Problem 1.01PP
Draw a component block diagram for each of the following feedback control systems,
(a)
The manual steering system of an automobile
(b)
Drebbel’s incubator
(c)
The water level controlled by a float and valve
(d)
Watt’s steam engine with fly-bail governor
In each case, indicate the location of the elements listed below and give the units associated with
each signal.
In each case, indicate the location of the elements listed below and give the units associated with
each signal.
• The process
• The process desired output signal
• The sensor
• The actuator
• The actuator output signal
• The controller
• The controller output signal
• The reference signal
• The error signal
Notice that in a number of cases the same physical device may perform more than one of these
functions.
Step-by-step solution
step 1 of 4
(A).
Manual Steering Systems
Step 2 of 4
(B)
Step 3 of 4
(c)
Set water
leveKRef Input)
(CorMleror
■ Actuator o/p)
Ftootvatve
Floating
(Sensor+contioller+Actualor)
Of float
Water
Tank
(process)
"Water Level Controller
Step 4 of 4
(D)
Fly-Ball Governor
Water level
Problem 1.02PP
Identify the physical principles and describe the operation of the thermostat in your home or
office.
Step-by-step solution
step 1 of 1
Thermostat is used as the sensing element for controlling the room temperature. It
performs the task o f automatic reduction o f error to zero, irrespectiTe o f the situation
created by disturbance.
Working Principle:
It contains a fluid which is able to e ^ a n d or contract due to
temperature change, which causes the sni^ - action of a switch that makes switchingON or OFF ofthe heat source.
Problem 1.03PP
A machine for making paper is diagrammed in Fig. There are two main parameters under
feedback control: the density of fibers as controlled by the consistency of the thick stock that
flows from the headbox onto the wire, and the moisture content of the final product that comes
out of the dryere. Stock from the machine chest is diluted by white water returning from under the
wire as controlled by a control valve {CV).A meter supplies a reading of the consistency. At the
“dry end” of the machine, there is a moisture sensor. Draw a block diagram and identify the nine
components listed in Problem part (d) for the following:
(a)
Control of consistency
(b)
Control of moisture
Figure A papermaking machine
Figure A papermaking machine
Watt’s steam engine with fly-ball governor
In each case, indicate the location of the elements listed below and give the units associated with
each signal.
• The process
• The process desired output signal
• The sensor
• The actuator
• The actuator output signal
• The controller
• The controller output signal
• The reference signal
• The error signal
Step-by-step solution
step 1 of 13
(a)
Step 2 of 13
Figure 1 shows the general block diagram to understand the process of the system. The nine
components listed below are identified in figure 1.
• Process
• Process desired output signal
• Sensor
• Actuator
• Actuator output signal
• Controller
• Controller output signal
• Reference signal
• Error signal
Actuating
signal
Actuator
ou^ut signal
Figure 1
step 3 of 13
Draw the block diagram for the figure 1.12 in the textbook to understand the process of control of
consistency.
Step 4 of 13 ^
Actuator
Error
Paper
consistency
Sensor
Figure 2
Step 5 of 13
Compare figure 1 and figure 2. The comparison shows the working of the paper making machine
>A/ifh the mainr ^nmnnnonfc in fho nrr»rocc nf nr*ntml nf nrinciotonnw
Step 6 of 13
The input reference signal is given to the controller and here the valve acts as an actuator for the
system. The valve initiates the mixing process and the consistency of the paper is monitored
according to time delay. The consistency checking meter checks the output of the process and
sends the signal to error detector. Here, the original required consistency Is checked with the
output received and checks for any variation in the output. The corresponding required
adjustments may be done in the controller to reduce the error rate, to improve the total output of
the process.
Step 7 of 13
Thus, the control of consistency is explained and the major components are identified.
Step 8 of 13
(b)
Step 9 of 13
Draw the block diagram for the figure 1.12 in the textbook to understand the process of control of
moisture.
Step 10 of 13
Paper
moisture
Sensor
Figures
step 11 of 13
Compare figure 1 and figure 3. The comparison shows the working of the paper making machine
with the major components in the process of control of moisture.
Step 12 of 13
Here, Dryer remains as a main process and the moisture is checked in the moisture checking
meter (sensor). The total output is compared in the error detector and required adjustments are
done by the controller to improve the accuracy.
Step 13 of 13
Thus, the control of moisture is explained and the major components are identified.
Problem 1.04PP
Many variables in the human body are under feedback control. For each of the following
controlled variables, draw a block diagram showing the process being controlled, the sensor that
measures the variable, the actuator that causes it to Increase and/or decrease, the information
path that completes the feedback path, and the disturbances that upset the variable. You may
need to consult an encyclopedia or textbook on human physiology for information on this
problem.
(a)
Blood pressure
(b)
Blood sugar concentration
(c)
Heart rate
(c)
Heart rate
(d)
Eye-pointing angle
(e)
Eye-pupil diameter
Step-by-step solution
step 1 of 2
(A)
.
(B) .
(C)
.
Block Diagram for Blood Pre ssure. Blood Sugar Concentration and Heart Bate
Control
Step 2 of 2
(D ).
rm *
12^T_*2?2!E!S.
Block Diagram o f Eye-Pointing Angle and Eye - P i ^ i l Diameter
Problem 1.05PP
Draw a block diagram of the components for temperature control in a refrigerator or automobile
air-conditioning system.
Step-by-step solution
step 1 of 2
Figure 1 shows the general block diagram to understand the temperature control of refrigerator
and in the automobile air-conditioning system.
ActuMor
Process
and in the automobile air-conditioning system.
Figure 1
Step 2 of 2
The required temperature signal is set in the thermostat, and then the controller actuates the
compressor for cooling process. The temperature sensor measures the temperature and
compares the required reference temperature with the measured temperature in the comparator.
When the required temperature is achieved, the controller stops the compressor and maintains
the temperature.
The temperature sensor checks the temperature of the system periodically and actuates the
controller if there is difference in the required reference temperature. This process is continued to
maintain the required temperature of the system thereby controlling the ON/OFF input to the
compressor.
Thus, the temperature control of refrigerator and in the automobile air-conditioning system is
explained.
Problem 1.06PP
Draw a block diagram of the components for an elevator-position control. Indicate how you would
measure the position of the elevator car. Consider a combined coarse and fine measurement
system. What accuracies do you suggest for each sensor? Your system should be able to correct
for the fact that in elevators for tall buildings there is significant cable stretch as a function of cab
load.
Step-by-step solution
step 1 of 2
The comoonents of elevator oosition control are shown in fiouret.
Step 1 of 2
The components of elevator position control are shown in figure!.
Controller
Actuator
Process
Floor
Sensor
Figure I
step 2 of 2
The input is given to the logic controller through comparator. The logic controller drives the
transmission mechanism of the elevator (Motor or hydraulic). The elevator moves up and down
and reaches the required floor position.
When the corresponding floor button is pressed, the controller reduces the speed of the motor to
stop the elevator in the respective floor. The electro switch acts as a sensor for coarse
measurement to measure the floor level. Accuracy level can be fixed in the sensor considering
the cable stretch due to cab load. This sensor enables to locate the elevator in the respective
floor accurately without any deviation in the measurements.
This complete mechanism forms a closed loop as in figure 1. The error detector compares the
reference input to the output of the closed loop to ensure the exact location of the elevator.
Hence, the block diagram of the elevator mechanism considering negligence is explained.
Problem 1.07PP
Feedback control requires being able to sense the variable being controlled. Because electrical
signals can be transmitted, amplified, and processed easily, often we want to have a sensor
whose output is a voltage or current proportional to the variable being measured. Describe a
sensor that would give an electrical output proportional to the following;
(a)
Temperature
(b)
Pressure
(c)
Liquid level
(d)
EJpwj^fJinJjiri ainnn a ninp (nr hlnnH alnnn an artprv^
(d)
Flow of liquid along a pipe (or blood along an artery)
(e)
Linear position (t) Rotational position
(g)
Linear velocity
(h)
Rotational speed
(i)
Translational acceleration
(j) Torque
Step-by-step solution
step 1 of 2
(A)
. TEMP M A T U R E : Thermocoi^lc.
(B)
. PRESSURE : Pressure gauge.
(C)
. U Q UlL? LEVEL : Bourdon tube and LV D T s combination.
(Py FLOW OF LIQUID ALONG PIPE FORCE : Any pres sure sensor can be used.
LINEAR POSITION : LVDT li n e a r Variable Differential Transformer).
Step 2 of 2
ROTATIONAL POSiriON : Potentiometer.
(G). U N EA R VELOCITY : Speedometer.
0 ^ . ROTATIONAL SPEED : Tachometer.
( 1 \ TRANSLATIONAL ACCLERATION : LVDT
( J y TO R Q U E: Combination o f Gear and Tachometer.
Problem 1.08PP
Each of the variables listed in Problem can be brought under feedback control. Describe an
actuator that could accept an electrical input and be used to control the variables listed. Give the
units of the actuator output signal.
Problem
Feedback control requires being able to sense the variable being controlled. Because electrical
signals can be transmitted, amplified, and processed easily, often we want to have a sensor
whose output is a voltage or current proportional to the variable being measured. Describe a
sensor that would give an electrical output proportional to the following;
(a) Temperature
(b) Pressure
(b) Pressure
(c)
Liquid level
(d)
Flow of liquid along a pipe (or blood along an artery)
(e)
Linear position (t) Rotational position
(g)
Linear velocity
(h)
Rotational speed
(i)
Translational acceleration
(j) Torque
Step-by-step solution
step 1 of 1 ^
An actuator amplifies the signal taken from the sensor. Any Electronic A n^lifier can be
used to do such. The Operational Amplifier is the most commonly used Actuator.
e .-
' OPAMP
OP AMP
Units o f the Actuator output signal are current (Anq^ere) or Voltage (Volts)
Problem 1.09PP
Feedback in Biology
(a) Negative Feedback in Biology. When a person is under long-term stress {say, a couple of
weeks before an exam!), hypothalamus (in the brain) secretes a hormone called Corticotropin
Releasing Factor (CRF) which binds to a receptor in the pituitary gland stimulating It to produce
Adrenocorticotropic hormone (ACTH), which in turn stimulates the adrenal cortex (outer part of
the adrenal glands) to release the stress hormone Glucocorticoid (GC). This In turn shuts down
(turns off the stress response) for both CRF and ACTH production by negative feedback via the
bloodstream until GC returns to Its normal level. Draw a block diagram of this closed-loop
system.
(b) Positive Feedback in Biology. This happens in some unique circumstances. Consider the
birth process of a baby. Pressure from the head of the baby going through the birth canal causes
(b) Positive Feedback in Biology. This happens in some unique circumstances. Consider the
birth process of a baby. Pressure from the head of the baby going through the birth canal causes
contractions via secretion of a hormone called oxytocin which causes more pressure which in
turn intensifies contractions. Once the baby is born, the system goes back to normal (negative
feedback). Draw a block diagram of this closed-loop system.
Step-by-step solution
step 1 of 9
(a)
Step 2 of 9
The block diagram to show an example of closed loop negative feedback in a biological system Is
shown in figure 1.
Step 3 of 9
Long
Nofinal
Gluco
oofticoids(CD)
Pituitary
glands
— ^L^H ypolhala m c«L |^^^
Mi
1
I
Adrenal
Glands
GCs
Stress response - off
^Stress response • off
Negatice feedback_______
Figure 1
Step 4 of 9
Here, ‘GD’ is the Glucocorticiods, ‘CRF’ is the Corticotrophin Releasing Factor, and ‘ACTH’ is the
Adrenocorticotropic hormone.
Step 5 of 9
Thus the block diagram of closed loop negative feedback In a biological system is drawn and
shown in figure 1.
Step 6 of 9
(b)
Step 7 of 9
The block diagram to show an example of closed loop positive feedback in a biological system Is
shown in figure 2.
Step 8 of 9
O j^tocin
Desired
Pressure
,
y ? y
Baby
head
Birth
canal
1 ^f
\
Pressure
1
^ Is
n (Contractions)
Positive feedback
Figure 2
Step 9 of 9
Thus the block diagram is drawn to show the closed loop positive feedback in a biological
system.
Problem 2.01 PP
Write the differential equations for the mechanical systems shown in Fig.. For Fig.(a) and (b),
state whether you think the system will eventually decay so that it has no motion at all, given that
there are nonzero initial conditions for both masses and there is no input; give a reason for your
answer.
Figure Mechanical systems
Nofnctioa
Nofrictkn
(I)
h
(I)
I—
^
w v —
—
I
Step-by-step solution
step 1 of 6
(a)
Refer to Figure 2.41 (a) in the text book for the block diagram of a mechanical system.
Draw the free body diagram of mass m ,.
^ *a*a
*1*1^ *1 *
Write the differential equation describing the system.
m,x, - - ^ x , -k^x^-k^x^
Step 2 of 6
Draw the free body diagram of mass
.
tz
Write the differential equation describing the system.
m^x^ ^-k^x^-k^x^-^k^x^-¥ k^y
+ *3 (*1 - ■»!)+*3
- .y ) = 0
Thus, the differential equations describing the system are.
(x,-JC j) s 0
There is friction that affects the motion of both the masses. Thus, the system decays to zero
motion for both the masses.
Step 3 of 6
(b)
Refer to Figure 2.41 (b) in the text book for the block diagram of a mechanical system.
Draw the free body diagram of mass
■
Write the differential equation describing the system.
( x j- x , )= 0
Step 4 of 6
Draw the free body diagram of mass
.
*2*1
Write the differential equation describing the system.
m^x^ - - b ^ x ^ - k ^ x ^ - k ^ x ^ + k ^
Al|X2 4'i^(x2~Xi)4'A^A^ = 0
Thus, the differential equations describing the system are.
+ it,x, +
(xj -
)* 0
Although friction affects only the motion of the left mass directly, continuing motion of the right
mass excites the left mass, and that interaction continues until all motion damps out.
Step 5 of 6
(c)
Refer to Figure 2.41 (c) in the text book for the block diagram of a mechanical system.
Draw the free body diagram of mass m ,.
|_ x .
^*1 ►tjX j
*1*1 * 2 *1 *
Write the differential equation describing the system.
m,x, *-i^ x , -ik,x, -A^x, +i^Xj+i\x^ = 0
m,jt, + 4| (x, - X,) + V i + *1 (*i - *2) = 0
Step 6 of 6
Draw the free body diagram of mass
iWj
s -k^x^ - A^Xj - AiXj + ^X| +
.
+F
+ i | X i+ i | ( x i- x i) + * 2 ( X 2 - x i) = F
Thus, the differential equations describing the system are.
n\Xy + 6 |(x ,-X j)+ A |X ,+ A ^ (x , - X j ) * 0
m ,x , + i|X , + A ^ ( i, - X |) + < : , ( x , - X |) = f
Problem 2.02PP
Write the differential equation for the mechanical system shown in Fig. State whether you think
the system wili eventuaily decay so that it has nomotion at ail, given that there are nonzero initial
conditions for both masses, and give a reason for your answer.
Figure Mechanical system
M
<1
Noftictioo
Step-by-step solution
Step-by-step solution
step 1 of 4
Refer to Figure 2.40 in the textbook.
Draw the free body diagram of left side mass, m j.
J— ,
K ^,
m^
lU X i- x ^
Figure 1
Step 2 of 4
Write the equation of motion for left side mass, m |.
m,
+ JCj (jcj - Xj
* 0
»Hj^+(Ar|+Ar,)aii-iC, x , + i ,{ x ,~ x ,) = 0
(1)
Draw the free body diagram of right side mass, m |.
m,
Figure 2
Step 3 of 4
Write the equation of motion for right side mass, m2 ■
m ,X 2 + J C 2 (x j-x ,)+ 6 ^ ( x j - xj)+JC ,;^ = 0
i^ ( i, - ii) = 0
(2)
Therefore, the differential equations for mechanical system are.
+ /T j)x ,-J ir2
- JITjX,+(AT, + Jirj)x2 +
( i j - x ,) s 0
step 4 of 4
The relative motion between x, and ji^ , decays to zero due to the damper. However, the two
masses continue oscillating together without decay, since there is no friction opposing that motion
and also no flexure of the end springs, and these are the essentials to maintain oscillation of the
two masses.
Thus, the system w ill finally decay, as the relative motion between x, and x^. decays to
zero.
Problem 2.03PP
Write the equations of motion for the double-pendulum system shown in Fig. Assume that the
displacement angles of the pendulums are small enough to ensure that the spring is always
horizontal. The pendulum rods are taken to be massless, of length /, and the springs are attached
three-fourths of the way down.
Step-by-step solution
step 1 of 6
Consider the circuit diagram.
r
m
m
Double pendulum system
Step 2 of 6 ^
Re-draw the circuit diagram.
^
/
I
1
^
/
/
m
Here there are two degrees of freedom
Step 3 of 6 ^
At any instant, let one of them be displayed by 6| and other by 6,.
Step 4 of 6
Writing about point ‘0’ the moment equation is.
Jt(sinO 2-sinO ,)cos0sm /^d,
Similarly, write the moment equation for other pendulu
-mg/sin6, - ^ ^ / j *(sin6,-sine|)cos82 =m/^e.
Step 5 of 6
Assume the angles are small, then sin0| s 0 |,
c o s 9 |S ]a n d c o s 0 2 s l.T h e
above equations are modified as.
* (e ,-e ,)= m /= 9 ,
Therefore, the above equation is moment equilibrium about the pivot point of the left pendulum.
Step 6 of 6 ^
Write the equation for moment equilibrium about the pivot point of the right pendulum.
* (6 ,-6 ,) - 0
Therefore, the above equation is moment equilibrium about the pivot point of the right pendulum.
Problem 2.04PP
Write the equations of motion of a pendulum consisting of a thin, 2 kg stick of length / suspended
from a pivot. How long should the rod be in order for the period to be exactly 1 sec? (The inertia I
of a thin stick about an end point is
Assume that 0 is small enough that sin Q ^
Q.)
Why do you think grandfather clocks are typically about 6 ft high?
Step-by-step solution
step 1 of 3
Consider the following equation for the sum of all external moments about the center of mass of
Consider the following equation for the sum of all external moments about the center of mass of
a body.
M ^ I a ...... (1)
Where.
/ is the body’s mass moment of inertia about its center of mass,
a is the angular acceleration of the body.
The coordinates and forces relationship is shown in Figure 1.
o
Figure 1
Step 2 of 3
From Figure 1, the equation of moment about point O is given below.
M o = -m g n ^ s m 0
........(2 )
—
Substitute
-m g
for /p in equation (2).
=
^ m /*tf+ mg X^ sin
=0
f l + ^ 8 i n f l = 0 ...... (3)
2i
Consider ^ is very small in equation (3).
g+ ^e = o
(4)
Consider the following general equation.
^+24^a>e+ai‘ f f = 0
(5)
Compare equation (4)and equation (5).
Step 3 of 3 ^
Consider the following equation for the period for a swing.
r=— ....(7)
Substitute equation (6)in equation (7).
2/
r - 2 > r / - = - ...... (8)
Substitute 2 for / and 9.81 for g in equation (8).
I 2x2
r = 2 ;rj
’'V 3 x 9.8I
= 2.31
= 2sec
Consider the following equation for the value of length.
/= - ? § -
(9)
Substitute 9.81 for g in equation (9).
3x9.81
/* 8^*
* 0.3727 m
Thus, the vaiue of length and period for a swing are [0.3727m]a
Hence, the 1 second for a swing from one side to the other. This pendulum is shorter because
the period is faster. But if the period had been 2 second, the pendulum length is 1.5 meters.
Problem 2.05PP
For the car suspension discussed in Example, plot the position of the car and the wheel after the
car hits a “unit bump”(that is. r is a unit step) using Matlab. Assume that m1 = 10 kg, m2 = 250
kg, Kw = 500,000 N/m, and Ks = 10,000 N/m. Find the value of b that you would prefer if you
were a passenger in the car.
Example A Two-Mass System: Suspension Model
Figure 1 shows an automobile suspension system. Write the equations of motion for the
automobile and wheel motion assuming one-dimensional vertical motion of one quarter of the car
mass above one wheel. A system consisting of one of the four wheel suspensions is usually
referred to as a quarter-car model. The system can be approximated by the simplified system
shown in Fig. 2 where two spring constants and a damping coefficient are defined. Assume that
the model is for a car with a mass of 1580 kg, including the four wheels, which have a mass of 20
kg each. By placing a known weight (an author) directly over a wheel and measuring the car?s
deflection, we find that ks = 130,000 N/m. Measuring the wheel?s deflection for the same applied
the model is for a car with a mass of 1580 kg, including the four wheels, which have a mass of 20
kg each. By placing a known weight (an author) directly over a wheel and measuring the car?s
deflection, we find that ks = 130,000 N/m. Measuring the wheel?s deflection for the same applied
weight, we find that Mv ~ 1,000,000 N/m. By using the step response data in Fig. 3(b) and
qualitatively observing that the car?s response to a step change matches the damping coefficient
curve for ^ = 0.7 in the figure, we conclude that b = 9800 N?sec/m.
Solution. The system can be approximated by the simplified system shown in Fig. 2. The
coordinates of the two masses, x and y, with the reference directions as shown, are the
displacements of the masses from their equilibrium conditions. The equilibrium positions are
offset from the springs? unstretched positions because of the force of gravity. The shock
absorber is represented in the schematic diagram by a dashpot symbol with friction constant b.
The magnitude of the force from the shock absorber is assumed to be proportional to the rate of
change of the relative displacement of the two masses?that is, the force =
The force of
gravity could be included in the free-body diagram; however, its effect is to produce a constant
offset of X and y. By defining x and y to be the distance from the equilibrium position, the need to
include the gravity forces is eliminated.
The force from the car suspension acts on both masses in proportion to their relative
displacement with spring constant ks. Figure 2.6 shows the free-body diagram of each mass.
Note that the forces from the spring on the two masses are equal in magnitude but act in
opposite directions, which is also the case for the damper. A positive displacement y of mass m2
will result in a force from the spring on m2 in the direction shown and a force from the spring on
m1 in the direction shown. However, a positive displacement x of mass m1 will result in a force
from the spring ks on m1 in the opposite direction to that drawn in Fig. 4, as indicated by the
minus X term for the spring force.
The lower spring kw represents the tire compressibility, for which there is insufficient damping
(velocity-dependent force) to warrant including a dashpot in the model. The force from this spring
is proportional to the distance the tire is compressed and the nominal equilibrium force would be
that required to support m1 and m2 against gravity. By defining x to be the distance from
equilibrium, a force will result if either the road surface has a bump (r changes from its
equilibrium value of zero) or the wheel bounces (x changes). The motion of the simplified car
over a bumpy road will result in a value of r(t) that is not constant.
As previously noted, there is a constant force of gravity acting on each mass; however, this force
has been omitted, as have been the equal and opposite forces from the springs. Gravitational
forces can always be omitted from vertical-spring mass systems (1) if the position coordinates
are defined from the equilibrium position that results when gravity is acting, and (2) if the spring
forces used in the analysis are actually the perturbation in spring forces from those forces acting
at equilibrium.
Figure 1 Automobile suspension
Figure 2 The quarter-car model?
T
Figure 3 Responses of second-order systems versus ^ ; (a)impulse responses; (b) step
responses
Figure 4 Free-body diagrams for suspension system
Ki-i
0
i^-r)
■r)
il
Applying Eq. (1) to each mass and noting that some forces on each mass are in the negative
(down) direction yields the system of equations
Some rearranging results in
F=m a,
(2.1)
4 (J -i)+ t,(y -i)-* ,(i-r )= « iii,
(18)
=
(19)
The most common source of error in writing equations for systems like these are sign emors. The
method for keeping the signs straight in the preceding development entailed mentally picturing
the displacement of the masses and drawing the resulting force in the direction that the
displacement would produce. Once you have obtained the equations for a system, a check on
the signs for systems that are obviously stable from physical reasoning can be quickly carried
out. As we will see when we study stability in Section 6 of Chapter 3, a stable system always has
the same signs on similar variables. For this system, Eq. (2) shows that the signs on the
X terms
are all positive, as they must be for stability. Likewise, the signs on the
and
and /te rm s
are all positive in Eq. (2).
1 + — (t-W
Ml
+ .^ d -y ) + ^ i = ^ r .
Mi
Ml
Mi
(1 1 0 )
(2 .1 1 )
9 + - ^ 0 - t ) + ^ ( y - x > = o .
M2
M2
The transfer function is obtained in a similar manner as before for zero initial conditions.
Substitutino s for d/dt in the differential eouations vields
J^X(I)+ 1— (Xd) - X(D) + .^(X (I) - X(D) + ^ X ( . ) = ^ « ( I) ,
Ml
Dll
Jfl]
Mi
^ n s )+ s — m ) - x m + — o r(s ) - x m = a
which, after some algebra and rearranging to eliminate X(s), yields the transfer function
I 'd )
^ (» + > )
* * + ( s r + ^ ) * " + ( ^ r + ^ + f e ) ‘" + ( ^ ) ' + i ^
(112)
To determine numerical values, we subtract the mass of the four wheels from the total car mass
of 1580 kg and divide by 4 to find that m2 = 375 kg. The wheel mass was measured directly to be
m1 = 20 kg. Therefore, the transfer function with the numerical values is
Y(s)
131e06(5 + 13.3)
* d ) “ J* + (516.1)s“ + (5.685«04)j“ + (I J (n < 0 ^ + l.733«Or
0.13)
We will see in Chapter 3 and later chaptero how this sort of transfer function will allow us to find
the response of the car body to inputs resulting from the car motion over a bumpy road.
Step-by-step solution
step 1 of 3
Consider the transfer function.
i'( 4
'»!'»!
4j
J a l 'U A A
1 m i« j
Where.
is the mass.
kg is the car deflection.
k^ is the applied weight.
b is the damping.
Write the MATLAB program,
m l = 10;
m2 = 250;
kw = 500000;
ks= 10000;
Bd = [1000 3000 4000 5000];
t = 0:0.01:2;
for i = 1:4
b = Bd(i):
A=[0 1 0 0;-( ks/ml + kw/ml ) -b/ml ks/ml b /m l;
0 0 0 1; ks/m2 b/m2 -ks/m2 -b/m2 ];
B=[0; kw/ml :0 ;0 ];
C=[1 0 0 0 ; 0 0 1 0 ];
D=0;
y=step(A.B.C.D,1,t):
subplot(2.2,i):
plot(t, y(:,1),t, y(:,2)):
legend(Wheer,'Car’):
title = sprintf('Response with b = % 4 .1 f, b );
end
Step 2 of 3
The output for the MATLAB code is given in figure 1.
Step 3 of 3
Thus, the smallest overshoot is with b is 5,000, and the jump in car position is the fastest with this
damping value.
From figure 1, ^ s 3,000 is the best compromise. Therefore, overshoot for lower values, and the
system gets too fast for larger values.
Problem 2.06PP
Write the equations of motion for a body of mass M suspended from a fixed point by a spring with
a constant k. Carefully define where the body’s displacement is zero.
Step-by-step solution
step 1 of 3
Displacement is Zero when forces are balanced
Hence
T
i-k x
No displecement
Step 2 of 3
^ e n M g = -la ,
^x&ere
S=
dt
=Acceleration due to gravity.
Then,
The displacement be comes Zero.
It is shown in the figure where displacement is zero.
Step 3 of 3
From the above figure,
kx+Mg = 0
I.', kx = -hK I
T
(di^koement)
Problem 2.07PP
Automobile manufacturers are contemplating building active suspension systems. The simplest
change is to make shock absorbers with a changeable damping, b(u1). It is also possible to
make a device to be placed in parallel with the springs that has the ability to supply an equal
force, u2. in opposite directions on the wheel axle and the car body.
(a)
Modily the equations of motion in Example to include such control inputs.
(b)
Is the resulting system linear?
(c)
Is it possible to use the force u2 to completely replace the springs and shock absorber? Is thi
a good idea?
Example A Two-Mass System: Suspension Model
a good idea?
Example A Two-Mass System: Suspension Model
Figure 1 shows an automobile suspension system. Write the equations of motion for the
automobile and wheel motion assuming one-dimensional vertical motion of one quarter of the car
mass above one wheel. A system consisting of one of the four wheel suspensions is usually
referred to as a quarter-car model. The system can be approximated by the simplified system
shown In Fig. 2 where two spring constants and a damping coefficient are defined. Assume that
the model is for a car with a mass of 1580 kg, including the four wheels, which have a mass of 20
kg each. By placing a known weight (an author) directly over a wheel and measuring the car?s
deflection, we find that ks = 130,000 N/m. Measuring the wheel?s deflection for the same applied
weight, we find that Mv ~ 1,000,000 N/m. By using the step response data in Fig. 3(b) and
qualitatively observing that the car?s response to a step change matches the damping coefficient
curve for ^ = 0.7 in the figure, we conclude that b = 9800 N?sec/m.
Solution. The system can be approximated by the simplified system shown in Fig. 2. The
coordinates of the two masses, x and y, with the reference directions as shown, are the
displacements of the masses from their equilibrium conditions. The equilibrium positions are
offset from the springs? unstretched positions because of the force of gravity. The shock
absorber is represented in the schematic diagram by a dashpot symbol with friction constant b.
The magnitude of the force from the shock absorber is assumed to be proportional to the rate of
change of the relative displacement of the two masses?that is, the force =
The force of
gravity could be included In the free-body diagram; however, its effect is to produce a constant
offset of X and y. By defining x and y to be the distance from the equilibrium position, the need to
include the gravity forces is eliminated.
The force from the car suspension acts on both masses in proportion to their relative
displacement with spring constant ks. Figure 2.6 shows the free-body diagram of each mass.
Note that the forces from the spring on the two masses are equal in magnitude but act in
opposite directions, which is also the case for the damper. A positive displacement y of mass m2
will result in a force from the spring on m2 in the direction shown and a force from the spring on
m1 in the direction shown. However, a positive displacement x of mass m1 will result in a force
from the spring ks on m1 in the opposite direction to that drawn in Fig. 4, as indicated by the
minus X term for the spring force.
The lower spring kw represents the tire compressibility, for which there is insufficient damping
(velocity-dependent force) to warrant including a dashpot in the model. The force from this spring
is proportional to the distance the tire Is compressed and the nominal equilibrium force would be
that required to support m1 and m2 against gravity. By defining x to be the distance from
equilibrium, a force will result if either the road surface has a bump (r changes from its
equilibrium value of zero) or the wheel bounces (x changes). The motion of the simplified car
over a bumpy road will result in a value of r(t) that is not constant.
As previously noted, there is a constant force of gravity acting on each mass; however, this force
has been omitted, as have been the equal and opposite forces from the springs. Gravitational
forces can always be omitted from vertical-spring mass systems (1) if the position coordinates
are defined from the equilibrium position that results when gravity is acting, and (2) if the spring
forces used in the analysis are actually the perturbation in spring forces from those forces acting
at equilibrium.
Figure 1 Automobile suspension
Figure 2 The quarter-car model?
U
Figure 3 Responses of second-order systems versus ^ ; (a)impulse responses; (b) step
responses
3^
z
m
s
Figure 4 Free-body diagrams for suspension system
lO-i
IJi-H
h^ -ii
Applying Eq. (1) to each mass and noting that some forces on each mass are in the negative
(down) direction yields the system of equations
Some rearranging results in
F=m a,
(2.1)
4 ( J - i) + t,( y - x ) - t,( i- r ) = « > i* ,
(2.8)
- f c O '- J ) - * ( » - «
(2.9)
The most common source of error in writing equations for systems like these are sign emors. The
method for keeping the signs straight In the preceding development entailed mentally picturing
the displacement of the masses and drawing the resulting force in the direction that the
displacement would produce. Once you have obtained the equations for a system, a check on
the signs for systems that are obviously stable from physical reasoning can be quickly carried
out. As we will see when we study stability In Section 6 of Chapter 3, a stable system always has
?
tKa oinne- nn
X terms
are all positive, as they must be for stability. Likewise, the signs on the
and /te rm s
are all positive In Eq. (2).
A ( t - y , + i.( x - y ) + t x = ^ r .
Ml
Mi
Mi
Mi
(2.10)
y + — 0 - t ) + ^ (y -x )= O .
M2
(2.11)
M2
The transfer function is obtained In a similar manner as before for zero initial conditions.
Substituting s for d/dt in the differential equations yields
j^ x c i) +
Mi
- r m + ^ ( X ( j ) - Y(sn+ ^ x w =
Mi
Mi
Mi
+ » — ()'(») - * ( ! ) ) + — ()'(j) - * ( » ) ) = a
which, after some algebra and rearranging to eliminate X(s), yields the transfer function
X M _____________________ ^ ( » + > )
«(»)
( 2 . 12)
To determine numerical values, we subtract the mass of the four wheels from the total car mass
of 1580 kg and divide by 4 to find that m2 = 375 kg. The wheel mass was measured directly to be
m l = 20 kg. Therefore, the transfer function with the numerical values is
r(s)
131e06(s + 13.3)
S(j) “ J* + (516.1)s“ + (5.685«04)j2 + (Ume06)s + l.733«07'
0.13)
We will see in Chapter 3 and later chaptero how this sort of transfer function will allow us to find
the response of the car body to inputs resulting from the car motion over a bumpy road.
Step-by-step solution
step 1 of 3
(A)
The FBD shows the a<iditioa of the variable force, \32 and shows 6 as in the FBD
of figure 2.5, However, here 6. Is a function o f the control variable ui. The forces
below are drawn in the direction that would result from a po sitive displac ement o f
^
O
"
-
* ) - ■K'.
M , y = - K , ( y - x ) - i { u ^ ) { y - i ) +U j
Step 2 of 3
The system is linear with respect to
because it is additive.
Step 3 of 3
(C)
It is technically possible. However it would take very high forces and thus a lot of
power and is therefore not done.
Problem 2.08PP
In many mechanical positioning systems there is flexibility between one part of the system and
another. An example is shown in Fig. 1 where there is flexibility of the solar panels. Figure 2
depicts such a situation, where a force u is applied to the mass M and another mass m is
connected to it. The coupling between the objects is often modeled by a spring constant k with a
damping coefficient b, although the actual situation is usually much more complicated than this,
(a) Write the equations of motion governing this system.
(b) Find the transfer function between the control input u and the output y.
Figure 1 Communications satellite Source: Courtesy Space Systems/Loral (SSL)
Figure 2 Schematic of a system with flexibility
Step-by-step solution
step 1 of 7
Sketch the schematic o f a sjrstem with flexibility.
Step 2 of 7 ^
Sketch the free body diagram o f the given figure.
iibc-y)
U
b ( f- y )
Step 3 of 7
>
X
K x -y )
Step 4 of 7 ^
From the free body diagrams, we get
mx = - J t ( x - i ( x - j^)
= u + ^ (x -
+ A ( i - J>)
Simplify further.
X H—
X H—
m
-k
—
X ------- y -------- y = 0
m
A .
m
_ jt
m
A.
1
X -------- x + V H-------- v + — V = — u
M
M
M
M
M
Thus, the equation o f motion governing the system is obtained as
- k ^ b . k
A . X H—
X H—
m
-k
—
X ------- y -------- y = 0
m
A .
m
.
m
^ A .
1
M
M
X -------- x + V H-------- v + — V = — u
M
M
M
Step 5 of 7 ^
(b)
Find ^
- k
Limlace tra n s f^ n o f the equations
\ . k
A. . *
X H— X H— X---- y ------ y = 0
m
m
-k
A .
M
M
m
-
m
it
^ A .
M
1
M
M
s^x+ - X + - s x - - r - - s r = 0
— X - — sX +
+ — Y + — ysY = — U
M
M
M
M
M
Rewrite in matrix form
(m ^ + b s + k
[ -(As + Jt)
^0^
-(b s + k )
Ms^ + bs + k ) [ Y ; [ u )
Step 6 of 7
Solve using the Cramer’s rule,
(m ^ + b s + k
0^
-(te + t)
u]
detl
T
det
+ As + it
- (As + it)
-(A s+ it)
A fs^+As+it
+ As + it
7 =
{^ms^ + As + itJ^Afe^ + As + itj - (As + it)^
Step 7 of 7
Simplify further.
y
^
ms^ + As + it
+ bs + k ^ (^Ms^ + b s + k'j - (As + k ^
+ As + it
U
mMs + (« + M')bs + (M + m) ks^
Problem 2.09PP
Modify the equation of motion for the cruise control in Example, Eq., so that it has a control law;
that is, let
u = K (v r-v ), (2.108)
where
vr= reference speed. (2.109)
/C= constant. (2.110)
This is a “proportionarcontrol law in which the difference between \^rand the actual speed is
used as a signal to speed the engine up or slow it down. Revise the equations of motion with vr
as the input and v as the output and find the transfer function. Assume that/n = 1500 kg and b =
70 N-sec/m, and find the response for a unit step In vr using Matlab. Using trial and error, find a
value of K that you think would result in a control system In which the actual speed converges as
quickly as possible to the reference speed with no objectionable behavior.
70 N-sec/m, and find the response for a unit step In vr using Matlab. Using trial and error, find a
value of K that you think would result in a control system In which the actual speed converges as
quickly as possible to the reference speed with no objectionable behavior.
Eq.
b
u
V H— V = —.
m
m
EXAMPLE 2.1
A Simple System; Cruise Control Model
1. Write die equedon of motion for the ipeed m l forward motion of
lhecaribow nm Fig.2.1 aasnmiiig that the en^ne imparta a force ■ ae
abown. Take the Laplace transform o f the reauiting differential eqoadon
and find the transfer function between the input« and the output v.
2. Use Matlab to 6nd the response o f the velocity of the car for the case
in which the inpntjumpe from being « = Oat time f = 0 to a constant
Hgui«2.1
Cruise control model
« B 500 N thereafter. Assume that the car a
visoous drag coeflkient, b B 50 N*secAn.
I « is 1000 kg and
Solution
1. Equations of motion: For sunpBdty we asanme that the rotmkNinl
inertia o f the wheeb b negligible and that there b ftiction retarding
the motion of the car that b proportional to the car’s speed with a
proportionality constant,
The car can then be approximated for
modeling poipoeea using the ftee-body diagram men m Fig. 2.2, which
defines coordinmes, shows all forces acting on the body (heavy lineaX
and imficates the accdetaCion (dashed line). The coordinate o f the car’a
positioo Xu the distaoce from the refotenoe line sbo«m and b chosen ao
that positive isto the r i ^ Note that in this case the meitial acceleration
u amply the second derivative o f x (that is, a * x) because the car
position b measured wkh respect to « inertial reference frame. The
equation of motion b found mung Eq. (2.1). The friction force acts
opposite to the direction motiMi; therefore k b drawn rqiposhe to the
direction o f positive motion m l eiketcd as a negative fioroe in Eq. (2.1).
T h eresu k b
( 2.2)
^
b.
M
X + —X = — .
03)
For the case of the automotive cruise control where the variable o f
interest b the speed, v ( ^ ) , the equbion o f motion becomes
V+ —V
* B —.
“
0 .4 )
H k sobtion of w ch an equation win be cowered in detail in Q u p te r 3;
however, the essence b that you aaaume a aolotionoftfae form v s
p v e a an input o f the form ■ b
Then, tince v b
the
differential equation chi be written aa^
03)
t::
Rgurt2,2
r>co body diagram for
The c* term C H ic e b out, and we find that
V _ -!■
yg
m
«.
.+ 1 '
06)
For reasons that will become efetf in Quqker 3, this b often written
using capital letters to signify that k b the‘Itunsforro'’o f the solution, or
vw _
i
*+
h'
a7)
This expresskm of die dUferential equation (2.4) b called the tra n ti’e r
function and wiD be used extensively m later chapters. Note that, in
essence, we have aubstkuled i fof dfdt in Eq. (Z4X T hb transfer Auh
tion serves as a mwh model that relates the c h ’ s velocity to the forces
propelling the car, that b , inputs from the accelerator pedal. Transfer
functions of a system will be used in later chapters to design feedback
corXrollers such as a cruise control device found in many modern cars.
X Time reaponae: The dynamics o fn system can be prescribed to Matlab
in terms of its transfer function as can be seen in the Matlab statemenb
below that implemeiits E<^ (2.7). The step fimetion in Madab calcnbles die time response o f a Ikictf system to ■ unit step input Bwuuie
the system b finear. the output for th b case can be multiplied by the
magnitude of the input mep to derive a step re9 onae o f any ampUtude.
Equivaleikly. ^ can be multiplied by die magnitude of the input step.
Stepresponsewtth Hatlab
9 S• (1/1000]/(s -f 5<V1000);
step(S00*9 s);
% setsupthem odetod«flnetlN
transfer ftincUon
% defines th e transfer function from
Eq. (2.7) with the numbers filled l a
% plots th e step response for u • 500.
calculate and plot the time reaponie o f velocity for an input step with
n 500>N magnitude. The Mep responae b shown m Fig. 2.3.
Rgure2.3
Response of the car
v e lo d ty toastepinti
Step-by-step solution
step 1 of 7
Consider the equation of motion.
V + — V ® — IT
Where,
u is the input
V is
the output
m is the mass
b is the friction force
Substitute J C (v ^ -v ) foru.
m
m
.
b
K
m
m
. b
K
V +— V+ — V=
m
m
K
m
K
— Vm
V + — V * — V ,------- V
A block diagram of the scheme is shown in Figure 1.
Step 2 of 7
u
1
m
. b .K
Step 3 of 7
Figure 1
Step 4 of 7
Consider the transfer function of the closed loop system.
y (s )
b
m
K (x )
K
m
S + — + —
Write the inputs for MATLAB.
K
m
den
I
m mj
Step 5 of 7
Consider K=100,200,1000,5000
Write the MATLAB program,
m = 1500;
b = 70;
k = [100 200 1000 5000];
hold on
t=0:0.2:50;
for i=1:length(k)
K=k(i);
num =K/m;
den = [1 b/m+K/m];
sys=tf(num,den);
y = step(sys,t);
plot(t,y)
grid
end
hold off
Step 6 of 7
The output for the MATLAB code is given in Figure 2.
Step 7 of 7
Thus, the larger the K is the better the performance with no objectionable behavior for any of the
cases. Increasing K also results in the need for higher acceleration is less obvious. In Figure 2,
for K - 1,000 , there is a response in 5 seconds and the steady state error is 5%.
Problem 2.1 OPP
Determine the dynamic equations for lateral motion of the robot in Fig. 1. Assume it has three
wheels with a single, steerable wheel in the front where the controller has direct control of the
rate of change of the steering angle, Usteer, with geometry as shown in Fig. 2. Assume the robot
is going in approximately a straight line and its angular deviation from that straight line is very
small. Also assume that the robot is traveling at a constant speed, Vo. The dynamic equations
relating the lateral velocity of the center of the robot as a result of commands in Usteer are
desired.
Figure 1 Robot for delivery of hospital supplies. Source: AP Images
Figure 2 Model for robot motion
Step-by-step solution
step 1 of 3
Refer FIGURE 2.46 in the textbook.
Consider the following equation for the sum of all external moments about the center of mass of
a body.
M ^ I a .......(1)
Where.
/ is the body’s mass moment of inertia about its center of mass.
a is the angular acceleration of the body.
Consider the following equation for the time rate of change of the steering wheel angle.
S, - U ^ .......(2)
Where.
is the control input.
The turning rate change with respect to x axis is shown in Figure 1.
Step 2 of 3
Consider the following equation for the carts turningrate of change with respect to x axis.
sin 9,
Where.
is nonzero.
L is the length of the wheel,
F^is the constant speed.
Consider
is small and rearrange the above equation.
(3)
The lateral motion as a function of y is shown in Figure 2.
Step 3 of 3
Consider the following equation for the actual change in the carts lateral position.
..... (4)
Take differentiation on both sides in equation (4).
y=
......(5)
Substitute equation (3)in equation (5).
..
v^e.
Take differentiation on both sides,
y » -2 —* • ......(o)
Substitute equation (1 )in equation (6).
(7)
Thus, there is no dynamics come into equation (7). Therefore, there was no need to invoke
equation (1).
Problem 2.11 PP
A first step toward a realistic model of an op-amp is given by the following equations and is
shown in Fig.:
10^
i+ = i-= 0 .
Figure Circuit for Problem
Find the transfer function of the simple amplification circuit shown using this model.
Step-by-step solution
step 1 of 2 >
Refer to Figure 2.43 in the circuit.
The output voltage is,
r^ = — [r,-r.]
5+ r
*
o)
The currents at inverting and noninverting terminals is,
I.-L - 0
Apply KirchhofTs current law at node V_.
y -V
—
—
y -V
— sfi-=0
R,
« ,)
\
K
R,
R,R^ )
R^R,
R„+R^
V=
V^+
R^+ R , -
^
r , ..... (2)
R^ + R , -
Step 2 of 2
The voltage at noninverting input terminal is,
> ;= o v
Substitute O V fo r y and ------^
v^= —
(o —
5+it^
“
-
R^+R/
10’
*/ y
s + lRi, + R^ “
s+lRj^+R^
f
Rm*Rf
o
r
y in equation (1).
J
<«’
R. y
s+ lJ^ +Rf “
s+lR^,+R^
K^LJ2L_5^ 1=.ioL_L^-^
j+
^
+
S + lR^+Ry "
r (s + I ) ( R , * R , ) + I O ^ R , )
•“[
iQ t
(/i„+Ji/)(j+l) J
R,
+
“
»'-[(*+i)(^+«,)+ioX]—io’«A
________ - 1 0 ’ i?,
K.
"(*+i)(J?.+«^)+ioX
[R^+R/J
*'*
» + i + i o ’ f — 5s— 1
Therefore, the transfer function of the circuit,
is
K.
j + 1 + 10’
Problem 2.12PP
Show that the op-amp connection shown in Fig. 1 results in Vout = Vin if the op-amp is ideal.
Give the transfer function if the op-amp has the nonideal transfer function of Problem.
Figure 1 Circuit
A first step toward a realistic model of an op-amp is given by the following equations and is
shown in Fig. 2:
A first step toward a realistic model of an op-amp is given by the following equations and is
shown in Fig. 2:
10^
i+ = i-= 0 .
Figure 2 Circuit
Find the transfer function of the simple amplification circuit shown using this model.
Step-by-step solution
step 1 of 3
Refer to the circuit diagram in Figure 2.38 in the textbook.
The voltage at the non-inverting terminal of the op-amp is,
v ,- V ^
The voltages at the inverting and non-inverting terminals of the op-amp are same for ideal opamp.
v. = n
The inverting node of the op-amp is directly connected to output node. So,
y^=ySubstitute
for v_ in the equation.
y^=yu
Hence, it is proved.
Step 2 of 3
The simplified op-amp circuit is shown in Figure 1.
Step 3 of 3
From Figure 1, the output voltage is,
= .4 (v .-v .)
Substitute
for
and
for v_ in the equation.
y ^ + A r^ = A V ^
( l + A ) V ^ = AVi.
y«
A
K.
1+ .4
Therefore, the transfer function.
of non-ideal op-amp is
A
\ +A
Problem 2.13PP
A common connection for a motor power amplifier is shown in Fig. The idea is to have the motor
current follow the input voltage, and the connection is calied a current amplifier. Assume that the
sense resistor rs is very smail compared with the feedback resistor R. and find the transfer
function from Vin to la. Also show the transfer function when R f=
Figure Op-amp circuit for Problem
Step-by-step solution
step 1 of 3
A common connection for a motor power an^lifier is given.
Sketch the given figure.
Step 2 of 3
At node A, write the voltage equation.
V L -0
^
R
At node B, write the voltage equation, with R^-^ R
,
^
0 -f%
0 -V .
R
R,
r
+
r/
,
‘
step 3 of 3 ^
The d3mamics o f the motor is modeled with negligible inductance as J^9^ +
We can rewrite as
4 , + * 4 = A'^.as
+
At the ou^ut, &om the equation
,
0 -r.
0 -r.
R
5,
,,
j , e , + b e , = K,i^
And the motor equation
Find To
V„ = l^R, + I^R^^■ K,
J^s -I- b
Substitute in the equation
R^
5 ,1 .* ■
y —0
R^
V —0 V —0
------ 1— = 0 and sinqjlify.
Rj
R
• J ,s ^ b )
R
This e^^ression shows that, in the steady state, when s —> 0 , the current is proportional to
the ii^u t voltage.
We know that the current anq^lifier has no feedback from the output voltage, in which
And thus, we get
U _
y>.
R
Thus, the transfer fimction is obtained as ^ - - M
U
l
R
Problem 2.14PP
An op-amp connection with feedback to both the negative and the positive terminals is shown in
Fig. 1. If the op-amp has the nonideal transfer function given in Problem, give the maximum
value possible for the positive feedback ratio, ' ’ = 7Tir' in terms of the negative feedback ratio,
N =
for the circuit to remain stable.
Figure 1 Op-amp circuit
A first step toward a realistic model of an op-amp is given by the following equations and is
shown in Fig. 2:
10^
i+ = i-= 0 .
Figure 2 Circuit for Problem
Find the transfer function of the simple amplification circuit shown using this model.
Step-by-step solution
step 1 of 4
Step 2 of 4
—
Clearly
E„
r+ R
E,
• [r „
r ,J L e ,+ e „ J
r]
V.
rt-R
Step 3 of 4
If the Op - Amp is non ideal then.
A(say)
l ,
V„ = A M
E ,+
I , r + R j j R (+ R k R
,^ R „
-AR,
E ,+ E „
I,
R ,+ R ,. R + rJ
Step 4 of 4
In the limiting case we put denominator = 0
We get 1+
R*,A
rA
R,4-Rj^
R+r
1+NA=PA
1=(P-N)
10’
's + 1
s + U C P - Al)10’
P oleisat[(P -N )10’ -l].
Equating it to Zero for limiting case,
,1+N.lO ’
We have P S 10'
Max value is |N+10'^
P ro b le m 2.15PP
Write the dynamic equations and find the transfer functions for the circuits shown in Fig..
(a) Passive lead circuit
(b)
Active iead circuit
(c)
Active lag circuit
(d)
Passive notch circuit
Figure (a) Passive iead; (b) active lead; (c) active lag; and (d) passive notch circuits
Figure (a) Passive iead; (b) active lead; (c) active lag; and (d) passive notch circuits
—W ^ ^ V V V - j —
-o * U
0 - i—v v \
I
t
vvnJ
- o
K/2> ^ 2 C
S-------------- A
A
o
S te p -b y -s te p s o lu tio n
(a)
Refer to Figure 2.48 (a) in the textbook for passive lead circuit.
Write the KirchhofTs current law equation at node y .
Q -+
(ri)
|>- = C B + -^ ii
(1)
Therefore, the dynamic equation is
Apply Laplace transform on both sides.
> { i ) = c » i ( i ) + — » (» )
/ X
Cl+ —
^
U
u(s)
Therefore, the transfer function. Z i £ l i
ii( s )
(b)
Refer to Figure 2.48 (b) in the textbook for active lead circuit.
Draw the circuit diagram with node voltages.
R
H
^
Figure 1
Write the KirchhofTs current law at node y .
From the circuit,
y - y
0 - r .
Substitute K i + ' ^ f ^ f o r y in equation (2).
“
.... (3)
R, “
Therefore, the dynamic equation is
— + - ^ V
[ r,
■*/
“
-
R, “
Apply Laplace transform on both sides of equation (3).
y ^ (s )= -c s K .(s )-± y ^ (s )
HI
fk W
K
4-ife]
/i,C
^
s J - L ^ - L ]
U c
R.C
*r
Therefore, the transfer function. U f ) :
K,
R,C)
U C
(C)
Refer to Figure 2.48 (c) in the textbook for active lag circuit.
Draw the circuit diagram with node voltages.
R^
R,
-VcMt
Write the KirchhofTs cument law equation at inverting terminal.
Z.
- ii.
K
Write the KirchhofTs cument law equation at node y .
Substitute - Z l k
K
for y .
S,
d l[ 1 1 , “
-)
K,
- ^ y ^ - ^ - c ^ K - c v „ ~ v , = o
c v «•
, . + ^0 = - c 0
^ v* , >- —
v, —
^ vW ,
0 M
0 0
c r „ + ^ = - c - ^ r „ - — ri+ -^ ^ ir„
R,
^
J !|j ^
(4)
Therefore, the dynamic equation is c v ^ , y ^ = ^ ^ v , . ±
“
It,
K
y^
.
It,.
Apply Laplace transform on both sides.
^C s +
K.
n .(4
til
■
/J,Cs+
Therefore, the transfer function. U f ) :
til
step 8 of 11 A
(< i)
Refer to Figure 2.48 (d) in the textbook for passive notch circuit.
Draw the circuit diagram with node voltages.
Write the KirchhofTs cument law equation at node P,.
Apply Laplace transform.
C t lK W - r . w ) + ^ + c i ( F ; w - F „(» )) = 0
= a K „W + c iK ,,( » )
Write the KirchhofTs cument law equation at node P,.
R
d t'
'
R
Apply Laplace transform.
R
+ 2 C s K ( i) +
— 2 < i_ i = 0
R
' '
Step 10 Of 11
From the circuit diagram.
Apply Laplace transform.
step 11 of 11
Substitute p;(« ) and ^ j( ^ ) in the equation.
f _ « ^
K .U )
(.2 ^ +2
RCV
2R O + 2
+ ____ I____ ]
2R + 2R^Cs)
s-HlJ
2R +2R‘
, ____ !____ ]
f
Therefore, the transfer function. U f ) :
2 ^0 +2
RCV
2RCS+2
2R+2R‘ C sJ
2R +2R‘
Problem 2.16PP
The very flexible circuit shown in Fig. is called a biquad because its transfer function can be
made to be the ratio of two second-order or quadratic poiynomiais. By selecting different values
for Ra. Rb, Rc, and Rd. the circuit can realize a low-pass, band-pass, high-pass, or band-reject
(notch) filter.
(a)
Show that if Ra = R and Rb = R c= R d=
the transfer function from Vin to Vout can be
written as the low-pass filter
A
where
A=
« i’
' RC*
R
(b)
Using the Matlab command step, compute and plot on the same graph the step responses fo
the biquad of Fig. for A = 2, wn = 2, and ^ = 0.1, 0.5, and 1.0.
Figure Op-amp biquad
Step-by-step solution
step 1 of 12
Refer Figure 2.52 in textbook.
Step 2 of 12
Consider the transfer function of the circuit
^ = -C K ,
(2)
R
=
...... (3)
2 i+ il+ iL + iiL ._ f i!.
^
R^ R,
(4)
R ^^~ R
Step 3 of 12
Take Laplace transform in Equation (1),{2) and (3).
......
^ = -CsV^
R
(6)
...... (7)
Step 4 of 12
Determine
and
Substitute —
in Equation (5) and (6).
for
in Equation (5).
(8)
U
) '
s.
■ ^ r ,+ a v ,= o
R
(9)
Step 5 of 12
From Equation (8) and (9) write the matrix form.
1
[-L .a - i l
Rj
R
’"*
0
t
R
Cs
R]-r
'
1
R
"Ji, ‘
1
— +Cs
0
^
j
1
- -
[ «
~SV^
S,
I
C s + — S+-rT
R,
R ^ [ rr,
Consider the value of F, and
C
( 10 )
y ,-
1
RR,
( 11)
y ,= -
c v + ^ j+ ^
R^
f i'
Step 6 of 12
Substitute
for
in Equation (4).
.L -
R
^
^
^
i-H h -kH "Substitute Equation (10) and (11).
R
{ R.
R ,) ^
R, '
R, “
I
r
I
V
'K
K
Rt
K?
}
'
f
f
c
1
y
‘
(
1
RR,
1 + l j
'
c
1
I
K ,”
C V + A j+ '
1 J?.
1,
1
*.
1
*
' n.
J
step 7 of 12
C
1
RR,
’ R
K
C V + -^ s+ -^
R^
'
R‘
c v + - ^ j+ - ^
«,
R\
I
f
RR,
c
I
I
=-R f - ' + ' 1
1 K Jt.) C V + A i + ^
1
C ij '+ A ^ + J ^
«‘ J
1
=- R
L J_±
RR ,* R^R^
4* + —^ 5 + ----
(RC)^
Step 8 of 12
Thus, the transfer function of the system is.
c'^
^ j’ +f '
R Rj
R^ R2
^ )
C'
y^
R,C
{R C f
Step 9 of 12
(a)
Substitute
for J^, co for J^, oo for J^and oo for R^.
1 1
R
~ RRR,
y^
Rfi (Rcy
RR,C’
R fi
(R c y
j?
_5_
(« C )’ i ’ + ^ j + l
_R
A.
y.
. (
12 )
(« C )’ i ’ + ^ i + l
A _
Thus, the transfer function of the system is
__________ ^
Step 10 Of 12
Compare Equation (12) and 2.111 in Textbook.
J ____ l_
RC
Step 11 of 12
(b)
Substitute 2 for A, 2 for
and 0.1 for g in Equation 2.111 in textbook,
write the MATLAB program.
A = 2;
wn = 2;
z = [0.1 0.5 1.0]:
hold on
for i = 1:3
num = [ A ];
den = [ 1Avn''2 2*z(i)/wn 1 ]
step( num, d e n )
end
hold on
Step 12 of 12
Figure 1 shows the output for the MATLAB program.
Figure 1
Thus, the step responses using MATLAB is shown in Figure 1.
J
Problem 2.17PP
Find the equations and transfer function for the biquad circuit of Fig. if Ra = R ,R d = R1. and Rb =
Rc=-^.
Figure Op-amp biquad
Step-by-step solution
step 1 of 4
Step 2 of 4
Given: R ,= R i,
R v= R ,=oo
R ,= R .
Using property o f Inverting Amplifier.
V,
RsC
'
^ = -1
V,
Writing Node Equations,
at V :
Step 3 of 4
Also ^ + ^ + ^ r i + s R j ) = 0
R,
R
R /
R,
Ri
R
V „+ V , = - ^ V „
V,
V ,=
RsC
V ,= R sC V,
R,
’I
R
Rs
J
1 1 ^ RsC +R R^^C ^’l
W
Rs
r
^
~ Ri
Step 4 of 4
[■r J+R’ sC+R’ R / C ? Y Vfc
' I
RRs
J R.
R R , V.,
’ R ^ j+ R ,R " s C + R ,R " R js " C ?
R , V_
V J._______ R
^-^2
^ R ,R j +R,R^ sC+R,R’ R js’ c '
-R, v
r
V „
-R R ,R ,+ R R ,+ R ,R ’ sC+R,R’ R ,s“C?'
Vi
R.
R iR j+ R iR ’ sC+R,R’ s^C?
R i ^ sV „ + R i ’ R ’ sC V „ + R i ^ ’ s^C ?V „+R R i R jV4
+R “ R jV j,+ R ’ R is C V ^ + R iR ^ js “ C ? V i= 0
Problem 2.18PP
The torque constant of a motor is the ratio of torque to current and is often given in ounce-inches
per ampere. (Ounce-inches have dimension force x distance, where an ounce is 1/16 of a
pound.) The electric constant of a motor is the ratio of back emf to speed and is often given in
volts per 1000 rpm. In consistent units, the two constants are the same for a given motor.
(a) Show that the units ounce-inches per ampere are proportional to volts per 1000 rpm by
reducing both to MKS (SI) units.
(b )A certain motor has a back emf of 25 V at 1000 rpm. What is its torque constant in ounceinches per ampere?
(c) What is the torque constant of the motor of part (b) in newton-meters per ampere?
(c) What is the torque constant of the motor of part (b) in newton-meters per ampere?
Step-by-step solution
step 1 of 4
(a)
9 .8 ^ x lk g =
2.205 Pounds
9 .8 N = 2 .2 0 5 x 16 Ounces
9.8 n = 35.28 Ounces
1 Ounce = 0.0283 kg x 9.8 “ = 0.277 N
s
H hch =2.54 cm = 0.0254 m
1 - O unce-inch = 7.036 x 10"’ N-m
1 —O u n ce-Inch per Amp =7.036 X 10"® N -m /a
(-)
Step 2 of 4
30
1 volt per 1000 amp = — xlQ*^ V o lt/ra d /s
n
rad.
= 9.55
= 9.55x10-®
c.rad
Nnn.s
c/s
1 V o ltp e rl0 0 0 fp m = 9 .5 5 x l0 ^
a
1 Ounce-inch/Amp 7.036 _
^
1 VoltperlOOOrpm 9.55
■M
Step 3 of 4
(B)
K , = 25 f — - — )
^
l^lOOOrpm)
Using equation (y)
1 V /1000 rpm = 1.358 Ounce - inch per Ampere
.'. Kj = 25 x l.3 5 8 = |33.9^ O unce-inch per Ampere
Step 4 of 4
(C)
.'. Kt = 33.96 Ounce - inch per Ampere
Using equation (a )
K t =33.96x7.036xlOr^N-m/A
= 239x10-® N-m/A
= |0.239N-m/A|
Problem 2.19PP
The electromechanical system shown in Fig. represents a simplified model of a capacitor
microphone. The system consists in part of a parallel plate capacitor connected into an electric
circuit. Capacitor plate a is rigidly fastened to the microphone frame. Sound waves pass through
the mouthpiece and exert a force fs (t) on plate b, which has mass M and is connected to the
frame by a set of springs and dampers. The capacitance C is a function of the distance x
between the plates, as follows;
where
£ = dielectric constant of the material between the plates,
A = surface area of the plates.
The charge q and the voltage e across the plates are related by
A = surface area of the plates.
The charge q and the voltage e across the plates are related by
» = C(i)e.
The electric field in turn produces the following force fe on the movable plate
that opposes its motion;
fe = 2eA'
(a) Write differential equations that describe the operation of this system. (It is acceptable to
leave in nonlinear form.)
(b) Can one get a linear model?
(c)
What is the output of the system?
Figure Simplified model for capacitor microphone
Step-by-step solution
step 1 of 6
W e h av e liie follow ing electxom edianical system :
Step 2 of 6 ^
G iv en th at capacitance C is a fooctioo o f th e distance x betw een d ie plates, a s follows:
C (x )= H ere. € is foe c b d ectiic constant o f d ie m aterial betw een die plates,
A is foe sur& ce a rea o f foe plates
T h e charge q a nd foe voltage « across foe plates are r d a te d b y
9 = C (x )«
T h e electric field in tu rn produces foe follow ing force
o n fo e m ovable p late foat
opposes its m otkm :
Step 3 of 6
T h e fie e b o d y diagram o f f o e c ^ n c i f o r p late (b )ii
b
- f
Step 4 of 6
T h e equation c fx notion fo r fo e above plate. 6 is
T h e equation f ix foe circuit p art is
at
( Since e is fo e
^
^ o fia c itiw o lta g e J
X
dt
d i‘
I S in c e i( /) = ^ j
eA
v = J 8 j+ l9 + .^
eA
T herefore d ie tw o coiqiled n o n linear
equations are
i & + B i+ K x + - ^ = / A t ) and
le A
" ’
v = R q -V L q + ^
Step 5 of 6 ^
(b)
W e cannot g e t a linear m odel because c£ d ie term s ^
differential equations.
Step 6 of 6
(c)
T h e output o f foe system is foe current |r ( t ) = g |
a n d qx p resen t in th e
Problem 2.20PP
A very typical problem of electromechanical position control is an electric motor driving a load
that has one dominant vibration mode. The problem arises in computer-disk-head control, reel-toreel tape drives, and many other applications. A schematic diagram is sketched in Fig. The motor
has an electrical constant Ke. a torque constant Kt, an armature inductance La. and a resistance
Ra. The rotor has an inertia J1 and a viscous friction B. The load has an inertia J2. The two
inertias are connected by a shaft with a spring constant k and an equivalent viscous damping b.
Write the equations of motion.
Figure Motor with a flexible load
R,
Step-by-step solution
step 1 of 2
Refer to Motor with a flexible load in Figure 2.51 in the textbook.
According to Newton’s second law of motion, the sum of torques acting on a rotational
mechanical system is zero. The sum of applied torque is equal to the sum of opposing torques on
a rotational system.
Write the equation of motion for a motor.
J A * B 0 ,+ b (i^ -e ,)+ iL { e ,-e ,)= T ,
J A = - B e ,- b ( g ,- e , ) - k ( e .- e , ) + T ,
(d
Here,
is the inertia of the rotor
^ is the viscous damping
g is the viscous friction
is the spring constant
is the motor torque
Write the equation of motion for a motor.
j A 2 = - b { A - e ,) - k ( e .- e . )
(2)
Step 2 of 2
Apply KirchhofTs voltage law.
Here,
is the electrical constant
It is known that, the output torque is directly proportional to the armature current.
T .= K ,l,
Here,
K , is the torque constant
Therefore, the equations of motion are.
Problem 2.21 PP
For the robot in Fig., assume you have command of the torque on a servo motor that is
connected to the drive wheels with gears that have a 2:1 ratio so that the torque on the wheels is
increased by a factor of 2 over that delivered by the servo. Determine the dynamic equations
relating the speed of the robot with respect to the torque command of the servo. Your equations
will require certain quantities, for example, mass of vehicle, inertia, and radius of the wheels.
Assume you have access to whatever you need.
Figure Robot for delivery of hospital supplies. Source: AP Images
Step-by-step solution
step 1 of 13
Refer Figure 2.45 in textbook.
Step 2 of 13
Assume that robot has no mass. So multiply the torque by a factor of 2.
Step 3 of 13
The motor must have a gear that is half the size of the gear attached to the wheel.
Let’s also assume there is no damping on the motor shaft.
Step 4 of 13
So, friction
and
are both zero.
Step 5 of 13
Consider the wheel attached to the robot
Where,
y^is the inertia of the drive wheel,
is the motor inertia,
is the wheel angular acceleration.
T is the commanded torque from the motor.
Step 6 of 13
The acceleration of the drive wheel is directly related to the acceleration of the robot and its other
wheels, provided there is no slippage. Add the rotational inertia of the two other wheels and the
inertia due to the translation of the cart plus the center of mass of the 3 wheels.
Step 7 of 13
Let the angular acceleration is
and assume the inertia is same as the drive wheel.
Step 8 of 13
Neglect the translation inertia of the system and write the equation
Step 9 of 13
Consider friction force.
f
Where,
is the mass of the cart plus all three wheels.
Step 10 of 13
Substitute r § for a.
Step 11 of 13
Consider angular inertia
Step 12 of 13
Substitute 2 for n.
( m „ r j + 3 7 ^ + 4 7 ^ ) ^ , = 2T^
Step 13 of 13
Thus, the dynamic equation relating the speed of the robot is
+ 3 7 ^ + 4 7 .) ^ ^ =27^
Problem 2.22PP
Using Fig., derive the transfer function between the applied torque, Tm. and the output. 62, for
the case when there is a spring attached to the output load. That is, there is a torque applied to
the output load. Ts. where Ts = -Ks92.
Figure (a) Geometry definitions and forces on teeth; (b) definitions for the dynamic analysis
“0 ^
Step-by-step solution
step 1 of 13
Refer Figure 2.35 in textbook.
Step 2 of 13
The force transmitted by the teeth of one gear is equal and opposite to the force applied to the
other gear as shown In Figure 2.35 (a) In textbook.
torque-ftwcexdistance
Step 3 of 13
Consider the torques applied to and from each shaft by the teeth
ZL.ZL
i =
/
Step 4 of 13
Consider the torque multiplication Is proportional to the radius of the gear
T, r,
r,
N,
Where,
N is the number of teeth.
n is the gear ratio.
Step 5 of 13
The velocity of the contact tooth of one gear is the same as the velocity of the tooth on the
opposite gear
velocity=o»’
Where.
€0 is the angular velocity.
r is the radius of the gear.
Step 6 of 13
The angles will change in proportion to the angular velocities
e, o,
Oi
AT,
N,
JV,
The servo motor output torque is
attached to gear 1. So the servo’s gear 1 is meshed with
gear 2 and the angle ^ Is position (body 2). The Inertia of gear 1 and all that is attached to body
1 is J^. The inertia of gear 2 and all that is attached to body 2 Is J^.
The friction ^ and
.
Step 7 of 13
Consider the equation of motion for body 1.
■ f A + W - T ,- T ,
(1)
Where,
T Is the reaction torque from gear 2 acting back on gear 1.
Step 8 of 13
The spring is only applied to the second rotational mass so torque only effects. Adding the spring
torque in body 2.
Step 9 of 13
Consider the equation of motion for body 2.
.........( 2 )
Where.
is the torque applied on gear 2 by the gear 1.
Step 10 of 13
Substitute
for ^ In Equation (1).
(3)
step 11 of 13
Substitute 7J for 7*, in Equation (2).
.......(4)
Step 12 of 13
Use Equation (3) and (4) the relationship in Equation 2.73 in textbook and eliminating 7J the two
equations
(y , + J,n')Si +
r ,( » )
(
i , +K,ff, =nT,
( j j + J , n ’ ) s ^ + ( b ,* b ,a ’ '^s + K ,
S te p 1 3 o f1 3
Thus, the transfer function of the system is
( J", +
Problem 2.23PP
A precision table-leveling scheme shown in Fig. relies on thermal expansion of actuators under
two corners to level the table by raising or lowering their respective corners. The parameters are
as follows:
Tact = actuator temperature,
Tamb = ambient air temperature.
Figure (a) Precision table kept level by actuators; (b) side view of one actuator
1 ,
(b)
R f= heat-flow coefficient between the actuator and the air,
C = thermal capacity of the actuator,
R = resistance of the heater.
Assume that (1) the actuator acts as a pure electric resistance, (2) the heat flow into the actuator
is proportional to the electric power input, and (3) the motion d is proportional to the difference
between Tacf and Tamb due to thermal expansion. Find the differential equations relating the
height of the actuator d versus the applied voltage vi.
Step-by-step solution
step 1 of 2
Step 2 of 2
Proportionality
d =
Power
551
In Put
R
And using equation
We get
HEAT
m l
R
2.73
C 7L +
cr„+
d
Rtqd . Bq.
Problem 2.24PP
An air conditioner supplies cold air at the same temperature to each room on the fourth floor of
the high-rise building shown in Fig.(a). The floor plan is shown in Fig.(b). The cold airflow
produces an equal amount of heat flow q out of each room. Write a set of differential equations
governing the temperature in each room, where
To = temperature outside the building,
Ro = resistance to heat flow through the outer walls,
Ri = resistance to heat flow through the inner walls.
Assume that (1) all rooms are perfect squares. (2) there is no heat flow through the floors or
ceilings, and (3) the temperature in each room is uniform throughout the room. Take advantage
of symmetry to reduce the number of differential equations to three.
Figure Building air-conditioning; (a) high-rise building; (b) floor plan of the fourth floor
ui syiiiineiiy (OTeuube me iiuinuei ui uiiieteiiiiai equauuiis lu iiiiee.
Figure Building air-conditioning; (a) high-rise building; (b) floor plan of the fourth floor
Step-by-step solution
step 1 of 2
Step 2 of 2
“C" Thermal cs^acity o f Air
*'1” Corners are equivalent
“2” Arc equivalent
Assuming; ^ >7J > ^ > 7 J
For * 3"
F o, -2 -
-
( 1)
=
--------- (2)
s,
F o r-r
=
RequideEquations are (l)& (2 )& (3 )
R,
+
Ro
-(3 )
Problem 2.25PP
For the two-tank fluid-flow system shown in Fig., find the differentialequations relating the flow
into the first tank to the flow out of the second tank.
Figure Two-tank fluid-flow system
Step-by-step solution
step 1 of 2
From the relation between the height of the water and mass flow rate, the continuity equations
are
(3)
M
Here,
^ Is the area of tank 1.
p is the density of water.
is the height of water in tank 1.
And,
= w -w ^
p A ji^ ■
...... (4)
PA i
Here,
is the area of tank 2.
p is the density of water.
Is the height of water in tank 2.
Step 2 of 2
From the relation between the pressure and outgoing mass flow rate.
(5)
(6)
I
I
Substitute
for w In equation (3).
A = -L f4 -i
I
Substitute
I
I
for w and —
i
f or
in equation (4).
p ^ W
(8 )
Therefore, the differential equations relating the flow into the first tank to the flow out of the
second tank are.
Problem 2.26PP
A laboratory experiment in the flow of water through two tanks is sketched in Fig. Assume that
Eq. describes flow through the equal-sized holes at points A, B, or C.
(a)
With holes at B and C. but none at A, write the equations of motion for this system in terms o
h1 and h2. Assume that when /?2 = 10 cm, the outflow is 200 g/min.
(b)
At h1 = 30 cm and h2= ^0 cm, compute a linearized model and the transfer function from
pump flow (in cubic-centimeters per minute) to h2.
(c)
Repeat parts (a) and (b) assuming hole B is closed and hole A is open. Assume that h3 = 20
cm, h i > 20 cm, and h2 < 20 cm.
Figure Two-tank fluid-flow system for Problem 2.26
cm, h i > 20 cm, and h2 < 20 cm.
Figure Two-tank fluid-flow system for Problem 2.26
Pump
Step-by-step solution
step 1 of 5
(a)
Step 2 of 5
Refer FIGURE 2.58 in the textbook.
The area of both tanks is A, the values given for the heights ensure that the water will flow is
below as.
(1)
( 2)
(3)
W ,-W ^ = pAh,
W ^-W g = p A h ^ ...... (4)
Consider the value of gravity.
g=981cm /sec*
slOOOdVsec^
Substitute 2 0 0 g /m iifo r
for ^a n d lOOOcn^sec* for g in equation (2).
200 g/inn = -!-[lg ia m /c c x lOOOcn^sec^ x lO c m p
1
T—
Jg = 2oo / — [lg ra m /c m * x l0 0 0 c m /s e c * x l0 c m J
1
J ts ^Igram /cm ^xlO O O cm /sec’ xlO cm
200g/60s
100
/g c m “ s>
J?s30 g^cm 2
30g ^cm
Step 3 of 5
(b)
Consider the following nonlinear equation from above information.
(5)
(6)
Consider the linearized equation.
(7)
..........( 8 )
Substitute equation (7) and equation (8) in equation (5).
1
/— 7Z— 7:— r —
Substitute 30 cm for
i
10 cm for Aj,, 100 cm2 for A, 1 for /?, IQOQcip/sfC^ for g and 30 for R
in equation (9).
-^(1)(I0O0)(30+5*i -1O-5^)h(1)(100)
(1)(100)(30)'
Sh,=
s k = - 2 ^ fi+ - L tfk — L t f t V J - K
’
30i, 40 ’ 40
( 10 )
100 "
Consider the total inflow equation.
W ^ = W ^ + S W ...... (11)
Consider the value of nomina! Inflow Is '^Iven below
10
=y>^cm V sec
Substitute equation (11) in equation (10).
’
1
40
3 0 l,
- - L a A .l+ - L ( » '
40
lo o ' ”
’
Substitute y - v ^ c m ’ /sec for
+ ^ » ')
’
.(12)
in equation (12).
(13)
S k— —
< S h ,-S h ,)+ — m
’
120 0' ’
100
Substitute equation (7) and equation (8) in equation (6).
.(14)
Substitute 30 cm for
10 cm for Aj,, 100 cm2 for A, 1 for p , lOOOcm/sec^ for g and 30 for R
in equation (14).
1
r ^ { l) ( im ) ( 3 0 + S h , - lO - S h j) (1 )(1 0 0 )(3 0 )'
'
^ ( l)( 1 0 0 0 )( 1 0 + 5 ^ )
(1)(100)(30)
1
(15)
From equation (13) and equation (15), holding the nominal flow rate maintains h i at equilibrium
but h2 will not stay at equilibrium. So, the constant term increasing h2.
Hence, the standard transfer function will not result.
Step 4 of 5
(c)
With hole B closed and hole A, the values given for the heights ensure that the water will flow is
below as.
W ^ - W ,= p A h ,
W ,-W ^ = pA h,
Consider the following nonlinear equation from above information.
(18)
.(19)
Substitute equation (7) in equation (18).
( 21 )
Substitute 30 cm for Aj^, 20 cm for A^, 100 cm2 for A, 1 for p , lOOOctn/sec^ for g and 30 for R
in equation (21).
S k ,= -
’
= - - L h + J -^ fc l+ J - W
30l,
20
( 22 )
100 *
Substitute equation (11) in equation (22).
^
(23)
= _ J _ | 1 + J _ ^ fc 1+J _ ( ( F
20 V 100^ " ■
^
Consider the value of nominal inflow is remc
301,
20
s k - —
’
J
100
+ — S W ...... (24)
600 ’ 100
Substitute equation (7) in equation (19).
(25)
Substitute 30 cm for Aj^, 20 cm for A^, 100 cm2 for A, 1 for p , lOOOctn/sec^ for g and 30 for R
in equation (25).
1
r^(l)(1000)(30+5A| -2 0 ) (1)(100)(30)'
'
(1)(100)(30)
^(l)(1000)(10+5^)
30l, 20
S k -—
’
600
30l. 20 ’ j
Sh,— — S h , ...... (26)
’
600
’
Take Laplace equation (24) and equation (26), the desired transfer function in cmYsec •
3 H ,{ s ) ..
. m (s)
7
T1
(27)
V 60oJ
f n ,i‘
pit
(28)
I, 600j
Step 5 of 5
Substitute equation (27) in equation (28).
1
^6003 f : 7 X j r , . ^ _ L )
l.
60o A
600j
S H j( s )
1
0.01
•Y
l. 600j
Convert the inflow unit in grams/min •
<?//;(»)_1 (0.0I)(60)
e w (s) ° 6 0 o r
1 V
r*6 0 o J
0.001
0.001
Thus, the desired transfer function is
("iJ
Problem 2.27PP
The equations for heating a house are given by Eqs. (1) and (2), and in a particular case can be
written with time in hoi/rs as
dt
= jr* — HLZlZe
R *
where
(a) C is the thermal capacity of the house, BTU/*F,
(b) Th is the temperature in the house. “ F,
(c) To is the temperature outside the house. *F.
(d) K is the heat rating of the furnace, = 90. 000 BTU/h,
(e) R is the thermal resistance. °F per BTU/h,
(f) u is the furnace switch. = 1 if the furnace is on and = 0 if the furnace is off.
It is measured that, with the outside temperature at 32°F and the house at 60°F, the furnace
(f) u is the furnace switch. = 1 if the furnace is on and = 0 if the furnace is off.
It is measured that, with the outside temperature at 32°F and the house at 60°F, the furnace
raises the temperature 2*F in six min (0.1 h). With the furnace off, the house temperature falls
2°F in 40 min. What are the values of C and R for the house?
Equation 1
q = ^ ( T i- T 2 ) ,
where
q = heat-energy flow, joules per second (J/sec)
R = thermal resistance, *C/J ? sec
T = temperature, “C.
Equation 2
t.y .
where C is the thermal capacity.
S te p -b y -s te p s o lu tio n
step 1 of 3
CdZ
dt
R
RTUlhr
90.000
Step 2 of 3
(0
7^ = 32^ F ,
7J=60"i?
u=l
^ = 2 < f F lh r
+ 2 0 c + — =90.000 -------R
(1)
'■ '
Step 3 of 3
(ii)
T,=32‘ F
.
T, = 60’ F
a=0
dt
= 2 F lh r
- 3 c + — = 9000
R
fiC = 2 8 / 3 --------
(2)
Putting (i) we get
i^ 5 ^ + 2 8 = 9 0 , 0 0 0 / i
5 = 2.385x10-® F lB T U jh r
C=
3.913x10® B T U l-F
Problem 3.01 PP
Find the Laplace transform of the following time functions:
(a )/(f)= 1 + 7 f
(b) fl;f) = 4 + 7f + f2 + 6(t). where 5(t) is the unit impulse function
(c) fl;0 = e - f + 2 e - 2 t + t e - 3 t
(d)
=
(e) m = sinh t
c C ontrol O f Dynamic Systems (7th Edition)
P ro b le m
Find the Laplace transfomi of the following time functions:
(a )/(f)= 1 + 7 f
(b) fl;f) = 4 + 7f + f2 + 6(t), where 5(t) is the unit impulse function
(c) f{f) = e - t + 2 e - 2 t+ te - 3 t
(d)
=
(e)
= sinh t
(e) f{f) = sinh t
S te p -b y -s te p s o lu tio n
step 1 of 12
(a)
Step 2 of 12
Consider the time function / ( 0 = l+ 7 r
Take Laplace transform.
1
7
s+7
Thus, the Laplace transfomi is
s+7
Step 3 of 12
(b)
Step 4 of 12
Consider the time function / ( r ) = 4 + 7 f + r * + ^ ( r )
step 5 of 12
Take Laplace transform.
s + \ s+2
(5+3)‘
Thus, the Laplace transfomi is
5+ 1 5+ 2
(5
+ 3 )*
Step 8 of 12
(d)
Step 9 of 12
Consider the time function.
« /’ +2/ + l
aiep iu or iz
Take Laplace transform.
i[ / ( / ) ] - L ( / ') + i( 2 / ) + L ( l)
2!
2
I
=-r + —+j s
s* + 2 s + 2
Problem 3.02PP
Find the Laplace transform of the following time functions:
(a )/(f)= 1 + 7 f
(b) fl;f) = 4 + 7f + f2 + 6(t). where 5(t) is the unit impulse function
(c) fl;0 = e - f + 2 e - 2 t + t e - 3 t
(d)
=
(e) f(f) = sinh /
(e) f{f) = sinh t
Step-by-step solution
step 1 of 12
(a)
Step 2 of 12
Consider the time function / ( 0 = l+ 7 r
Take Laplace transform.
1
7
s+7
s+7
Thus, the Laplace transfomi is
Step 3 of 12
(b)
Step 4 of 12
Consider the time function / ( r ) = 4 + 7 f + / * + ^ ( r )
step 5 of 12
Take Laplace transform.
4 7 2 !,
= - + — + — +1
s s
s
«’ +45* + 7j + 2
+7^+2
Thus, the Laplace transfomi is
Step 6 of 12
(c)
Step 7 of 12
Consider the time function f { l ) = e " + 2 e ' ^ +16'”
Take Laplace transform.
1
2
1
- + -------+ s + \ s + 2 (5 + 3 )'
Thus, the Laplace transfomi is
5+1 5+2
(5 +3 )*
Step 8 of 12
(d)
Step 9 of 12
Consider the time function.
«/’ +2/+l
aiep iu or iz
Take Laplace transform.
i[ / ( / ) ] - L ( / ') + i( 2 / ) + L ( l)
2!
2
I
=- + — + r
j
s
s * + 2 i+ 2
Thus, the Laplace transfomi is 5*+25+2
Step 11 of 12
(e)
Step 12 of 12
Consider the time function.
/(/) * s in h r
Take Laplace transform.
1
*5 *-l
Thus, the Laplace transfomi is
5 ^ -1
Problem 3.03PP
Find the Laplace transform of the following time functions:
(a) f{t) = 4 cos 6f
(b) ^t) = sin 3f + 2 cos 3f + e -t sin 3f
(c) f{f) = t2 + e -2 t sin 3t
Step-by-step solution
Step-by-step solution
step 1 of 6
(a)
Step 2 of 6
Consider the time function / ( / ) = 4 c o s 6 f
Consider
£(cos<tf)=- j ---;•
S + fl
Take Laplace transform.
s4s^+ 3 6
Thus, the Laplace transfomi is
Step 3 of 6
(b)
Step 4 of 6 ^
Consider the time function / ( / ) =
Consider
£(smo/)g
sin3f+2cos3/+e*'$in3/ •
,•
S +<r
Take Laplace transform.
l(/(r)) =i(sin3/)+i(2cos3r)+i(«‘'sin3/)
3
2s
3
-^ + -:
j'+9r + s'+9
(s+l)’+9
Thus, the Laplace transfomi is
2s
s'+9 s'+9 (s+l)"+9
Step 5 of 6
(C )
step 6 of 6
Consider the time function / ( » ) = « ’ +«■“ sin 3»
Take Laplace transform.
£[/(/)]■ L(/*)+£(«■*sin3/)
2!
3
*- r +
s’ (s+2)’ +9
=A
3
°s’ '^(s+2)’+9
2
Thus, the Laplace transfomi is - r +
s’ (s+2)’ +9
Problem 3.04PP
Find the Laplace transform of the following time functions:
(a) f{t) = / sin /
(b) f{t) = t cos 3f
(c) f{f) = t e - t + 2 t c o s t
(d) f[t) = ts\n 3 / - 2 f c o s f
(e)/(0=1(fJ + 2/cos 2f
(e) /(0 = 1(f)+ 2 / cos 2f
Step-by-step solution
step 1 of 9
(a)
Consider the time function,
Apply the Laplace transfomi on both the sides of the time function.
Since there are two variables in the above function we need to use the multiplication by time
Laplace transform property. In order to do so. we solve the given time function by separating
them.
First, consider ^ ^ /^ = s in r and find the Laplace transform of it.
Step 2 of 9
Now. find the Laplace transform for the whole function.
^ {« (0 } = —
G (>)
f ( . » - . l) ( 0 ) - ( l) ( 2 . ) l
(.-.I)-
J
2s
( s - . lf
2s
j‘
+ 2j ’ +1
Therefore, the Laplace transfomi of the time function, /s in / is
s* + 2s^ + \
Step 3 of 9
(b)
Consider the time function, /]^;r^= to a B 3 f
Apply the Laplace transfomi on both the sides of the time function. £ [ f ( i ) ) = £ { tc o s 2 i\
Since there are two variables in the above function we need to use the multiplication by time
Laplace transfomi property. In order to do so. we solve the given time function by separating
them.
First, consider g ^/^ = cos3/ and find the Laplace transform of it.
Step 4 of 9
Now. find the Laplace transform for the whole function.
£ { ,g ij) } = - ± G ( s )
-s ‘ -9 + 2 s‘
° j ' +18s’ +81
s ^ -9
s'+\Zs'*% \
Therefore, the Laplace transfomi of the time function, /cos3/ is
4 ^ -9
Step 5 of 9
(C)
Consider the time function,
+ 2 tc o B t
Appiy the Lapiace transfomi on both the sides of the time function.
/:{/(/)} = £ { le " +2fcost}
F{s) = £ { le " } +2r {/cos/}
1
( s ‘ + 1 ) - s (2 j )
r-2
(*+ > r
__ i_
-2
s’ - 2 i ' + I
Step 6 of 9
Further simplification gives.
1 -s ’
F (s )
{s ^ iy
____1 , ^(^‘ - 0
~ { s ^ l f (s'+lf
(s’ + 1 ) '+ 2 ( s + 1)’ (s* - 1)
(s+l)’(s*+l)'
(s*+2j’ +1)+(j^+2s+l)(2s*- 2)
(s“+2s+l)(i* +2s’ +l)
^ s * + 2 s ^ + l + 2 s * - 2 s * + 4 s * - 4 s + 2 s '- 2
s‘ + 2s*+s’ + 2s*+4i’ + 2s+s*+ 2i“ + 1
3s* + 4 s ’ + 2 j " - 4 s - I
" j ‘ + 2s’ +3s*+4s’ +3s’ + 2j + 1
Therefore, the Laplace transform of the time function, te^ + 2 tc o s t '®
3 s *+ 4 s’ + 2 s ^ - 4 s - l
Step 7 of 9
(d)
Consider the time function,
Apply the Laplace transfomi on both the sides of the time function.
/■ { /{ /)} =
COS/ }
F ( j ) = ^ { t s in y ) ~ 2 jC {/c o s /}
,
j
( 2 £ L + 2 [ ( £ i 1)z ! E ) 1
r+2
**
j ’ + l- 2 s '
c-^21
6s
2(»’ - l )
(s’ + 9 ) '" ( s '+ 1)'
Therefore, the Laplace transfomi of the time function, /s in 3 /- 2 /c o s / is
6s
2 ( s '- l)
( s '+ 9 y
( s 't l) ^
Step 8 of 9
(e)
Consider the time function,
i ( /^ 4 '2 / oob2 /
Appiy the Lapiace transfomi on both the sides of the time function.
£ { f { i ) } = £ {l(l)+ 2te os2l}
F ( s ) = / '{ l ( / ) } + 2 r { / c o s 2 / }
=
1 -2
=
1 -2
( s* + 4 ) - ( 2 j ) 5
s '+ 4 - 2 s ’
( s '. 4 f
Step 9 of 9
Further simplification yields,
j( j* + 4 ) *
^ s*+Bs^ + \ 6 + 2 s ^ - ^
s {s ^ + 4 f
s * + 2 s ^ + S s ^ -S s + l6
s (s ^ + 4 )"
Therefore, the Laplace transfomi of the time function, l( /)+ 2 /c o s 2 / is
5 *+ 2 j ’ + 8 s* - 8 s +16
s(s^ +A '^
Problem 3.05PP
Find the Laplace transform of the following time functions (* denotes convolution):
(a) ^ t) = sin t sin 7t
(b) fl:0 = s in 2 f+ 7 c o s 2 f
(c) f{f) = {sin t)A
(^ ) / ( O
= / c o s(^ — t )
sinr d r
0
(^) f i t )
= J
cos(f -
T)
sinxdx
0
Step-by-step solution
(a)
Consider the time function
/ ( r ) = s io / s in 7 r
(1)
Consider the equation.
s u u > /s in 6 /K -^ c o s (|a -6 |/)—
...... (2)
Compare Equation (1) and (2) and find a and b.
a = hb = 7
Substitute 1 for a and 7 for b in Equation (2).
/(0 =^cos(|l-7|»)-icos(|l+7|f)
=-cos6f--cos8r
2
2
Take Laplace transform.
= i [ ^ ______ f _ l
2U'+36 i*+64j
I4«
( j’ +36)(j*+64)
14f
Thus, the Laplace transfomi is
+ 3 6 )^5 ^+ 6 4 )
(b)
Consider the time function.
/(r)=sin*/+7cos^/
(3)
Consider the trigonometric fomiula.
. 2 . l-cos2/
(4)
2
cos'f *-l+cos2r
(5)
Substitute Equation (4) and (5) in Equation (3).
=4+3cos2/
Take Laplace transform.
■ KHA)
7^^+16
s (s “ + 4 )
Thus, the Laplace transfomi is
75*+ 16
r(j*+4)
(c)
Consider that division by time is equivalent to integration in the frequency domain
F { s ) = ] e - f ( l) d l
Jf (j)* =J je -“ / { l) d l
Interchange the order of integration.
Consider the time function.
/ ( 0 = ^
= lan‘'(oo)-tan"'(5)
- f - t a n - 'M
Thus, the Laplace transfomi is
step 10 of 11
(d)
Consider the time function.
/(/) s $ in /* s in /..... (6)
Take Laplace transform.
L ( s m ( ) = - jL '
' 5 +1
(7)
Substitute Equation (7) in Equation (6).
1
5 *+ 2 5 *+ !
1
Thus, the Laplace transfomi is
5* + 2j * + 1
Step 11 of 11
(e)
Consider the time function.
/ ( / ) = J cos(/ - r)sin rdr
0
Take Laplace transform.
cos(f-r)sinr</rj
i(/(7))=i(cos(/)*sid(/))
L (c o s t)= - ^
(9)
L (s m ()= -jL -
(1 0 )
f +1
'
' 5
(8)
+1
Substitute Equation (9) and (10) in Equation (8).
i * + 2 i’ + l
Thus, the Laplace transfomi is
Problem 3.06PP
Given that the Laplace transform of /(/) is F(s), find the Laplace transfonn of the following:
(a) g(t) = f{t) cos t
tn
(b)
gV) = fff(r )d rd ti
Step-by-step solution
step 1 of 2
Step 1 of 2
(a)
g(t)-f(t)cost
Usingproperty L[x(t)y(t)]-^X(s)*Y(s),
fe)
•F (s)
Step 2 of 2 ^
Usingtheintegrationpropertyoflaplacetransforms»
L \£ K x )d t]
cks> - ! ^ = --------- ^
8
Problem 3.07PP
Find the time function corresponding to each of the following Laplace transforms using partialfraction expansions:
(a)
F(s) -
(b)
f W — s(s+j)(s+5)
3i+2
(c) F ( l) =
i'+ i+ lO
3F+6S+6
(d) F(s) =
(i+ l)(i^
' W + 6t i+ 1 0 )
(j+ D d ^+ d a + lO )
_1
? + l6
(e) F(s)
(f)
2(i+3)
F(s) =
(J+1)(J^+16)
(g) m
= ^
F(s) =
(h)
4
?44
F(s) =
(i)
e -‘
(i) F (4 ) = 7 ^
Step-by-step solution
Step 1 of 16
(a)
Consider the Laplace transform equation.
Take partial fraction expansion and rewrite F (.r)F (s ).? -+ J -
(1)
J + 1
S
Use the cover up method and find a and b ■
* 4 u ,= - i
Substitute i for a and _ ] for b
equation (1).
F ( i) = i — L
5+ 1
S
Take Inverse Laplace Transform.
i- ( F ( 5 ) ) = r '( i) - r '( ^ ]
Thus, the time function of
^
is
Step 2 of 16
(b)
Consider the Laplace transform equation
^
5{5 + 1)( j + 5)
Take partial fraction expansion and rewrite F ( ^ ) •
a,
Use the cover up method and find
^and
c.
(5 + 1)(5 + 5)u
•
1
__5
"~4
- _ 5_ |
s (5 + \y ^
I
"4
Step 3 of 16
Substitute i for a , —— for b ^nd — for c in equation (2).
4
4
F ( f ) s ------- 2 _ + _ l _
s
s+ 1
f+ 5
Take Inverse Laplace Transform.
f
' 5 '
L - '( F ( 5 ) ) = r ' i
j '
_4_ + r ‘ _ L
5+ 1
5+5
- r '
Thus, the time function of
«( j + I)(5 + 5 )
Step 4 of 16
(c)
Consider the Laplace transform equation
F (5 ).
' '
5’ +25 + 10
Take partial fraction expansion and rewrite F ( j ) -
3(5 + 1)
1
~(5+1)’ + 3 '
(5+1)'+3*
Take Inverse Laplace Transform.
/ ( / ) = 3e"'cos3f l ( / ) - i c ‘ 's in 3 r l( / )
3j + 2
Thus, the time function of
j
s ^ * 2 s * l0
3e"'cos 3 /1( / ) - —c"* sin 3 /1( /)
Step 5 of 16
(d)
Consider the Laplace transform equation.
=
' '
___
(5 + l)(5 ’
+ 10)
+65
Take partial fraction expansion and rewrite F ( s ) =
,3)
s*\
j +6^+10
Find the value of a.
( 3 5 ’ + 6 s + 6 )|
3
‘' ” (5 "+ 6 5 + 1 0 ) ''~ '“ 5
Substitute — for a in equation (3) and equate numerators.
(s + i)
fa + e
3 s *+ 6 s + 6
( s *+ 6 s + 10)
( s + 2 ) ( s * + 6 f+ 1 0 )
^ 6 + | j s * + ^ ^ + c + 6 j s + ( c + 6 ) = 3 s *+ 6 s + 6
Step 6 of 16
Equate powers of s to find ^ and c.
“
y
c+ 6 -6
CmO
3
12
Substitute j
^
^ in equation (3).
12
3
F (5 ) =
( 5 + 1)
y '
( 5^ + 65 + 10)
3
12
<
( 5 + 1)
5
5 ( 5^ + 65 + 10)
Step 7 of 16
Take Inverse Laplace Transform
>
3
r '( F ( 5 ) ) = r '
( 5 + 1)
[ 5
( 5 ^+ 6 5 + 1 0 ) ]
=[|e'''(0-ye'*s‘»(')l(0+y«'*<»s{')l(0]
3s*+6s+6
Thus, the time function of
(5 + 1)(5=+65 + 10)
Step 8 of 16
(e)
Consider the Laplace transform equation.
Take partial fraction expansion and rewrite F ( s ) -
Take Inverse Laplace Transform.
i- '( F ( 5 ) ) = i - '
l 4 ( 5 '+ 4 ‘ )
/(/) =lsin4/1(5)
Thus, the time function of
—sin4/l(r)
s*+16
Step 9 of 16
(f)
Consider the Laplace transform equation.
^
^
(5 + 1 )(5 *+ 1 6 )
Take partial fraction expansion and rewrite F ( s ) bs+c
F (5 ) =
(5 + 1)
( 5^ + 16)
+ 3) .
4
2 (5
(4)
Step 10 of 16
Substitute j y
17
^ in equation (4) and equate numerators.
bs+c
2 (s + 3 )
(s+l) (j* +16^ (s+l)(j* +16)
^ ^ + * j 4 ’ + ( * + b)4 + [ ^ + ' j = 25 + 6
Equate powers of s to find ^ and c.
64
-+ c =6
_38
^ " l7
*+ c = 2
Step 11 of 16
4
4
38
Substitute — for a , ------for A and — for c In equation (4).
17
17
17
4
4
38
-T ^^+ T
F f 5 l = - ! 2 - + - 1 2 ___ n
( 5 + 1)
(5 -+ 1 6 )
4
4
19
E _____ u l
(s + 1 ) ( j^ + 4 * )
4
3 4 ( j^ + 4 * )
Take Inverse Laplace Transform.
f
i- '( F ( 5 ) ) = i- '
4
>
r 4 . '1
i f 19
4
17 - r '
17^
+C
(5 + 1)
(5»+4»)
[ 3 4 ( 5 '+ 4 = ) J
^
/
/ W = f | e '''( ') - ^ « » 4 < l( 0 + ^ s in 4 / l( / )
2 ( 5 + 3)
Thus, the time function of
(5 + 1)(5*+16)
jy C o s 4 / l( / ) + ^ s in 4 / l( r )
Step 12 of 16
(g)
Consider the Laplace transform equation.
F (5 )= ^
Take partial fraction expansion and rewrite ^ ( 5)F ( 5 ) = l+ i
5 5
Take Inverse Laplace Transform.
r '( F ( 5 ) ) = r '( i) + L - ( l]
/(/) = ( i+ /) i( /)
Thus, the time function o
<f ^
is |(1 + /)1 (/)|.
step 13 of 16
(h )
Consider the Laplace transform equation.
^ W = 7
Take Inverse Laplace Transform.
rH' ( F ( 5 ) ) = r ' [ l )
/( ') = ^ '( 0
120
''
Thus, the time function of — is
120
1(0
step 14 of 16
(i)
Consider the Laplace transform equation.
Take partial fraction expansion and rewrite ^ ( 5)-
'
^ 5+1
- i5 + l
5'+ 2 5 + 2
5'- 2 5 + 2
(5
+ 1)’
+1
( 5 - 1 )*+ !
Step 15 of 16
Take Inverse Laplace Transform
(. 5 + 1.) .- - ' - ' '
i- '( F ( 5 ) ) = £T'
'( 4 - O - I 4 '
- r
(5 + 1) ^ + !
( 5 - l ) '+ l
/ ( / ) = 5-* c o s ( / ) - ~ ( e - ' s in ( / ) ) - « ' c o s ( / ) - ~ ( e ’ s in (/))
= tf"'c o s (f)—j( - e " 's in ( / ) + c o s ( / ) « " ') - e 'o o s ( / ) + ^ ( c ' s in (/)+ c o s (/)e ')
= - e < K ( / ) |2 l^ jl( ,) + 8 i„ ( , ) |z l^ j|( ,)
Therefore, / ( / ) = - c o s ( /) s in h ( r ) l( /) + s iii( r ) c o s h ( /) l( /)
5
Thus, the time function of - j l — is |-c o s ( /)s in h ( /) l( /) + s in ( /) c o s h ( /) l( /) |
Step 16 of 16
G)
Consider the Laplace transform equation.
F (s )-f
Take Inverse Laplace Transform.
i- ( F ( 5 ) ) = r '( ^ ]
/ ( / ) . ( /- l) 9 ( /- l)
Here,
is Heaviside step function.
Thus, the time function of £ — is l ( » - i) g ( < - i) l
Problem 3.08PP
Find the time function corresponding to each of the following Laplace transforms:
(b)
=
(0
=
(d)
=
(f) F (s ) =
W iW
(g) F(.s) = taa ' ( i )
Step-by-step solution
step 1 of 12
(a)
Consider the Laplace transform equation.
Take partial fraction expansion and rewrite F (.r)r^/ \
O
b
C
Use the cover up method and find a , ^and
'
c.
tU='
( ,+ iy
c = iu = - i
- 7 U
.
* -l
Step 2 Of 12
Substitute 1 for a . • ] for ^and . ] for c in equation (1).
Take Inverse Laplace Transform.
/ ( 0 = ( i- e - '- « - ') i( 0
1
Thus, the time function c
Step 3 of 12
(b )
Consider the Laplace transform equation.
Take partial fraction expansion and rewrite F (.r)-
Ff$)=— fl±£±l—
1
^“ 1
Take Inverse Laplace Transform.
r'(F (.))= r'(^ ]
/ ( ') = ^ 'W
Thus, the time function of
is
Step 4 of 12
(c)
Consider the Laplace transform equation.
. .
2 ( j* + j+ l)
--------- 5 -i
Take partial fraction expansion and rewrite F ( 4)r^/ \ O
b
C
'" W 'T +sT T n T + T T T T T
(»+l) (i+l)’
(2)
2 (5 ^ + 5 + 1),
j+ iy
2^5*
+ lV
V .= -2
2 ( j* + j+ l)
1
2(2 j + 1 ) 5 - 2 ( j * + « + 1)
u .
step 5 of 12
Substitute 2 for 0 < —2 for ^ and 0 for c in equation (2).
=./ \
2
0
-2
Take Inverse Laplace Transform.
r '( F W
) = r 'g ) . r '[ ^ ]
/ ( / ) = (2-2/d-')l(0
2 (5 * + 4 + 1)
Thus, the time function of - i --------- ^ i
j( j+ l) *
Step 6 of 12
(d)
Consider the Laplace transform equation.
Take partial fraction expansion and rewrite ^ ( 4)F (s )^
As -¥B
Cs + D
j* - 2
j" + 2
3 1 1 1
—j + — —s —
. 4 ___ 2 . 4 ___ 2
■ 5^-2
+2
Step 7 of 12
Take Inverse Laplace Transform
(3-5 +-n
i-'(F(j))=r' 4 2 + r‘ 4 2
*'-2
i’ +2
^
J
)
= ^ s ‘* * ( ' / 2 ' ) + ^ c o s h ( , ^ / ) - ^ s i n ( , ^ ( ) + i c o s ( V 2 f )
Thus, the time function of j * + 5 + 2 i_
s *-A
^ s in h ( , ^ / ) + lc o s h ( , ^ / ) - ^ s in ( ,^ » ) + ic o s ( . / 2 ^ )
Step 8 of 12
(e)
Consider the Laplace transform equation.
j, , .
2(^-f2 )(^H -S )’
'
( i + !) ( * = + 4 ) ’
Take partial fraction expansion and rewrite ^ ( 5)-
a.
o=(j+l)F(j)|,_,
Find the value of
2S
>1.280
Find the value of
d ^ ( s - 2 j) ^ F ( s ) \^ ,
^ -8 3 -3 9 y
20
*-4,150-;i.950
Step 9 of 12
Find the value of e.
e^dr
- -4 .1 5 0 + y l.9 5 0
Find the value of J}.
^ -1 2 8 -5 7 9 j
200
= -0 .6 4 -y 2 .8 9 5
c.
Find the value of
c -b *
= - 0 .6 4 + y 2.895
Substitute 1,280 for a , -0 .6 4 -> 2 .8 9 5 for 6 , -0.64+>2.895 for c . -4.1 50-> 1.950 for
d and -4.150+>1.950 for e in equation (3).
1-280 ^-0.64->2.895 ^-0.64+>2.895 ^^.150->1.950 ^-4.150+>1.950
F(
s+\ *
S -2 J
^
S + 2J
{ s - 2 jy
*
{ s * 2 j) ‘
Step 10 Of 12
Take Inverse Laplace Transform.
1.28«-' + /[(-4 .1 5 0 + y i.9 5 )« -^ -(4 .1 5 0 + y i.9 5 )e ^ ]1
/ ( ') ■
-(0 .6 4 -;2 .8 9 5 )e -^ “ -( 0 .6 4 + ;2 .8 9 5 )e '”
|
fl.28«-' + /[(-4.150e--'“ + yl.95«-^“ )- ( 4 .I 5 0 « '’' + yl.95e'“ )]l
|- ( 0 . 6 4 e - ^ - y 2 .8 9 5 e - ^ ) - ( 0 .6 4 « ^ + J 2 M S e ‘ ^ )
_ |l.2 8 « - ' -/4.150e-^“ +
J
-« 4 .1 5 0 e ^ -yrt.O S e'” !
~ |- 0 .6 4 « - ^ + y 2 .8 9 5 e -F '-0 .6 4 e ^ -y 2 .8 9 5 e '”
J
Rearrange the equation.
[l.28«^ -/4.150(e-^“ + « '“ ) + y/1.95(«-^
'^“ )1
”|-0.64(«-F' +*>“)+y2.895(«-''“ -e^)
r
1.28*'' -/4.150(2cos(2/))+yrt.95(-y2sin(2/))
>(')
“|-0.64(2cos(2/))+;2.895(-y2sin(2/))
1.28*- -/4.1S0(2cos(2/))-/rt.95(2sin(2/))
”|-0.64(2cos(2/))-/2.895(2sin(2/))
={l.28e-'-»8.3cos(2»)+3.9/sin(2/)-1.28cos(2/)+5.79(sin(2/))}l(/)
Rearrange the equation.
/(/) =[l.28e-'+3.9/sin(2/)+5.79siii(2/)-8.3/cos(2/)-1.28cos(2/)]l(l)
2(i +2)(j+5)'
-------- ^ is
(*+!)(*’ +4)
|[l.28g-*+3.9<sm(2/)+5.79sin(2<)-8.3/c<»(2()-1.28cos(2<)]l(>)|.
Thus, the time function of
Step 11 of 12
(f)
Consider the Laplace transform equation.
Take partial fraction expansion and rewrite F ( s ) 5*
(*
1
+ 0
(»
+ ')
Take Inverse Laplace Transform.
r'(F(*))=r
W
- 'f
(.= +■ )'
/ ( 0 = ( / c o s » ) l( / )
Thus, the time function of -------- ^
15* +11
is ( /c o s / ) l( /) •
*----------------- *
Step 12 Of 12
(g)
Consider the Laplace transform equation.
F(.)=tan-[i)
Rewrite F ( 5) in terms of series.
^W=7-37^57-Take Inverse Laplace Transform.
r'(F(*))=r'[t»-g]]
°'^'(7"37‘"57""]
/ ( 0 = i- 3 ( - ^ -
sin(/)
•w
Thus, the time function of tan"' -
■ 0 '
is
s in (/)
>(0
Problem 3.09PP
Solve the following ODEs using Laplace transforms;
(a) 5 ( 0 + 5 ( 0 + 3 y (0 = 0; y (0 ) = 1, y (0 ) = 2
(b) 5 ( 0 - 2 5 (0 + 4 y ( 0 = 0 ; y ( 0 ) = 1. 5 ((0 = 2
(a) 5 ( 0 + j ( 0 = sin<; y (0 ) = 1, 5 (0 ) = 2
(d) 5 ( 0 + 3 y (0 = sin »; y (0 ) = I , 5 (0 ) = 2
(e) 5 ( 0 + 2 j ( 0 = «!:.
i^ 9 W 2^-
(e) 5 ( 0 + 2 5 (0 = e*; y (0 ) = 1, 5 (0 ) = 2
m 5 ( 0 + y ( i ) = t; y ( 0 ) = 1. 5 ( 0 ) = - 1
S te p -b y -s te p s o lu tio n
step 1 of 19
(a)
Consider the equation is 5 ( 0 '* '5 ( t ) + 3 y ( r ) = O
Apply Laplace Transform to the equation.
[ s » l '( » ) - s y ( 0 ) - 5 ( 0 ) ] + [ j l '( s ) - y ( 0 ) ] + 3 r ( s ) = 0
Substitute 1 for y(0 )a n d 2 for 5 ( 0 ) 'h tbe equation.
[j*r(j)-j-2]+[»i'(j)-i]+3y(»)=o
(s*+ff+3)K(j)*j+3
j+3
! '( * ) =
i^ + j+ 3
i+ 3
'FPf]R^
step 2 of 19
Apply partial fractions.
s+ 3
„ l . y
Determine the value of
:7 n
(1)
.
i+ 3
- 1- 4
..
(/v ri)
.llif
■(w n)
step 3 of 19
Determine the value of g .
1
(
.V T T ^ i
.V iT 'l
(-y fH )
i- iM .
-2 — 1 .
■ (-y v n )
Step 4 of 19
Substitute the value of .4 and ^ in equation (1).
y(s)
step 5 of 19
Apply inverse Laplace transform
cL
■ ;? ii
2y
? cos---Ju,1—vsin--5 . VfT,le -i*
=e‘
2
2
V ll
Therefore, the expression for H O is.
y (0 = c » '[ c o s [ : | ', ] - ^ s in [ 4 ^ , ] n ( 0 ]
Step 6 of 19
(b)
Consider the equation is 5 ( t ) - 2 5 ( / ) + 4 y ( / ) = 0
Apply Laplace Transform to the equation.
[ s * y ( s ) - 4»-(0 ) - 5 ( 0 ) ] - 2 [ j l ' ( s ) - y ( 0 ) ] + 4 r ( s ) = 0
Substitute 1 for y^Q)and 2 for 5 ( 0 ) in the equation.
[ s 'y ( i ) - 4 - 2 ] - 2 [ s i'( 4 ) - i ] + 4 y ( 4 ) = o
y ( j) [ i'- 2 4 + 4 ] - j
’'( * ) = 7 d j7 4
i- i+ i
(a -O ^ (^ )
—f
\
s -l
*”*
(
^
(2)
step 7 of 19
Apply Inverse Laplace Transform. Use the following fomiula;
f ---- — ^ — r l = e~*cosbt
r '
Therefore,
y ( 0 = « 'c o s ^ + ^ « 's in ^
HO is.
Therefore, the expression for
y ( t) = e fc o S 'j3 l+ -^ e fs m - j3 t
Step 8 of 19
(c )
Consider the equation is 5 ( t ) + 5 ( 0 “ ®^^
Taking Laplace Transform
[ s ' y ( j ) - j y ( 0) - 5 ( 0) ] + [ j r ( j ) - y ( 0) ] =
^
Substitute 1 for y^Q)snd 2 for 5 ( 0 ) in the equation.
[ s - y ( s ) - s - 2 ]+ 4 y ( s ) - i= p ^
y ( j) ( i'+ s ) = p | j^ + ( 4 + 3 )
(4+3)(s’ +i)+1
4(i+l)(s’ +l)
y (s );
4 *+ 3 s * + 4 + 4
4(s+l)(s^+l)
A
B
C t+ D
= — + ------ ; ------------
S
J+1
» +1
Step 9 of 19
Apply partial fraction.
j’
+ 3 j ’ + » + 4 = ^ ( » + 1 ) ( j ' + i ) + & ( 4 ’ + i ) + ( C s + D ) » ( j + 1 ) ...... (3)
Determine the value of
.
Substitute ^ = 0 ih equation (3).
y4(0 + l) ( 0 ’ + l ) = 4
A=A
Step 10 of 19
Determine the value of g .
Substitute 5 s -1 in equation (3).
B (- l) ( ( - l) * + l) = ( - l) ‘ + 3 ( - l) * - l+ 4
-2B = - l + 3 - l + 4
2
Step 11 of 19
Determine the value of Q .
Compare 5* coefficients in equation (3).
\= A + B + C
1 = 4 --+ C
2
C=l - i
2
^ = T
Compare ^coefficients in equation (3).
\= A + B + D
l = 4 - |+ D
0 = 1- -
Substitute the values of AyB ^C iD in y ( * ) expression.
j,, , 4 2.5 0.S4+0.S
i
4’ + !
4+1
=l_ i4 .0 .5 ^ .0 .5
4+ 1
4
4+1
'
4+1
Step 12 Of 19
Apply Inverse Laplace transform.
Use the following formula;
= s in i(
i') - '
‘ " ( H
=cosbl
‘" ( T r y ) "
Therefore,
y(») = 4ii(r)-2.5«"'-0.5cos/-0.5sin/
Therefore, the expression for y ( ' ) is,
|y(0=4if(f)-2.5c"*-0.5oosf-0.5sigf|.
Step 13 of 19
(b)
Consider the equation is 5 ( t ) + 3 y ( / ) = s in f.
Apply Laplace Transform.
[4 “ r ( 4 ) - 4 p ( o ) - 5 ( o ) ] + 3 y ( 4 ) = p | ^
Substitute 1 for y(0 )a n d 2 for 5 ( 0 ) 'b Ibe equation.
[ 4 'y ( 4 ) - 4 - 2 ] + 3 y ( 4 ) = p l^
(4 ^ + 3 )y ( 4 ) = ^ + ( 2 + 4 )
2
1
s
_ i ( 4 ^ + 3 )-(4 ^ + 1)
2
4
2 (4 ^+ 3 )(4 - + 1)
_ i ___ 1
(4‘ + ( V 3 f )
^_I_
s
step 14 of 19
Apply inverse Laplace transform.
Use the following formula;
s in i(
cosbt
y ( » ) = ^ ^ s in /- ^ s iii^ j+ ^ s iii- \^ r + c o s x ^
Therefore, the expression for
HO is.
} < » = j siiw — UsinV3< + ^ s i n ^ + c o s ^
step 15 of 19
(e)
Consider the equation is; 5 ( » ) + 2 5 ( /) = e '
Apply Laplace Transform.
[4’ y ( 4 ) - 4 y ( 0 )- 5 ( 0 )]+ 2 (4 r (4 )-y (0 )) = j ^
Substitute 1 for y(0 )a n d 2 for 5 ( 0 ) 'b the equation.
4*y(4) + 24y(4) = — +4 + 4
' 4 -1
y (4 )=
(4 + 4 ) ( 4 - l ) + l
( 4 - l ) ( 4 + 2)4
4 * + 3 4 -3
( 4 -1 )( 4 + 2)4
A ^ B ^ C
4
4 -1
4+2
4 * + 3 4 - 3 = 4 ( 4 - 1 ) ( 4 + 2 ) + B 4 (4 + 2 )+ C s( 4 - 1 ) .
step 16 of 19
Apply partial fractions.
Determine the value of
.
Substitute 4 s 0 In equation (4).
-3 = 4 (-1 )(2 )
- f
Determine the value of g .
Substitute 4 = 1 In equation (4).
l+ 3 - 3 = f i( l + 2)
Determine the value of C ■
Substitute 4 = -2 In equation (4).
4 - 6 - 3 = C ( - 2 ) ( - 2 -1 )
-5 = 6C
y (,) = L 5 ^ 0 ^ _ 0 ^
4
' '
4 -1
4+ 2
= ( 1 .5 ) i+ ( 0 .3 3 ) ^ - ( 0 .8 3 3 ) ^
Step 17 of 19
Apply Inverse Laplace transform.
Use the following formula;
t
1= H 0
Therefore,
y{l) = 1 .5 ii(/)+0 .3 V -0.833«-“
Therefore, the expression for y ( 0 is,
|y(>) = 1,5ii (<)+0.33e' - 0.833e-4‘ |
Step 18 of 19 A
(f)
Consider the equation is y ( j ) + y { t ) = t
Apply Laplace transform on both sides.
[4 ’ y ( 4 ) - 4 p ( o ) - 5 ( o ) ] + y ( 4 ) = ^
Substitute 1 for y ( 0 ) and
for 5 ( 0 ) 'b Ibe equation.
[4 > y ( 4 ) - 4 + i]+ y ( 4 ) = l
(4*+l)y(4) = ^ - l +
4
y ( 4 ) = - r y i ---- r + - ^
^ ' 4^ ( 4^ + ! ) 4 * + l
4_____ l _
■ 4^ ( 4 ’ + 1)
1
__1_
4’
4’ + l
4
4’ + l * 4 ’ + l
4 '+ l
1
4’ + !
= -L -2 -!-+ -J 4’
4’ + l 4^+1
Step 19 of 19
Apply inverse Laplace transform.
Use the following formula;
r'
c'
U+4^
|=sini(
r' ( * ^|=cos4f
U + i’;
Therefore,
y(»)=/-2sin»+cosf
Therefore, the expression for y ( ' ) is,
|y ( t ) = / - 2 s in f + c o s f |.
(4)
Problem 3.1 OPP
Using the convolution integral, find the step response of the system whose impulse response is
given below and shown in Fig.;
» < 0.
0
Figure Impulse response
Step-by-step solution
step 1 of 5
Refer to impulse response in Figure 3.47 in the textbook.
The function is,
' ^
\0
(< 0
The expression for convolution is,
j'(/)= J A(r)ii{/-r)rfr
(1)
-«e
0
/
«
-«
0
f
It is known that,
/.
fl
/S O
"W=|o
/<o
And,
,
,
fl
' [0
TS/
otherwise
Step 2 of 5
The shifted unit step function is.
Step 3 of 5 ^
Case {1):
The multiplication of shifted unit step function shown in Figure 1 with
A(r) becomes zero for
/£ 0
Recall equation (1).
)’,(!)= J /l(T)u(l-T)tfT
=0
Since, there is no overlap between the two functions
u(/—r) and *(r) and the output is zero.
Case (2):
Find the output of the system for the / ^ 0.
— ( r + l) * Case (3):
Find the output of the system for the / * oo,
=0
step 4 of 5 ^
MATLAB code to plot the step response of the system:
» t=0:0.01:10;
» y=-(t+l).*exp(-t):
» plot(t,y)
» title{'step response')
» xlabel('t{sec)')
» ylabel('y(t)')
Step 5 of 5
The step response of the system is shown in Figure 2.
step response
t(sec)
Figure 2: Step response of the system
Hence, the required step response is obtained using the convolution integral.
Problem 3.11 PP
Using the convolution integral, find the step response of the system whose impulse response is
given below and shown in Fig;
0,
r < 0 and / > 2 .
Figure Impulse response
-I
-OS
0
OS
1
15
2
25
3
1106 (nc)
Step-by-step solution
step 1 of 6
Given impulse response o f the system is
^
ri, o ^ t ^ 2
|0 , £ < 0 and t > 2
The given impulse response is
*(/)
-1 -0.5 0 0 5
I t.5 2 2.5 3
lunc(soc)
Step 2 of 6 ^
There are three cases to consider as shown in the following figure.
w
For the case £ ^ 0, the situation is illustrated in the below Sgure.
There is no overlc^ between the two functions u (£ - r ) and h (r) and the output
is zero, that is y i (i) = 0.
Step 3 of 6
(b)
For the case 0 £ £^ 2, the situation is displayed in the below Bgure and shows
partial overls^. The o u ^ u t ofthe system is
s£
Step 4 of 6 ^
(c)
For the case £ ^ 2, the situation is displayed in the below figure and shows
total overU^. The output of the system is
^ W = f* A W « (£ -T )^ T
s2
Step 5 of 6
The figures to illustrate convolution are
u(t)
b
bet)
i(sec)
(
uCt)
2
t(stc)
2
t(s a :}
b
ll(T)
T (sec)
•
h(T)ii(l-t)
h(T)
1
x(sec)
t
2
xfsecj
h(T)u(l-T)^
1
b (i)
^ t- j)
step 6 of 6
The output of the system is the con^osite o f the three segments computed above a
shown in the following figure.
Thus, the required step response is obtained.
Problem 3 .1 2PP
Consider the standard second-order system
G(s) = ^ + 2^Q)nS +
(a)
Write the Laplace transform of the signal in Fig.
(b)
What is the transform of the output if this signal is appiied to Gfs^?
(c)
Find the output of the system for the input shown in Fig.
Figure Plot of input signai
t “ (')
I
I
I
Figure Plot of input signai
«(0
1
2
3
Tim e (sec)
Step-by-step solution
Step 1 of 8
(a)
The input signai is.
Define the signal,
.
u{l) = t ; 0 £ / £ l
=1
; l£ ( £ 2
= - /;2 S /S 3
Step 2 of 8
Write the formula for the Laplace transform. F W of a function, / ( ' )
Define the signal,
!/( /) » /
.
;0 £ /£ l
= -/;2 S rS 3
Step 2 of 8
Write the formula for the Laplace transform. F W of a function, / ( ' )
f/t\
Ri ih^titiitp thp fi inrtinn
The transform of the output Y ( j ) is.
Y{s) = G{s)U(s)
3— p ------ ^ f o r < ? (j)a n d
S
* ( i_ « - * _ « - » ' + e - > ') + i( - 3 e - * ' + 3e-*') f
U (s)
a i ^ [ l - e ' * - ( l + 3 5 ) e ^ + ( l + 3 j)g '* * ]
5*
+ 2^01^ +
)
j'( i " + 2 < « v j + « ^ )
Thus, the transform of the output is
f l^ [l- c ~ * - F ( l + 3 5 )(g ~ ^ -e ~ ^ )]
step 4 of 8
(c)
The transform of the output I ' M
is,
< t^ [l-g ~ * + ( l + 3 s ) ( g ~ ^ - g ~ ^ ) ]
S* ( j * + 2^0f,5 +
)
of
Consider the transform, K ( j ) ■ ,> ,------ ---------- tt .
' j " ( i ’ +2fflV > + “ 2)
Apply inverse Laplace transform.
ait
s^[s^+2Ca>^ + a i)
s+ C
O A
+ -T + s
+
+
Step 5 of 8
Equate the coefficients on both sides.
B + D ^O
A + C + 2 ^ o )^ D —0
2 C (o ,A + o iD ^ 0
Aoi = a i
A=l
Determine D.
2 ( a ,( - l) + a iD = 0
b
= -K
Determine C.
l+ C + 2 < f l» ,^ = 0
C = - ( 2 f * + l)
Problem 3.13PP
A rotating load is connected to a field-controlled DC motor with negligible field Inductance.A test
results in the output load reaching a speed of 1 rad/sec within 1/2 sec when a constant input of
lOOVis applied to the motor terminals. The output steady-state speed from the same test is found
to be 2 rad/sec. Determine the transfer function 0 W ,of the motor.
Step-by-step solution
ste p 1 of 6
The equation of motion for a DC motor is.
oiep I ui o
The equation of motion for a DC motor is,
J A + m . - k j . ...... (1)
+
(2)
Since, there is negligible filed Inductance,
sQ .
From equation (2).
RJ. = v ,- K .d .
<3>
Step 2 of 6
for
in equation (1).
R . j. e . * R jb 0 ,= K . v ,- K j:, e .
Apply Laplace transform both sides.
R .J j^ e . ( s ) * R J b s e .( s ) = K . y , ( ! ) - K . K j , e . ( ! )
+
(
*
, *
g .W
+
(s) = K .V , (j )
K.
v ^ (j) ■ R ,J,s'*(R J> + K ,K ,)s
K.
K J.
KJ.
v ^ (s )
......( .,
5 ( f+ a )
Where.
IC.
R Jb + K ^K ,
V.
V .
= — ss_ and
----- s—
Step 3 of 6
The applied Input voltage Is,
V j,(/)
= 100V
Apply Laplace transform.
100
The output load reaching a speed of 2 lad/swithin — secTherefore,
t f f r l = 2rad/s
100
Substitute iiiii for Vy. ( 5 ) in equation (4).
s{s+ a)
100
s
lOOK
S0{s)
•(5)
s{s+ a)
Step 4 of 6
Apply Final value theorem to equation (5).
'
<-*A
*(i)-s
lim
2=
1 00 ^
s Hm.s
\0 0 K
( j+ « )
100 ^
(0 + a)
lOOK
-*2
• ( 6)
From equation (5).
iooa:
a
a
. ^ ( j+ a )
'W =
_ioo/:|~i
1
a
5 + flJ
L*
j
Step 5 of 6
Apply Inverse Laplace transform on both sides.
1= 2 ( 1 - * - )
l- e - * 0 J
e - " - 0 .5
Ine"* = ln0.5
-0.693
fl 1
0 = 1.39
From equation (6).
lOOK
-=2
1.39
a: B 0.0278
Step 6 of 6
Substitute 1.39 for a and 0.0278 for K ^ equation (4).
Vy(s)
0.0278
s ( s + 1.39)
Therefore, the transfer function of the motor. — “
is
0.0278
!(*-H .39)
Problem 3.14PP
A simplified sketch of a computer tape drive is given in Fig.
(a)
Write the equations of motion in terms of the parameters listed below. K and B represent the
spring constant and the damping of tape stretch, respectively, and o)1 and oj2 are angular
velocities. A positive current applied to the DC motor will provide a torque on the capstan in the
clockwise direction as shown by the arrow. Find the value of current that just cancels the force, F,
then eliminate the constant current and its balancing force, F. from your equations. Assume
positive angular velocities of the two wheels are in the directions shown by the arrows.
J1 = 5 X 10-5 kg m2, motor and capstan inertia,
= 1 X 10-2 N m sec, motor damping,
SI
n = 2 X 10-2 m,
/C/ = 3 X 10-2 N•m/A, motor-torque constant,
/C/ = 3 X 10-2 N•m/A, motor-torque constant,
/C= 2 x 1 0 4 N/m,
S = 20 N/m sec,
r2 = 2 x 1 0 -2 m,
J2 = 2 x 1 0 -5 kg m2,
82 = 2 X 10-2 N m sec, viscous damping, idler,
F = 6 N, constant force,
i i = tape velocity m/sec {variable to be controlled).
(b)
Find the transfer function from motor current to the tape position.
(c)
Find the poles and zeros of the transfer function in part (b).
(d)
Use Matlab to find the response of x1 to a step in ia.
Figure Tape drive schematic
Step-by-step solution
step 1 of 8
(a)
Consider the electrical system in Figure 3.50 from the text book.
The electrical equation of the system is.
Take Laplace transform on both sides.
L .sI.{s )* R .I.(s ) = V .{s)-K ^ 0 ,{s)
The mechanical equation describing the system is,
=A y,
Apply Laplace transform on both sides.
The transfer function of the motor system is,
V (s)
, I
JC.JC.'l
v + i
K
j( r j+ l)
Here,
K m
K,
B R ,^ K ,K ,
BR, + K,K,
Step 2 of 8
Consider the mechanical system in Figure 3.50 from the text book.
Write the equations describing the system.
Apply Laplace transform on both sides.
Bs [ x ^ ( ! ) - x , ( j ) ] + i c [ j r , ( j ) - . s r , ( * ) ] = f ( s)
(5s+A:)[jr,(i)-Ar,(j)]=F(s)
(i)
The mechanical equation describing the system is,
—F
Apply Laplace transform on both sides.
( j ) + B ^s^ (s ) + F ( j ) = 0
+ B ^ ) $ j{ s ) + F { s ) ^ 0
F {s) = - ( J ^ '+ B ^ ) e ,{ s )
+ Bjsj02{^)
Substitute
/ »• .
IF /_ \
IF / - \ T _
F {s) in equation (1).
f r Ji , B *\/» / * \
step 3 of 8
ation between linear and angular displacements is,
9,
“ 'A U )
= r,02(s)
te » ;^,(i)fo r AT,(i)ancl r,^ ^(^)fo r J f , ( j ) in equation (2).
i equations of motion describing the system i
Step 4 of 8
(b)
Consider the equations of motion describing the system.
( jy * B , s ) 0 , ( s ) = K J .(s )
Substitute ' “ ^ '‘"■^for A ( ^ ) .
r,
S‘^»stitute ^ _ ^ _ j / . ( , ) f o r
The transfer function of the system is.
I,(s )
V -t-g ^
Bs + K
’
r,r2 K ,(B s * K )
{jfS ^ +
+ BjS+r^Bs +
r,r^K,{Bs + K)
s {J iS +
(
1
+
+
X tj)
step 5 of 8
Substitute 2x10"*
•^1’ IxlO"^^®*^
2x10"*
3x10"*
K,.2QtoT B . 2 x l 0^ f°'’ /C, 5 x l0 "* f° ''
2x10'^^°'^ / j and 2x10"*^°'^
inthetransferfunction.
X , ( s ) _________ ( 2 x l0 - ^ ) ( 2 x l0 - » ) ( 3 x l0 - » ) ( 2 0 ^ + 2 x l 0 * )
2 x l 0-’j " + 2 x l 0- '( l + 2 x l 0-’ ) i +
/.( » )
i ( 5 x l O " ’i + l x lO '’ )
( 2 x l 0* )( 2 x l 0-*)
1 2 x l 0' ‘ ( 2a s + 2 x l 0‘ )
i ( 5 x l 0 ’j + l x l 0 - ’ ) ( 2 x l 0 - V + 1 .0 2 i+ 4 0 0 )
2 4 0 x l0 - » ( n - 1 0 0 0 )
■ ( 5 x l 0 - ’ ) ( 2 x l 0 - ‘ ) i ( s + 2 0 0 ) ( j '+ 5 1 0 0 s + 2 x l 0 ’ )
2 .4 x ltf( n - 1 0 0 0 )
i ( i + 2 0 0 ) ( i’ + 5 1 0 0 s + 2 x l0 ’ )
Thus, the transfer function from the motor current to tape position is.
2 f ,( i)
2 . 4 x i y ( j + l0 0 0 )
/,( s ) ” * ( i + 2 0 0 ) ( l’ + 5 1 0 0 s + 2 x l 0 ’ )
Step 6 of 8
(c)
Consider the transfer function,
X ,(s )
2 . 4 x l t f ( i + 1000)
/ , ( j ) " j ( j + 2 0 0 ) ( j ' + 5100s + 2 x l 0 ’ )
Use MATLAB to find the roots of the denominator polynomial.
» ro o ts ([1 5100 2E7])
ans =
1.Oe+03 *
-2.5500 + 3.6739i
-2.5500 - 3.6739i
Thus, the poles of the transfer function are |0 ,-2 0 0 ,2 5 0 0 ± y 3 6 73.9| and zero of the transfer
function is |»1Q00|.
Step 7 of 8
(d)
Enter the following code in MATLAB to find the step response of the system.
» num=2.4E5*[1 1000];
» den=conv([1 200 0],[1 5100 2E7]);
» sys=tf(num,den):
» sys1=feedback(sys,1):
» step(sysl)
Step 8 of 8 ^
The following is the MATLAB output for step response:
Figure 1
Thus, the response to unit step input is plotted.
Problem 3.15PP
For the system in Fig., compute the transfer function from the motor voltage to position 02.
Figure Motor with a flexible load
i.
S te p -b y -s te p s o lu tio n
step 1 of 4
Step 1 of 4
Consider the electrical system in Figure 2.54 from the text book.
Apply KirchhofTs voltage law.
Apply Laplace transform on both sides.
The mechanical equation describing the system is,
J ,0 ,+ b (e ,-0 ,)+ k (e ,-0 ,)+ B $ ,= K ,i,
Apply Laplace transform on both sides.
[ V + ( B + i ) j + * ] « l ( i ) - [ f a + * ] « , ( i ) = A : , / , ( » ) ...... (2)
Step 2 of 4
The mechanical equation describing the load is,
Apply Laplace transform on both sides.
J2S^$2 (^ )
[
^^502 (<s)
+ its+
(-r)] * k \$ 2 ( j ) - ^1 ( j ) ] = 0
(s)-[its+ k\9^ (s)= 0
[to+ k]0f (s)= [/jj*+ to+*]
(s)
3
=
, ,
Step 3 of 4
Recall equation (2).
+ ( B + 6 ) s + * ] ^, ( j ) - [ t o + Jfc] j (s ) = K ,I^ (s )
substitute W
Z M
M
f o r / (,).
Substitute
for
L ^ + R^
Jt
to + *
J
^
{ [ V + ( B + 4 ) i + * ] ( i ^ + « . ) + A : , J t , s } ( j , i ’ +bs+k)-
{ b s + k f{ L ^ + R,) ___________________________________
'iW
L^+R ,
(V + J ? ,)(fa + * )
Step 4 of 4
On further simplification;
bK,s+kK,
K {‘ )
k { L ^ + R ,)+ K ,K ^
J'
'
( 4 y + 2 M j+ * ’ ) ( V + i ^ )
bK,s+kK,
V
/ + [ V i + ( ® + * ) i. ]» ‘ +
( y ,j* + 4 s + 4 ) -
[ ( 5 + 4 ) i ^ + * t , + A T ,A :,]j+ iM ,
L , i V + ( 2 i . 6 * + *(LJc‘ *2RJ>k)s+RJc‘
bKs+kK,
V
/ + [V i+ ( fi+ * ) i.]» ’ +
(y,»’ + fts + * )-
[ ( B + 4 ) J ^ , + * t . + X , 4 : , ] j+ * ) ^
*[LJ c' *2R Jbk)s*R J^
i.4 V
Therefore, the transfer function form the motor voltage to position 0 l( s ) i:
bK,s+kK,
« ,( j)
K i‘ )
1V
i*’
+ [V
i + ( f i + 4 ) i .] * ’
+
( j^ + b s + k ) -
|[ ( 5 + 4 ) « . +kL. + K,K,']s+kR ,
LJ>W +{2LJ>k+RJ^)s‘ + (L J :‘ t-2R,bk)s+RJ[^
Problem 3.16PP
Compute the transfer function for the two-tank system in Fig. with holes at A and C.
Figure Two-tank fluid-flow system
Step-by-step solution
step 1 of 1
We have
A h ,-
A h ,( S ) -
AA^ + 6ct^ - 2 0
6a
Problem 3 .1 7PP
For a second-order system with transfer function
G(s) = .
5^ + s — 2*
determine the following;
(a)
The DC gain;
(b)
The final value to a unit-step input.
Step-by-step solution
Step-by-step solution
step 1 of 2
(b)
Determine the poles of the system.
i^ + j- 2 - O
Therefore ^ = 1,-2
Thus, the system has an unstable pole, so the final value theorem Is not applicable.
Step 2 of 2
(a)
Consider a second order system with transfer function.
G (* ) = - r^
s + S -2
Consider the DC gain formula.
D C g a in -G (5 )|,j,
Substitute 0 for s.
C (0) = 4
Thus, the DC gain of the system is
ED
Thus, the DC gain is not defined for an unstable system so the output of the system is
unbounded.
Problem 3.18PP
Consider the continuous rolling mill depicted in Fig. Suppose that the motion of the adjustable
roller has a damping coefficient b, and that the force exerted by the rolled material on the
adjustable roller is proportional to the material’s change in thickness: Fs = c ( T - x). Suppose
further that the DC motor has a torque constant Kt and a back emf constant Ke, and that the
rack-and-pinion has effective radius of R.
(a) What are the inputs to this system? The output?
(b)Without neglecting the effects of gravity on the adjustable roller, draw a block diagram of the
system that explicitly shows the following quantities: Vs (s). 10 {s^, F(s) (the force the motor exerts
on the adjustable roller), and X(s).
(c)
Simplify your block diagram as much as possible while still identifying each output and input
separately.
Figure Continuous rolling mill
Step-by-step solution
step 1 of 8
(a)
Refer to Figure 3.51 in the textbook.
From the Figure 3.51, the inputs to the system are.
Input voltage
thickness
and gravity {M g Y
The output to the system is.
output thickness ^ x) •
Therefore, the inputs to the system are.
Input voltage(v,(/)), thickness(r)andgravity(A^)
[ouq)utthickness(x)]■
and the output to the system is
Step 2 of 8
(b)
Write the equation of motion for adjustable roller.
na = c {T -x )-m g -b x -F ^
Apply Laplace transform.
(m$’ + & y + c ) j ir ( s ) + ^ ( s ) + ^ ^ —^ - 0
(1)
Write the equation for torque in rack and pinion.
T ^ = KF.
.........( 2 )
Write the expression for torque of a motor.
Here,
K , Is the torque constant.
I f is the field current.
Substitute
for
N K .If
in equation (2).
(3)
Step 3 of 8
From the dc motor circuit, the loop equation is.
Apply Laplace transform on both sides.
(^>
It is known that,
v, ( O
= a :.0
Apply Laplace transform.
V ^ {s )^ K ^ {s )
(5)
And.
e {s )R
jr ( i)
N X {s)
eW
Step 4 of 8
Substitute
fbr e W In equation (5).
for
in equation (4).
------------------------------R ,* L js
...
I. { s ) ~
* ------ —
<®>
R .+ L .S
Step 5 of 8
Substitute
M ir j
* I j for ^
in equation (1).
( m j* + f o + c ) A '( s ) + - ^ ^ ^ ^ i. ^
=0
sK^N ,
Substitute
for
/
\ / X N K ,If
(ms* +fo+c)jr(s)+—
m g -c T
(7)
R .* L js
Step 6 of 8
Draw the block diagram from the equation (7).
Figure 1
Step 7 of 8
(c)
1
In Figure 1, the blocks ------------
N K ,I f
and---------—
are in cascade. Calculate the equivalent block.
*
N K ,If
Move the summing point ahead of a block
G(s) as shown In Figure 2.
Figure 2
Step 8 of 8
Simplify the feedback loop in Figure 2.
f
] ( __
^__ ]
J v in y ^+ A y + C /
r(j).
(
m ,l,
Y
1
Y A T .y .> |
1+
m ,ifR
R^
+
+As+c)+(Affl:,//)(A:,iVs)
Draw the simplified block diagram.
-<■(*)
Problem 3.19PP
Consider the block diagram shown in Fig.. Note that ai and bi are constants. Compute the
transfer function for this system. This special structure is called the “control canonical fonn” and
will be discussed further in Chapter 7.
Figure Block diagram
Step-by-step solution
step 1 of 7
Refer to Figure 3.19 in the textbook for block diagram.
Form Figure 3.19 in the textbook, the transfer function of positive feedback ioop is,
l- G W W ( j)
The reductions of positive feedback loop with G (s ) = i and
=
will be,
s
1
rM —
TTT
I '
1
's + a ,
step 2 of 7 ^
Shift the Pick-off point at j|f| to the right past the second integrator to get b^s+b 2 asshownin
Figure 1.
Figure 1
Step 3 of 7 ^
From Figure 1, the transfer function of positive feedback loop is.
r ( , ) --------1 - G ( f) tf( j)
The reductions of positive feedback loop with
f—
G(5 ) = [— ^— I—ahd
will be.
1 -
r(*)
'- f —
1 -
U )
1
+ 0 ^ + 02
step 4 of 7
Shift the Pick-off point at X 2
right past the third integrator to get
shown in Figure 2.
Figure 2
Step 5 of 7
From Figure 2, the transfer function of positive feedback loop is.
r fj^ = —
..
' ' 1- G { s ) H ( s )
The reductions of positive feedback loop with
be.
f___!___
+ 0^3 + 0 2 ) 3
r(i)=
5 ( 5*+< 1,5 + 02)
ff(ff*+<I,5 + <^)
1
5^5^+0,5 + <l2) + <l}
1
5*+0,5^+025 + ^
Step 6 of 7
Draw the reduced block diagram
y (5 )o -
1
+<1,5 +<^5 + <l3
—5------- 5-------------S
( V + *a )s + ^
Figure 3
Step 7 of 7
From Figure 3, the transfer function is,
1/ ( 5)
------- 2~-----------^^5’ +0,S*+025 + < ^ J ''^
btS^+bi3+bj
5’ +0,5*+<^5 + <^
Therefore, the transfer function. I l £ i i
G(s)
b,s^ + bjS+bj
3 + O fS + 023 + 0^
"11»)
as
Problem 3.20PP
Find the transfer functions for the block diagrams in Fig.
Figure Block diagrams
Step-by-step solution
step 1 of 19
(fli)
T h e fbUowiiig is Ihe givefi blodc diagram :
Step 2 of 19
F ro m F ig m e l.
T h e negative un ity feedback tra n sfe r fim ction is.
q
i+ q ( i)
I
l+ G W j
q
i+q
step 3 of 19
F ^ u r e 2 is d ie reduction d iagram o f F ig u re l.
Step 4 of 19 ^
F rom F igure 2.
T h e transfer function is.
R
1+G i
q + q ,(i+ q )
i+q
^ q + q + q q
i+ q
Therefore, d ie tra n sfe r function o f th e g iv en b lo ck diagram is
R~
l+ O j
Step 5 of 19
(b)
T h e follow ing is d ie given blo d c diagram :
F ig u re s
Step 6 of 19
F irst fitMltfig feedbadc paths a nd deten n in e tra n sfe r functions. F ig u re 4 is d ie
m odified blodc diagram o f F igure 3.
Step 7 of 19
T ransfer fo nctioa fo r first feedback is H, = — ^ _
i+qq
and
T ransfer function fo r second feedback is H ^ = — — — .
* I+ G 4G 5
Step 8 of 19 ^
F igure 5 is th e reduction diagram o f F ig u re 4.
Cascade path
F igure 5
Step 9 of 19
W e sinyilify d ie b lo ck diagram b y red u cin g d ie parallel cofnbination o£ d ie
controller pa th (cascade padi). T he resultant is show n in F igure 6.
G .G jG .G .
/Jo-
+ (^
( l+ q C jX l+ G .G ,)
F ig u re d
Step 10 of 19
F rom th e above block diagram , th e transfer function becom es
r
qqq,G.
R
( l+ G iG ,) ( l+ G ,G 5)
+0,
’
qqq.q
r+q
1+G^G2 + G 4G5 + G i 020^0^
+ - K ^ (l+ G iG 3 + G 4G5 + 6 ^0 , 0405 )
1+ Q G j + G 4GS+QG 3G4GS
Therefore, th e tra n sfe r function o f th e g iv en b lo ck diagram is
~
r \ __________ 1 + G ^ G a + G 4 G ^ + G ^G aG 4Q 5___________
Step 11 of 19
(c)
G iven b lo ck diagram is
F ig u re ?
Step 12 of 19 ^
F irst fitidtfig feedbadc padis a nd deten n in e tra n sfe r functions. F ig u re 8 is d ie
m odified blodc diagram o f F igure 7.
Feedbackl
Feedbaclc2
F ig u re s
Step 13 of 19
T ransfer function fo r first feedback is Hi = - ^ - and
1+G ,
T ransfer fo nctioa fo r second feedbadc is
'
l+ G ,
.
Step 14 of 19 ^
F ^ u ie 9 is d ie red u cd o n diagram o f F ig u re 8.
Cascade path2
Cascade pathl
F igure 9
Step 15 of 19 ^
W e sim plify d ie blo d c diagram b y red u cin g d ie parallel com bination o£ d ie
contrcdler pa th (cascade path).
R esuttant equation fo r cascade p a d il is O iOiOj
1+G ,
R esultant equation fo r cascade p a d il is
G4GS
I+ G 4 ’
Step 16 of 19 ^
F ^ u r e 10 is th e reduction d ia ^ a m o f F ig u re 9.
F igure 10
Step 17 of 19 ^
F rom th e above block diagram , th e sum m ing equation becom es
. G.G jG, ^
1+G.
(1+G,)(1+G.)
^ G .G ,G ,( l+ G ,)+ q C ^ q G ,G ,
(l+ Q X l+ G ,)
. g|GsG,(i+Gi)+qGiqq,Gi
l-^G ^+ G ^+ G ^G t
Step 18 of 19 ^
F ^ u r e 11 is th e reduction d ia ^ a m o f F ig u re 10.
Step 19 of 19
F rom th e above block diagram , d ie tran sfer function becom es
r
.X
G ,G ,G .(l+ Q ,)+ G i< ^C ^Q .G , ^
1+G 2+G 4+G ^G,
G^G^Gf ( 1+ G 2)+G^G 2G^G4Gj + G 7(l+ G ^ + G 4 + ^ 0 , )
i+q+q+qq
Therefore, d ie tra n sfe r funetkm o f th e g iv en b lo d c diagram is
r
G ,G .G ,(l+ G ^ )+ G j 0 ,G3G , G i+ Q , ( l + q i+ 0 4 + q ,Q .)
R
1+G 2+G 4+G ^G i
Problem 3.21 PP
Find the transfer functions for the block diagrams in Fig., using the ideas of block-diagram
simplification. The special structure in Fig.(b) is called the “observer canonical form” and will be
discussed in Chapter 7.
Figure Block diagrams
S te p -b y -s te p s o lu tio n
step 1 of 14
(a)
Refer to Figure 3.54 (a) in the text book for the block diagram.
Redraw the block diagram by reducing the feedback loops.
Figure 1
Here.
G,
o; =
Step 2 of 14 ^
Add fonvard path to
and draw the reduced block diagram.
Figure 2
Step 3 of 14
Determine the transfer function of the system, — .
Jt
g |( l- K ^ )
Substitute — ^ —
for GC bbd — —*— for G f .
G ,(l-G ,g , + G ,)(l-G ,g ,)
( i- G , H , ) ( l - G ^ , ) + G ,( \- G ,H ,* G ,) G ,
G ,{ l-G ,H ,)(i-G ,H ,) + G ,G ,{\-G ,H ,)
'(} -G ,H ,)( l-G ,H ,)* G ,G fi,* G ,G ,(l- G ,H ,)
Thus, the transfer function of the system is
G ,(l-G ,g,)(l-G ,tf.)-bG ,G ,(l-G ,g,)
{i-G ,H ,){l-G ,H ,)+ G ,G ,G , + G ,G ,(l-G ,H ,)
Step 4 of 14
(b)
Refer to Figure 3.54 (b) in the text book for the block diagram.
Simplily the circuit by moving the block ^ before the summing point. By moving the block
becomes
and the total block becomes
, it
The blocks below the summing point
—» and a, form a feedback loop and the total block value becomes
Draw the reduced block diagram.
Figure 3
Step 5 of 14
Repeat moving the blocks before the summing point for b^+bfS to get
value becomes
points,
•¥b^s)s and the total
above the summing point. The blocks below the summing
and — are in series and the total value becomes —; -------r . This block forms s
feedback loop with
and the total value becomes
Draw the reduced block diagram.
Figure 4
Step 6 of 14
Similarly, the blocks
1
and — are in series and equivalent value is
. This block forms a feedback loop with
and the total value of the block
I
below the summing point becomes
s ^ * a ,s ^ + a ^ + a .
Draw the reduced block diagram.
-or
b j + ( b 2 + b ^ s )s
s ^ + O jS ^ + O j S + a j
Figure 5
Step 7 of 14
Determine the transfer function of the system, — .
Jt
_
biS^+bjS + bj
bjS + b ^ + bj
Thus, the transfer function — of the system is
Step 8 of 14
(c)
Refer to Figure 3.54 (c) in the text book for the block diagram.
The blocks above the summing points are reduced by shifting the blocks left to the summing
point and the blocks below the summing points are simplified by reducing the innermost loop first
and proceed further.
Draw the reduced block diagram.
Figure 6
Step 9 of 14
The resultant value of blocks — and a, which are forming a feedback loop is --------. Now shift
S
* + <*!
the block b^
to the summing point. The resultant value at the top of the summing point is
* j( 4 + a , )
The blocks — !— and — are in series and the resulting block forms a feedback loop with a,.
s+a,
s
1
The resultant value is
• . Now shift the block b^ left to the summing point. The
blocks b^, ^ ( 5 + a |).e n d
are summing above the summing pointsand the
resultant value is
The blocks below the summing points,
^
• and i . are in series.
4 (4 + a ,)+ < ^
,
Step 10 of 14
Draw the reduced block diagram.
Figure 7
Step 11 of 14
The resultant value below the summing point is,
1
__
1
1+
(
1
]
s{s' + a,!,5 +
s^ + a ^ s ^+ a ^+ a ^
Draw the reduced block diagram.
R
1
b . + i^ ( s + a . \+ h ( s ^ + a s + a ,\
o-
-o r
Figure 8
Step 12 of 14
Determine the transfer function of the system
^5^+(a,6^-f
Y
+ a^b^ + b^
j -F
Thus, the transfer function — of the system is
4i5* + (0 |i| + ^ ) j- F a , 62+ a 2^ +4^
J +0,5
+
Step 13 of 14
(d)
Refer to Figure 3.54 (d) in the text book for the block diagram.
Redraw the block diagram by reducing the feedback loops.
Step 14 of 14
Determine the transfer function of the system, — .
R
AB’ + D
r ~ u g {ab*+ d )
for g '.
UBH
A
\.\+ b h ] ' “
l+ G
J
AB + D {\* B H )
AB+BHD+D
\ + BH + ABG+DG + BHDG
Thus, the transfer function — of the system is
AB+BHD+D
[+ BH + ABG+ DG +BHDG
Problem 3.22PP
Use block-diagram algebra to determine the transfer function between R(s) and Y(s) in Fig.
Figure Block diagram
Step-by-step solution
Step-by-step solution
step 1 of 4
Refer to Figure 3.55 in the textbook for block diagram.
The blocks
and G, are having the same common feedback loop through the block /fjT h e
resultant blocks are given as follows;
The resultant block for the feedback loop
and
is,
The resultant block for the feedback loop Gy and
\*G M y
Draw the simplified block diagram.
-QO,
Step-by-step solution
step 1 of 4
Refer to Figure 3.55 in the textbook for block diagram.
The blocks G^ and Gy are having the same common feedback loop through the block /fjT h e
resultant blocks are given as follows;
The resultant block for the feedback loop G^ and Hy is,
l* G ^ H y
The resultant block for the feedback loop Gy and Hy is.
\*G M y
Draw the simplified block diagram.
-QDraw the simplified block diagram.
!• ( ,)
Problem 3.23PP
Find the transfer functions for the block diagrams in Fig., using Mason’s rule.
Figure Block diagrams
Step-by-step solution
Step 1 of 16
(a)
Refer to Figure 3.54 (a) in the text book for the block diagram.
Redraw the block diagram by reducing the feedback loops.
Figure 1
Here.
o; =
Step 2 of 16
Draw the signal flow graph.
There are two fonvard paths and two loops.
Step 3 of 16 ^
Write the formula for the transfer function using Mason’s gain formula.
y,
P i+ j> .
Here.
p^ and
are the forward path gains
/^ and / j are the loop gains
Determine the fon/vard path gains.
p ,= C ,< ^
f t = G,
Substitute —
---- for G *.
G,G,
P i'
Step 4 of 16
Determine the loop gains.
/, = -G|C^G,’
/ 2 = - c ,g ;
Substitute — ^ — for ( j! and — —*— for G f .
I-G jf f ,
’
1 - G ,ff ,
’
'
■ [I-G .W J I.I-G .W J
_____________________
( l- G ,/f ,) ( l- G .f f ,)
l2 = -G,G",
.
g ,g .
\-G ,H ,
Step 5 of 16
Determine the transfer function.
l-G ,H ,
1+
'
{ l-G ,H ,)( \-G ,H ,)
l-G ,H ,
rG ,G , + G , ( l- G , W Q l ( l - G . f f , )
'( l-G ,W ,)(l-G /f,) + G,G,G. + G ,G ,(l-G ,//0
G ,G .( I- G .W ,)^ ■ G ,(1 -G .W .)(I-G ,W ,)
■(i - g ,« ,) ( i - g ,w,) + g ,g .G3+g ,g ,( i - g,//,)
Thus, the transfer function — using Mason’s gain formula is
G ,G 3(l-G ,//.)+G ,(l-G ,tf3)(l-G ,g,)
( l- G 3 g :) ( l- G , g .) + G ,G ,G , - t - G , G , ( l- G . W , )
Step 6 of 16
(b)
Refer to Figure 3.54 (b) in the text book for the block diagram.
Construct signal flow graph for the block diagram.
Figure 3
There are three forward paths and three loops.
Step 7 of 16 ^
Write the formula for the transfer function using Mason’s gain formula.
Y
p ,* p ,* p .
Here,
P\ ’ P i
P i^^^ ^be fonvard path gains
/j, /^and ly are the loop gains
Determine the fonvard path gains.
a
= -A
s
1 .
f t = -!■ * !
ft- ^ is
Step 8 of 16
Determine the loop gains.
,
1
,
1
I , - ----vO,
,
1
Determine the transfer function.
y . ft+ ft+ p .
R
•i
11
1 1
1
1
1
",
j +fl,5 +ajj+<ij
Thus, the transfer function — of the system is
byS ‘¥ b ^ -¥ b j
Step 9 of 16
(c)
Refer to Figure 3.54 (c) in the text book for the block diagram.
Construct signal flow graph for the block diagram.
There are three fonvard paths and three loops.
Step 10 of 16 ^
Write the formula for the transfer function using Mason’s gain formula.
y ^ ft+ ft+ ft
Pi ’ P i
P i^^^ ^be fonvard path gains
/|, /^and ly are the loop gains
Step 11 of 16
Determine the fonvard path gains.
ft - 4 ^
S te p 1 2 o f1 6
Determine the loop gains.
,
1
,
1
I , - ----vO,
,
1
Determine the transfer function.
y _ Pl± £ i ± £ l
R ~ l-l,-l,-l,
b,[^s* + a,s+ay)'*'by(s'*-a^)+by
s ^ + O fS ^ + a ^ + a ^
* { a ^ l\+ b y )s + a ^ + a ^ b 2 + l^
+OyS+ai
Thus, the transfer function — of the system is
+ (a,6^ + 4^) j + ajAj +
s +a^s +ayS+ay
Step 13 of 16 ^
(d)
Refer to Figure 3.54 (d) in the text book for the block diagram.
Construct signal flow graph for the block diagram.
5 '* -
\ + BH
There are two fonvard paths and two loops.
Step 14 of 16 ^
Write the formula for the transfer function using Mason’s gain formula.
y
ft+ft
R ~ l-l,-l,
Here.
p, and py are the fonvard path gains
/i and iy are the loop gains
Determine the fonvard path gains.
p, = AB'
Pi = D
for
\ + BH
A *
AB
UBH
Step 15 of 16
Determine the loop gains.
l,= -A B ’G
l^= -D G
Substitute
for
UBH
AB*G
ABC
l + B ff
Step 16 of 16
Determine the transfer function.
y
ft+ft
R ~ l-l,-l^
AB
*D
J+ B H
\-¥BH
AB-^BHD + D
^ \- ^ B H ^ A B G ^ D 0 - ¥ B H D G
Thus, the transfer function — of the system is
AB+BHD+D
[+ BH + ABG + DG+BHDG
+ 4^
Problem 3.24PP
UseMason’s rule to determine the transfer function between R(s) and Y(s) in Fig.
Figure Block diagram
Step-by-step solution
Step-by-step solution
step 1 of 10
(b )
Construct signal flow graph from Figure 3.52 in textbook.
-H j
-H a
Figure 1
Step 2 of 10
From Figure 1, the fon/vard path gains are 2 and loop path gain is €
Consider Mason’s rule for Figure 1.
Determine the first Fonvard path gains from Figure 1.
G 2 _________________G 4
G i
Ge
Figure 2
Step 3 of 10 ^
Refer Figure 2 and write the equation for first fonvard path gain.
Pi =
...... (1)
Thus, the first fonvard path gain is G f i 2G^G^
Determine the Second Fonvard path gains from Figure 1.
G i
Ge
Gs
G i
Figures
Step 4 of 10
From Figure 3, write the second fonvard path gain.
P 2 = G iG jG jG j...... (2)
Thus, the second fonvard path gain is G,G)G5G^.
Determine the first loop path gain from Figure 1.
-H i
Figure 4
Step 5 of 10
From Figure 4, write the first loop path gain.
/,= - G ,G jG ,G * //,
(3)
Thus, the first loop path gain is
Determine the second loop path gain from Figure 1.
-H a
Figure 5
Step 6 of 10
From Figure 5, write the second loop path gain.
I2 =-G ,G ,G ^G ,i¥,
(4)
Thus, the second loop path gain is -G jG jG ^G j/f^
Determine the third loop path gain from Figure 1.
-H i
Gi
Gs
Figure 6
Step 7 of 10
From Figure 6 , write the third loop path gain.
/, = -G ,G ,G ,G * /f,
(5)
Thus, the third loop path gain is -GjGjGjG^Z/j
Determine the fourth loop path gain from Figure 1.
G\
Step 8 of 10
From Figure 7, write the third loop path gain.
U =-G ,G ,G jG */f^
(6)
Thus, the fourth loop path gain is -G fijG ^G JH ^
Determine fifth loop path gain from Figure 1.
Step 9 of 10
From Figure 8 , write the fifth loop path gain.
.......(7)
Thus, the fifth loop path gain is —G4H 2
Determine the sixth loop path gain from Figure 1.
-H
2
Step 10 Of 10 ^
From Figure 9, write the fifth loop path gain.
G , / 7 j...... (8)
Thus, the sixth loop path gain is —GjJVj
Consider mason’s gain formula.
P ,* P l
(9)
Where,
p^,P 2 is the fonvard path gains.
is the loop path gains.
Substitute Equation {1),(2),(3),(4),{5),(6),(7) and (8 ) in Equation (9).
y(» )_____________________ g,G,G.G.+G,G,G.G.
I+ G 1G2G4G4/ / } ^-G iG j^tG ^/f^+G |G )G jG 4/f) + G|G}G5G4/ f 4 -\-G^H2 -¥G^H2
_______________ G |G ,(G ,G , + G ,G ,)_____________
1+ / / , ( G , + G ,) + G ,G ,( G ,G , + G ,G ,) ( / /, + f f .
Thus, the transfer function of the system is
G |G ,(G ,G , + G ,G ,)
Problem 3.25PP
For the electric circuit shown in Fig., find the following:
(a) The time-domain equation relating i(t) and v\ {t) ;
(b) The time-domain equation relating i(t) and v2 {t) ;
(c)
Assuming all initial conditions are zero, the transfer function V2 {s)
1^1 (s^ and the damping ratio ^ and undamped natural frequency ojn of the system;
(d)
The values of R that will result in v2 (fj having an overshoot of no more than 25%, assuming
v1 {t) is a unit step. L = 10 mH, and C = 4 fJF.
(d) The values of R that will result in v2 {t) having an overshoot of no more than 25%, assuming
v1 {t) is a unit step. L = 10 mH, and C = 4 fJF.
Figure Block diagrams
Step-by-step solution
step 1 of 10
The following is the given electric circuit:
L
R
Step 2 of 10
(a)
Apply KirchhofTs current law to the input loop.
v . ( o = i ^ + « < ( o 4 r '( '> *
Thus, the time domain equation relating
and V |(r)is ,
step 3 of 10
(b)
From the circuit in Figure 1, the voltage across the capacitor is,
Thus, the time domain equation relating
and v ^ {t) is.
Step 4 of 10
(c)
Apply KirchhofTs current law to the input loop.
Apply Laplace transform with zero initial conditions.
y,{s) = s U ( s ) + R I(s )+ -^ I( s )
The capacitor voltage is.
Apply Laplace transform with zero initial conditions.
( 2)
/ ( i ) = C tF ,(i)
step 5 of 10
Substitute equation (2) in equation (1).
Rearrange the terms to obtain the transfer function.
s*L+sR+—
C
7L t s^
~2+ s —
r + —n
L IC )
J £ _
R
^
*
Thus, the transfer function.
V ( 5)
( of the system is
J£ _
f '. M
step 6 of 10
Compare the characteristic equation
R
1
+ s — h— with the standard second-order
L
characteristic expression. 5*
LC
•
2^i». - j
—
LC
= of
"
Thus, the undamped natural frequency,
1
of the system is
T ie
Step 7 of 10
Substitute
j
for a>^ in 2^e>,
sLC
*
1 lC ~ L
RyfZC
2L
_ R
[C
' i ' l l
Thus, the damping ratio, ^ of the system is R
C
2 Vl
Step 8 of 10
(d)
The maximum peak overshoot is.
Where,
^ is the damping ratio which lies between 0 and 1.
For maximum overshoot of 25%,
0 .2 s = e ~ ^
—13863 = —
Step 9 of 10
Determine the value of ^ .
1.3863 s
C
1.3863
V > -« "
*
- S — = 0.441
f = 0.441^1
Square the equation on both sides,
< -* = 0 .4 4 l’ ( l - { - * )
f ’ = 0 . 1 9 5 ( l - f “)
= 0 .1 9 5 - 0 .1 9 5 f’
1 .1 9 5 f’ = 0.195
Step 10 of 10
Simplify further to obtain ^ .
0195
^
1.195
-0 .1 6 3 2
^• = 0.4
Write the expression for damping factor, ^ .
f-lJ!
The condition to be satisfied is,
* J ^ > 0 .4
2>Il
Substitute 4 x 1 0 “*^°*^ C 3 ^d 10x10“*^°*^ X . ^be equation.
4x10-*
-[■
2 V 10x10-“
1
> 0 .4
^ ( 0 .0 2 ) > 0 .4
j, > ( 0 . 4 ) ( 2 )
R> 40Q
Thus, the values of R that will result in more than 25% overshoot are |/t > 4 0 q ] .
Problem 3.26PP
For the unity feedback system shown in Fig., specify the gain K of the proportional controller so
that the output y(t) has an overshoot of no more than 10% in response to a unit step.
Figure Unity feedback system
« *)o
Step-by-step solution
step 1 of 3
Step 1 of 3
T h e fbUowiiig is tfie b lo ck diagram
a u n ity feedback system :
Step 2 of 3
T h e tran sfer function o f th e system is.
K
s(s+ 2 )
i( i + 2 )
r(j)
R (s)=
K
~ s (s + 2 )+ K
K
^+2s+K
T h e tnaxiim im p eak overshoot is»
W h ere ^ is th e da n ^n n g ratio vrfu d i lies betw een 0
F o r m axim um overshoot o f 10%,
1.
M.
0 .1 = 8 ^
Solve fo r ^
2 .3 0 2 6 =
Q
23026
- r ^ — = 0.7329
^ = 0 .T 3 2 9 4 \-C
Square o n b o th sides,
^ = 0 . 7 3 2 9 '( l- ^ )
^ ’ = 0 . 5 3 7 ( 1 - ^ “)
^ = 0 .5 3 7 -0 .5 3 7 ^ ^
1.537^^ = 0.537
Sim plify fiirther to obtam ^
^
0-537
^
1.537
^ = 0 .3 4 9 4
^ = 0 .5 9
Step 3 of 3 ^
C otn p ate ^ + 2 s + K w ife th e standard second-order aqw essioii.
2^<n=2
K=at
Substitute 0.59 for ^ in
2^01=2
2 (0 .5 9 )0 ,. = 2
1
0.59
^ 1 .6 9 5
T h e constant K is,
K =ei
= 1.695'
= 2.87
T hus, fee gatn K o f fe e proportional controller to h a v e a n overshoot o f m o re than 10% is
Io < j : < z 8 ^ .
Problem 3.27PP
For the unity feedback system shown in Fig., specify the gain and pole location of the
compensator so that the overall closed-loop response to a unitstep input has an overshoot of no
more than 25%, and a 1% settling time of no more than 0.1 sec. Verily your design using Matlab.
Figure Unity feedback system
C om pensator
Phot
Step-by-step solution
Step-by-step solution
ste p 1 of 14 ^
Consider the unity feedback system shown in Figure 1.
C o m p e n s a to r
P la n t
K
100
5+a
*+ 2 S
y ( j)
r
Figure 1
Step 2 of 14
Find the transfer function for the system shown in Figure 1.
________ lo o y
'( j+ o ) ( s + 2 5 ) + 1 0 0 X
___________ lOOK^__________
% ’ + »(o + 2 5 )+ (25a+ 100A :)
Step 3 of 14 ^
Write the standard form of transfer function for the second order system.
^ --------=■
s* + 2 ( ;6 }^ + e i
Compare the denominator term: ^*+ j(a+25)+ (25fl+ 100A T ) with the denominator in the
+2^Q>^+es^
standard form:
Compare the coefficient of s terms.
2^< » .= (a+ 2 5 )
Compare the coefficient of constant terms.
« ^ = (2 5 o + 1 0 0 X )
Step 4 of 14
Consider the following data:
The peak overshoot,
The settling time,
£ 0.25
^0.1 sec
Write the formula for settling time.
Re-write this expression.
_
4.6
f ® .= Substitute Q.l sec for t, in this expression.
.
4.6
46
b
Step 5 of 14 ^
in the expression; 2^(8), = (a + 2 5 )
Substitute 46 for
2(46) = (a+ 2 S )
0 + 2 5 = 92
0 = 67
Write the formuia for peak overshoot.
4 /, = e ^
Take natural logarithm on both sides of this expression.
b>M =
7 ’^
Square on both sides of this expression.
1
rr'
.
Simplify further.
1
« ‘ + ( \ a M , 'f
,
I
Substitute OJ25 for
I
in this expression.
(l"0-25)’
’ ~ \ ) r ’ +(ln0.25)’
= 0.4
Step 6 of 14
Substitute 0.4 for ^ in the expression: <i«i|.=46
0.4a>.=46
® '° 0 .4
®, = 115
Substitute 115 for or^ and 67 for o in the expression:
=(2So+100/C)
115= = 25(67)+100Jf
100/: = 13225-1675
11550
100
K = 115.5
Thus, the gain of the system,
is [ i i l s l
Step 7 of 14
Determine the pole location of the compensator.
Clearly, from Figure 1, the compensator has a pole aX s ^ - a
for a in the expression: s ^ - a
Substitute
s = -6 7
Thus, the pole location of the compensator is [5 a - ^ 7 | .
Step 8 of 14
Consider the transfer function:
iooa:
+ s (n + 2 5 )+ (2 5 a + 1 0 0 /:)
Substitute 115.5 for Jf,an d 67 for a in this transfer function.
100(115.5)
» W = :t
i ’ + s(6 7 + 2 5)+ (25 (67)+100(115.5))
11550
” i * + 9 2 j+ 1 3 2 2 5
Write the expression for transfer function.
y (» )
11550
« ( s ) “ * '+ 9 2 8 + 1 3 2 2 5
Step 9 of 14 ^
The input is unit step. That is r ( , ) = « (/)
Find the Laplace transfonn of unit step input.
M *)= i
. X
y(s)
11550
in the transferfunction: —r 4 = “ s-----------------' ’
R(s) *'+928+13225
1
Substitute —for
*
r{ s )
1
11550
“ * '+ 9 2 i+ 1 3 2 2 5
y ( s ) = _____ 11550______
^ '
* (* '+92*+ 13225)
Step 10 of 14 A
Consider the following function:
"550
' '
*(*'+928+13225)
Use partial fraction approach to detennine the response H O
11550
Bs+C
J
* (* '+ 9 2 * + 1 3 2 2 5 )” *
* '+ 9 2 * + 13225
11550 = X (* '+ 9 2 * + 1 3 2 2 5 )+ * (« * + C )
11550 = .4 * '+ 9 2 .4 * + 13 2 2 5 . 1 + & '+ C8
11550 = 4 * '+ B * '+ 9 2 .4 * + C 8+ 1 3 2 2 5 4
Simplify further.
11 550 = ( 4 + f l ) * ' + ( 9 2 4 + C ) * + 1 3 2 2 5 4
Compare the coefficients of 5*
Compare the coefficients of s
924< + C = 0
Compare the coefficients of constant tenris.
1322544 = 11550
v4 = 0.873
Substitute 0.873 for
in the expression: 92*4 + C = 0
92(0.873)+ C = 0
C = -80.316
Substitute 0.873 for ^ in the expression: *4<(>SsO
0.873+B = 0
5 = -0.873
Step 11 of 14
Simplify further.
,
Bs+C
, 4
^ * '° * ^ * '+ 9 2 * + 13225
Substitute 0.873 for
-0.873 for
and -$0,316 for C >n this expression.
w V - 0-873 ^ (-0 .8 73)^-80.316
s
^ j*-h92j-Fl3225
0.873
s
0.873s
+ 13225
0.873s
s *+92 j
0.873
s
( s + 4 6 ) +(105.4)
80.316
+ 13225
80.316
s *+92 j
(s + 4 6 ) +(105.4)
Step 12 of 14
Consider the following Laplace transform pairs.
------- -------- "
»c~* QO&bt
(s + a ) ^ + 6 *
b
LT ^
« "* s in 6 /
(s + a ) ^ + d *
Apply inverse Laplace transform on both sides of the following expression.
y / j V _ f 0-873^ f
*
\ r 80.316Y
0-873S
105.4
J [(*+ 4 6 )*+ (1 0 5 .4 )'J I 105.4 J [(* + 4 6 )’ +(105.4)'
,-if0.873'|
' I
*
0.873*
J
1
[(*+ 4 6 )'+ (1 0 5 .4 )'J
i- { y ( * ) } =
80.316)
. 105.4
J
/
105.4
1
[(* + 4 6 )'+ (1 0 5 .4 )'J
0.873tt(»)-0.873e-“ cos(105.4»)'
ste p 13 of 14 ^
Simplify further.
>'(/) = 0.873i((/)-0.873e-“ cos(105.4»)-0.762e'“ sin(105.4»)
Write a MATLAB code to get the plot of H O
» t = 0:0.01:0.14;
» y = 0.873 - 0.873.*exp(-46.*t).*cos(105.4.*t)0.762.*exp(-46*t).*sin(105.4.*t);
» plot(t,y);
The plot of output ^ ( / ) is shown in Figure 2.
Figure 2
Verify this design using MATLAB.
Consider the following function:
"550
' '
* ( * '+ 9 2 8 + 1 3 2 2 5 )
Write the MATLAB code for this function.
» numH = [11550]:
» denH = [1 92 13225 Oj;
» sysH = tf(numH,denH);
» impulse(sysH):
The result of this MATALB code is a plot shown in Figure 3.
Step 14 of 14
Hence, the plot obtained in MATLAB is similar to the plot obtained by theoretical calculations.
Thus, the design is verified using MATLAB.
Problem 3.28PP
Suppose you desire the peak time of a given second-order system to be less than tp.. Draw the
region in the s-plane that corresponds to values of the poles that meet the specification tp < tp.
Step-by-step solution
step 1 of 2
Second order sjrstem.
G=j.
2
s = - ^ o ^ ± c v 7 i7
_ n _
n
'*
o c iV r?
S=-5o)„±0>4
=>— < v
*p
*p
step 2 of 2
-IX + -
Problem 3.29PP
A certain servomechanism system has dynamics dominated by a pair of complex poles and no
finite zeros. The time-domain specifications on the rise time
(tr), percent overshoot {Mp). and settling time (fs) are given by
tr< 0.6 sec,
M p< 17%.
ts < 9.2 sec.
(a)
Sketch the region in the s-plane where the poles could be placed so that the system will mee
all three specifications.
(b)
Indicate on your sketch the specific locations (denoted by
that will have the smallest rise
time and also meet the settling time specification exactly.
(b)
Indicate on your sketch the specific locations (denoted by
that will have the smallest rise
time and also meet the settling time specification exactly.
Step-by-step solution
step 1 of 7
A certain servo mechanism system has dynamics dominated by a pair of complex poles and no
finite zeros.
The time domain specifications on the rise time (fr). percent overshoot (Mp), and the settling time
(fs) are given by t, 5 0.6 s,
^ 17%, /, £ 9.2 s.
(a)
Consider the percentage peak overshoot is
^ 17%.
Write the expression for percentage peak overshoot.
xl00%
Substitute
17%for
17%*
%PO
tha equation.
xl00%
0M =e
K ..1 7 ) = - ^
-1.7719*-
Step 2 of 7
Apply squaring on both sides of the above equation.
(9 .8 6 % fM
Ra ^
3.1398*^
V
3.1398-13.009f’
=0.24135
f =0.5
Step 3 of 7
Determine the damping angle.
=cos"'(0.5)
=60®
Determine the real part of dominant pole of complex pole.
Adjecentside
Hypothesis
cos60® Adjecentside
Hypothesis
1_ Adjecentside
2 Hypothesis
Adjecent$ide(a)=l and Hypothesis=2The dominant pole is.
s = a ± Jto
-l±yi.732
Step 4 of 7
Write the expression for rise time.
Substitute
0.6sfor
/j. and
0.5 for
^ in the equation.
J y ll- 0 .5 ^
0.6
.
O.S
®,>/l-0.5*
)T-tan"' (>/3)
'iT j
step 5 of 7
Simplify the equation further,
2x
0.6 -
2x
3
m
-4
]
rads/s
Step 6 of 7
Sketch the region in the j —plane.
Step 7 of 7
(b)
Sketch the region in the j —plane.
Figure 2
The specific location is denoted by x
that meets the given specifications.
figure 2 that will have the smallest rise time and also
Problem 3.30PP
A feedback system has the following response specifications;
• Percent overshoot Wp < 16%
• Settling time ts < 6.9 sec
• Rise time fr < 1.8 sec
(a) Sketch the region of acceptable closed-loop poles in the s-plane for the system, assuming the
transfer function can be approximated as simple second order.
(b) What Is the expected overshoot if the rise time and settling time specifications are met
exactly?
Step-by-step solution
step 1 of 5
(a)
Write the formula for settling time.
Here.
is the settling time,
<r is the negative part of the pole.
Substitute 5.9 for
.
6.9=± i
°6.9
•
0.66
Step 2 of 5
Write the formula for rise time.
.
1.8
Here.
r^is the rise time.
<p^is the undamped natural frequency.
Substitute 1.8 for /
1.8 = M
0>M
:*1
step 3 of 5
Refer Figure 3.24 in the text book, the value of damping ratio ^ is
0.5for maximum overshoot
of
Refer Figure 3.20 in textbook. For damping ratio ( f ) ° f 0 .5 , the location of the pole is 30®.
Draw the sketch of s- plane as shown In Figure 1.
Figure 1
Thus, the simple second order acceptable closed loop poles in the s plane shown in Figure 1.
sten 4 of 5
(b)
Refer Figure 1, locate the points where
and
line meet ^x)whlch Is the damping ratio. So
damping ratio ^ is 0.66.
Write the formula for overshoot.
M =e'
Substitute
0.66 for f •
-irriU*
=0.063
Step 5 of 5
Find the value of percentage of
Percentageof
x 100
Substitute 0.063 for A/,
Percentageof =0.063x100
=6.3%
Thus, the percentage overshoot
is |6.3%l •
Problem 3.31 PP
Suppose you are to design a unity feedback controller for a first-order plant depicted in Fig.1. (As
you will learn in Chapter 4, the configuration shown is referred to as a proportional-integral
controller.) You are to design the controller so that the closed-loop poles lie within the shaded
regions shown in Fig.2.
Figure 1 Unity feedback system
Figure 2 Desired closed-loop pole locations
tlraCs)
i r
Figure 2 Desired closed-loop pole locations
(a)
What values of a)n and ^ correspond to the shaded regions in Fig. 2?
(A simple estimate from the figure is sufficient.)
(b)
Let Ka= a = 2 . Find values for K and Kl so that the poles of the closed-loop system lie within
the shaded regions.
(c)
Prove that no matter what the values of Ka and a are, the controller provides enough flexibili
to place the poles anywhere in the complex (left-half) plane.
Step-by-step solution
step 1 of 7
Refer to Figure 3.56 for unity feedback system.
From the block diagram, the transfer function is.
I
J
s(s+ a)*{s+ K ,){IC K ,)
s(s+ a )
s’ +sa+ sKK , + KK,K,
K K ,(s + K ,)
s ‘ + s{a + K K ,)+ K K ,K ,
Step 2 of 7
(a)
Refer to Figure 3.57 for desired closed loop pole location.
Choose roots that lie in the center of the shaded region.
5=-3±J2
It is known that,
s*
±
-3±y2 - -fffl. ±
Compare on both sides.
=2
•ii'- i) -
4 -1
a , = J \3
ffl =3.6
Therefore, the value of
correspond to the shaded region is 2.6^0>, ^4.6 -
Step 3 of 7
Redraw the Figure 3.57.
Step 4 of 7
Step 5 of 7
^2 - s i n ^
sO.948
Therefore, the value of ^ correspond to the shaded region is |0.554^^S0.948[-
Step 6 of 7
(b)
From the transfer function, the characteristic equation is.
s ^ + s ( a + K K , ) + K K ^ ,= 0
Substitute 2 for a and I f o r K ^ .
s ^ + s ( 2 + 2 K ) + 2 K K ,= 0 ...... (1)
Choose roots that lie in the center of the shaded region.
s , j= - 3 ± J 2
Calculate the characteristic equation.
( j+ ( 3 + ; 2 ) ) ( i+ ( 3 - y 2 ) ) = 0
»*+6* + 13 = 0
(2)
Compare equation (1) with equation (2).
2-¥2K = 6
2K = 4
K =2
And.
2 K K ,= \i
2(2)K ,-13
13
K ,= -
13
Therefore, the values of K m i K , are 2 and — respectively.
_____ 4
Step 7 of 7
(c)
For the closed loop pole positions found in part (b), in the equation (1), the value of fc can be
chosen to make the coefficient of s take on any value. For this value of
a value of AT, can
be chosen so that the quantity K K ,K ^ takes on any value desired. This implies that the pole
can be placed anywhere in the complex plane.
Problem 3.32PP
The open-loop transfer function of a unity feedback system is
C(j) =
s(f-l-2)
The desired system response to a step input is specified as peak time fp = 1 sec and overshoot
Mp = 5%.
(a) Determine whether both specifications can be met simultaneously by selecting the right value
ofK.
(b) Sketch the associated region in the s-plane where both specifications are met, and indicate
what root locations are possible for some likely values of K.
what root locations are possible for some likely values of K.
(c) Relax the specifications in part (a) by the same factor and pick a suitable value for K, and use
Matlab to verily that the new specifications are satisfied.
S te p -b y -s te p s o lu tio n
step 1 of 10
Consider the open-loop transfer function of a unity feedback system.
<?(*)=
5(5 + 2)
The desired system response to a step input is specified by,
s (sec and
The peak time,
The overshoot, A/ =5%
Step 2 of 10
(a) Determine the transfer function as follows;
g (^ )
1+G(j)
K
__51+25_
K
s‘ * 2 s * K
The general expression for the transfer function of a second order system is - s + 2 (a ^ + a ;
Equate the coefficients of powers of s and we get,
2 -2 ^ 0 ,
(1)
Step 3 of 10
Consider the given overshoot value to solve the damping factor as follows;
100
100
=0.05
=-2.9957
( - < ) '= ( - 2 .9 9 5 7 ) '( l- f » )
8.9742
9.87
1 -f’
<•’ =0.909(l-<-')
Further simplification yields,
-0.909 -0.909C*
C’ +0.909C* = 0.909
a 0.909
^ ” 2.909
-0.69
Step 4 of 10
We know that the peak time is given e
0 > . ^
Evaluate the natural frequency by substituting the given peak time value.
/. - 1 sec
1-
=*
Substitute the value of damping coefficient in this equation to get.
-0.4761
X
^ 0.5239
=4.34
Now substitute the values of C and O)^ which are calculated earlier in the respective equations
for C and
given in equation (1) to see whether these two values are equal.
2 = 2Ca,,
(0.69)(4.34) = I
2.99
From the result it is evident that the values do not satisfy the equation. Hence, both of these
specifications cannot be met simultaneously by selecting the right value of K.
Step 5 of 10
(b) Assume that the two specifications are met for the following conditions.
8 /, = (0.05)r
/, = (lsec)r
Here, r is the relaxation factor.
On the other hand, the overshoot and peak time equations respectively are given by.
(2)
e ^ = (0 .0 5 )r
= ( ls e c ) r ......(3 )
From the part (a) we have the following conditions:
0>. = J k ......(4)
(5)
Step 6 of 10
Replace C nnd
values in equation (3) with the values obtained from equations (4) and (5).
Therefore,
s ( l $ec)r
■ ( 6)
K -\
Step 7 of 10 ^
Replace the C value in equation (2) with the value obtained from equation (5). Therefore,
(0.05) r = e ^
Further simplification yields.
(0 .0 5 )r =
i^^^^
*725 ^ o .0 5 r
Substitute the r value from equation (6). Therefore,
5 " '- 0 . 0 5 r
From the trial and error method, the value of r is,
r-2 .2 1
Step 8 of 10
Simplify equation (6) to find the expression for k •
r = -
X
K -l
a:
= 4 + i
Substitute the r value in this equation for the value of K:
K = i+ ^
( 2.21)^
9.86
V-----4.88
= 3.02
Substitute this value of K in the given conditions for
and tp as follows;
8 /,= (0 .0 5 )r
= (2.21)(0.05)
= 0.11
»,.(lsec)r
= (2.21)(1)
=2.21sec
We have calculated two specifications Af^ and
, now sketch these two root locations on the
s-plane and derive some root locations that are likely the values for K.
Step 9 of 10
(c)
We have the value of K from part (b). Use this value iC*3.02 ^nd the following MATLAB
code for plotting to verify the new specification.
K=3.02;
num=[K]:
den=[1,2, K];
sys=tf(num,den);
t=0:0.01:7;
y=step(sys,t):
Plot(t,y);
yss = dcgain(sys);
Mp = {max(y) - yss)*100;
% Finding maximum overshoot
msg_overshoot = sprintf(.Max overshoot = %3.2f%%., Mp);
% Finding peak time
idx = max(.nd(y==(max(y))));
tp = t(idx):
msg_peaktime = sprintf(.Peak time = %3.2f., tp);
xlabel{.Time (sec).);
ylabel{.y(t).);
msg_title = sprintf(.Step Response with K=%3.2f.,K);
title(msg_title);
text(1.1,0.3, msg_overshoot);
Step 10 of 10
The following is the screen shot of the MATLAB:
s te p R e sp o n se w ith K=3.02
Figure 2
Problem 3.33PP
A simple mechanical system is shown in Fig., The parameters are k = spring constant, b =
viscous friction constant, m = mass. A step of 2 N force is applied as F = 2 * ^{t) and the
resulting step response is shown in Fig. What are the values of the system parameters k, b. and
m7
Figure Mechanical system
■
*
U
u '
No b ictio
(■)
I
r~
Step-by-step solution
step 1 of 13
Refer Figure 3.58(a) in textbook to identify the direction of the spring forces on the object and
draw the free body diagram as shown in Figure 1.
Step 2 of 13
Figure 1
step 3 of 13
Consider the Newton’s law of equations of motion for any mechanical system.
F = n m ...... (1)
Where.
F is the vector sum of all forces applied to each body in a system
a is the vector acceleration of each body with respect to an inertial reference
frame.
m is the mass of the body.
Step 4 of 13
Refer figure 1 and write the equation of motion.
F = n a + b i+ tc
(2)
Step 5 of 13
Determine the transfer function of the system.
-G (s)
f( j)
Rewrite Equation
Eque
(2).
1
I
f i b
ff s*+— s +—
\
m
mj
2 b
k
5 *+ —j + —
m
G(s)-.
m
2 b
k
r + —4 + —
■(3)
Thus, the transfer function of the system is
2 b
k
s^+ — s+ —
m
m
Step 6 of 13
Substitute 0 for s in Equation (3).
G(0) = f
Step 7 of 13
Refer figure 3.58(b) in textbook.
2 G (0 ) = 0.1
k^20
Thus, the value of the system parameter k is I
Step 8 of 13
Consider characteristics equation is
...... (4)
Compare Equation (3) and (4).
fflj - - .......(5)
m
2 S 6 > . = - ...... (6)
m
Consider general formula for rise time.
.......(7)
Where.
is the rise time.
is the undamped natural frequency.
Step 9 of 13
Refer Figure 3.58(b) in the textbook.
t, v l s e c
Substitute 1 sector
in Equation (7).
i= M
Thus, the undamped natural frequency is 1.8.
Step 10 of 13
Step 11 of 13
Substitute 20 for k and 1.8 for
in Equation (5).
1.8^ = “
m
m«6.17
Thus, the value of the system parameter m is |5 , | 7 | .
Step 12 of 13
Consider general formula for maximum overshoot.
y (i,)-y H
(8)
Where,
is the peak time.
Refer Figure 3.58(b) in the textbook.
;-(/,) = 0.113
^ (o o )s O .l
Substitute 0.113 for
and 0.1 for ^^oo)in Equation (8).
0.113-0.1
0.1
>0.13
Thus, the maximum overshoot is 0.13.
Step 13 of 13
Refer Figure 3.24 in textbook. The damping ratio (? ) is 0.5
Substitute 0.5 for ^ .1 .8 for
and 6.17 form in Equation (6).
m
b=
= 2(0.5)(1.8)(6.17)
= 12.06
Thus, the value of the system parameter b is ||2.Q6| •
Problem 3.34PP
A mechanical system is shown in Fig. The mass M = 20 kg and thecontrol force, u, is proportional
to the reference input, u = Ar.
(a) Derive the transfer function from R to Y.
(b) Determine the values of the parameters k, b, A such that the system has a rise time of fr = 1
sec and overshoot of Mp = 16%, and zero-steady-state error to a step in r.
Figure Simple mechanical system
t i i z
p
-
Step-by-step solution
step 1 of 7
(a)
Refer Figure 3.59 in textbook to identify the direction of the spring forces on the object and draw
the free body diagram shown in Figure 1.
U
Step 2 of 7
Write the formula for Newton’s law of equations of motion for any mechanical system.
F = n m ..... (1)
Here.
p is the vector sum of all forces applied to each body in a system
a is the vector acceleration of each body with respect to an inertial reference frame,
m is the mass of the body.
Step 3 of 7
Refer Figure 1 and write the equation of motion.
u = A ^ + b y + f y ..... (2)
Write the expression for transfer function of the system.
R{s)
^ >
Since, the input is step response with amplitude x ■Therefore, the value of 1 /(5 ) is
Take Laplace transform for equation (2).
l/ ( s ) = ( A & ^ + t e + * ) l'( 5 )
Substitute , « ( i ) f o r 1/ ( 5 ).
(JI6’ + * » + * ) } '( » ) = . M ( j )
hts^+bs*k
A
4/fj’+-^5+
V Aa
A
M
)
T{s) =
A
M
2 A
k
4 + — 4 + ----
M
hi
A_
Thus, the transfer function of the system is
M
i b
^ ^
k
M
Step 4 of 7
(b)
Substitute 0 for s in equation (3).
r(o )= -^
M
A
Consider
r(o)=i
(4)
Consider the unit step input. So, the value of ^ is ).
Substitute i for x io equation (4).
Step 5 of 7
Consider the general form of characteristics equation.
»4+2{»»,i+a^
(5)
Compare denominator of equation (3) and equation (5).
(6)
-(/)
Step 6 of 7
Write the formula for rise time.
, = l* H e r e ,
is the rise time.
is the undamped natural frequency.
Substitute 1sec for
i= M
Substitute ( fo r it and l.S for
in equation (6 ).
A / s 0.30$
Step 7 of 7
Refer Figure 3.24 in textbook. The value of damping ratio (? ) is 0.5 for maximum overshoot of
16%
Substitute 0.5fot ? . I.S fot t»|, and 0.308 for Jt/ in equation (7).
2 x 0 .S x l.g =
0.308
4 = 0.554
Thus, the vaiue of the system parameter Jt. .,4 .and ^ is [)]. [ijand |Q.5S4| respectiveiy.
Problem 3.35PP
The equations of motion for the DC motor shown in Fig. were given in Eqs. (1-2) as
Assume that
Jm = 0.01 kgm2,
b = 0.001 N-msec,
Ke = 0.02 Vsec,
/C/=0.02Nm /A,
Ra = 10Q.
(a) Find the transfer function between the applied voltage va and the motor speed 9m-
(a)
Find the transfer function between the applied voltage va and the motor speed Om-
(b)
What is the steady-state speed of the motor after a voitage va = 10 V has been applied?
(c)
Find the transfer function between the applied voltage va and the shaft angle dm.
(d)
Suppose feedback is added to the system in part (c) so that it becomes a position servo
device such that the applied voltage is given by
va = K( dr - dm)
where K is the feedback gain. Find the transfer function between 0rand dm.
(e)
What is the maximum value of K that can be used if an overshoot M< 20% is desired?
(f)
What values of K will provide a rise time of less than 4 sec? (Ignore the Mp constraint.)
(g)
Use Matlab to plot the step response of the position servo system for values of the gain K =
0.5,1. and 2. Find the overshoot and rise time for each of the three step responses by examining
your plots. Are the plots consistent with your calculations in parts (e) and (f)?
Figure Sketch of a DC motor
S t e p - b y - s t e p s o lu t io n
step 1 of 20
(a)
Step 2 of 20
Consider the equation of motion for the DC motor.
Take Laplace transform.
(1)
Rewrite Equation (1).
S 0 .(s )
R,
K(s)
n.
b
’T 7
K .K ,)
)
t
K
J.R.
I
J. J.R.)
Substitute0.01 for J , ,0.001 fo rij, 0.02for JC,,0.02tbr JC,and10for R^.
0.02
‘ R .U )
s)
0.01x10
0 ^
0.02 x 0.02
r
0.01
0.01x10
J
0.2
5 + 0 .1 0 4
5 g ,(5 )
0.2
(2)
K .( i) “ » + 0 .1 0 4
Step 3 of 20
0.2
Thus, the transfer function of the system is
f + 0 .1 0 4
Step 4 of 20
(b)
Step 5 of 20
Consider the final value theorem.
(*)!» »
Step 6 of 20
Rewrite Equation (2).
5 (5 + 0.104)1”
Substitute 10 for
r .
•
» (1 0 )(0 .2 ).
’ 5 (5 + 0 .1 0 4 )'”
_^
° 0.104
-1 9 .2 3
Step 7 of 20
Thus, the final value theorem is ||9,2?|-
Step 8 of 20
(c)
Step 9 of 20
Rewrite Equation (2).
g,(5)
0.2
r .( 5 )
5(5 + 0 . 104)
(3)
Thus, the transfer function between the applied voltage
and the shaft angle
is
0.2
j( j+ 0 .1 0 4 ) ^
Step 10 of 20
(d)
Step 11 of 20
Rewrite Equation (3).
’
5 (5
+ 0 .1 0 4 )
Step 12 of 20
Substitute
for J'
0 .2 ^ - R
,)
5 ( 5 + 0 . 104)
'
0.2AT
0 ,(i)
(4)
5’ +0.1045 + 0.2A:
Thus, the transfer function of the system is
0.2K
5’+0.1045+0.2A:
step 13 of 20
(e)
Consider general formula for maximum overshoot.
M ,= e ^
Substitute 0.2 for
.
=e ^
02
Take natural log on both sides.
log, 0.2 * loge'*'
- 1 .6 0 9 = * ^
( V i V ) ( l - 6 0 9 ) = 5»
Square on both sides.
2.S8]
Step 14 of 20
Thus, the damping ratio is 0.455
Step 15 of 20
Consider characteristics equation is ^
...... (5)
Compare Equation (4) and (5).
2 ^ . = 0.104
0.104
2ff
Substitute 0.455 for g .
0.104
•
2 (0 .4 5 5 )
= 0.1142
Thus, the undamped natural frequency is 0.1142.
Compare Equation (4) and (5).
= 0 .2 ^: ...... (6)
Substitute 0.1142 for
.
0 .1 1 4 * * 0 .2 /:
0.0129
/:= 0.2
- 0 .0 6
This, the maximum value o f i s
I
Step 16 of 20
(f)
Consider general formula for rise time.
...... (7)
Where,
is the rise time.
is the undamped natural frequency.
Substitute 4 sec for / in Equation (7).
4=M
® ,= 0 .4 5
Thus, the undamped natural frequency is 0.45.
Substitute 0.45 for
in Equation (6).
0.45* = 0 . 2 /:
a:
=1.01
Thus, the value of the system parameter K is IT o il
Step 17 of 20
(g)
Consider the characteristic equation.
S* + 2 ^ ^ + 6>1
(8)
Compare Equation (4) and (8).
5,(5) 5’+0.1045+«^
Step 18 of 20
Write the MATLAB program,
close all
K1=[0.5 1.0 2.0 6.5e-2];
t=0:0.01:150;
for i=1:1:length(K1)
K = K1(i);
titleText = sprintf('K= % 1 .4 fK );
wn = sqrt(0.2*K);
num=wn''2;
den=[1 0.104 wn''2]:
zeta=0.104/2/wn;
sys = tf(num, den);
y= step(sys, t);
% Finding maximum overshoot
if zeta < 1
Mp = (max(y)-1)*100;
overshootText = sprintf('Max overshoot = %3.2f %', Mp);
else
overshootText = sprintf('No overshoot');
end
% Finding rise time
idx_01 = max(find(y<0.1)):
idx_09 = min(find(y>0.9));
t_r = t(idx_09)-t(idx_01);
risetimeText = sprintf('Rise time = %3.2f sec', t_r);
% Plotting
subplot(3,2,i):
piot(t,y);
grid on;
title(titleText);
text( 0.5, 0.3, overshootText);
text( 0.5, 0.1, risetimeText);
end
% Function for computing rise time
fun tr = risetime(t,y)
% normalize y to 1:
y = y/y{length{y));
idxl = min(find(y>=0.1));
idx2 = min(find(y>=0.9));
if -isempty(idx1) & ~isempty(idx2);
tr = t(idx2)-t(idx1);
else
tr = 0
end
Step 19 of 20
The output for the MATLAB code is given in figure 1.
■ /V M
”-
IM
IM
Figure 1
Step 20 of 20
Thus for the value of 8: < 0 .0 6 the value of M ,< 2 0 % and if
4 seconds which is proved using Matlab and shown in Figure 1.
1,01 the rise time is less than
Problem 3.36PP
You wish to control the elevation of the satellite-tracking antenna shown in Fig. 1 and Fig. 2. The
antenna and drive parts have a moment of inertia J and a damping B ; these arise to some
extent from bearing and aerodynamic friction, but mostly from the back emf of the DC drive
motor. The equations of motion are
J0 + B$=Te,
where Tc is the torque from the drive motor. Assume that
J = 600,000 kg m2 B = 20,000 N m sec.
(a) Find the transfer function between the applied torque Tc and the antenna angle 6.
(b) Suppose the applied torque is computed so that 9 tracks a reference command 9r according
to the feedback law
(b) Suppose the applied torque is computed so that 9 tracks a reference command 9r according
to the feedback law
Tc = K ( d r -d ),
where K is the feedback gain. Find the transfer function between 9rand 9.
(c)
What is the maximum value of K that can be used if you wish to have an overshoot Mp <
10%?
(d)
What values of K will provide a rise time of less than 80 sec? (Ignore the Mp constraint.)
(e)
Use Matlab to plot the step response of the antenna system lor K = 200,400,1000, and
2000. Find the overshoot and rise time of the four step responses by examining your plots. Do
the plots to confirm your calculations in parts (c) and (d)7
Figure 1 Satellite-tracking antenna Source: Courtesy Space Systems/Loral
Step-by-step solution
step 1 of 8
Given motion equation
J 9 + B 0 = T,
In L ^ la c e domain this circuit is as below
Js^S(s)+Bs&(s) = %{s)
_____ (1)
s (J s + B ) ............
T,[s)
Given values
/ s 600000kg.m^ and f s 20000N.m.sec
Now equation (1) becomes
f W
___________1
T ;(s )
s
S (s)
(6 0 0 0 0 0 s + 2 0 0 0 0 )
1.67x10-*
Step 2 of 8
(b)
Given T , = K { 0 ,- e )
Now equation (1) becomes
^
^ [6 ,(s )-6 ( s )]
1
s[Js+ B )
£l(s) [ s ( Js+ f i ) ] + ^ :6 (s ) = r e , (s)
K
6 (e) _
6, ( b) ” [ b (J s + 6 ) ] + ^
6 (e) _
e, ( e )
K
[ E ( 600000E +
6 ( e)
20000) ] +
jr
1 6 7 x 1 0 -* ^ -
+1.67x10-* a:
step 3 of 8
(c)
T akt
^
6 , ( e)
■___
[ e (Je + 6 )]+ A T
K
^ '(s ) _______ J
(IF!
By con^aring this with standard equation we can get
B
and 2 ^ ^ =
■J
(2)
Step 4 of 8
We know that
0.\ = e
2.3026=.
(2.3026)*
+(2.3026)"
i = 0.591155......... (3)
Step 5 of 8
Now from equations (2) and (3) we can write
0.591155 =
6*
K =-
(2 x0.591155)V
K =
__________
20000^___________
(2x0.591155)" (600000)
20000"
K =
(2x0.591155) (600000)
a: = 476.93
|AT <476.931
Step 6 of 8
(<5
t. = H < 8 0
1.8
^ O L > ----
80
IF 1.8
^80
=>|A:> 303.751
step 7 of 8
w
fig u r e ;
s u b p l o t ( 2 , 2 , 1) ;
K=200;
n u m = [K /6 0 0 0 0 0 ] ;
d e n u m = [l 1 / 3 0 K / 6 0 0 0 0 0 ] ;
s t e p ( n u m , denum )
a x is ( [0 ,4 0 0 ,0 ,1 .4 ]) ;a x is
g r id ;
s u b p l o t ( 2 , 2 , 2) ;
K=400;
n u m = [K /6 0 0 0 0 0 ] ;
d e n u m = [l 1 / 3 0 K / 6 0 0 0 0 0 ] ;
s t e p ( n u m , denum )
a x is ([0 ,4 0 0 , 0 , 1 . 4 ] ) ; a x is
g r id ;
s u b p l o t ( 2 , 2 , 3) ;
K=1000;
n u m = [K /6 0 0 0 0 0 ];
d e n u m = [l 1 / 3 0 K / 6 0 0 0 0 0 ] ;
s te p ( n u m ,d e n u m )
a x i s ( [ 0 , 4 0 0 , 0, 1 . 4 ] ) ; a x i s
g r id ;
s u b p l o t ( 2 , 2 , 4) ;
K=2000;
n u m = [K /6 0 0 0 0 0 ];
d e n u m = [l 1 / 3 0 K / 6 0 0 0 0 0 ] ;
s te p ( n u m ,d e n u m )
a x i s ( [ 0 , 4 0 0 , 0, 1 . 4 ] ) ; a x i s
g r id ;
(* s q u a re *) ;
( * s q iia r e * ) ;
(* s q u a re *) ;
(* s q u a re * ) ;
Step 8 of 8
LF
.
f - ....
............
200
bM(MC)
,
.....
...
step 5 of 5
^, = 1.5
0 , s l . 2 , T s O.94
Problem 3.38PP
In aircraft control systems, an ideal pitch response {qo) versus a pitch command {qc) is described
by the transfer function
Qa(s) _
raig(i + l/ r )
The actual aircraft response is more complicated than this ideal transfer function; nevertheless,
the ideal model is used as a guide for autopilot design.
Assume that fris the desired rise time and that
1.789
l - l l
T ~ tr*
t=0.89.
Show that this ideal response possesses a fast settling time and minimal overshoot by plotting
the step response for tr = 0.8,1.0,1.2, and 1.5 sec.
Step-by-step solution
step 1 of 5
For tp = 0.8
= 2.23
= 0 .5
^ = 0.89
t
Step 2 of 5
Step 3 of 5
=1.789
s r = 0.625
Step 4 of 5
r, = 1.2
=1.5. T = 0.75
Step 5 of 5
I, = 1.5
0 , = 1.2. t =0.94
1 \
0.5 \
' 1 V
; 1/
1
1.6
.
•.
Problem 3.39PP
Approximate each of the transfer functions given below with a second-order transfer function.
c. m =
(O.Sm -1)(»+D
( 0 J 5 J + 1)(0 .9 5 j + 1)( j 2 + j + !)■
<hW =
( 0 . 5 i+ l ) ( i+ l)
(OJSj + 1)(0.95j + 1)(j2 + O J 1 + 1)’
(0.9SI+ 1)(0.05j + 1)(i 2 + » + !)■
C4(J) =
C5W =
C5W =
(0.5s + l) ( l + l )
(O.S5f + 1)(0.0Sj + l)(a 2 + i+ I)’
(0.5 l+ 1)(0.02j + 1)
( O j S j + 1)(0 .9 5 j + 1){s^ + j + 1 ) ‘
f ifyuMu t i)
( 0 5 5 s + l) ( 0 . 9 5 s + l ) ( f i + S + l ) ‘
Step-by-step solution
step 1 of 8
Consider the transfer function.
(0.55 + 1)( j + 1 )
G .W =' 7( 0 . 5 5 j
+ 1 ) ( 0 .9 5 j + l ) ( i ’ + J + 1)
(1)
Rewrite Equation (1).
( 0 .5 j ^ + 1 .5 s + | )
G i(f) =
(0 .5 2 2 5 i’ + 1.5 j + 1 )( j ‘ + i + 1)
1
(»’ + i + l )
Thus, the approximated standard second order system <?, ( j ) is
Step 2 of 8
Consider the transfer function.
( 0 .5 g + l ) ( 5 + l )
Gi W = 7
’ (0.55 j + 1)(0.95 j + l ) ( j * + 0.2 j + 1)
(2)
Rewrite Equation (2).
(0 .5 j’ + I .5 i + l)
G ,( i) =
(0 .5 2 2 5 i’ +1.5* + l ) ( j ’ + 0.2 j + 1)
I
( j ’ + 0 .2 i+ l)
Thus, the approximated standard second order system <7^ (s) is
( j ' + 0.2j + | )
Step 3 of 8
Consider the transfer function
( - 0 .5 » + l) ( » + l)
G ,(* ) =
’ ( 0 .9 5 » + l)( 0 .0 5 i+ l ) ( i ’ + i + l)
(3)
Rewrite Equation (3).
(-O .S» + l ) ( i - H )
G .W =
(0 .9 5 s+ l)(0 .05i + 1)(«* +* + 1)
[v (» + l ) « (0.954 + 1)]
Step 4 of 8
Step 5 of 8
In Equation (3), (0.054+1) is the real pole in the right half plane, which increases
in the
homogeneous response. This leads the system to be unstable.
' '
( j ’ +4 + l)
Thus, the approximated standard serxrnd order system G, ( 4 ) i
( -0 .5 4 + 1)
( 4“ +4 + 1)
Step 6 of 8
Consider the transfer function.
(0.54+1)(4+1)
G .( * ) = 7
’ (0 .5 5 4 + 1 )(0 .0 5 4 + 1 )(4 ® + 4 + 1 )
■(4)
Rewrite Equation (4).
G .( 4 ) ________ ,
^
'
(0.554 + 1)(0.054 + 1)(4‘ + 4 + 1)
[ v (0 .5 4 + 1 )» (0 .5 5 4 + 1 )]
In equation (4), (0.054 + 1) is the extra zero in the equation. This increases the overshoot of the
response. So the equation can be approximated to second order system by neglecting
(0.054 + 1)
(filL
<7+f4^*T
" ' ( 4'+ 4 + 1)
Thus, the approximated by standard second order system <7, ( 4) i
(4 + 1)
( 4* + 4 + 1)
Step 7 of 8
Consider the transfer function.
G ,(4)=
(0.54 + I)(0.024 + I)
' (0.554 + 1)(0.954 + 1)(4’ + 4 + 1)
-(5)
Rewrite Equation (5).
G ,(4)=
(0.54 + 0 (0 .0 2 4 + 1 )
(0.554 + 0 (0 .9 5 4 + 0 ( 4“ + 4 + 1)
G ,( 4 ) .
( 0.954 + 0 ( 4’ + 4 + 1)
[ v (O .54+ 1)«(O .554+ 0]
Step 8 of 8
In equation (5), (0.954 + 1) is the extra pole in the equation. This increases the rise time of the
response. So the equation [cannot be approximated! to second order system as there is extra
pole ( 0.954 + 1) in the system.
Problem 3.40PP
A system has the closed-loop transfer function
m
_________ 2700(1-1-25)_________
( l+ l) ( s + 4 S ) ( l + 60)(l2 + gs+2 5)’
*(J)
where /? is a step of size 7.
(a) Give an expression for the form of the output time history as a sum of terms showing the
shape of each component of the response.
(b) Give an estimate of the settling time of this step response.
S t e p - b y - s t e p s o lu t io n
S t e p - b y - s t e p s o lu t io n
step 1 of 5
Consider the closed loop transfer function.
2 ,7 0 0 (j-i-2 5 )
T (s) =
(1)
( i+ l)( s + 4 5 ) ( i+ 6 0 ) ( j= + 8 s + 2 5 )
Step 2 of 5
Consider the general formula closed loop transfer function.
Substitute 0 for s in Equation (1).
r(o)
2 ,7 0 0 (2 5 )
°(1 )(4 S )(6 0 )(2 5 )
Step 3 of 5
Rewrite Equation (1).
2.700(^ -t-25)
r ( j) =
( i+ l)( s + 4 5 ) (i+ 6 0 )( j’ + 8s+ 2 5 )
(>»(*))
7
Substitute — for J?(^)
s
2 ,7 0 0 (^ h»25)
( i + l) ( j+ 4 5 ) ( i+ 6 0 ) ( » ’ + 8s ^
+ 25)
(7 )
' '
Rewrite y ( ^ ) -
B
C
s+ 4 5
s+ 6 0
„/ V 7
s
i+ 1
Ds + E
(s+ 4 )’ +9
Assign D s + E - F ^^d take Inverse Laplace Transform.
y(t) = l ( t ) * A e - + B e ^ + C « - * + F e *' sin ( 3 /+ ^ )
Thus, the time function ^ ( r ) is |7 (/) + ,4^^'
+Cg~*“ + f t '* s i n ( 3 t + ^ ) |
Step 4 of 5
(b)
Settling time is set by the first pole at -1.
Consider general formula for settling time.
t,=—
.... (2 )
a
Where,
is the settling time
cr is the negative real part of the pole.
Substitute 1 for <r.
'<=T
« 4 .6 s e c
Thus, the settling time by the first pole t, is S I
Step 5 of 5
Consider the settling time is set by the second pole at -45.
Substitute 45 for <r in Equation (2).
'
45
sO .lO sec
Thus, the settling time by the second pole
is | o jo | .
Problem 3.41 PP
Consider the system shown in Fig, where
Figure Unity feedback system
OC»)
-o m
Find K, z, and p so that the closed-loop system has a 10% overshoot to a step input and s
settling time of 1.5 sec (1 % criterion).
Step-by-step solution
Step-by-step solution
step 1 of 6
Refer to the block diagram of problem 3.41 in the text book.
The transfer function of the closed loop system is,
y (^ )
G M oM
«(s)
l + G(i)D(i)
Substitute —
^ for G ( j ) and
K {s+ z)
1
K (j)
j
for G (5 )-
(5 + 3)
s
+p
K (s^-z)
« ( j + 3 ) s +p
1
K {s + z)
5 ( j + 3 ) { j + /» )+ A T (j+ r )
Step 2 of 6
The general fonn of the second order system with an extra pole is,
H {s )-
a^p
( j + p ) ( 5* + 2^0)^ +
)
Choose 2 s 3 that results in pole-zero cancellation to reduce the transfer function into general
form.
y (^ )
^ ( j+ 3 )
j ( f + 3 ) ( j + />) + A r(5 + 3 )
K
s{s + p )+ K
K
s^ + ps + K
Compare the denominator with the general second-order transfer function denominator.
2 fa , = p
ai =K
Step 3 of 6
The specifications of the closed-loop system are 10% overshoot and 1.5 s of settling time.
Write the formula for peak overshoot.
Equate the overshoot to 0.1.
0.1
=« ^
- = ln (0 .l)
= 2.3026
* * ^ ' = 2 .3 0 2 6 ^ (1 -^ *)
= 0 .3 495
<• = 0.591
Step 4 of 6
Write the formula for the settling time, t, .
Substitute 1.5 s for
15=
and 0.591 for ^ .
{O.S9\)a>,
B5.189rad/s
Step 5 of 6
Determine K by substituting 5.189 rad/s for
in the relation.
K = oi
= 5.189’
= 26.93
Determine the value of p by substituting 5.189 rad/s for
and 0.591 for ^ in the relation.
p = 240,
= ( 2 )( 0 .5 9 I ) (5 .I8 9 )
= 6.13
Thus, the value o f /< is 126.931 •
The value of z is [ ^ .
The value of p is |6.13|.
Step 6 of 6
Verify the specifications by drawing the step response of the closed loop system using MATLAB.
K=26.93;
p=6.13;
num= K;
den=[1 p k]:
sys=tf(num,den);
step(sys)
Obtain the step response of the system from MATLAB.
Hence, the specifications are met.
Problem 3.42PP
Sketch the step response of a system with the transfer function G(s) = s/2 + 1
^ (J /4 0 + l)[(l/4 )2 + s /4 + l]‘
Justify your answer on the basis of the locations of the poles and zeros. {Do not find the inverse
Laplace transform.) Then compare your answer with the step response computed using Matlab.
Step-by-step solution
ste p 1 of 2
Step 1 of 2
c?w = r+ 1
1.40
^
( s + 4 0 ) ( s' + 4 e + 1 6 )
3 2 0 (s+ 2 )
(s + 4 0 )[s^ + 4 s+ 1 6 )
Step response Y (s) =
320 ( g + 2)
s {s + 40)
+ 4 s +16)
3 20(s+ 2)
s (s+ 4 0 ) ( s + 2 + J3.46) ( s + 2 - ^3.46)
By observing the location po les, we can say that the system is limitedly stable with
S=0
Problem 3.43PP
A closed-loop transfer function Is given below;
(A) +'] H+ *] [rfr +']
[ ( i) ^ + ( 4 ) + *] [ ( A) ^ +
+
'
]
[® fe + *1
Estimate the percent overshoot, Mp, and the transient settling time, ts, for this system.
Step-by-step solution
step 1 of 4
Step 1 of 4
Consider the closed loop transfer function.
J E
d
i h
i l i i H
Simplify H ( s )
7+1
H {s )^
4;
s+2
— 2 ___
16
4
s+2
=______ 2______
16(s ’ + 4 j + 16)
s+2
» w =
2
(1)
1 6 ( i" + 4 i + 16)
From Equation {1) write the second order system.
Step 2 of 4
Consider the characteristic equation.
y (» )= .
—
5' ' S * + 2 ^ ^ + 6>1
<2>
Where.
g is the damping ratio
a>^ Is the undamped natural frequency
Compare Equation (1) and (2).
2g(&, = 4 ...... (3)
-4
Substitute 4 for
in Equation (3).
2 f(4 ) = 4
2
^=i
= 0.5
Thus, the damping ratio g is 0.5.
Step 3 of 4
Consider general formula for settling time
,.= il
f® .
Substitute 0.5 for ^ and 4 for
'
0 .5x4
= 2.3sec
Thus, the settling time
is [2.3|.
Step 4 of 4
Consider general formula for overshoot.
J l/, = e ^
Substitute 0.5 for g .
A f,= e ^
-1.570
-g fU U
= 0.163
Thus, the percentage overshoot
is | | 6 .3 i | .
Problem 3.44PP
A transfer function. G(s), is given below:
G(s) =
[(iro)^
(iro) + *]
[(to)^+(to) + *] [3 + *] [(iro)^+ (iro) + *]
If a step input is applied to this plant, what do you estimate the rise-time, settling time, and
overshoot to be? Give a brief statement of your reasons in each case.
Step-by-step solution
step 1 of 4
Step 1 of 4
Consider the transfer function.
...
(1)
Rewrite Equation (1).
—
(j’ +j+io,ooo)
10, 000 ^______
G(i
>
— ( i '+ 1 0 t + IO O )i(i+ 5 ) —
( j ^ + IO j + 1 0 ,0 0 0 )
100 '
^5'
M 0, 000 '
'
500(i’+i+10,000)
' (*’ +15*’ + 150s + 5 0 0 )(j* + 10s + 10, 000)
G(s)=
500(s’ +s+10,000)
(s^+15s"+150s+500)(s"+10s+10,000)
(2)
Step 2 of 4 ^
In second order system it is easy to find the damping ratio using the classical method. However,
it is not easy to follow the behavior of a higher order system by the same classical method. So
evaluate the problem using MATLAB code.
Step 3 of 4
Refer Equation (2) and write the MATLAB program.
num=500*[1 1 10000];den =conv([1 15 150 500],[1 10 10000]);step{num,den)
Figure 1 shows the output for the MATLAB program.
Figure 1
Step 4 of 4
Consider the values from Figure 1.
Maximum overshoot Rise time Settling time
(M ,)%
0
( t , ) sec
0.348
0.915
Table 1
Thus, the percentage maximum overshoot
time
is 10.9151-
is
, Rise time
is |Q.348| and Settling
Problem 3.45PP
Three closed-loop transfer functions are given below:
I'M
^
« (j) -
.
I'M
*M
2
- ( i2 + 2 i + 2)’
2(» + 3)
2(lZ + 2 l+ 2 )'
'
i ;m _
m
_
_________6
(i + 3)(j 2 + 2 j + 2)'
In each case, provide estimates of the rise time, settling time, and percent overshoot to a unitstep input in r.
i^ ta n -K w -e ta n e n liitin n
Step-by-step solution
Consider the closed loop transfer function.
« (s )
( s * + 2 j + 2)
^ '
(»= + 2 j + 2 )
' '
i? W = l
Consider the characteristic equation.
(2)
Here,
^ is the damping ratio,
is the undamped natural frequency.
Compare equation (1) and (2).
2 s * » ,= 2 ...... (3)
=2
(4)
Consider general formula for rise time.
1.8
—
...... (5)
Here,
is the rise time.
Substitute >12 for
in equation (5).
, _ l -8
'- - li
=1.272sec
Thus, the rise time / is 11.2721.
Consider general formula for settling time.
..... (6)
Substitute ■ n for
in equation (3).
2 f> j2 ~ 2
1
^ ° l2
= 0.707
Thus, the damping ratio g is 0.707.
Substitute 0.707 for g and ■Ji for
in equation (6).
0.707xV2
=4.6sec
Thus, the settling time t, is m i
Consider general formula for overshoot.
Substitute 0.707 for f .
-gxO.707
= 0.0432
Thus, the percentage overshoot
is m u
Consider the closed loop transfer function.
d /..\
' '
2 ( j ’ + 2 i + 2)
^ ^
S (s ) = l
...... , , ,
' '
2 (j’ +2s + 2)
From equation (7), write the second order term.
» *+ 2 t+ 2 = 0
Consider the characteristic equation.
o>.
5’ + 24»»,f + ffl’
■(B)
Here,
^ is the damping ratio,
is the undamped natural frequency.
Compare equation (7) and (8).
2««». = 2 ..... (9)
e i= 2
r - ...... (10)
a , = >n
Consider general formula for rise time.
1.8
............( 11)
Here,
/^ is the rise time.
Substitute >12 for
in equation (11).
, _ l -8
'- - l2
=1.272sec
Thus, the rise time /^ is I1.2721.
Consider general formula for settling time.
..... ( 12)
Substitute ■ n for
in equation (9).
2 f> j2 ~ 2
1
^ ° l2
= 0.707
Thus, the damping ratio g is 0.707.
Substitute 0.707 for g and ■Ji for
in equation (12).
0.707xV2
=4.6sec
Thus, the settling time t, is E U
step 8 of 11 ^
Consider general formula for overshoot.
3 /, = . ^
Substitute 0.707 for g .
-gxO.707
J iW
^
= 0.0432
Thus, the percentage overshoot
is E m
J o f ll ^
In second order system it is easy to find the damping ratio using the classical method. However,
it is not easy to follow the behavior of a third order system by the same classical method. So
evaluate the problem using MATLAB code.
Consider the closed loop transfer function.
-
6
( s + 3 ) ( s * + 2 s + 2)
6
Y {s )^
R(s)
( j + 3 ) ( s^ + 2j + 2 )
+ 3 )^
+ 2j + 2 J
(13)
Refer equation (13) and write the MATLAB program.
num=[6]:
den=conv([1 3], [1 2 2]);
step(num, den), grid
T = tf{num, den)
damp(T)
Step 10 of 11
Figure 1 shows the output for the MATLAB program.
Figure 1
step 11 of 11
Consider the values from Figure 1.
Maximum overshoot Rise time Settling time
(M ,)%
3.65
( / , ) sec
1.69
( '. ) sec
4.54
Thus, the percentage maximum overshoot M^ \s |3.6Sl ■Rise time /^ is
time /. is 14.541
and Settling
are given below:
3 .4 6 P P
ransfer function(s) will meet an overshoot specitication of IWp < 5%?
:ransfer ftjnction(s) will meet a rise time specification of tr< 0.5 sec?
ransfer function(s) will meet a settling time specification of fs < 2.5 sec?
40
2 + 4i + 40)’
40
+ I)(j2 + 41 + 40)’
120
-i-^tr.f2^4v4.4m ’
+ l)(P + 4 i + 40)’
120
+ 3)(s? + 4 l+ 4 0 )’
20(1 + 2)
+ l)(s? + 4 l+ 4 0 )’
3«040/40I<j 2 + 1+401)
2 + 4 l + 40)(l2 + i + 901)’
Step-by-step solution
step 1 of 19
Step 2 of 19
le transfer function of the system.
40
(1)
y ^ + 4 j+ 4 0 )
Step 3 of 19
re characteristic equation
_£S_
+ 2 fa ^ + a l
(2)
amping ratio
undamped natural frequency
Equation (1) and (2).
......(3)
(4)
eneral formula for rise time.
....(5)
3e time.
6.32 for
in Equation (5).
sec
isetime
is [0 .2 8 4 1
Step 4 of 19
eneral formula for settling time.
.... (6)
6.32 for
in Equation (3).
-4
2
6.32
• 0.316
lamping ratio g is 0.316.
0.316 for g and 6.32 for
in Equation (6).
L6___
■x6.32
ec
lettlingtime
is |2.3Q|
Step 5 of 19
eneral formula for overshoot.
0.316 for g .
>^j«
ruii*
;i4
lercentage overshoot 3 /^ is 13541
Step 6 of 19
ition (1) and write the MATLAB program.
I 40];
den), grid
□ obtains the step response plot
den)
lows the output for the MATtAB program.
Step 7 of 19
le values from Figure 1.
overshoot Rise time Settling time
( '/ )
0.213
1.77
Id values are tabulated in table 2.
overshoot Rise time Settling time
( '/ )
<0.5
<2.5
from table 1 and 2. the following obsen/ations are made for the transfer ft
quation (1).
rved K
) % is 35.1 but it should be less than 5%. Hence the equation
Maximum peak overshoot condition.
rved
is 0.213 sec but it should be less than 0.5 sec. Hence the equr
d rise time condition.
rved
J is 1.77 but it should be less than 2.5 sec. Hence the equation s
sttling time condition.
Step 8 of 19
irder system it is easy to find the damping ratio using the classical metho
ry to follow the behavior of a third order system by the same classical me
le problem using MATtAB code.
Step 9 of 19
Step 10 of 19
le transfer function of the system.
40
(7)
t-l)( l* + 4 j+ 4 0 )
Step 11 of 19
rtion (7) and write the MATtAB program.
'([1 1],[1
4 401);
den), grid
den)
Step 12 of 19
le values from Figure 2.
overshoot Rise time Settling time
( / , ) sec
( '. ) sec
2.1
3.99
Id values are tabulated in table 4.
overshoot Rise time Settling time
( / , ) sec
( '. ) sec
<0.5
<2.5
from table 3 and 4. the following obsen/ations are made for the transfer ft
quation (7).
rved K
) % is 0 but it should be less than 5%. Hence the equation sat
aximum peak overshoot condition.
rved
is 2.1 sec but it should be less than 0.5 sec. Hence the equatic
required rise time condition.
rved
J is 3.99 but it should be less than 2.5 sec . Hence the equation
required Settling time condition.
Step 13 of 19
le transfer function of the system.
120
(8)
l+ 3 )(l^ + 4 i+ 4 0 )
Ition (8) and write the MATtAB program.
'([1 3],[1
4 40]);
den), grid
den)
Step 14 of 19
le values from Figure 3.
overshoot Rise time Settling time
( '/ )
0.431
1.53
Id value is from the given condition is tabulated in table 6.
overshoot Rise time Settling time
( '/ )
<0.5
<2.5
from table 5 and 6. the following obsen/ations are made for the transfer ft
quation (8).
rved K
) % is 0.869 but it should be less than 5% . Hence the equatio
d Maximum peak overshoot condition.
rved
is 0.431 sec but it should be less than 0.5 sec. Hence the equr
d rise time condition.
rved
is 1.53 but it should be less than 2.5 sec . Hence the equation
sttling time condition.
Step 15 of 19
le transfer function of the system.
20(i + 2)
(9)
j + l ) ( l '+ 4 i + 4 0 )
Ition (9) and write the MATtAB program.
01;
'(]1 1],[1
4 40]);
, grid
den)
lows the output for the MATtAB program
Step 16 of 19
le values from Figure 4.
overshoot Rise time Settling time
( / , ) sec
( '. ) sec
1.42
3.32
Id value is from the given condition is tabulated in table 6.
overshoot Rise time Settling time
( '/ )
<0.5
<2.5
from table 7 and 8. the following obsen/ations are made for the transfer ft
quation (9).
rved K
) % is 0 but it should be less than 5% . Hence the equation sa
aximum peak overshoot condition.
rved
is 1.42 sec but it should be less than 0.5 sec. Hence the equat
e required rise time condition.
rved
J is 3.32 but it should be less than 2.5 sec . Hence the equation
e required Settling time condition.
Step 17 of 19
le transfer function of the system.
3 6 ,040
4 0 l(i" + j+ 4 0 l)
j ‘ + 4 i+ 4 0 )( i^ + s + 9 0 l)
89.87
s‘ + 4 s + 4 0 ) ( i ’ + i + 9 0 l ) ( i ’ + 1 + 4 0 I )
89.87
s‘ + 4 s + 4 0 ) ( j ’ + i + 9 0 l ) ( i ’ + 1 + 4 0 l )
89.87
j ’ + 4 i + 4 0 ) (j* + 2 s’ + l,303s^ + l ,3 0 2 i + 3 6 l,3 0 l)
89.87
1 * + 4 i + 4 0 ) ( i ' + 2 i ’ + 1303s’ + 13021+361301)
( 10 )
Ition (10) and write the MATtAB program.
'(]1 4 40],[1 2 1303 1302 361301]);
den), grid
den)
lows the output for the MATtAB program
Step 18 of 19
le values from Figure 5.
overshoot Rise time Settling time
( / , ) sec
0.113
( '. ) sec
4.72
Id value is from the given condition is tabulated in table 6.
overshoot Rise time Settling time
( '/ )
<0.5
<2.5
from table 9 and 10. the following obsen/ations are made for the transfer
quation (10).
rved K
) % is 46.8 but it should be less than 5%. Hence the equation
required Maximum peak overshoot condition.
Step 19 of 19
red
is 0.113 sec but it should be less than 0.5 sec. Hence the equati
d rise time condition.
rved
is 4.72 but it should be less than 2.5 sec. Hence the equation (
required Settling time condition.
r.?ir.1s(a) Sketch the unit-step responses for G1(sJ and G2(s^, paying close attention to the transient
part of the response.
(b) Explain the difference in the behavior of the two responses as it relates to the zero locations.
(c) Consider a stable, strictly proper system (that is, m zeros and n poles, where m < n). Let y(t)
denote the step response of the system. The step response is said to have an undershoot if it
initially starts off in the “wrong” direction. Prove that a stable, strictly proper system has an
undershoot if and only if its transfer function has an odd number of real RHP zeros.
(c) Consider a stable, strictly proper system (that is, m zeros and n poles, where m < n). Let y(t)
denote the step response of the system. The step response is said to have an undershoot if it
initially starts off in the “wrong” direction. Prove that a stable, strictly proper system has an
undershoot if and only if its transfer function has an odd number of real RHP zeros.
Step-by-step solution
s te p 1 of 18
(a)
The two non-minimum phase system are.
G ,W =
’
( j + l ) ( i + 2)
(s + l ) ( i + 2 ) ( j + 3)
The unit step response for Gy ( j ) i'
2 ( j- l)
- .
—r —
^ r
for ( 7, ( e l in the above equation.
( j + I ) ( * + 2)
i;w =
i ( i + l ) ( l + 2)
Step 2 of 18
Apply partial fractions to simplify the function Yy(^ ) •
!;(,)= _ + — .+ — _
S 5+1 5+2
(1)
Use residue method to calculate coefficient x ■
_
I
2 ( .- l)
( s + 1 )(j + 2 ) L
•L
2 (0 -0
( 0 + l)(0 + 2 )
' 2
Step 3 of 18
Use residue method to calculate coefficient g .
0 (^ + 2 )JL .
_
i(£ z O |
5(54
2(-<-0
- ( - l) ( - l+ 2 )
-1
Step 4 of 18
Use residue method to calculate coefficient CC = ( s * 2 ) Y ,{ s l^
= (* + 2 )f
2 (^-0
*(*+oL
2( - 2 - 0
{-2 )(-2 + 0
' 2
Step 5 of 18
Recall equation (1).
\
A
B
C
5
5+1
5+2
i; ( j) = - + — -+ — -
Substitute ( fo r X ’ - 4 for ^ ,a n d 3 for ^ in the above equation.
' '
5
5+1
1
4
5
5+1
5+2
^ 3
5+2
It is known that.
i - ' i . i
‘■ 'f c ) " '
Apply Inverse Laplace transform.
3 ',(/) = l- 4 e ''+ 3 e '* '
Step 6 of 18
MATLAB code to sketch the unit step response for Gy ( 5)
» t=0:0.1:10;
» yl=l-4.*exp(-t)+3.*exp(-2.*t);
» plot(t,y1)
» title{'Step response for G1 (t)’)
» xlabel('time(s)')
» ylabel('y1 (t)')
Step 7 of 18
The sketch of unit response for Gy ( 5) is shown in Figure 1.
S t ^ respoase f o r G l ( t )
tim e (s )
Figure 1
Therefore, the sketch of unit response for Gy ( 5) is shown in Figure 1.
Step 8 of 18
The unit step response for <72( 5) is.
(5 + I)(5 + 2)(5 + 3)
r(A '
for ( 72( 5) in the above equation.
2 ( » - I) (.- 2 )
s ( j + l ) ( i+ 2 ) ( * + 3 )
Apply partial fractions to simplify the function
V/ \
A
B
C
D
5
5+1
5 +2
5+3
( 5 ).
,r>\
y ,( i) = - + — -+ — - + — -
(2)
Step 9 of 18
Use residue method to calculate coefficient x ■
f l) (i+ 2 ) (j+ 3 )
2 ( » - l) ( ^ - 2 )
(5 + 1){5 + 2)(5 + 3)
3 (0 -l)(0 -2 )
( 0 + l) ( 0 + 2 ) ( 0 + 3 )
3 (-l)(-2 )
(1 )(2 )(3 )
-I
Step 10 of 18
Use residue method to calculate coefficient g .
= ( j+ l)
3 ( ^ - l) ( » - 2 )L . 1
5(5 + 1){5 + 2)(5
_ 3 M K fz 2 )
5 (5 + 2 )(5 + 3)
3 (-1 -1 )(-1 -2 )
- ( - l ) ( - l + 2 )(-l+ 3 )
- ii
‘ -2
—9
s te p 11 of 18
Use residue method to calculate coefficient C-
- ( . i J
1
3 ( ^ - l) ( ^ - 2 ) |
5(54
» + ') ( ^ + 3 ) L ,
3 ( -2 -l)(-2 -2 )
(_ 2 )(-2 + 1 )(-2 + 3 )
_36
' 2
-1 8
Step 12 of 18
Use residue method to calculate coefficient / ) .
f= ( * + 3 ) j;M L _
-(X I3 )[
1
3 ( » - l) ( ^ - 2 )
i ( i + l ) ( j + 2 ) [_ j
3 (-3 -1 )(-3 -2 )
(_ 3 )(-3 + 1 )(-3 + 2 )
"-6
— 10
Step 13 of 18
Recall equation (2).
A
B
C
D
5
5+1
5 +2
5+3
K ( 5 ) * — + -----+ -------- + -------
-1 0 for C ill the above equation.
Substitute ( fo r x ^ - 9 for q , 18 for
5
5+1
1
9
5
5+1
5 +2
^ 18
5+2
5+3
10
5+3
It is known that.
i - ' i . i
‘■ 'f c ) " '
Apply Inverse Laplace transform.
3'2(/) = l- 9 e - ' + I85‘ *'-1 0 e **'
Step 14 of 18
MATLAB code to sketch the unit step response for G^ { 5) ^
» t=0:0.1:10;
»y2=1-9.*exp(-t)+18.*exp{-2.*t)-10.*exp(-3.*t);
» plot(t,y2)
» title{'Step response for G2(t)')
» xlabel('time(s)')
» ylabel('y2(t)')
Step 15 of 18
The sketch of unit response for ^ 2 ( 5 ) is shown in Figure 2.
S t ^ respoase fo r G 2(t)
^1-------------------- ^--------------------
Problem 3.48PP
Find the relationships for the impulse response and the step response corresponding to Eq. for
the cases where
(a) the roots are repeated.
(b) the roots are both real. Express your answers in terms of hyperbolic functions (sinh, cosh) to
best show the properties of the system response.
(c) the value of the damping coefficient,
Is negative.
HW =
HW =
_________£5_________
(i + { < b,)2 + <o5 ( 1 - !: 2 ) '
Step-by-step solution
step 1 of 5
w
Li this case we have ^ = 1 ■
Find H(s).
m
-
U [.)
( s + a i)
Step 2 of 5 ^
For the impiilse response U(s) = 1, we get
k{t) = aj'te-"-'.
We can then integrate the in^ulse response to obtain the step response.
Alternatively, for a unit step input, U (s) = —
By applying inverse Laplace we get
Step 3 of 5
(b)
Fin<)H(s).
H [s ):
'U W
H {s ) = --------------------
^ -------------------
- where |^] > 1
For the impulse response U{s) = 1.
F ind^(0.
*(0 -
- lj
h{t)=
step 4 of 5
We can then integrate the in^ulse response to obtain the unit step response.
Alternatively, for a unit step ii^ut,
) = - and using partial &action expansion, &nd
m _________ 1_________
r(r)
__________1_________
' I
Simplify further.
1+
Y(s) =
1
j'( s ) = i+ —
!—
. - f i. '
____ 1
1
24^
r ( s ) = i + — r.:----- . - f " '
^ ‘
2 4 i> - \
Find^Cf).
Thus, we get H O = ! - « ■ *
Step 5 of 5 ^
Now we have the remaining case where C ^ negative and | ^ < 1. since we
already dealt with the case of | ^ > 1 in the previous part The impulse response and the
step responses are exactly the same, that is ^(<) - —= ^ ^ = e ~ ^ sin(<P^), except now C
k jl- C
is negative and the e^ onential terms become imboimded and the system is unstable.
Problem 3.49PP
Consider the following second-order system with an extra pole;
«W = ;
( * + P )(^ + 2f
Show that the unit-step response is
y(0 = 1 +Ae~^ +Be-^* an(coat - e),
where
A=
«3 -2 < a i« p + p 2 ’
P
-{
P -{« *
(a)
Which term dominates y(t) as p gets large?
(b)
Give approximate values for A and B for small values of p.
(c)
Which term dominates as p gets small? (Small with respect to what?)
(d)
Using the preceding explicit expression for y(t) or the step command in Matlab, and assuming
that ojnr = 1 and ^ = 0.7, plot the step response of the preceding system for several values of p
ranging from very small to very large. At what point does the extra pole cease to have much
effect on the system response?
Step-by-step solution
step 1 of 5
H (s) =
(s+p)(s“ +2?a^s+i!^’ )
°>»P
Y (s)
s(s+p)(s“ +25ti>i,s+ti>«’ )
Y (s )= ^ + ^ +
Step 2 of 5
Solving, w e get
K i= l
K j= A
V
And
2
2 ^ ( K ’ )(p=-2?avP+aw’ )
<
9 -^
y f t ) = l +Ae'*‘ +Be’* sin(cDat-0)
^ Q)
(1)
''
Step 3 of 5
(a)
I * Term domnates
Step 4 of 5
(b)
“ 2^co,p
P
Step 5 of 5
(c)
(last term becomes small as P becomes small with respect to H ^term
Problem 3.50PP
Consider the second-order unity DC gain system with an extra zero.
H (A =
a»g<J + 2)
-I- 2i<on5 + ai§)
(a) Show that the unit-step response for the system is given by
'c(IE((iy( + A ),
y (0 =
where
f t =UD
(b) Derive an expression for the step response overshoot. Mp, of this system.
(c) For a given value of overshoot. Mp, how do we solve for ^and ojn7
Step-by-step solution
step 1 of 7
Consider the second order unity DC gain S3rstem with an extra zero.
z ( e^ +
<z?'l
w
We write the transfer fimction in partial fic tio n form, as
^
2 .
^
^
^
We know that^(^) =
2 .
I *.2 ■
.
z dt
Step 2 of 7
Find.y(i).
-^ = = c o s (® rfi - ^ ) + - ^ = = s in ( a y - ^
— 1^-1+ ijc o s (a > ,( - ^ ) + - ^ 1 — s in (® ,i -
^ ( 0 = 1-
Where
-1
_ L _
Simplify further.
^ (0 = 1
g3
^ ~
- (^ + A )]
Where
I ca^,-z
Step 3 of 7
We combine the last two terms in the argument o f the cosine term, and find
s /? +
.
P - tan"
\
Hence it is proved that
Step 4 of 7
(b)
we know that — ^—- = 0.
At peak time
dt
a
^ [./z’ + a>; - 2 ;a i,) cos (a>,f - fi,) ■
^
+ a j - 2gai, )sin (a>^ - /?)
cos(iV-A-4) = o
Therefore, we get
.1 __ C
4 = tan"
Step 5 of 7
F in d ^ + 4 .
_z.
f l + 4 = tan"
I
“i
J
Thus, we g ett, = —
'
JM
Find
= —^z^ - z ^ +
Step 6 of 7
Thus, we get
1
(
tan"^
z - ^
\
3
M
, = - ^ z ^ - l - a^e
Step 7 of 7
(c)
For a given
Overshoot Mp, the values o f
and ^ have to be found by trial and error. In
general, they will be different than the standard second order system values unless
z is large that is the zero is far away.
Problem 3.51 PP
The block diagram of an autopilot designed to maintain the pitch attitude d of an aircraft is shown
in Fig. The transfer function relating the elevator angle 5e and the pitch attitude d is
e(5)
W j)
5 0 (i+ l)(5 + 2 )
(sZ + 5 s + 4 0 )(l2 + 0.031+0.06)’
where d is the pitch attitude in degrees and 6e is the elevator angle in degrees.
The autopilot controller uses the pitch attitude error e to adjust the elevator according to the
transfer function
E(s)
j+ 1 0
Using Matlab, find a value of K that will provide an overshoot of less than 10% and a rise time
faster than 0.5 sec for a unit-step change in d r . After examining the step response of the system
for various values of K. comment on the difficulty associated with making rise time and overshoot
uoii ly ivieiuaL;; iiTiu a voiUc vj1 r\‘ u lai' win ^ivjviuc ai i uvci si luui ui icso u lai i i u /o ai ivj a i isc uiiic
faster than 0.5 sec for a unit-step change in d r . After examining the step response of the system
for various values of K. comment on the difficulty associated with making rise time and overshoot
measurements for complicated systems.
Figure Block diagram of autopilot
lAifcn
■
D ^)
Step-by-step solution
step 1 of 7
Consider the following block diagram of autopilot designed to maintain the pitch attitude $ of an
aircraft:
Figure 1; Block diagram of autopilot.
Step 2 of 7
Consider the transfer function relating the elevator angle, S, in degrees and the pitch attitude,
B in degrees.
5D (^+l)(^+2)_____ ^
+a0S s+O €6j
The autopilot controller uses the pitch attitude error e to adjust the elevator according to the
following transfer function.
= D{s)
4^+3)
s+ W
Step 3 of 7
The error signal £ ( i ) is.
The transfer function of the system is.
g ( i)
G (i)0 (i)
6 l,( i) ° l + G (i)D (i)
S 0 ( i+ l) (i+ 2 )
J
(i^ + 5 s + 4 0 )(j* + 0 .0 3 s+ 0 .0 6 )J l, i+ lO
5 0 (i+ l)(i+ 2 )
^ s ’ + 5 i+ 4 0 )(* ’ + 0 .0 3 j+ 0 .0 6 )Jt.
50K{s*
soa
: ( j + i )( j
(
J
V A :(i3 -3 n
J
+ 10 )
+ 2 )( j + 3)
[ (( j ^+
* + 5 j + 4 0 )( j ^+0.03
*
j + 0.06)( j + 10)
j
J
5 0 g ( j + l ) ( j + 2 ) ( j+ 3 )
1+
+ 5 j + 4 0 )( j ’ +0.03 j + 0 .0 6 )( j + 10) ]
Step 4 of 7
Simplify this equation further and we get.
J
5 0 y ( j + l ) ( j + 2 ) ( j+ 3 )
(
[ ( j ' + 5 j + 4 0 )(j’ +0.03 j + 0.06)( j + 10)J
e,(s)
+ 5j + 4 0 )( j =+0.03 j + 0 .0 6 )( j + 10)+50^C( j + 1)( j + 2 )( j + 3 )
(j "
j’
+ 5 j + 4 0 ( j ’ +0.03 j + 0 .0 6 )( j +10)
5 0 y ( j + l ) ( j + 2 )(j+ 3 )
+ 5j + 4 0 )( j =+0.03 j + 0 .0 6 )( j + 1 0 )+ 5 0 K ( j + 1)( j + 2 ) ( j + 3 )
” (j '
SOA: ( j ’ + 3j ’ + 2j * + 6j + j ' + 3 j + 2j + 6 )
7 ( j ’ + 15j '+ 9 0 j + 4 0 0 )( j ’ +0.03 j +0.06)
[+50A : ( j ’ + 3j ' + 2j ’ + S j + j ’ + 3j + 2j + 6 )
50Ar(j’ +ds’ + l U + 6 )
' r j ’ +15.03J*+(50AT+90.51)j’ + (3 0 0 /C + 4 0 3 .6 )j’ 1
[
J
+(550A:+17.4) j +(3 0 0 A :+ 2 4 )
Step 5 of 7
The output must be normalized to the final value of
for easy computation of the overshoot
and rise-time. The next step is to find the value of K that will provide an overshoot of less than
10% and a rise time faster than 0.5 sec for a unit-step change in
The following is the MATLAB function for computing the rise time:
%Function for computing rise time
function tr = risteime(t,y)
% normalization of y to 1
y=y/y(length{y));
idxl = min(find(y>=0.1))
idx2 = min(find(y>=0.9))
if -isempty{idx1) & ~isempty(idx2)
tr= t{idx2)-t(idx1);
else
tr = 0
end
Enter the following commands to plot the pitch attitude $ and also to find the peak overshoot
and rise time.
% Program to find peak overshoot and rise time and also to plot the pitch attitude
clear all
close all
K1 =[3.5 3.0 0.3 0.03];
t=0:0.01:150;
for i=1:1:length(K1)
K = K1(i);
titleText = sprintf('K=%1.4f ,K);
wn = sqrt(0.2 *K);
num= wn''2;
den=[10.104wn^2];
zeta=0.104/2/wn;
sys = tf(num, den);
y= step(sys, t);
% Finding maximum overshoot
if zeta ,1
Mp = {max(y)-1)*100:
overshootText = sprintf('Max overshoot = %3.2f, Mp);
else
overshootText = sprintf('No overshoot')
end
% Finding rise time
idx_01 = max{find{y<0.1)):
idx_09 = min(find(y>0.9));
t_r=t(idx_09)-t(idx_01);
risetimeText = sprintf('Rise time = %3.2f sec', t_r);
% Plotting the pitch attitude
subplot(3,2,i):
Plot(t,y);
grid on;
title(titleText);
text( 0.5, 0.3, overshootText);
text( 0.5, 0.1, risetimeText);
end
Step 6 of 7
The following are the plots of pitch attitude verses time for various values of X For a : = 3.5:
K= 3.5000
1.5
0.5
Max overs hoot= 17. >1
Rise time = 0.11 sei
0
50
100
Time (sec)
150
Figure 2: Step response of autopilot system for K —Z.S
Step 7 of 7
For K = 3
K= 3.0000
1.5
0.5
Max overs hoot =12.
Rise time = 0.12 se :
0
50
100
Time (sec)
150
Figure 3; Step response of autopilot system for K —i For a: = 0.3:
K= 0.3000
0
50
100
Time (sec)
Figure 4; Step response of autopilot system for K —0 3 For a: = 0.03:
K= 0.0300
150
Problem 3.52PP
A measure of the degree of instability in an unstable aircraft response is the amount of time it
takes for the amplitude of the time response to double (see Fig), given some nonzero initial
condition.
(a) For a first-order system, show that the tim e to double is
bi2
where p is the pole location in the RHP.
(b)
For a second-order system (with two complex poles in the RHP), show that
lo2
*2 = —;:— ■
fai2
*2 = —;:— ■
Figure Time to double Time Amplitude 2A
Anqditude
______
I te w
L
*2
»
Step-by-step solution
step 1 of 4
w
For the 1* order system.
Let the Transfer Function is 0 ( e ) = ———
s—p
By taking Inverse L ^ lace Transform we can get
g {t)= k e ’‘
Now consider the initial time as /g
And consider at t = tgthe an^litude of the response is as ‘A’
Then,
g ( g = t e < ' = A ............. (1)
Step 2 of 4
Now consider that at time t s ^ the an^litude is a s ' 2A’
Then,
g W = f e * = 2 A ............. (2)
By taking the ration of equation (2) to equation (1), we can get
fe"
2 A _ te *
A “ te ”
_2
p
Given that the increased time as
Therefore
ln 2
= ti-to
Where
Step 3 of 4 ^
(b)
For the second order system,
_a
Let the Transfer Function as O fs) = -*----- -=-------- r
Then, the respons e is given by taking the Inverse Ls^lace Transform o f 0{s) a
g (<) = go ^
A tt
^ sin [a ^ +cos"‘ i )
let the amplitude of the response be 'A'.
Then, theresponeis g (^ ) = gn ^
s in fa t^ +cos“* ^)= A
Where cos"* ^ = cos"* |^ |+ ^
Therefore g ( < b )= -g o -4 — -s in (® A + cos"“|{|) = A ........ (3)
iH if
Step 4 of 4
At t s
let the amplitude of the response be 2A'.
Then, tile respone is g (O = So ^
^ ^ (<Vi
Where cos'* ^ = cos”* |^|+?r
Therefore g(t^) = - g , ^
^ ^ sin
From the equation (3) |g(A))| = “ So
+ cos"* |^|) = 2 A ........ (4)
= A ..
•(5)
By taking ratio of equation (6) to equation (5), we can get
2A
wfc-41
Here a -= \{\a ,= -{a ^
Therefore
Problem 3.53PP
Suppose that unity feedback is to be applied around the listed open-loop systems. Use Routh’s
stability criterion to determine whether the resulting closed-loop systems will be stable.
(b)
= ^
W
«” <»> =
Step-by-step solution
Step-by-step solution
(a)
The transfer function of open-loop system is.
' '
+25^+35+4)
First, find the characteristic equation.
I+ ^ G (s) = 0
5 ( i ’ + 2 j* + 3 » + 4 )
j*+ 2 j*+ 3 j’
+8s +8 = 0
To determine the Routh array, first arrange the coefficients of the characteristic polynomial in two
rows, beginning with first and second coefficients and followed by the even numbered and oddnumbered coefficients.
Write the Routh array for the polynomial.
1 3
8
2 8
i* :
a b
i* :
c
d
Evaluate the variables a,b,cnoAd 2 x 3 -8 x l
_ 6 -8
2
=-l
,
2 x 8 - lx 0
16
° 2
Sa-2b
-1
-8 -1 6
-1
-2 4
d=b=^
According to Routh array a system is stable If and only If all the elements in the first column of
the Routh array are positive.
From the obtained Routh array, it is clear that the sign changes twice in the first column, so there
are two poles not in the left hand side of the plane, hence the given closed-loop system Is
unstable.
Step 5 of 11
(b)
The transfer function of open-loop system is.
XG(s)
S* ( j + l)
First, find the characteristic equation.
I + ^ G ( j) = 0
5*( j + I ) + 2 ( 5 + 4 ) = 0
5’ + 5 * + 2 j + 8 » 0
To determine the Routh array, first arrange the coefficients of the characteristic polynomial in two
rows, beginning with first and second coefficients and followed by the even numbered and oddnumbered coefficients.
Write the Routh array for the polynomial.
1 2
1 8
-6
8
>7 of 11
According to Routh array a system is stable If and only If all the elements in the first column of
the Routh array are positive.
From the obtained Routh array, it is clear that the sign changes twice in the first column, so there
are two poles not in the left hand side of the plane, hence the given closed-loop system Is
unstable.
(c)
The transfer function of open-loop system is.
First, find the characteristic equation.
I-flC G (s ) = 0
4 ( j ’ + 2 j * + j +1)
1+ - ; V ; ----- ;----------{ = 0
s‘ (s’ +2s‘ - s - l )
s‘ (s’ +2s‘ - s - l) + 4 ( s ’ +2s‘ + s + l) = 0
j’
+ 2s* + 3 j ’ + 7 j ’ + 4 j + 4 - 0
To determine the Routh array, first arrange the coefficients of the characteristic polynomial in two
rows, beginning with first and second coefficients and followed by the even numbered and oddnumbered coefficients.
Write the Routh array for the polynomial.
1 3 4
2
* :
7
4
<4 “ i
i* :
*1
step 10 of 11
Solving for C| and
we get,
30+2
15
_32
" l5
4 =4
Step 11 of 11
According to Routh array a system is stable If and only If all the elements in the first column of
the Routh array are positive.
From the obtained Routh array, it is clear that the sign changes twice in the first column, so there
are two poles not in the left hand side of the plane, hence the given closed-loop system Is
unstable.
Problem 3.54PP
Use Routh’s stability criterion to determine how many roots with positive real parts the following
equations have;
(a) s4 + 8s3 + 32s2 + 80s + 100 = 0
(b) s5 + 10s4 + 30s3 + 80s2 + 344s + 480 = 0
(c)
s4 + 2s3 + 7s2 - 2s + 8 = 0
(d)
S3 + s2 + 20s + 78 = 0
(d)
S3 + s2 + 20s + 78 = 0
(e)
s4 + 6s2 + 25 = 0
Step-by-step solution
step 1 of 5
(a)
The characteristic equation is,
* * + &t’ + 3 2 i’ + 80s + 1 0 0 = 0
To determine the Routh array, first arrange the coefficients of the characteristic polynomial in two
rows, beginning with first and second coefficients and followed by the even numbered and oddnumbered coefficients.
Write the Routh array for the polynomial.
s*:
1
32
s’ :
8
80
22
100
100
s' : 43.6
s":
100
Since there are no sign changes in the first column of Routh array, the number of roots with the
positive real parts is zero.
Step 2 of 5 ^
(b)
The characteristic equation is,
s* +1 Oj ^ + 30s’ + 80s^ + 3 4 4 s + 4 8 0 = 0
To determine the Routh array, first arrange the coefficients of the characteristic polynomial in two
rows, beginning with first and second coefficients and followed by the even numbered and oddnumbered coefficients.
Write the Routh array for the polynomial.
1
30
344
10
80
480
s’ : 22
296
s’ : - 5 4 .5
480
s':
49 0
s’ : 4 80
Since there are two sign changes in the first column of Routh array, the number of roots with the
positive real parts is two.
Step 3 of 5
(c)
The characteristic equation is,
s*+ 2s’ + 7s’ - 2 s + 8 = 0
To determine the Routh array, first arrange the coefficients of the characteristic polynomial in two
rows, beginning with first and second coefficients and followed by the even numbered and oddnumbered coefficients.
Write the Routh array for the polynomial.
1 7
8
2
-2
j’ ;
8
8
»>:
-4
j* :
8
Since there are two sign changes in the first column of Routh array, the number of roots with the
positive real parts is two.
Step 4 of 5
(d)
The characteristic equation is,
*’ + i ’ + 2 0 * + 78 = 0
To determine the Routh array, first arrange the coefficients of the characteristic polynomial in two
rows, beginning with first and second coefficients and followed by the even numbered and oddnumbered coefficients.
Write the Routh array for the polynomial.
1* :
1 20
1 78
-5 8
78
Since there are two si'^n Ghan*^es in the first c-oiui 1 nf Rniith arrai/ tha niimhpr nf mnt« u/ith thp
positive real parts a two.
Step 5 of 5 ^
(e)
The characteristic equation is,
sV
6 j“ + 2 5 -0
5 * + Os’ +
+ O r+ 2 5 = 0
To determine the Routh array, first arrange the coefficients of the characteristic polynomial in two
rows, beginning with first and second coefficients and followed by the even numbered and oddnumbered coefficients.
Write the Routh array for the polynomial.
t’ :
1
6
i* :
4
12
3
i* :
12-
i* :
25
25
25
100
-2 1 3
From the characteristic equation, it is clear that there two coefficients missing so there are roots
outside of the LHP. Since there are two sign changes in the first column of Routh array, the
number of roots with the positive real parts is two.
Problem 3.55PP
Find the range of K for which all the roots of the following polynomial are in the LHP:
s5 + 5s4 + 10s3 + 10s2 + 5s + /< = 0.
Use Matlab to verify your answer by plotting the roots of the polynomial in the s-plane for various
values of K.
Step-by-step solution
step 1 of 2
s’ +Ss’ +lOs’ +lOs’ SrHtN)
oiep Iui ^
8^+58^+10s^+10s^5s^^c=0
1
10
5
5
10
k
8
-5
j+ (k-25)+ 80
+8
■1
-8
L ,.( fc - 2 5 ) ^ + 8 0 ( k - 2 5 ) l
k-25+80
*
k
Step 2 of 2
— - i( k - 2 5 ) > 0
k+55 5^
^
64k+j(k-25)(lc4-55) <0
3201rtk’ +30k-(25x55) <0
=>k^+350k-1375 <0
■ |-353.85>k>3.85l
Problem 3.56PP
The transfer function of a typical tape-drive system is given by
__________________________
j[(j + OJ)(j+ l)(j2 + 0 ^ + 4 ) ] ’
where time is measured in miiiiseconds. Using Routh’s stability criterion, determine the range of
K for which this system is stabie when the characteristic equation is 1 + KG(s) = 0.
Step-by-step solution
step 1 of 6
Step 1 of 6
Consider the transfer function of a typical tape drive system.
, ,
__________g ( j + 4)__________
s
+ 0.5)( j + 1)(
+ 0.4j + 4)J
Step 2 of 6
Consider characteristic equation.
l + ^T G (j) = 0 .......(2)
Step 3 of 6
Substitute Equation (1) in Equation (2).
,
K (s+ 4)
1+ f
^
~Q
4|^(»+ 0 .5 )( 4 + l ) ( s + 0 .4 s + 4 ) ]
j [ ( j + 0 . 5 ) ( j + l) ( » ’ + 0 . 4 s + 4 ) ] + ^ r ( j + 4 ) = 0
s ’ + l.9 s ‘ + 5 .ls ’ + 6 .6 j ’ + 2 s + j A :+ 4A: = 0
s ’ + 1.9s‘ + 5 .1 i’ + 6 .2 s ’ + ( 2 + A : ) i + 4 X = 0
Step 4 of 6
Thus, the characteristic equation is
+1.9s*+5.L $^+ 6.2»^ + ( 2 + J l)4+4AT = 0
Apply Routh array for this polynomial
Step 5 of 6
2 +K
5.1
1.9
6.2
4K
F ig u re 1
Step 6 of 6
The system is stable if the equation satisfies the following conditions;
• All the terms in the first column of the Routh’s array should have a positive sign.
• The first column of Routh’s array should not posses any sign change.
From the above statement, the stability conditions are,
/:+ 3 .6 3 > 0
^ > -3 .6 3
And
- 8 .4 3 < ^ : < 0.78
Thus, the stability condition is |0<J^T < 0.781-
the conditions on the system parameters (a,
^ f f ^ ) to guarantee closed-loop
system stability.
Figure Magnetic levitation system
a
Re
(f+ p )
Step-by-step solution
Step 1 of 6
Consider transfer function of the system.
Step-by-step solution
step 1 of 6
Consider transfer function of the system.
21
-y+p
K,
s+ p
j '- o “
{s+ p )(s‘ - a ‘ )+ K K ,{s+ z)
__________ K K ,{ s + z ) __________
j*+ps’ +(A3Ti, - o’ +JCKjZ- po*
K K ,{s* z)
Thus, the transfer function of the system is
5^ + ps^ + ( k K ^ - < ^ ] s + K K qZ - p<^
Step 2 of 6
Step-by-step solution
step 1 of 6
Consider transfer function of the system.
21
5+p
j+ p
s ^ -a *
j’ -o ’
( » + p ) ( j’ - f l ’ ) + A ^ r ,( i+ z )
__________ K K , { s + z ) __________
j * + ps’ + (A3Tj - o’ ) z + JCKjZ- pc’
Thus, the transfer function of the system is
___________ K K ^(s*z) ___________
+ ps^ + ( K K ^ - i^ ) s + K K f^ z - p ( ^
Step 2 of 6
KK(gz-pd^
K K a z -p d i
Figure I
Step 6 of 6
The system is stable if the equation satisfies the following condition.
• All the terms in the first column of the Routh’s array should have same sign.
• The first column of Routh’s array should not posses any sign change.
• All the terms in the first column of the Routh’s array should be greater than zero.
Therefore, for stability all the elements in the first column to be positive and we obtain the
following constraints.
From the above statement, write the stability conditions.
p> 0
K K ^ p - K K ^ > 0 ifK > 0 ^ p > z
K K ^-p a ‘ > 0 i/K > 0 = z > -^
KK^
Therefore, for stability all the elements in the first column to be positive system condition on the
system parameters is \ p >zland
,V 2 KK,
Problem 3.58PP
Consider the system shown in Fig.
Figure Control system
A
J(f+ I)
(a) Compute the closed-loop characteristic equation.
(b) For what values of (7,
is the system stable? H int: An approximate answer may be found
using
i+ h '
for the pure delay. As an alternative, you could use the computer Matlab (SImulink) to simulate
the system or to find the roots of the system’s characteristic equation for various values of 7 and
A.
Step-by-step solution
step 1 of 2
(a)
Y (s)
s(s+ l)
1+
Ae^
s(s+ l)
s(s+ l)+ A e'^
CH Equation:
s (s+ l)4 A (l-T s) = 0
(By taking the ^proxim ation of exponential term)
s^+s-ATs+A=0
|s^+(l-AT)s+A=0|
Step 2 of 2
0>)
1
A
1-AT
0
A
1-A T>0
A >0
T <—
A
A >0 & T <-
Problem 3.59PP
Modify the Routh criterion so that it applies to the case in which ali the poies are to be to the left
of - a when a > 0. Apply the modified test to the polynomial
S3 + (6 + K)s2 + ( 5 + 6K)s + 5K = 0.
finding those values of K for which all poles have a real part less than -1.
Step-by-step solution
step 1 of 1
Shift the root to origin and then
-
Routh’s Criteria.
ste p 1 o n
Shift the root to origin and then
Replacing s by s-a.
For the given Equation
Routh’s Criteria.
( s - lf +(64*;)(s-l)“ +(5+6k)(s-l)+3k=0
=>(s=-l+3s-3s’ )+ (sH l-2s)(64k)+ (5+ 6fc)(!-l)+ 5k= 0
=> s’+ s' (-343c«) +s (3-12-2k+5+€k) -l+€4k-5.61d-5k=0
s*+(lri-3)!^+s(4M )=0
s’
1
s’
Icf3
s’ 4k-4
s"
0
0
Problem 3.60PP
Suppose the characteristic polynomial of a given closed-loop system is computed to be
s4 + (11 +K2)s3 + (121 +K■\)s2 + {K^ + K1K2 + 110K2 + 210)s + 11/C1 +100 = 0.
Find constraints on the two gains K^ and K2 that guarantee a stable closedloop system, and plot
the allowable region(s) in the (/Cl, K2) plane.You may wish to use the computer to help solve this
problem.
Step-by-step solution
step 1 of 3
Step 1 of 3
The characteristic polynomial of the closed loop system is.
j^ + ( ii+ A : 2 ) j’ + ( i2 i+ ic , y + ( x , + A : ,A :2 + iio ^ : 2 + 2 io ) i+ iiK |+ 10 0=0
Write the Routh array for the polynomial.
s*-.
I
i i + a:2
s^:
I2 I+ £ ,
HAT,+100
ac, + a:|A:2+iiojc 2+21 o
o
a
s :
IIA:,+100
b
iia :,
+ ioo
Where.
^
( ii+ A r 2 ) ( i2 i+ A :i) - ( A :i+ A :|A :2 + iiO A :2 + 2 io ) ( i)
l I + ATj
133i + i i a :,+ 1 2 ia :2 + a :, a : 2 - a : i -AT,AC;- 1 1 0 ^ 2 - 2 1 0
11+AC2
10AC,+MAC;+ 1121
II + AC2
And,
^
a(AC| + AC|AC2 + 110AC2+2l0)-(llAC, + 100)(ll+A:2)
a
+AC,AC2 + iioa :2 + 210)-(1 lAC,+100)(11 +AC2 )
) ___________________________________________
l l + AT;
V.
io / : i+ n A : a + ii 2 i
11+ ^2
( ioa: , -1- 11 ^ 2
+ ^ 1^ 2 -*-iioa :2-(-21o) - ( i i + a:2)(iia :,+ ioo)( i i -i- a:2)
io a : , + i i
/ : 2 + i i 21
step 2 of 3
For a stable system, first column elements of Routh array must be greater than zero.
Therefore,
iu a
:2 > o
And,
ii/:,-i-ioo>o
\\K t> -m
-1 0 0
Kt>-
11
Therefore, the two gain of the system are AC,
'
11
and AC2 > -11
^
Step 3 of 3
Draw the allowable region in the ( ac„ a :2 ) plane.
i 1C2
3
10 0
A llo w a b le re g io n
3
50
•lO O /ld
-150
-120
-90
-AO
-30
^
0
....................................................
1
i
-50
-100
1
1
1
1
-150
•200
1
1
1
Hence, the plot is drawn.
-250
30
60
90
120
Problem 3.61 PP
Overhead electric power lines sometimes experience a low-frequency, highamplitude vertical
oscillation, or gallop, during winter storms when the line conductors become covered with ice. In
the presence of wind, this ice can assume aerodynamic lift and drag forces that result in a gallop
up to several meters In amplitude. Large-amplitude gallop can cause clashing conductors and
structural damage to the line support structures caused by the large dynamic loads. These
effects in turn can lead to power outages. Assume that the line conductor Is a rigid rod,
constrained to vertical motion only, and suspended by springs and dampers as shown in Fig. A
simple model of this conductor galloping is
(»2+,2)l/2
Where
m = mass of conductor,
ir“fnr’c \/ortinal riicnla^om ont
m = mass of conductor,
y = conductor’s vertical displacement,
D = aerodynamic drag force,
L = aerodynamic lift force,
V=wind velocity,
a = aerodynamic angle of attack = —
T = conductor tension,
n = number of harmonic frequencies,
/ = length of conductor.
Assume that L(0) = 0 and D(0) =D0 {a constant), and linearize the equation around the value y =
^ =0. Use Routh’s stability criterion to show that galloping can occur whenever
^+D b<0.
Figure Electric power-line conductor
Step-by-step solution
step 1 of 2
D (a )!!^ -H a )V
dx
dx^
1
\ dD d a
_, .
D ( a ) j: ,- L ( a ) r
Step 2 of 2
Now
So,
-1
1
-ao
da
da
m
D { a ) x j- L { a ) V
Problem 4.01 PP
If S is the sensitivity of the unity feedback system to changes in the plant transfer function and T
is the transfer function from reference to output, show that S + T =
Step-by-step solution
step 1 of 1 ^
Given that S is the sensitivity o f the unity fe edback system to changes in the plant transfer
function and T is the transfer function &om reference to output.
Given that S is the sensitivity o f the unity fe edback sjrstem to changes in the plant transfer
function and T is the transfer function from reference to output.
We know that
^
^
S+T= — 5
1 + 01D,i
s + r = 1 + GfDd
1 + GZ)„
T=
GD.
| g + r = i|
1 + ODd
Problem 4.02PP
We define the sensitivity of a transfer function G to one of its parameters k as the ratio of percent
change in G to percent change in k.
^
d G jG
d \n G
k dG
d t/T “
d to t “
G ^ '
* “
The purpose of this problem is to examine the effect of feedback on sensitivity. In particular, we
would like to compare the topologies shown in Fig. for connecting three amplifier stages with a
gain of - K into a single amplifier with a gain of -10.
(a)
For each topology in Fig., compute j8/ so that if /C= 10, V = -10R.
(b)
For each topology, compute
when G =
(Use the respective j8/values found ii
part(a).) Which case is the least sensitive?
(b) For each topology, compute
when G =
(Use the respective j8/values found in
part(a).) Which case is the least sensitive?
(c)
Compute the sensitivities of the systems in Fig.(b, c) to j82 and /S3. Using your results,
comment on the relative need for precision in sensors and actuators.
Figure Three-amplifier topologies
Step-by-step solution
step 1 of 14
(a)
Refer to Figure 4.24 (a) in the text book.
The transfer function is.
r = - fi,K ‘R
Compare above equation with y = -10Jtf i, K ’ = 10
Substitute 10 for Kin the above equation.
10’
10- ’
= 0.01
^
=
Therefore, the value of
is |o,q i |-
Step 2 of 14
Refer to Figure 4.24 (b) in the text book.
Determine the transfer function of the system.
Compare above equation with y = -10Jt^10
( i + j cPai J)
^
.V io
l + JTA
K
1+JCA =
2.1544
Substitute 10 for Kin the above equation.
10
2.1544
1+ 1 0 ^ = 4 .6 4 2
^ = 0 .3 6 4 2
Therefore, the value of ^
is |Q.3642l-
Step 3 of 14
Refer to Figure 4.24 (c) in the text book.
Determine the transfer function of the system.
Y
R
(-iC )(-A :)(-iC )
\- { - K ) ( - K ) ( - K ) P ,
K'
r= -
\+K%
Compare above equation with y = -10Jtjt ’
= 10
i+ ic ’A
i+ jc ’A "
10
‘
a :’’ A - o. ia : ’ - i
jc
0.1k ’ - 1
A = Substitute 10 for Kin the above equation.
A = -
10*
_ 9 9 -l
1000
"
-0 .0 9 9
Therefore, the value of ^
is |0.Q99|.
Step 4 of 14
(b)
Determine the sensitivity,
,0
.
dG/G
dK/K
Step 5 of 14
Refer to Figure 4.24 (a) in the text book.
The transfer function is.
i= /j( - js r ) ( - A : ) { - A r )
G = -p ,K '
Determine the sensitivity.
-3
Therefore, the sensitivity
is [ | ] .
Step 6 of 14
Refer to Figure 4.24 (b) in the text book.
The transfer function is.
R ( i + A T A A ' + ^ A J U + J^AJ
G=( i+ jfA ) ’
dG
dK
dG
dK
d
AC’
dK .(1 + AC>^)’ .
-(1 + ACA)’ (3AC’ )
( i + a:A )‘
- 3 a: ’
Step 7 of 14
Determine the sensitivity.
S?
K
7
{
-3AT’
1
a:’
I
( i+ ^ A )’J
i+ « A
Substitute 10 for K and 0.3642 for 0^ in the above equation.
5?-
1+(10)(0.3642)
3
1+3.642
*0.646
Therefore, the sensitivity 5 ^ is |Q.646l •
Step 8 of 14
Refer to Figure 4.24 (c) in the text book.
The transfer function is.
Y
( - a : ) ( - a: ) ( - a: )
R
\- { - K ) ( - K ) { - K ) P ,
K'
G= -
l+K%
dG
dK
d_
'd K
dK
_ ^ 1
1+ a
a: ’/
’a J
- ( i + a : ’ >», ) 3AT’ + a :’ ( o + 3a : ’ ; ^ )
-3 K ^ -3 K ’P ,+ 3 K ‘P,
( l + JC’A ) '
- 3 a: ’
7
+ a: ’ a )’
i
Step 9 of 14
Determine the sensitivity.
S?
H m
-3AT’
K’
3
i + a:’a
Substitute 10 for Kand 0.099 for
in the above equation.
1+(10)’ (0.099)
3
1+99
= 0.03
Therefore, the sensitivity 5 ^ is [q^03 |.
The topology in (c) is least sensitive.
Step 10 of 14
(c)
Determii
Here,
c=-
Step 11 of 14
Refer to Figure 4.24 (b) in the text book.
The transfer function is,
R
( i + a t a A ' + ^ a J U + j^ a J
AT’
(1+ ATA)’
Write the expression for the sensitivity of the system in Figure 4.24 (b) to
■
^ J P A d G
dG
PPx
d
AC’
dp. (l+ACA)’ .
dG
= a : ’ [ ( - 3 ) ( i + a : a )- ’- ' ( a : ) ]
3AT*
0 + a : a )‘
S te p 1 2 o f1 4
Determine the sensitivity,
.
, JpA dG
\ G ) d f i,
3AT’
AT’
i
+
a: a
1
)‘ J
(1 + ATA)’
3ATA
l + Kp,
Substitute 10 for Kand 0.3642 for
.
3 ( 1 0 ) ( 0 .3 6 4 2 )
*
1 + ( 1 0 ) ( 0 .3 6 4 2 )
in the above equation.
1 0 .9 2 6
4 .6 4 2
- 2 .3 5 4
Therefore, the sensitivity 5 ^ is |~ 2 .3 5 4 l •
Step 13 of 14
Refer to Figure 4.24 (c) in the text book.
The transfer function is.
Y
R
(- K ) ( - K ) ( - K )
l- ( - K ) { - K ) ( - K ) P ,
K'
\+K%
G = —
Write the expression for the sensitivity of the system in Figure 4.24 (c) to
*
.
lo J d A
dG
Calculate ------.
r fA
dG
</ r
dp,
AT’
1
~ d p X \* K ^ P ,\
= - a: ’ ( - i ) ( i +
a: ’a
) " " ( a: ’ )
K‘
(i +
a : ’a
)'
Step 14 of 14
Determine the sensitivity, 5 ^ .
, j p
'
^
^
\G ) d p ,
A
AC’
AC‘
(1+AC’A ) ’
1 + ac’ a J
i
+
a: ’a
Substitute 10 for Kand 0.099 for 0^ in the above equation.
j,.
*
(1 0 )’ (0 .0 9 9 )
1 + ( 1 0 ) ’ ( 0 .0 9 9 )
99
1+99
.-0 .9 9
Therefore, the sensitivity 5 ^ is |-Q .99|.
The results indicate that the closed loop system is much sensitive to errors in the feedback path
than that in the fonward path.
Hence, sensors should have more precision than actuators.
Problem 4.03PP
Compare the two structures shown in Fig. with respect to sensitivity to changes in the overall
gain due to changes in the amplifier gain. Use the relation
as the measure. Select H1 and H2 so that the nominal system outputs satisfy
F1 = F2, and assume /CH1 > 0.
Figure Block diagrams
I h p '-
— { !]—
^
T—
Q
—
---------- 1
Step-by-step solution
step 1 of 10
It is given diat.
T h e eatpressio n t o m easure s e n sitiv i^ is,
r fh iF
S=-
FdK
W here.
K , th e feed forw ard gain
, th e o u ^ nit the system
iS, fe e m easure o f sensitivity
Step 2 of 10 ^
T h e U o d c d i a g r m o f l h e first structure is fe o w n in F igure 1.
F igure 1
Step 3 of 10 ^
T h e U o d c diagram o f th e second structure is show n in F ig u re 2.
F igure 2
Step 4 of 10
A ssu m in g fe e intetm ediate outyut from fe e felt b lo ck in F i ^ n e 1. C alculate th e outyut
th e b lo d c diagram in F ^ u r e 3.
F ig u re s
Step 5 of 10 ^
C alculate th e g am o f fe e b lo ck left blo d c in F ig u re 3.
7 ^ = -? ^
' \+ K H ^
(1)
C a l c u l i th e ga in o f fe e blo d c r ^ i t U o d c in F ig u re 3.
(F ,-F ^ ,)K = F ,
F ^ = F ^ {y + K H ^ )
7 ? = - ^
' 1 + ja r ,
C)
T hus, com bine equation (1 ) a n d ^ ) to obtain fee iityut outp u t relation.
p - ____ K
(
SK ^
.... ®
Step 6 of 10 ^
C alculate fe e g am o f fe e b lo ck d io w n in F ^ u r e 2.
( 7 J - 7 r ^ ,) ( ^ : ) ( J i:) = F ,
7ac’ = 7 rj(i+ J i:*7 7 j)
.......
step 7 of 10
n a It i» i^vcu uiot, luc MMpuw wi
»
th e r e la tio n b e t w e e n .? ] a n d ? 2 -
7!?77j = (l+ JE H i)“ - I
'
'
(5)
K
Step 8 of 10
T h e m easure o f sensitivity d ue to changes in fe e am plifier g a in is,
'
F [d K
d
f—
[ l+ K
(
K
H j
J» ]
( (l+ ^ ,)^ (7 i)"| 2 K {l+ X H ,f-2 {l+ K H ,){H ,)(K ^ )
J[
[
.
(i+Jsar,)’ (2A:+2is:’7r, -2 k ^h )
"
W(l+7SSf,)‘
T hus, th e m easure o f sensitivity d ue to changes in fe e a n ^ lifie r g a in fo r fee first b lo d c is,
S j = — 2—
'
....... (fi)
l+TSK
Step 9 of 10
T h e m easure o f sensitivity d ue to changes in fe e a m p litW g a in is,
'
7^1 liK
{ (l+JC ^7r,)(77)Y 2 K (1 + K ^ H ^ )- ( 2 K H ^ ) { k ^ )
2K
( K ) { l+ K ^ H ^ )
T hus, th e m easure o f sensitivity d ue to c f a a n ^ in fe e antylifier g a in f iv fee first b lo ck is,
2
S f.
l+ K ^ H ^
tu te
te fe
Substitu
feevalue
?2
fromequation(^)infeisequationtoobtainthe
2
_______ 2_______
i+ ( i: ^ 7 r j‘ + 2 K H ;)
2
s f =
2
■m
Step 10 Of 10 ^
H e n c ^ from equations (6) a nd (7) observe fea t, fe e sensitivity
seco n d b lock is related to sensitivity o f first blo ck by.
g a in c h a n ^ fo r fee
T herefine, as J i? ] is g reater fe a n zero, w e can o ra c h id e fea t fe e second b lo d c is less
sensitive
fe e first blodc.
Problem 4.04PP
A unity feedback control system has the open-loop transfer function
(a) Compute the sensitivity of the closed-loop transfer function to changes in the parameter A.
(b) Compute the sensitivity of the closed-loop transfer function to changes in the parameter a.
(c)
If the unity gain in the feedback changes to a value of
^ 1, compute the sensitivity of the
closed-loop transfer function with respect to f3.
Step-by-step solution
step 1 of 5
A uni^ feedback system has the open-bop transfer functbn
(a)
Find the cbsed loop transfer hmction r(s) v^dch is given by
r(s) =
^ ' 1 + G (s)
s ( s + a)
1+
a (s + a)
r ( a ) = - 5 — ---------
+ as + A
^ ^
Step 2 of 5
We know that
<JT(s) _ s^ + as + A - A
(s“ + as + A f
dT{s) _
^
^ + as
(s’ + OB
The sensitivity of the cbsed loop transfer hmction is given by
^ _ A ^
^ TdA'
Find the sensitivity.
^
^
S^ + os
^°T d A
r i- _
■*
A
(7 7 ^ 7 7 1 7
s (5 + a )
s(s + a) +
Thus, tile sensitivity of tiie cbsed b op transfer fimction to chaises in tiie parameter A is obtained
-I- a)
s(s + a ) + A
(b)
We know that
Fmd —
aa
dT
-sA
(s“ + os + il)
step 3 of 5
The sensitivity of tiie cbsed loop transfer fimction is given by dSf =
T da
Find tiie sensitivity.
•
Tda
^
“
a(s^ + a s + . j )
-a l
^
(7 7 ^ 7 ^ 7
s ( s + a ) + j1
Thus, the sensitivity of the cbsed bop transfer fimctionto chaises in the parameters is obtained
sH =
s ( s + a) + j4
Step 4 of 5
(0
Find T(s)
r(s) =
l + yS a W
dT
-O ^ s f
[1 + ^ ' ( s ) 7
f id T
j?(l + /Sffj
O
Tdp~
-0 “
(1 +
Find the sensitivity of the cbsed b op transfer function.
-P A
rf
5(5 + a)
- fiA
! + ■
(s + a)
Step 5 of 5
- fiA
Sf = s (s + a ) +
Thus, tiie sensitivity is obtained as s ; =
-p A
s ( s + a) + PA
Problem 4.05PP
Compute the equation for the system error for the feedback system shown in Fig.
Figure Closed-loop system with sensor dynamics. R = reference, U = control. Y = output, V =
sensor noise
Step-by-step solution
Step-by-step solution
step 1 of 2
Refer Figure 4.5 in the textbook for the closed loop system with sensor dynamics.
Consider the equation for mass of the car.
- w
\* D G H
\* D G H
- ^ v
\ + DGH
(1)
Write the output of the error detector from Figure 4.5.
E ^ R - Y ..... (2)
Step 2 of 2
Substitute equation (1) in (2).
G
UDGH
UDGH
UDGH
(X-^DGH)R-FDGR
G
„
DGH „
“---------------------------- -- ■
-tY +
V
l+ D G H
R + D G H R -F D G R
l + DGH
l +D G (H -F )
l +DGH
l+ D G H
G
W
l + DGH
G
-< -T
"
W
l+ D G H
l +DGH
+ - ^ V
l + DGH
+ - ^ V
l+ D G H
Hence, the system error equation is.
\->rDGH
1+ DGtf
X ^D G H
Problem 4.06PP
Consider the DC motor control system with rate (tachometer) feedback shown in Fig.(a).
(a)
Find values for K and
so that the system of Fig.(b) has the same transfer function as the
system of Fig.(a).
(b)
Determine the system type with respect to tracking dr and compute the system Kv in terms o
parameter fC and
(c)
.
Does the addition of tachometer feedback with positive kt increase or decrease Kv?
Figure Control system
Step-by-step solution
step 1 of 7
(a)
Refer to Figure 4.25 (a) in the textbook.
The simplified circuit of Figure 4.25 (a) is shown in Figure 1.
Figure 1
Step 2 of 7
Calculate the closed loop transfer function.
K K „K ^
(')
KKJC,
~ * [ s ( i+ r ^ ) + / a : j f c ^ ] + A M : . jr ,
KKJC.
+KKJcJa + KKJC,
■
K K ,K ,
(1)
step 3 of 7
Refer to Figure 4.25 (b) in the textbook.
Calculate the closed loop transfer function from the Figure 4.25 (b).
K'
i ( l + r ^ ) + A T '( l + * »
9
K'
(2)
Step 4 of 7
Compare the equations (1) and (2).
trt _ K K .K .
*
And
K% = KK.k,
Substitute
'
KKJC.
i f o r jc 'in K% = KKJc,
K.
Therefore, the values of j f ' and
are
k
'
K,
Step 5 of 7
(b)
The inner loop in Figure 4.25 (a) is.
!(* ) =
AA.
j( i+ r , i) + A a r , t , i
KK.
s (1 + t, s + K K J .)
KK,
s '0 + T . ^ i- K K . k , )
From the
^e denominator of the tr
transfer function, it is clear that one pole exists at the origin.
Therefore,
jre, the type of the syst
system is 1.
Hence, the type of the system is jT y p e l
Step 6 of 7
The value of K , is.
JCy s lim4Z.(4)
KK.
s lim ^
i-** 5 ( i + r . i + i a : . t , )
—
KK.
s lim
■ - - i+ r , s + / a r , * ,
KK.
1 + X K jt,
( since fiom equation (3) w e have^
[K 'k ; = K K J ,
)
KV,
K .=
k,{l+K'k;)
Therefore, the value of K , in terms of j f ' and
is
Step 7 of 7
(C)
The velocity error constant is.
Kk;
■(3)
From equation (3). it is clear that,
is inversely proportional to k,-
That is.
The addition of tachometer feedback with positive k, Idecrwisfsl AT, ■
Problem 4.07PP
A block diagram of a control system is shown in Fig.
(a)
If r is a step function and the system is closed-loop stable, what is the steady-state tracking
(b)
What is the system type?
(c)
What is the steady-state error to a ramp velocity 5.0
K2 = 2 and K^ is adjusted to give s
system step overshoot of 17%?
Figure Closed-loop system
Step-by-step solution
step 1 of 5
Refer to the figure of problem 4-7P in the text book.
Determine the transfer function of the system.
Determine the transfer function of the inner feedback loop.
s ^ + { \0 + K 2 ) s
Determine the transfer function of the system.
I M - 0 5—
for T,(s).
i ’ + ( io + A :,)i
3 ) ______ 1
y (4
(^^-10)
■ i i Q 5^ .( » ^ 3 )
»
(i+io) j=+(io+is:j)j
______________ O.SK,(s + i) _____________
” ( i + 1 0 )[j’ + (I0 + ^ j ) s ] + 0.5^:, (» + 3)
________________ 0.5J:,(5 + 3)_______________
i ’ + ( 2 0 + a: , ) j ’ + ( i o o + i o ^ :j + o . 5 A : , ) i + i . 5 i : |
Step 2 of 5
Determine the tracking error.
£ ( s ) = 0 .5 « ( i) - 0 .5 ir ( i)
- 0 .5 [ « ( j) - > r ( s ) ]
M
=0 5 l- r M l
J !(.)
i? (* )J
_______________ Q -iK ,(s * 3 ) _______________
j ’ + ( 2 0 + « ,) j* + ( 1 0 0 + 1 0 A r j+ 0 .5 A :,) s + 1.5Ar,
£M
= 0.5 1 -
R{s)
+ (2 0 +
E (s) = 0.5 1 -
R (s)
O.SK,{s + i)
K j)s^ + (1 0 0 + 1 0 ^ :, + 0 . 5 K ,) i + 1.5AT,
0 .5 3 :|(t + 3)
j ’ + ( 2 0 + ^ r j) j^ + ( io o + iO A :i+ o .5 A r ,) i+ i.5 i:,
For a unit step input,
■
____________________ 0 . 5 K , { S * 3 ) ______________]
'
■[
l
i ’ + ( 2 o + i : , ) i= + ( i o o + i o ir , + o . 5 * : , ) i + i. 5 i: , J »
step 3 of 5
Determine the steady state tracking error.
= lt o i£ ( j)
„
r
—>
0 . 5 £ , ( i + 3)
L
i ’ + ( 2 0 + A :,) s ’ + (io o + iO A :,+ o .5 A r,)j+ i.5 A :,
0 .S £ ,( ^ + 3)
'l ™ ® - ^ [ '” , ’ + ( 2 0 + iC ,) j" + (100 + IOAr, + 0 .5 £ ,)i+ 1 .5 A :,
0 .5 x 3 x 0 .5 £ |
1.5£,
= 0.5
Thus, the steady state tracking error is |Q,5| .
Step 4 of 5
(b)
The transfer function of the system is,
y (^ )
£ (s)
_______________ o .5a :i ( j + 3 )
s ’ + (2 0 + A:2 ) ? + ( 1 0 0 + 1 0 A :, +
0 .5 £ ,) j + 1.5A:,
______________________ 0 .5 A :,( j + 3)___________________
“ j * [ j ’ + ( 2 0 + £ 2 ) i * + (1 0 0 + 1 0 £ j + 0 .5 £ | ) j + 1 3 A : ,]
Thus, the system type is
|type0|-
Step 5 of 5
(c)
The tracking error is.
,,
r _________ O.S£,(ft-3)_________ 1 , ,
■ [ ' ” i ’ + ( 2 0 + £ j ) j ^ + ( 1 0 0 + l 0 £ j + 0 .5 A : , ) i + 1.5A:,J
For a ramp input,
£ ( j ) = 0.5
—
s
0 .5 £ ,( n -3 )
[
i* + ( 2 0 + £ j) s * + ( 1 0 0 + 1 0 £ j+ 0 .5 £ ,) j + 1 .5 £ ,J j'
Determine the steady state tracking error.
« . = lim j£ (j)
= lim^O.S
0.5
= lim—
0 .5 £ |(^ + 3)
i ’ + ( 2 0 + + (100 + 10£, + 0 .5 £ ,)s + 1.5£, J j '
0 .5 £ |(s + 3)
i* + ( 2 0 + £ j ) i ’ + (1 0 0 + 1 0 £ j+ 0 .5 £ ,)j + 1.5£,
Thus, the steady state error to a ramp velocity is 0
Observe that the steady state error to a ramp input is independent of
and
.
Problem 4 .0 8 P P
A standard feedback control block diagram is shown in Fig. with the values
(f4U0)'
0 and compute the transfer function from R to Y.
(a) Let
(b)
Let R = 0 and compute the transfer function from IVto Y.
(c)
What is the tracking emor if R is a unit-step input and W e 0?
(d)
What is the tracking error if R is a unit-ramp input and W e 0?
(d)
What is the tracking error if R is a unit-ramp input and W e 0?
(e)
What is the system type and the corresponding error coefficient?
Figure Closed-loop system with sensor dynamics. R = reference, U = control. Y = output, V =
sensor noise
Step-by-step solution
ste p 1 of 6
(a)
Consider the gain of the plant.
G(s) = i
(1)
Consider the gain of the controller.
- ^ (^ + 0 ...... (2)
Consider the gain of sensor.
Conside
100
(3)
; + l0 0
Refer from Figure 4.5 in fhe textbook and write the transfer function equation with respect to zero
IVvaiue.
y (» )
p„(4
« (i)
) g (4)
(4)
l+ D „ ( i) G ( i) W ( 4 )
Substitute equations (1), (2) and (3) in equation (4).
( ^ ]( ;)
rw
A(s
( M
)
j* ( s + 1 0 0 )+ 2 (j+ l)(1 0 0 )
i '( 4 + 100)
2 (4 -H )( h -100)
j ’ ( i+ 1 0 0 ) + 2 ( j + l)(100)
,5)
R{s)
4’ + 100i’ + 2004+200
Hence, the transfer function of the system from R to Y is
y (4 )
2(4 + 1)(4 + 100)
« (4 )
4“ + 1004“ +2004 + 200
Step 2 of 6
(b)
Refer from Figure 4.5 in the textbook and write the transfer function equation —
—*-4^ \iwith respect
W {s )
to zero Rvalue.
l i f l .
(^ (i)
(6)
l+ £ > „ ( s ) G ( 4 ) « ( i)
Substitute equations (1), (2) and (3) in equation (6).
(i)
I 'M
100
W( s)
1+
^ A ^ + io o J
a
j
’ (4 + 1 0 0 )+ 2 ( j +1)(100)
4’ ( l + 100)
j ( 4 + l0 0 )
»’ ( i + 100)+2(4 + l)(l0 0 )
K (4 )_
4 (4 ^1 00)_____
B '( i)
i ’ +100j*+20as+200
Hence, the transfer function of the system from Wto Yis
I 'M
(V(4)
4 ( 4 + 100)
4’ + 1004“ + 2004 + 200
Step 3 of 6
(C)
Calculate the —“
with respect to zero Wvalue.
Ig -'-'-w ...."
2(44-1)(4-I-100)
From equation (5). substitute
for r ( 4 ) in equation (8).
4*+1004*+ 2004+200
2(4+ 1)(4 + I00)
R (s )
4 *+ 1004“ + 2004 + 200
2 ( 4“ + 1014+100)
° ' “ 7 + io o 7 + 2 o o 4 + 2 o o
4“ +1 004 “ +2004 + 200 - 24 ’ - 2024 - 200
4“ + 1004“ + 2004 + 200
4’ + 984“ -2 4
V
+ 1004“+2004 + 200
Step 4 of 6
Calculate the value of E{s) in unit step input.
Substitute —for J?($) in equation (9).
s
4*+984=-24
(10)
^ W = -5 4’ + 1004“ + 2004 + 2001,
Calculate the ^ ( o o ) = H m 4 £ ( 4 ) from equation (10).
, .
,.
4*+984“ - 2 4
fO
, ™ * ^ + i o 0 4 ' + 2004 + 2 0 o [ 4 j
^ (*)= 0
(11)
Hence, the value of tracking emorat unit step input with zero Wvalue is |zero|.
Step 5 of 6
(d)
Calculate the value of tracking error at unit ramp input with zero Wvalue.
From equation (5). substitute __ 2(5 + 1) ( j +100)
for r ( 4 ) in equation (8).
5^+1005 *+ 2005+200
Calculate the value of E{s) in unit step input.
Substitute -Lfor J?(5 ) in equation (9).
5
5*+985*-25
£ ( 4) = -
(\
( 12)
7)
4’ + 1004’ +2004 + 2001
=(,
Calculate the e , ^ ( « ) = lim 4£(4) from equation (12).
,
,
,.
4“ +984“ - 2 4
( 'I'l
^ < ” ^ ° ! a % »+1004“ + 2004 + 20ol 7 j
■ lim
4’ + 984“ -2 4
4’ +1004’ + 2004 + 2 0 0 U
4“ +984 - 2
— 4“ +1004’ + 2004 + 200
s lim
(13)
Hence, the value of tracking emorat unit ramp Input with zero IVvalue Is
100
Step 6 of 6
(e)
Determine the velocity emor coefficient value K, from 5^ ^00)
Write the general formula for K, ■
(14)
k— H
Substitute equation (13) in equation (14).
a:
»-
1
1
100
A:, = 1 0 0 se c -'...... (15)
Determine the type of the system __ 2(4 + l)(4 + 100)__
4*+ 1004“ + 2004 + 200
In the transfer function, the numerator part having j^term and denominator part have 5*term.
So it is [Type-Isyslem| and velocity error coefficient is ||0Q sfc~'| .
Step-by-step solutiOdPP
ow = 3—
=«
showing a lead compensation in the feedback path. What is the value of the velocity error
coefficient, Kv7
Figure Closed-loop system with sensor dynamics. R = reference, U = control. Y = output, V =
sensor noise
QlW
A
*w
Step-by-step solution
step 1 of 7
(a)
Refer from Figure 4.5 in the textbook and write the transfer function equation with respect to zero
Wvalue.
T ( i) = f ( 4
(1)
l+ D „ W G W f f( * )
Calculate the —“ with respect to zero W value.
R (s )
I g - '- '- w
......
Substitute equations (1) in equation (2).
R {s )
‘^ ^ ^ h *D A s )C (s )H {s )
£ (£ l
1+ D M G { s ) H { s ) - F { s ) D J s ) C ( s )
R {s )~
l + A ,( * ) G ( * ) //( j)
f i+ A , W G W w M - F M A , M G M '|
<3>
step 2 of 7
Calculate the value of tracking error at unit ramp input with zero Wvalue.
Substitute -L fo r J?($) in equation (3).
s
f i + q , ( ^ ) g ( . ) w ( ^ ) - f ( » ) q , M g m Y 1 'I
h £>,(,)g(,)«m
Calculate the
}[7)
( « ) = l i m j £ ( i ) from equation (4),
JU^J
l+ Z ) „ ( * ) G W /fW
=
...... (5,
l+ D ^ ( j) G ( j) H ( j)
JW
Hence, the value of steady-state tracking error at unit ramp input is
\ + D A ^ ) G U ) flU ) - F ( s ) D A s ) G ( s ) '
Step 3 of 7
(b)
Consider the gain G (^)h as single pole origin in the s plane.
<?(5) = j G ,(5)
(6)
Consider the gain of the controller is.
Z ) ^ ( j) = 0 .7 3 ...... (7)
Calculate the value of tracking error at unit step input with zero Wvalue.
Substitute —for
in equation (3).
r i+ A , M G ( » ) g ( » ) - F ( . ) o „ M G M ^ i
^
’ [
l +DAs)G{s)H(s)
Js
Calculate the ^ ( o o ) = H m s F (i) from equation (8),
^
Js
l + G „ (s )G W tfW
l + G , W G W i/ W
J
Step 4 of 7
Substitute equations (6) and (7) in equation (9).
1+ 0 .7 3 i G, ( j ) W ( j ) - F ( i )0 .7 3 i G, ( j
« ii* H = S S
l+ 0 . 7 3 iG , ( s ) H ( j)
'» + 0 . 7 3 G |( j ) W ( j ) - F ( ^ ) 0 . 7 3 G |( j ) '
foo) s lim
l+ 0 .7 3 G ,( 5 )//( y )
lim f1^1
i+ 0 .7 3 G | ( j ) W ( i ) - F ( i ) 0 . 7 3 G | (;
iM j.
l+ 0 .7 3 G ,( s )//( s )
.(10)
Hence, from above equation, requirement on the value of H ( j)s u c h that the system will remain
type-l system are:
^ ^o o )= 0 w h e n
is equal to 1 and /f^ ^ ^ is also equal to one.
Hence, the DC gain of the system H { s ) is |unity|.
Step 5 of 7
(c)
Consider the gain of the plant.
1
GW =
(11)
5(5 + 1)*
Consider the gain of sensor.
H(s)
2.755+1
(12)
0.365+1
Substitute equations (7), (11) and (12) in equation (5).
1+ 0 .73 -
1
2.755+1
j ( j + i y 0 .3 6 I+ I
- F ( i) 0 .7 3 l i t l l
H = li2
1+0 73—
i( ^ + l) M 0 - 3 f e + l)
t ( t + i y ( 0 . 3 f o + l)+ 0 .7 3 (2 .7 5 i + l ) - F ( j ) 0 . 7 3 ( 0 . 3 f a + l)
j ( i + lf( 0 .3 f a + l)
«*p\ /
^
i ( j + l f (0.3 fa + l) + 0 .7 3 ( 2 .7 5 j+ l)
j ( i + l) '( 0 . 3 f a + l)
* ( s + 1)’ ( 0 .3 & + 1 )+ 0 .7 3 (2 .7 5 * )
__ (11
• 5(5 + 1)* (0.3 6 5 + 1 )+ 0 .7 3 (2 .7 5 5 + 1 )U
Step 6 of 7 ^
Consider the value of F ( s ) is equal to 1 in above equation.
'* (* + 1 )’ (0.36s + 1 ) + 0 .7 3 (2 .7 5 * )+ 0 1
..73V
- 0 .7 3 ( 0 .3 6 * ) - 0 .7 3 ________
— I
* (* + 1)‘ ( 0 .3 6 * + 1)+ 0 .7 3 (2 .7 5 * + 1 )
a
_ | . ^ r * (* + I)^ (0 .3 6 * + 1 ) + 0 .7 3 (2 .7 5 * )-0 .7 3 (0 .3 6 * ) Y i ’|
* (* + l)’ (0
( 0 .3 6 * + l)+ 0 .7 3 ( 2 .7 5 * + l)
JU J
r ( i + l)‘ (0 .3 6 * + l)+ 0 .7 3 ( 2 .7 5 ) -0 .7 3 (0 .3 6 ) ')
’
*(* + l)"(0 .3 6 * + l)+ 0 .7 3 (2 .7 5 * + l)
J
r l + 0 .7 3 ( 2 .7 5 ) -0 .7 3 ( 0 .3 6 ) 'l
~[
«-33
J
« - , ( * ) = 3-7599
Step 7 of 7
Determine the velocity emor coefficient value J^T^from 5^ ( o o ) .
Write the general formula for K , ■
(15)
k— H
Substitute equation (14) in equation (15).
1
3.7599
a:
- 0 .2 6 6 s « - '
Hence, the value of velocity coefficient value is |j^^sQ .266sfC~'| ■
a
Problem 4.1 OPP
Consider the system shown in Fig., where
(a)
Prove that if the system is stable, it is capable of tracking a sinusoidal reference input r= sin
wof with zero steady-state error. {Hint: Look at the transfer function from R to E and consider the
gain at cuo.)
(b)
Use Routh’s criterion to find the range of K such that the closed-loop system remains stable
wo = 1 and a = 0.25.
Figure Control system
cControl O f Dynamic Systems
(7th Edition)
Problem
Consider the system shown in Fig., where
Dc{s) = K
(a) Prove that if the system is stable, it is capable of tracking a sinusoidal reference input r= sin
wof with zero steady-state error. {Hint: Look at the transfer function from R to E and consider the
gain at wo.)
(b) Use Routh’s criterion to find the range of K such that the closed-loop system remains stable if
wo = 1 and a = 0.25.
Figure Control system
(b)
S.S.E=limsE(s)
s’ (s + l)(s ’ +<D,’ )
lim
= 0
■-*" (s’ +£0,’ )s(s+ l)+ K (s+ a)’ (s’ +00,’ )
Step 3 of 3
(c)
Characteristic equatioii is
s ([s^+ lj (rH ) +K (s+0.25)^ =0
i.e.,
s*-h ’ + s’
(K+1)+ s (1+0.5K)+(0.25)’ k =0
Using Routh's criterion for stability.
s*
1
K+1
s’
1
m ).5 K
s’
05K
(0.25)’ K
s‘
O5K+0.875
0
s"
(0 2 5 )’ K
(0.25)’ K
0
For stabili^, as all the first column elements should be greater than 0,
Problem 4.11 PP
Consider the system shown in Fig., which represents control of the angle of a pendulum that has
no damping.
(a) What condition must Dc(s) satisfy so that the system can track a ramp reference input with
constant steady-state error?
(b) For a transfer function Dc(s) that stabilizes the system and satisfies the condition in part(a),
find the class of disturbances w(t) that the system can reject with zero steady-state error.
Figure Control system
cControl O f Dynamic Systems
(7th Edition)
Problem
Consider the system shown in Fig., which represents control of the angle of a pendulum that has
no damping.
(a) What condition must Dc(s) satisfy so that the system can track a ramp reference input with
constant steady-state error?
(b) For a transfer function Dc(s) that stabilizes the system and satisfies the condition in part(a),
find the class of disturbances w(t) that the system can reject with zero steady-state error.
Figure Control system
j * + a:
(1)
step 2 of 6
The reduced block diagram is shown in Figure 1.
w
Step 3 of 6
From Figure 1, the closed loop transfer function is.
Substitute — !— for r ( s ) in the above equation.
A M
_ ! _
W (s]
step 4 of 6
Determine the expression for error signal.
=*M-
A M
i ’ + A : + o . ( j)
' '
i'+ A : + z ) . ( j )
w (,)
Problem 4 .1 2PP
A unity feedback system has the overall transfer function
R(s)
'
+
+
Give the system type and corresponding error constant for tracking polynomial reference inputs
in terms of ^ and ojn.
Step-by-step solution
step 1 of 2
Step 1 of 2
- ^ (0 -
t m
^
V
R (s)
E (0 = E (0 -Y (0
=
j R (0
s +2^cd^ s+ cch^
S .S .E -lim sE (s)
By observing the above function.we can say &at &e system is of [Type l\.
Step 2 of 2
■S.S.E.=lim
= 2i
25
Problem 4 .1 3PP
Consider the second-order system
ds)
? + 2 f7 + T *
We would like to add a transfer function of the fonn
Dc(s) =
in series with G(s) in s
unity feedback structure.
(a) Ignoring stability for the moment, what are the constraints on K, a, and b so that the system is
Type 1?
(b) What are the constraints placed on K. a, and b so that the system is both stable and Type 1?
(c) What are the constraints on a and b so that the system is both Type 1 and remains stable for
(c) What are the constraints on a and b so that the system is both Type 1 and remains stable for
every positive value for K?
Step-by-step solution
step 1 of 5
s"+2?s+l
D (s ) =
s+b
Y (s) _________ K (s+a)
E (s)
(s“+2?s+l)(s+b)+K (s+a)
( ! “ + 2 ? !+ l)(s+ b )
E = R -Y = . , \
----- -R (s)
(s“ +2? s+ l)(s+ b)+ K (s+ a)
Step 2 of 5
a.
, .
S.S.E.=lim sE(s)=
sfs+b)fs^+2^s+ll
, .
^
R (0
(s^+ 2^+ l) ( s ^ ) +K (s+a)
Fortype -1 systemJb=0, K ^ O , a ^ O
Step 3 of 5
b.
For the given system,
b=0;
R ( s ) = |.
s
and the system shoul d be of type -1.
S.S.E.=limsEfs)
i-»0
' '
s"(s"+2?s+l)
1
= lim -------- ^------------ ------ X- r
*-»«(s^+24s+l)s+K(s+a) s^
Char, equation = s^+2^^+(K+l)s+Kat*0
Step 4 of 5
For Stability,
1
K+1
s’
Ka
s‘
s"
0
Ka
(K+1) - E i > 0
25
=>Ka<2^(K+l)
Ka>0,
The constraints are | b=0,Ka>0, K a<2^(K +l)|
Step 5 of 5
c.
A sK >0, r^>0|
>1 ^
a<
2^(K +1)
|0<a<2^(K +l)|
Problem 4.14PP
Consider the system shown in Fig.(a).
(a)
What is the system type? Compute the steady-state tracking error due to a ramp input r(t) =
rof\{t).
(b)
For the modified system with a feed forward path shown in Fig.(b), give the vaiue of H f so the
system is Type 2 for reference inputs and compute the Ka in this case.
(c)
Is the resulting Type 2 property of this system robust with respect to changes in Hf. that is, wi
the system remain Type 2 if H f changes slightly?
Figure Control system
Figure Control system
A
»o— ^ ( i ) -
-o r
V
HfS
Step-by-step solution
step 1 of 4
T (0 = s (ts+ l) +A
R (s)= ^
E ( .) = R - Y = ^ !i^
s ( ts+ 1)+ A
S.S.E. = lim sE (s)=lim f
*-.0
'
* - * 0 s ( t s + l) 4 A U v
S .S .E .= ^
____ A
System is of [Type -1[
Step 2 of 4
b.
From block dis^am , we get
(R E , -Y+RHf s) A=Ys
A + s(ts+ l)
A + s ( ts+1)
Niim erator= ts^+s -Hf As+A (1-H,)
Step 3 of 4
Fortype-2sy8tem, itm usthavetw o zeros at s ^ .
=>(l-H fA )=0
and[ ^ ^
S. S.E. =lim sE (s) =lim
*-»o
A
.
,-----x f -^1
»-»<• s ( ts+1)+A
Ka
Step 4 of 4 ^
c.
system wouldbecome |T]rpe ^
Problem 4 .1 5PP
A controller for a satellite attitude control with transfer function G=1/s2 has been designed with c
unity feedback structure and has the transfer function
DcW = - ! ^ .
(a) Find the system type for reference tracking and the corresponding error constant for this
system.
(b )if a disturbance torque adds to the control so that the input to the process \s u + w. what is the
system type and corresponding error constant with respect to disturbance rejection?
Step-by-step solution
Step-by-step solution
step 1 of 8
(a)
The transfer function of the controller for a satellite attitude control is.
c=4The transfer function of unity feedback structure is.
10 ( 1 + 2 )
£ ) (.)
s+5
In order to determine a system for reference tracking, first identify the poles, and then determine
the type of system and the finally evaluate the error constants.
By looking at the transfer function, the system has two poles at 5 = 0 Thus, the system type is [Type 2 |.
Step 2 of 8
The closed loop transfer function is.
r(s )
C(s)D(s)
« (s)
1 + G (j)c (s )
1-h
10(^ + 2)
^ * ( « + 5 ) + I 0 ( f + 2)
Step 3 of 8
Calculate the steady state error.
s^(s+S)+W(s
f
s’ (s + S)
+2)i’J
l']
™ [ j > (, + 5 )+ 1 0 (5 + 2 )5 ’ J
= lim f- 7 ^ -----— > |_ * '( i + 5 ) + 1 0 ( i + 2 ) J
0+5
(0 ) (0 + 5 )+ 1 0 (0 + 2 )
___ 5 _
°
0+20
• 0.25
Step 4 of 8
Determine error constant i f .
jf . - —
0.25
•4
=4
Therefore, the error constant. K
is 0
Step 5 of 8
(b)
When a disturbance torque adds to the control, it acts as a disturbance input. For a disturbance
input, the poles at 5 s 0 ^re usually after the Input. Hence the system type is Type 0.
Thus, the system type is |t ^
Step 6 of 8
Determine the transfer function with disturbance feedback.
IW - I
ir(s )
G
I + GD
=1—
j+ 5
• 1—
j ' ( i + 5 )+ IO (j+ 2 )
f^ (i+ 5 )+ 1 0 (i + 2 ) -» + 5
i ’ ( i + 5 )+ 1 0 (* + 2 )
( » '- l ) ( j + 5 ) + I O ( i + 2)
i ’ (j+ 5 )+ 1 0 (i+ 2 )
Step 7 of 8
Calculate the steady state error.
*lim
'( j * - l ) ( s + 5 ) + 1 0 ( j + 2 )
j ’ ( i+ 5 )+ 1 0 (j+ 2 )
(0 -l)(0 + 5 )+ 1 0 (0 + 2 )
(0 )(0 + 5 )+ 1 0 (0 + 2 )
-5 + 2 0
20
_15
'2 0
3
°4
Step 8 of 8
Determine error constant K .
1
3
1
4
1 + a: .
'
3
4 -3
3
Therefore, the error constant,
I
Problem 4 .1 6PP
A compensated motor position control system is shown in Fig. Assume that the sensor dynamics
are H(s) = 1.
Figure Control system
(a)
Can the system track a step reference input rwith zero steady-state error? If yes. give the
value of the velocity constant.
(a)
Can the system track a step reference input rwith zero steady-state emor? If yes. give the
value of the velocity constant.
(b)
Can the system reject a step disturbance wwith zero steady-state en^or? If yes. give the valu
of the velocity constant.
(c)
Compute the sensitivity of the closed-loop transfer function to changes in the plant pole at -2
(d)
In some instances there are dynamics in the sensor. Repeat parts (a) to (c) for |W(*) =
and compare the corresponding velocity constants.
Step-by-step solution
ste p 1 of 8
Refer to Figure 4.29 in the textbook.
(a)
From the block diagram, it is clear that the system is Type 1 with
y (4
R (s )
0 (4
l+ H ( s ) G ( s )
y (« )
R (s )
g (4
1 + G ( j)
Write the expression for the error signal.
' '
i +a ( s )
'• ’
(2)
Step 2 of 8
, .
1 6 0 (j+ 4 )
Substitute i for A ^^jand —^
— 30)^°'"
£W =-
, .
1 6 0 (j+ 4 )
1+
s(i +2)(*+30)
(j+2)(i+30)
i( i+ 2 ) ( i+ 3 0 ) + 1 6 0 ( i+ 4 )
Now. evaluate the steady-state system error.
e„ - liin
----- ^
= lin ii f
'-*• ^^s(5+2)(5+30)+160(j+4)J
sO
Therefore, the system can track a step input with zero steady-state emor.
Step 3 of 8
Determine the value of velocity constant.
= H m j(7(5)
( ^ j( j+ 2 ) ( j+ 3 0 ) J
(
1 6 0 (j+ 4 )
^
™[(j -f2)(j +30)J
(»60)(4 )
- (2 )(3 0 )
=10.67
Therefore, the value of the velocity constant,
is |1Q.67|-
Step 4 of 8
(b)
The system is Type 0 with respect to the disturbance and has the steady-state emor. Find the
transfer function for 7 ( 4 ) •
1
4(4 + 2 )
y (^ )= -
r i60(4+4)Y
1+
I\
1
]
IL«r«+2^l
■- / /
\- — /
I
4(4 + 2 )
4(4 + 2 ) (4 + 3 0 ) + I6 0 ( 4 + 4 )
4(4 + 2 )(4 + 30)
4+ 3 0
4(4+2)(4+30)+160(4+4)
Determine the steady-state error to a disturbance input.
y.
*-*® (^4(4 + 2)(4+30)+160(4+4)J
Hence, the system cannot reject a constant disturbance.
Step 5 of 8
(c)
In order to compute the sensitivity of the closed-loop transfer function to changes in the plant
pole at _2. Now by determine the transfer function.
l6 0 (^ + 4 )
r(^ )= «(« + >4)(s + 3 0 )+ I6 0 ( 2 + 4 )
Where. A was inserted for the pole at the nominal value of 2.
Now evaluate the sensitivity of the system.
AST
T S A .......(3)
Step 6 of 8
Apply partial differentiation to the transfer function with respect to A.
ST
a(
160(5+44)
160(5+
'I
04 “ a 4 [(^5(5
5 (5 ++y4()(5
) (5 + 330)
0 )+
+ I6 0 (5 + 4 )J
160(5 + 4 )(5 )(5 + 30)
[5 (5 + 4 )(5 + 3 0 )+ 1 6 0 (5 + 4 ) ]’
o r.
Substitute — value in equation (3).
SA
4 [5 (5 + 4 )(5 + 3 0 )+ 1 6 0 (5 + 4 )][1 6 0 (5 + 4 )(5 )(5 + 3 0 )]
[1 6 0 ( 5 + 4 ) ][5 ( 5 + 4 )(5 + 30)+160(5 + 4 ) ] ‘
"
4 5 (5
+ 30)
5(5 + 4 )(5 + 3 0 )+ 1 6 0 (5 + 4)
25(5 + 3 0 )
(Since, 4 = 2)
5(5 + 2 )(5 + 3 0 )+ 1 6 0 (5 + 4 )
Therefore, the sensitivity of the closed-loop transfer function is
25 ( 5 + 30 )
5(5 + 2)(5 + 30)+160(5 + 4)
Step 7 of 8
(<i)
Consider the following information.
f f( 5 ) =
20
4+20
Determine the system type by computing ^ = Q a n d at the value, H = l , then the system is Type
1 with respect to the reference input.
The new transfer function is.
f f (4 ).l-
4 + 20
Step 8 of 8
The velocity feedback J ^ w ill also change. Hence, evaluate the new expression for the error.
*W = [i-r(4 > (4
[ « l( j+ 4 )
^
< ( 5 + 2 ) ( 5 + % ) + ie D ( 5 + 4 )
*(4
5 (j+ 2 y j+ 3 Q )+ 1 6 0 (j+ 4 )-ia i(j+ 4 )
5(5+2)(5+30) + M 0 (5 + 4 )
R [s]
5 ( 5 + 2 ) ( 5 + 3 0 ( 5 + a » )- 5 ( l6 D X 5 + 4 ) )
,
5 ( f+ 2 )i[5 + 3 0 ( s + 2 0 ) + 2 0 (lie D ]( 5 + 4 ))
'
Evaluate the velocity constant from the emor expression.
30 x2
K = 2.63
- 22.86
Therefore, the velocity constant.
is 122.86! •
Problem 4 .1 7PP
The general unity feedback system shown in Fig. has disturbance inputs w1, w2, and w3 and is
asymptotically stable. Also,
* ' ' '' j'>
Vi nn f2* .i(r *- +■p~ii) ’
'
« ^ n rJ i(» + w )'
Show that the system is of Type 0, Type /1. and Type (/1 + 12) with respect to disturbance inputs
w1, w2. and vv3 respectively.
Figure Single input-single output unity feedback system with disturbance inputs
Step-by-step solution
step 1 of 5
Draw the diagram of single input-single output unity feedback system with disturbance inputs.
w.
w .
Figure 1
Step 2 of 5
Consider the following data;
Step 3 of 5 ^
Write the expression for output from the block diagram shown in Figure 1.
y /A _
y / jX
Substitute
for G ,( i) a n d
for a , ( s )
1 + G , W G , W '^ ^ ^
)+
Here.
t U 'n 3 i ( * + / > i i ) J U ^ n ; ,i
(*+Pa)JJ
^ are the poles and zeros of the
and ^
not at the origin.
Step 4 of 5
Now evaluate the steady-state error.
(
= Iiin
y
g ,g ,[n ,(z 4 -z ,)]» ^ (z )
[ j * + i ' ’ I l ,( s + p ,) + / :,A :, n ,( i + z ,) J
= liin .
Analyzing this system, we see that there are no poles at the origin, hence its Type 0 system.
Step 5 of 5
Write the expression for output for disturbance
■
l + G |( j ) G j ( j
r A r , n ; , ( z + Z a ) '|
i .i‘ r c , ( i + P a ) J
1,
W ,{s)
n;,(z+z„)Y K, n;,
{ k , n:, (j+z„))(a:, nj, (»+zj,))
(s* n ; , ( z + p „ ) ) j ^ n 2 , ( z + P a )
A T ,r n ,(» + ? .,)V n ,(z + f» „ )„ , , ,
In this expression, in the denominator A (z) is the characteristic polynomial. Now evaluating the
steady-state error.
.(ifL O i.
By analyzing this system, we see that the system is of Type
Write the expression for output for disturbance
■
i^ A s )
I 'M -
l+ G ,( i) G ,( i)
(F .W
l- f
f f ^ , n : , ( » + r , , ) Y A : , n ; , ( z t r , , ) ^^
TC, (j+p„)Ji*' nj, ( s + P u ) } )
(» *n ;i(j+ p „ ))(» * -n ;,(z + p „ ))+ (A :,n ;,(j+ z „ ))(A :,n 5 i(z + z a ))
( j ' n ; , ( j + p „ ) ) ( i ' ’ n ; , ( » + P j, ) )
------------A W
In this expression, in the denominator A (z) is the characteristic polynomial. Now evaluate the
steady-state error.
in , A
By analyzing this system, we see that the system is of Type k * k
Problem 4 .1 8PP
One possible representation of an automobile speed-control system with integral control is shown
in Fig.
(a) With a zero reference velocity input vc = 0, find the transfer function relating the output speed
v to the wind disturbance w.
(b) What is the steady-state response of v if w is a unit-ramp function?.
(c)
What type is this system in relation to reference inputs? What is the value of the
corresponding error constant?
WVWtjat.la.thj?iv/ie aodJ^rreRnnnriinn prrnr mnstant nf this sustpm in rplptinn tn trankinn thp
(d)
What is the type and corresponding error constant of this system in relation to tracking the
disturbance w?
Figure System using integral control
Step-by-step solution
step 1 of 7
Refer to Figure 4.31 in the textbook.
From the block diagram in Figure 4.31,
y h .v * w = { - + ^ + iC \v
s
\m
V=
,
s
)
. . K + - , ------ ---------------W .......(1)
+kyms+mkfk2 *
•¥mk^k^-¥kjnis
Step 2 of 7
(a)
The reference input voltage, f ' s 0 V
Substitute 0 V for V in equation (1).
K ( j ) = - ----- ^
-------- ( 0 ) + ^ ------- — ----------I F f*
^ ^ s* ^+ k^ms+
^ MK>^ mkykj
wttk t
' ‘
’ r ^ f t t ^ tr
F (,) =
W (s )
P (j)
ms
( j)
+ m k ^ + mk^k^
Thus, the transfer function relating the output speed to the wind disturbance is
f P ( j)
(2)
+ ky/ns+ k^k^m
Step 3 of 7
(b)
The wind disturbance is a ramp function,
'
s
The steady state response of the velocity v is.
Substitute the value of K (f)fro m equation (2) in this equation.
(
= lim
*-»o
s lim
ms
Y » ;^ 1
s^ +kyms+k^k2mj
k^k^
Thus, the steady state value of the velocity v is V
step 4 of 7
(c)
Calculate the emor in system with respect to reference inputs.
£ W = n ( i) - f 'W
s* ■¥k^ms-¥mk^k2
-------- V , ( j )
\ s * •^kyms^mkyk^)
E (s ) =
' ^
K (s)
s^+k^ms^
The error of the system can be compared with a system of unit gain feedback and having e
nk^k^
fonward gain of <7( j ) * - 5— — .A s the gain <7( 4 ) has a pole at origin.
s ^ k^frts
The positional constant of the system is,
J S r,.B m G W
mk^k2
s lim
Step 5 of 7 ^
The steady state error of the system for the unit step input.
e„ = lim— ^-r-r y. ( 5)
lim ^
/ | ’\
)UJ
l + K m G (j)v
1
1 + /:,
sO
step 6 of 7
The velocity constant of the system is.
JS:, = limiG(j)
n kikj
s lim j»-»• s +kjm s
Now. the steady state error of the system for the unit ramp input.
s li m — —
r
1
^
l + lim 5 G (^)
(^ )v s * j
1
s + K^
^ J_
K
k^k^
As steady state error for unit step input is zero and for the ramp input is — . the system is a
Type 1 system.
Therefore, the velocity emor constant is
_____
Step 7 of 7
(d)
Calculate the emor in system with respect to disturbance w .
E ( ,) = W { s )-V (s )
5* •i-kyms'^k^k^m» '( 4
s*■^k,na’¥ k , k , m - m s \ „ , t \
[
The transfer function from the disturbance input 1^ ( 5 ) to the emor £ ( * ) is,
£ ( j)
*+ k^m s+ ktkim ~ m s
fV {s )
5^+
^m s+ m ft,j^
^+k^ms+mk^k2
J
The steady state error for a unit step disturbance input is,
K lim f£ ( 5 )
*-»® ^
5^ +kyms+mk^k2
j
s *kfms*mk^k2
J\sJ
Iju il
+ k^ms+ mk^k^
' '
I
J
kjkjm
As the steady state emor for a unit step disturbance is non-zero this is a Type 0 system.
Therefore, the value of the error constant is Q].
Problem 4 .1 9PP
For the feedback system shown in Fig., find the value of a that will make the system Type 1 for K
= 5. Give the corresponding velocity constant. Show that the system is not robust by using this
value of a and computing the tracking error e = r - y to a step reference for /C= 4 and K = 6.
Figure Control system
Step-by-step solution
5tep^y-step~so1ution
step 1 of 3
Y ( s) = - 5 ^ R ( s)
S+2+K
AsK=5,
S.S.E.=lim sE(s)=lim s(R -y )
c^O
' '
*-»0 '
(s+7)
'
ix R (s )
for type-1 system, ( 7 - 5 a ) ^
7
OF —
5
— = S.S .E .= K .
7
=7
Step 2 of 3
for
:
2
s+F- i
s+2— K
E(s)=R-y= t____ 5 L
___55
s+6
S+2+K
s f s + lj
S.S.E.=lim sE (s)=lim ^
c-kO
*-»0
c-kO
' •^ *-»0
(s)
-jj :
g+6
'• ^
system becomes type-0
1
with S.S.E.=15
Step 3 of 3
for K=6 :
S + 2 --K
E (s)= R -Y = ------ 5 _ |
_5
S+2+K
‘
S.S.E.=lim s£fs)= lim
<-kd
' •
<-k0
s+8
■ H )
R (0
> System becomes type-0 witti S.S.E.=—
System is not robust.
Problem 4.20PP
Suppose you are given the system depicted in Fig.(a) where the plant parameter a is subject to
variations.
(a)
Find G(s) so that the system shown in Fig.(b) has the same transfer function from r to y as th
system in Fig.(a).
Assume that a = 1 is the nominal value of the plant parameter. What are the system type and
(b)
the error constant in this case?
(c)
Now assume that a = ^ + 5a, where 6a is some perturbation to the plant parameter. What are
the system type and the error constant for the perturbed system?
Figure Control system
the system type and the error constant for the perturbed system?
Figure Control system
Step-by-step solution
step 1 of 8
(a)
Refer to the system in Figure 4.35(a) in the textbook.
Consider the expression for left side loop.
1
=X
J(s+a)
4 R * X - 4 Y = {s+ a )X
x
^
: ^
^
...... (1,
Step 2 of 8
Consider the expression for right side loop.
[ jr + ;ir ( s + o ) ] = iy
X {s+ a + l) = sY
„
sY
(s+ a + I)
(2)
Step 3 of 8
Equate both the equations (1) and (2).
4 {R -Y )
sY
( i+ a + 1 )
4 ( /e - y )( s + o + l) = j K ( j+ a - l)
4 / i( « + a + l) - 4 r ( 5 + a + I) = 5 y ( « + a - l)
4 /t( j+ a + l) = y [ s ( j+ f l- l) + 4 ( j+ a + l) ]
Y
R
4 ( g 4 -g 4 -l)
^ ( ^ + f l- l) + 4 ( i+ g + l)
The closed-loop transfer function of the system is.
r ( j)
4 (5 -t-g + i)
R (s)
4 (* + f l - l ) + 4 ( 5 + g + l )
(3)
step 4 of 8
The general fonri of the closed-loop transfer function is.
y (4
« (s)
g (^)
1+ G ( j )
(4)
Here C {4 ) is the open-loop transfer function.
Find the open-loop transfer function <7(j)From equations (3) and (4),
5 ( i+ f l- l)+ 4 ( 5 + g + l)
1 + G (j)
(l + G (s ))4 (5 + a + l) = G ( j ) [ j( 5 + a - l) + 4 ( s + a + l ) ]
4 ( j+ g + l)» G ( j) [j( 5 + fl-l) + 4 (5 + g + l) -4 ( s + a + l) ]
4 ( s + a + l) = G ( j)[ j( 5 + f l- ! )]
(5)
Thus, the open-loop transfer function C {4 ) is
4 ( x + g + l)
Step 5 of 8
(b)
Assume that a ^ \ To define the system type open loop transfer function is considered.
Recall equation (5).
Substitute 1 for a in the equation.
' '
s (* + l- l)
4 ( i + 2)
i(s )
4 ( i + 2)
There are two poles at origin, hence the system is a type-2 system.
Thus, the system type is
.
Step 6 of 8
Find the error constant.
For a type-2 system, the error constant exists for parabolic input only and the error constant is
zero for step and ramp inputs.
The acceleration error constant is.
it. ■U ni5^G (5)
* j-»0
' '
s U m [4 ^ 'i'8 ]
= [4 (0 )+ 8 ]
=8
Thus, the error constant is
Step 7 of 8
(c)
Assume that j s U d a
Here Sa is the some perturbation to the plant parameter.
To define the system type open loop transfer function is considered.
Recall equation (5).
Substitute 1 + for a in the equation.
s{s+ Sa)
There is only one pole at origin, hence the system is a type-1 system.
Thus, the system type is [ |] .
Step 8 of 8
Find the error constant.
For a type-1 system, the error constant exists for ramp input only and the error constant is zero
for step input and infinite parabola input.
The acceleration error constant is.
= lim j
f4 (5 + 2 + ^ g )
(5+&j) J
^ |- 4 ( 2 ^ j
4 (2 + ^
Sa
Thus, the error constant is
4 ( 2 + ^ f l)
Sa
Problem 4.21 PP
Two feedback systems are shown in Fig.
(a) Determine values for K^. K2, and K3 so that
(i) both systems exhibit zero steady-state error to step inputs (that is. both are Type 1). and
(ii) their static velocity constant /Cv = 1 when KO = 1.
(b) Suppose KQ undergoes a small perturbation: KQ -*
+ 5KQ. What effect does this have on
the system type in each case? Which system has a type which is robust? Which system do you
think would be preferred?
Figure Two feedback systems
h p ''
^
l- p '’
Step-by-step solution
step 1 of 4
(a)
Refer from Figure 4.36 (a) in the textbook and write the emor detector output equation.
E ^ R -Y
E ( s U - ^
^ ^ 1 + G (s )
(1)
Substitute the gain of the system from Figure 4.36 (a) in the textbook for G (« ) in equation (1).
^ '
A s^+s+K ^,
(2)
' '
Determine the value of velocity error coefficient K , ■
Write the general formula for velocity error coefficient.
Calculate the ^ . ^ ( o o ) = U in s £ ( j) from equation (2).
,
.
'b j-p V
)
..
i( 4 s + l )
1
4 s^ + s + K JC ^s'
(4 s + \)
^ H = l i 2 ’ -1^ 4z sr‘ + s + K ^ i
(o o )s — !— .
’ K .K ,
(4)
F orJS :.= l , find the value of
(oo) ■ — .
(® )
(5)
Substitute equations (5) in equation (3).
=
....... (6)
Hence, from equation (6), the value of
is equal to |pnel •
Step 2 of 4
Refer from Figure 4.36 (a) in the textbook and write the emor detector output equation.
E s R —Y
Substitute the gain of the system from Figure 4.36 (b) in the textbook for
in equation (7).
--------------i 4 £ ± i £ _ i d £ ± y j t ( , )
------------ R(s)
4J + 1+ C A
4s + l
(8)
Calculate the
from equation (8).
4 s + i + a :, a :„
^
, ,
4 j+ i+ ir A ( i- A : ,)
( * ) = ------------- “
4 1 + i + a :,* :.
, X l i M
M
Consider the value of
e .« ,H = 0
J
(9)
^ (» ) •
(10)
Equate equations (9) and (10),
U K ,K ,( \- K ,)
i+ A r , x . ( i - ^ : , ) = o
(11)
For AT^ « i , modify equation (11).
i + i :,( i - a: ,) = o
......(1 2 )
Step 3 of 4
Determine the value of velocity error coefficient K ,
Calculate the
'
from equation (8).
’
4s + i + a:, k ,
,-M
■—
*’
4 s + i+ ^ r ,^ :.
s
^ ^ 1 ^ K ,K ,(\-K ,)
• lim — S
.■M
4
pH
^
■(13)
'
F o r j r . = l , find the value of
4
■.......... ....... (14)
UK,
' ^
,H
Substitute equations (14) in equation (3).
\U K ,\
a: . *
UK,
For AT, a 1. find the value of K ,a: , =
3 ........ (15)
Substitute equations (15) in equation (12).
i+ 3 ( i- * r j) = o
.(16)
K p -- 1
«
3
Hence, from equations (15), and (16). the value of ATjis
and the value of AT, is [ ^ .
(b)
Consider the value of A^.
K ^ ^ K ^ + 8 K ^ ...... (17)
Calculate the ^ * p H = ' ™ » ^ W
‘i. * p l
)
from equation (2).
‘™ * 4 i ’ + i+ ( A r .+ S A r ,) iir , j
.(18)
Substitute AT| value in equation (18).
•b»pv
I
« P 4 i ' + * + ( * ; , + 8 a:„)
e .« p H = 0
(19)
Hence, the Figure 4.36 (a) is regardless of K^ value.
Step 4 of 4
Substitute equations (17) in equation (9).
,
.
i
+ a: , ( a: . + 5 a: . ) ( i -
t _ ( < » ) = ---------------------4 '^
■»»p V
)
i
+
a: , ( a: , +
,
8 a: , )
a: , )
...... (20)
Substitute 1 for A^ . 4/3 for K2, and 3 for K3 in equation (20).
l - p 3 ( H - 5 i C . ) [ l - |]
^ » pH
1+3(1+8A:,]
-& K .
Hence, the Figure 4.36 (b) is robust. So, the designers favor to [choosesystem (a ) than
because of output variation is huge with small variation in input changes.
Problem 4.22PP
You are given the system shown in Fig., where the feedback gain jS is subject to variations. You
are to design a controller for this system so that the output y(t) accurately tracks the reference
input r(t).
(a) Let f3 =1. You are given the following three options for the controller Dci(s):
Choose the controller (including particular values for the controller constants) that will result in e
Type 1 system with a steady-state error to a unit reference ramp of less than
.
(b) Next, suppose that there is some attenuation in the feedback path that is modeled by p = 0.9.
Find the steady-state error due to a ramp input for your choice of Dei = (s) in part (a).
(d ; iNexi, suppose inai iiieie is some uuenuauun in uie leeuuauK paui inai is iiiuueieu oy p = u.s.
Find the steady-state error due to a ramp input for your choice of Dei = (s) in part (a).
(c) If
= 0.9, what is the system type for part (b)? What are the values of the appropriate error
constant?
Figure Control system
X
S te p -b y -s te p s o lu tio n
step 1 of 8
Refer to Figure 4.35 in the textbook.
(a)
In order to choose a controller that will result in Type 1 system, first evaluate the transfer function
and then determine the steady-state error.
First choose
as an integrator in the loop and evaluate the transfer function.
( ( i + l) ( i+ 1 0 )
10____ Y * » ^ * * ' j
[(.-m) U io)A
' * ^ ( ( , + l)(“ + 1 0 ) ) ( ^ )
1 0 (V + *,)
_
5(s + l) ( ^ + IO)
1 0 (A ^ + Jfc,)
1+ A
j ( 5 + !)(« +10)
Step 2 of 8
Now evaluate the system error.
io(*^+*,)
5 ( j + l ) ( j + I0 )
1—
10(*,5 + * , )
j( 5 + l) ( « + 10)^
j ( j + l ) ( « + 10)
1-
j ( j + l) ( i+ 1 0 ) + 1 0 ( * , j+ * , ) / ?
j ( j + l) ( 5 + 10)
_r
r
(
io ( v + * , )
¥
4 ^ ^ i)( » - M o )
]] 1
( i ( * + l) ( j + 1 0 )J (j(j+ l)(j+ 1 0 )+ 1 0 (* ,s+ * ,)/jJ J « ’
fi
1 0 (V + *,)
[
i(s + l) (5 + 10)+ 10(t,^+t,))»
) 1
p(^+i)(»+io)+io(V+*,)/>-io(V+*,)~| 1
[
j ( i + l ) ( s + 1 0 )+ 1 0 (* ^ + * ,)^
(1)
step 3 of 8
Substitute ^ = t in equation (1).
[
i ( i + l) ( s + 1 0 ) + 1 0 ( * ^ + t , ) ( l )
s (5 + l ) ( j + 10)
Js'
^ I
[^j(5+l)(i +10)+10(*^+*,)Js*
■t
Now apply final value theorem to find the steady state emor.
=
---------] j .
^^j(i + l)(s + 1 0 )+ 1 0 (t^ + * ,)J i
f
s lim
( i + l ) ( i + 10)
)
[ i ( i + l)(i+ 1 0 )+ I0 (* ,i+ * ,) J
^10
‘ lOJfe,
...... (2)
Step 4 of 8
From equation (2), it is clear that
2 1 0 which meets the steady state specifications. The next
step is to determine the stability, and see if all the poles are on the open left half plane, use the
Routh criterion
The characteristic equation of the closed-loop poles is,
t i l s ’ 4 - ia s 4 - l o ( i ^ + 1^) = 0
Step 5 of 8
To determine the Routh array, we first arrange the coefficients of the characteristic polynomial in
two rows, beginning with first and second coefficients and followed by the even numbered and
odd-numbered coefficients.
Write the Routh array,
s’ :
1
f’ :
11
KM,
1 1 0 (ltl:,)-l(l^
11
s*:
lo t.
From the Routh array, it is clear that
> 0 and l l ( l + * , ) > 0 for the system to be stable.
Step 6 of 8
(b)
The value of feedback gain, f i = 0.9
The input is,
Substitute f i = 0.9 in equation (1).
f ^ (^ + i)( » * io )- ^ io (V + * ,)^ -io (V + * ,)] j .
^
(
J '■
s ( i + l) ( s + 1 0 ) + 1 0 ( * , f + t , ) ^
i ( s + l ) ( s + l 0 ) + 1 0 ( * , s + t , ) ( 0 . 9 ) - 1 0 ( t , s + t , ) j^ 1 j
i(i +l)(j +10)+10(*,s+t,)(0.9)
s (j
+1)(s + 10)+9(*, s + * ,)-1 0 (1 , s + 1 ,)Y i ^
i(s +l)(s+10)+9(*,s+*,)
l)(j +10)+9t,s+9*,-10*,j
s(s+l)(s+10)+9(t,s+*,)
JU’J
s (s + l ) ( j + 1 0 ) + 9 t , s + 9 * ,- 1 0 * , s - 1 0 * , Y I j
JTfil (s{s
f +l)(s+10)+9(*,s+i,)Jl.s’J
V 1)
^
Step 7 of 8
Apply final value theorem to find the steady state error.
e . = lim s [£ ( j) ]
|^5 (5 + l)(j+ 1 0 )+ 9 (A :^5 + ifc ,)J L 5 *y
Therefore, the steady-state emor due to ramp input is 0
Since the steady-state error due to ramp input is infinite, the system is no longer Type 1.
Step 8 of 8
(c)
The value of feedback gain, f i = 0.9
The input is,
—
s
Calculate the steady state error.
e .- to s [£ ( j) ]
^ 4 (4 + l)(j + 10) + 9(jfc^J+*;) j\s)
j { 5 + l ) ( j + 1 0 )-ik ^ j-ifc ,
■ lim
j (j
+ 1 )( j + 10)+9(A :, j + * , )
9k,
=- i
9
From error formula we see the system is Type 0. For the system Type 0 the error for step position
1
l+ K ,
9
l + K^
\ + K ,^ -9
a:
= -1 0
Therefore, the appropriate value for error constant.
is |» | q | .
Problem 4.23PP
Consider the system shown in Fig.
(a)
Find the transfer function from the reference input to the tracking error.
(b)
For this system to respond to inputs of the form r(t) = fn1 {t) (Where n < q ) with zero steady-
state error, what constraint is piaced on the open-loop poles p1, p2. • • •. pg?
Figure Control system
(«+;»iXi+ f t ) " ' C*+/>«)
Step-by-step solution
step 1 of 2
Y (0 _
G
R (s )
1+G
f E (s )= R (s )-Y (s )
,E (s )_
n ( ^ ‘)
step 2 of 2
S.S.E.=limsEfs)
t-»o
' ''
n ( '+ P i )
A nd
„i
sn(s+pi)
»i_______ is required to be equal to zero.
Hm--------------------
For this condition to be True, at least (n + 1) zeros out o f *q’ should present at the
origia
Problem 4.24PP
Consider the system shown in Fig.
(a)
Compute the transfer function from R(s) to E(s) and determine the steady-state error {ess) fo
a unit-step reference input signal, and a unit-ramp reference input signal.
(b)
Determine the locations of the closed-loop poles of the system..
(c)
Select the system parameters {k, kP, kl) such that the closed-loop system has damping
coefficient ^ = 0.707 and ojn=
What percent overshoot would you expect in y(t) for unit-step
reference input?
fdXJFinrlthft,trackinn prmr <;innpl a.<?a fiinrtinn nf timp
(d)
if thp rpfprpnnp inniit tn thp svRfpm
Find the tracking error signal as a function of time, e(t), if the reference input to the system,
r(t), is a unit-ramp.
(e)
How can we select the PI controller parameters {kP, kl) to ensure that the amplitude of the
transient tracking error, |effj|, from part (d) is small?.
(f)
What is the transient behavior of the tracking error, e(t), for a unit-ramp reference input if the
magnitude of the integral gain, kl. is very large? Does the unit-ramp response have an overshoot
in that case?
Figure Control system diagram
GmtfDller
Haat
(*.+ 4 )-S L
i.
Step-by-step solution
step 1 of 10
(a)
Refer from Figure 4.39 in the textbook and write the error detector output equation.
E = R -Y
I
E{s) = -
> « (*)
(1)
1+ G(5)D c (»)
Substitute the plant gain of the system and controller gain from Figure 4.39 in the textbook for
G {.) in equation (1).
1
E {s )= 1+
(‘- 4 )
1+
s ^ + k jM + k ,k
J ? (i)
R(s)
■(2)
s* + k ^ + k , k
Hence, the transfer function of the system from J?(;)to £ ( j ) is £ W = .
R{s)
+ k ^ + k ,k
Step 2 of 10 ^
Calculate the
=
from equation (2).
-2
1
»-»» s ^ + k fk s + k jk s
s .« H = o
(3)
Hence, the steady-state error for the unit step input is signal is |e_ ^ (oo) = 0 |.
Step 3 of 10 ^
Calculate the
from equation (2).
= lim -
*-*® s^+ k p k s + k fk
= 0 ...... (4)
Hence, the steady-state error for the unit ramp input is signal is |e
^ (oo) = 0 |.
Step 4 of 10
(b)
Write the denominator part of the equation (2).
s ^ + k ^ + k , k » 0 ...... (5)
Find the roots of the above equation.
■■
2
Hence, the roots of the closed loop poles from the transfer function from J?(5 )to
is
- k ^ ± y l{ k ^ y - 4 ( k ,k )
Step 5 of 10
(c)
Write the characteristic equation of the second order system.
5’ -f 2Co ^ - K i> / » 0 ...... (6)
Equate the equations (5) and (6).
s * + k fk s + k fk ^ s ^ + 2 f y a ^ + t o * ...... (7)
Substitute 0.707 for ^ and 1 for <o^ in equation (7).
j ^ + * ^ + * , * = j '+ 2 ( 0 . 7 0 7 ) ( l ) f + ( l ) ’
* = »’ + 2 ( 0 . 7 0 7 ) i + I .......(8)
Write the coefficients of the equation.
* , * = ! ...... (9)
* ,Jt = 2 (0 .7 0 7 )
*,Jt = 1.414......(10)
Write the transfer function of the entire system.
K (,)_
R(s)
0 {s )D A s )
1+G (*)D c ( 4
Step 6 of 10 ^
Substitute the plant gain of the system and controller gain from Figure 4.39 in the textbook for
in equation (11).
iif i.M l
« (•)
s^ + k.jk+ k,k
Y (s)_ k ( k ^ * k ,) .....
■i-kpks+kk,
Substitute equations (9) and (10) in equation (12).
i* + l.4 1 4 . f+ l
...... ,t3 )
Determine the peak overshoot value from damping ratio value
and natural frequency value
(«>.)
Write the general formula for percent of peak overshoot.
W , = e”* ''^ x l 0 0
(14)
Substitute 0.707 for ^ and 1 for <o^ in equation (14).
% A /,= e -^ 4 ” ’l ' ' ' i ^ x l 0 0
M , = 4 3 % ...... (15)
Hence, the value of peak overshoot is \M ^ =4.3% |.
Step 7 of 10
(d)
Determine the tracking emor signal as a function of time e ( /) at unit ramp signal.
Write £ ( j) fr o m equation (2).
R {s ) ..... ( 1 6 )
£ (4 ) = ^
s ^ + k fk s + k ,k
Substitute -L fo r
in equation (16).
-^kfks-ykfk
+ k ^ + k ,k
E(s)~
.(17)
Consider the general form of inverse Laplace transform.
r ' f - ----- 1 — - | = « '* 's i ii i<
(18)
Apply inverse Laplace transform in equation (17).
e(,)=
^ Sin
(19)
V 4 tt,-(tt,y
Hence, the value of tracking error signal as a function of time e (/)a t unit ramp signal is
^ 4 tt,- ( « , ) ’
2
'
^
/
Step 8 of 10
(e)
Write the small value of |e(/)| from the equation (19).
< > h- (* * ,) ’ » 0
4 tt, » ( * * , ) ’
^ > 0
2
(20)
...... (21)
Step 9 of 10 ^
Hence, In PI controller the relation between parameters ifc^and ifej,must be satisfy the
4 j M : , » ( J M : ^ f a n d ^ > 0 condition.
Step 10 of 10
(f)
Determine the transient behavior of the tracking emor for unit ramp input from equation (19).
. fj4tt,-(**,)’
8in J
s
2
O
V.
M
'
- (* * ,) ’ .
_
r —n
From equation (22), the overshoot time is finite and rather small for practical purposes.
Hence, the relation must satisfy the condition |4jbfc^
Problem 4.25PP
A linear ODE model of the DC motor with negligible armature inductance {La = 0) and with a
disturbance torque wwas given earlier in the chapter; it is restated here, in slightly different form.
= Mb +
Ki
^W,
Ki
where dm is measured In radians. Dividing through by the coefficient o f’^ , w e obtain
Sm+ 0\4m — ^Mfl +
where
„
KtKe
.
K,
^
I
With rotating potentiometers, it is possible to measure the positioning error between 6 and the
reference angle 0ror e = d re f- 0/n.With a tachometer we can measure the motor speed 6mWith rotating potentiometers, it is possible to measure the positioning error between d and the
reference angle drox e = dref - 0/n.With a tachometer we can measure the motor speed
Consider using feedback of the error e and the motor speed ^
in the form
where K and TD are controller gains to be determined.
(a)
Draw a block diagram of the resulting feedback system showing both dm and ^
as variables
in the diagram representing the motor.
(b)
Suppose the numbers work out so that a1 = 65. bO = 200, and cO = 10. If there is no load
torque (w = 0), what speed {in rpm) results from va = 100 V?
(c)
Using the parameter values given in part (b), let the control be D = kP + kDs and find kP and
kD so that, using the results of Chapter 3, a step change in dref with zero load torque results in a
transient that has an approximately 17% overshoot and that settles to within 5% of steady state
in less than 0.05 sec.
(d)
Derive an expression for the steady-state error to a reference angle input and compute its
value for your design In part (c) assuming dref= 1 rad.
(e)
Derive an expression for the steady-state error to a constant disturbance torque when dref=
and compute Its value for your design in part (c) assuming w = 1.0.
Step-by-step solution
step 1 of 5
Sketch the block diagram of the resulting feedback sjrstem.
Thus, the required block diagram is sketched.
Step 2 of 5
(b)
If V a = constant the system is in steady state
&= ^K -
Find
6 which is given by
^
2 0 0 x 1 0 0 60 rad s-’
p= -
65
27T rpm
0 = 2938 rpm
Thus, we get ^ = 2938 rpm .
Step 3 of 5
(c)
Find —
0
0y
Conq>aring with standard second order equation
0
aj
0^
= \1% we get
When
^ '= 0 .5 .
= 0.05 s
.
Find
= 0.05
k = 120
Thus, on comparing the coefficients, we get
i : = 72|
t; =
3 . 8 xlQT^I
step 4 of 5
(d)
Find the steady state error.
£ (s) = 0 ^ -0
s ( s + a .T ^ IC b .\
s^ + s(a^ +
For^,.«
1
5
we get
s „ = lim s 5 (s)
K=o|
Step 5 of 5
(e)
We know that
0
Qi
s* +
TjfKbft) + Kbfi
Find the response to torque.
0 „ = lim s0{s)
"
j-»o
'■ '
0 „ = lim s0{s)
"
j-*o
'
Problem 4.26PP
We wish to design an automatic speed control for an automobile. Assume that (1) the car has a
mass /n of 1000 kg, (2) the accelerator is the control U and supplies a force on the automobile of
10 N per degree of accelerator motion, and (3) air drag provides a friction force proportional to
velocity of 10 N • sec/m .
(a)
Obtain the transfer function from control input U to the velocity of the automobile.
(b)
Assume the velocity changes are given by
s+0,02
«+ao2
where V is given in meters per second. U is in degrees, and W is the percent grade of the road.
Design a proportional control law U = -kP V that will maintain a velocity error of less than 1 m/sec
in the presence of a constant 2% grade.
where V is given in meters per second. U is in degrees, and W is the percent grade of the road.
Design a proportional control law U = -kP V that will maintain a velocity error of less than 1 m/sec
in the presence of a constant 2% grade.
(c) Discuss what advantage (if any) integral control would have for this problem.
(d) Assuming that pure integral control (that is, no proportional tenn) is advantageous, select the
feedback gain so that the roots have critical damping (C= 1)•
Step-by-step solution
step 1 of 6
(a)
Consider the mass of the car.
ffl=l,000kg... (1)
Consider the accelerator is control U and supplies a force on the automobile.
=10Nperdegreeofacelerator....(2)
Consider the air drag provides a friction force proportional to velocity.
D=10N.sec/m...... (3)
Consider the relationship between the mass, acceleration and force.
m r = X ;^ ’
n a = K ^ - D x ...... (4)
Where.
u is displacement,
X is velocity,
x is acceleration.
Modify equation (4).
m v - K j u - D v ...... (5)
Where.
Vis velocity,
vis acceleration.
Apply Laplace transform in equation (5).
m sy (s) = K ,U {s)-D V {s)
m sy (s )+ D y (s ) = /C.U(s)
(ms + D )y (s ) = /C,U(s)
1/ (s)
(6)
(»« + D )
Calculate the transfer function of the automobile system.
Substitute equations (1), (2) and (3) in equation (6).
10
U (s)
(1,000s-M O )
l'( s )
1,000(0.01)
I / ( s ) ° l,0 0 0 (s +0.01)
y (s)
0.01
(7)
u ( s ) ” (i+ 0 .0 1 )
Hence, the transfer function of the automobile system is
y (s )_
t/( s )
0.01
( s + 0 .0 1 )
Step 2 of 6
(b)
Draw the general block diagram of automobile speed control system.
fy
Figure 1
Step 3 of 6
Write the output equation of error detector from Figure 1.
£ (s ) = » ;(s )-F (s )
f
M
J
£ ( s ) = i; ( s ) -
(8)
1+—^
1+—^
s+ a
V.
s+ a
Consider the velocity changes of the system.
LF(s)+
F ( i ) . — !—
' ’
j+ 0 .0 2
' ’
j+ 0 .0 2
HW
.......(9)
Step 4 of 6
Compare equation (8) from equation (9) and find the parameters a, A and B. Substitute a, A and
6 in equation (8).
0.0S
S+0-02
£ (s ) = i; ( s ) -
1+
»,
s+0-02
*1+-
G (,)
s+0.02
J+ 0 .0 2
0.05
= > ;(*)-
t + 0 .0 2 + t,
1.
> s + 0 . 0 2 + t,
s+ 0.02
^ 7
s+ 0 .0 2
;
( s + 0 .0 2 + i t,) F ,,( s ) - < : ,F ,,( s ) - 0 .0 5 G ( s )
S + 0 .0 2 + * ,
j,, ■
(s + 0 .0 2 ) i;(s ) -0 .0 S C ( s )
^ ’
J + 0 .0 2 + * ,
. ( 10)
Consider the velocity error of less than 1m/sec in the presence of a constant 2% grade. Assume
F ^ (j) is equal to zero.
Calculate the
= l i m j £ ( j ) — from equation (10).
| « . ( s / ^ ) | = Ums
- 0 .0 5
- 0.1
J+0.02+A ,
s lim
-
0.1
( 11)
0 .0 2 + * ,
The value of
-
0.1
0 .0 2 + * ,
2
J + 0 .0 2 + * , j
( jf ^ ) |
lesser than 1.
<1
* , > 0 . 0 8 .......(12)
Hence, the proportional constant value is * , > 0.0^
Step 5 of 6
(c)
The clear advantage of integral control is
[zerosteadystateeiTor| for step input.
Step 6 of 6
(d)
Calculate integral constant value.
Consider the value of proportional constant.
■(13)
* ,S -
Substitute equation (13) in equation (10).
( » + 0 .0 2 ) l( ,( 5 ) - 0 .0 5 g ( ^ )
i+ 0 . 0 2 + ^
, ■
j(j+ 0 .0 2 ) F ,,( i) - 0 .0 S jG ( t)
.(14)
j ’ + 0 .0 2 s + t,
Write the general characteristics equation of the second order system.
* ’ +2t«>,4+<i>,’ - 0 ...... (15)
Compare the equation (15) with denominator of equation (14).
* ’ + 25 o>,j+<a.’ . » ’ + 0 .02 i + * , ...... (16)
Write the damping ratio at critical damped condition.
C = 1 ...... (17)
Determine *^ from equations (16) and (17).
j * + 2 « ) ^ + c i ) / s j * + 0 .0 2 j + * ;
2<o,=0.02
<0, =0.01
*,= 0 .0 0 0 1
Hence, the value of Integral constant value is [*, - O.OOOI].
Problem 4.27PP
Consider the automobile speed control system depicted in Fig.
(a)
Find the transfer functions from W(s) and from R(s) to Y(s).
(b)
Assume that the desired speed is a constant reference r, so that
j{(j) = &
Assume that the road is level, so w(t) = 0. Compute values of the gains kP, Hr, and Hy to
guarantee that
Include both the open-loop (assuming Hy = 0) and feedback cases {Hy ^ 0) in your discussion.
(c)
Repeat part (b) assuming that a constant grade disturbance
(c)
Repeat part (b) assuming that a constant grade disturbance
W (s) = ^
^
is present in
is present in
addition to the reference input. In particular, find the variation in speed due to the grade change
for both the feed forward and feedback cases. Use your results to explain (1) why feedback
control is necessary and (2) how the gain kP should be chosen to reduce steady-state error.
(d)
Assume that w(t) = 0 and that the gain A undergoes the perturbation A + 5A. Determine the
error in speed due to the gain change for both the feed fon/vard and feedback cases. How should
the gains be chosen in this case to reduce the effects of 6A7
Figure Automobile speed-control system
^— 0 S te p -b y -s te p s o lu tio n
step 1 of 8
(a)
Refer from Figure 4.40 in the textbook and write the transfer function from IP^^Jand from J?(5 )
to y ( i ) .
k .—
H.
■R(s)
r ( ,) =
U k . — H,
' s*a ‘
\+ k .— H,
j+ a *
y ( j ) = ------------------- — « ( , ) ......................................(1)
^ ' s+ a+ k^AH , ' ’ s+a+k„AH ,
’
Hence, the transfers function from JP(j)andfrom J?(f)to y ( * ) is
i'W =
s + a + k ,A H ,
' '
s + a + k ,A H ,
' '
Step 2 of 8
(b)
Consider the value of road level.
w { s ) = 0 ...... (2)
Consider the gain of the feedback path.
H , = 0 ...... (3)
Consider the value of reference input.
S
The feedback gain value is considered as zero means the system belong to open loop system.
Consider the feed fonvard ( H , = 0 )
Calculate the value of lim « y ( j)fro m equation (1).
liin fK (f)s lim j[------- —
------ ^ ( j ) + —
r-«
w
^s+a+k^A N ^
' ^
s+ a+ k^A H ^
(5)
' ^
Substitute 0 for /f^and 0 for i r ( s ) in equation (5).
Kmj y (g) = Iim 5 ^ -^ ^ ^
j ....... (6)
Substitute equation (4) in equation (5).
Jk H
I i m 5 y ( 5 ) « l i m - ^ ^ ^ r , ...... (7)
Ak H
Consider the value of |im
^ ' j- .
•-* s + a ‘
lim— -—
*-“> s+ a
r, =r.
( 8)
Write the value of jyj,from equation (8).
■(9)
A k,
Hence, the value of //^ in feed forward system is
k = —
'
AH.
Ak^
zero value of f f = 0
step 3 of 8
Calculate the value of
at feedback path.
Calculate the value of Hm>’(f)from equation (1).
B
, V
A k ,ff,
Km y ( / ) s lim s ---------£ ---------iy ( s ) + ---------— — R (s ]
MA ' '
m O s+a+ k^A H y
' ' s*a ^k^A M ^ ' ^
( 10 )
Substitute equation (2) in equation (10).
^^ r
lim y ( / ) s lim s ------------------- R(.
^
\_s+ a+ k^A H y '
(11)
Substitute equation (4) in equation (1
lim y (/) = Ii m[—
f
—
12 )
. (
k ^ [s + a + k ,A H ,
Consider the value of Km
AkM ,
Km
■(13)
s+a+ k^A H ^ * j
Write the value of H^trom equation (13).
(14)
Hence, the value of H ,\n feedback system is:
'
Ak^
and the value of
at H 960
H ^ A -A H ,
Step 4 of 8
(c)
Consider the value of disturbance.
.(15)
s
Calculate the value of K m ^ ^(f)fro rn equation (1)
! ™ y r ( ') = 'i2 ^
JU
——
s+a+k,
^ '
B
, ,
AkH ,
j i n i y , ( / ) = s ------------- fy(s)+ ------ ‘- ^ R ( s )
a * k ,A H , ' ’ a * k ,A H , ' ’
*(*)]
.(16)
Substitute equation (9) in equation (16).
Ak.
B
, ,
' Ak,
— - — i y ( i ) + -----------a + k AH
' ' a + k AH
lim .) - ,( /) = » --------------- ^ ( 4 ) + --------------a+ k,A H , ^ ^ a+ k,A H ,
Substitute 0 for
, .
' ’
.(17)
' ^
in equation (17).
.(18)
Substitute equations (4) and (15) in equation (18).
! ™ y ^ ^ W = [ f « 4 + f 'o ]
B
U m > -,(0 = -« 4 + < i
B
«)>(«>) = — Wo
(19)
(20)
Hence, no possibilities available to [reduce the effect o f ftishirbanc^ in feed fonward system.
So the feedback control is not necessary in this system.
Step 5 of 8
Calculate the value of H*ity_^(f)from equation (1).
B
lim y ^ f f ) = lim 5 -
—— ^ w l
S + a + ki,A
, H,
------ --— r ( » ) + - d M : - j t ( $ ) ]
[a + * ,y W ,
' ' a+ k,A H , ' ' J
^B
/V
Ak„H,
J
w,+
r,
(21)
Substitute equation (14) in equation (21).
A k ^ U
^
■■■ W„ + ---------— }
.
B
a+k.AH,,
-----------H5, + ------- 2-----L ,
.(23)
« y > (“ ') = :a * k ,A H ,
Hence, effect of disturbance can managed by chooses
[largevalii^ of
k^.
Step 6 of 8
(d )
In feedfonward system, calculate the y y (oo) from equation (1).
y,(oo) = lim j[------- —
------ ^ W + —
f
ys+ a+ k^A H ^
A s [
B
?
(o o )= [------ ------
'
\a * k ,A H , •
— /f(s)l
s+a+ k^A M ^
^ 'J
Ak„H,
a + k, A H ,'
/V
^ '( ” )
^
^ '
a+ k,A H , "
.(24)
Substitute equation (2) in equation (24).
,
A kM ,
,
Substitute 0 for Hj,and A+SA^^^
, ^ ( „ ) . id ± 5 :4 * c 5 L r ,
in equation (25).
(26)
Step 7 of 8
Substitute equation (9) in equation (26).
R fH '
...... (27)
Hence, from the above equation concludes that |10% o f errorl is raised in the A value, the
same of % error is tracking.
Step 8 of 8
In feedback system, calculate the y ^ (oo) from equation (1).
,
r
B
’ \a * k ,A H , •
Ak,H ,
a+ k,A H , "
,
(28)
Substitute A+ S A ^^^ ^ in equation (28).
,
,
r
B
( A + & A )k ,H .
1
(29)
Substitute equation (2) in equation (29).
( A * iA ) k , H ,
Ra H ' -
a+ (A+ SA )k,H , "
(30)
Substitute equation (14) in equation (29).
(y4 + &4)Jt,
a + (A + S A )k ^ ffy
f
(^+ & 4 )
a + k ,A H A
&A
BA
Hence, the equation concludes that the tracking parameter variation is reduced by
of
[largev a lii^
Problem 4.28PP
Consider the multivariable system shown in Fig. Assume that the system is stable. Find the
transfer functions from each disturbance input to each output and determine the steady-state
values of y1 and y2 for constant disturbances. We define a multivariable system to be type k with
respect to polynomial inputs at wi if the steady-state value of every output is zero for any
combination of inputs of degree less than k and at least one input is a nonzero constant for an
input of degree k. What is the system type with respect to disturbance rejection at vv1 and at w2?
Figure Multivariable system
Step-by-step solution
step 1 of 4
Sketch the given figure o f the multi variable system.
-o n
step 2 of 4
We know that
„
1
_
s
—
~T~.-----T7
^i = T-
5 + ff + l
s(s + 1) «
~r~,----- T7 ^
s +s+\
e + ff + 1
Consider that there is constant disturbance condition.
Obtain the es^ression for the diff^erent variables in the equation
t V „ + ( s + \)W :„
+ S+1
Step 3 of 4
Let U2 be the signal coi^ling both the sjrstems.
Thus we get
(s + l ) ( j ^ - y a ) + s ( s + l ) » ;
ff + S + 1
Obtain the eg ressio n for Y2 .
y
S.
’ TTlT+l
I (^ + 1 )^ .
+3s+ 2
( s + l ) ’ ( -ira ) + s ( s + l ) V a
(s’ + 3s + 2 )(s’ + s + 1)
Step 4 of 4 ^
Obtain the different system types with respect to disturbances:
with respect to
Type 1
with respect to
Type 1
y 2 with respect to Wj
Type 1
y^ with respect to
Type 0
Problem 4.29PP
The transfer functions of speed control for a magnetic tape-drive system are shown in Fig. The
speed sensor is fast enough that its dynamics can be neglected and the diagram shows the
equivalent unity feedback system.
(a)
Assuming the reference is zero, what is the steady-state error due to a step disturbance
torque of 1 N m? What must the amplifier gain K be in order to make the steady-state error ess <
0.01 rad/sec?
(b)
Piot the roots of the closed-loop system in the complex plane and accurately sketch the time
response of the output for a step reference input using the gain K computed in part (a).
(c)
Plot the region in the complex plane of acceptable closed-loop poles corresponding to the
(c)
Plot the region in the complex plane of acceptable closed-loop poles corresponding to the
specifications of a 1% settling time of ts < 0.1 sec and an overshoot Mp < 5%.
(d)
Give values for kP and kD for a PD controller, which will meet the specifications.
(e)
How would the disturbance-induced steady-state error change with the new control scheme i
part (d)? How could the steady-state error to a disturbance torque be eliminated entirely?
Figure Speed-control system for a magnetic tape drive
/-aiokra*
S t e p - b y - s t e p s o lu t io n
Sketch the given figure.
Figure 1
(a)
Consider the reference is zero and the torque is 1 N.m.
^ 0.01nd/s•
The given steady-state error is
Determine the transfer function — for disturbance.
W
1
Js + b
0.5j + lj
{J s + b
Determine the steady-state emor,
(stepin
•
-------
Thus, the steady state error with a step due to step disturbance is
I + 1 0 *.
£ 0.01 we get AT, ^ 9.9
Thus, the amplifier gain
lAT^ lOl. to make the steady state error
^ 0.01
nd/s•
(b)
r(s )
Find the expression for
*.
10
1
y ( j)
+b'0.5s + 1)
Substitute 10 for
1 for ^ and 0.1 for J in the equation.
10(10) .
0 .5 J -H o . i j + i
y ( j)
l^ O .b + 1
0.5j + i J
100
(O.U+1)(0.5j + 1)+100
______
lOO_______
"(O.I)(0.5)[(j +10)(j +2)+2000]
2000
5^ + 12.^+20 + 2000
‘)(W I
5^ +
125+ 2020
Compare the expression with the standard system response and find
Thus,
and^
weget s V2020
2V 2^
Approximate the values to the nearest whole number to get a>^ s 44.94, ^ s0.13Hence, the roots are undesirable because, the damping is too low and there is high overshoot.
The roots of the closed loop are plotted and sketched.
4>
•3
I
Figure 2
The time response of the system for a step input is sketched.
I
1
Figure 3
(c)
Given that there is 1 % of settling time of
We know that for
^ 0.
1secand an over shoot of
i 0.1, we have ^ ^ 46-
Similarly for
S 0.05, ^ ^ 0.7
Step 8 of 11 ^
The region of acceptable closed loop poles is sketched in the complex plane.
Lines ofdam ping and
(d )
For a PD controller larger values of
and^
This can be achieved by increasing
and adding derivative fed back.
finri
are required.
y ( i)
' ^
lOAT,_L
y (» )
0 . 5 i- f l J s * b
w k
, ( k ^ + \)
(o .5 j + i) ( y * + ft)
200X.
y ( j)
n ,(s )
+ ( 1 2 + 2 0 O * :,* r„ ) i + 20(1 + 1 0 *:,)
Hence, the transfer function is
r(s )
200 * : ,
a ,(s )
s* + 12 + 2 0 0 * :,* :.) s + 20(1 + 10* : , )
By choosing the suitable values for Kp and KD any values for
ojj,, and^
can be achieved.
Step 10 of 11
(e)
Find the transfer function — for the disturbance.
1
Js + b
1
IT
J s + ft
io * : , ( * : „ f + i)
(0.5» + 1 )
20(0.5t +1)
ly ” j ’ + ( i 2 + 200 * : , * : . ) j + 20(1 + 10* : , )
The error when a step input is given is e,(step m W) = —
10*:,
Step 11 of 11
The derivative feedback affects the transient response only.
To eliminate the steady state error an integrator is added to the loop..
This can be represented by replacing AT^with AT^ +
in the fon/vard loop and still keeping PD
control in the feedback loop.
Thus, we get
y
20(0 .S s + l ) i
ly " s’ + (1 2 + 200 * : , * : . )s ’ + (2 0 + 200 * : , + 200 * : , * : . )s + 200 * :,
Hence, the steady state error to the disturbance will be [c , (step in W) = O] Thus, the steady state error to the disturbance torque is eliminated entirely.
Problem 4.30PP
Consider the system shown in Fig with PI control,
(a)
Determine the transfer function from R to Y.
(b)
Determine the transfer function from W to Y.
(c)
What are the system type and error constant with respect to reference tracking?
(d)
What are the system type and error constant with respect to disturbance rejection?
Figure Control system
Figure Control system
777T
Step-by-step solution
step 1 of 8
(a)
Refer to block diagram in Figure 4.43 in the textbook.
To calculate the transfer function from r to y , equate disturbance rejection to zero. The
modified block diagram is shown in Figure 1.
Step 2 of 8
Calculate the closed loop transfer function from R to Y.
10
I t S)
[
s
JU -
R(s)
+ S + 20J
10
y + f+ 2 o j
(V-t-t,)(io)
5 ( i * + j + 2 0 )+ ( + it;) (10)
io ( V + * ,)
»’ + j ’ +20s + 10*,i + 10*,
I0 ( * ,s + t ,)
»’ + f ’ + 1 0 (*,+ 2 )f+ IO *,
Therefore, the closed loop transfer function from R to Y is
s’ + i ’ + 1 0 (* ,+ 2 )i + 10t,
Step 3 of 8
(b)
To calculate the transfer function from fp to y , equate reference tracking to zero. The modified
block diagram is shown in Figure 2.
J
‘
10
i'+ i+ 2 0
V +*/
s
Figure 1
Step 4 of 8
Calculate the closed loop transfer function from jp to Y.
R(s)
1+
10
(“
)(7+ S + 2 0 J
10
, _______ s i + j + 2 0 _ _ ^
j ( j * + s + 2 0 ) + 1 0 ( i , j + i, )
i ( i ’ +s+20)
IQs
s’ + j ’ + 20s+10*^ + l0*,
lOs
” s’ + 4 ’ + 1 0 ( t ,+ 2 ) s + 10t,
Therefore, the closed loop transfer function from W to Y is
\0s
s’ + s ‘ + 1 0 ( t, + 2 ) j + 10t,
Step 5 of 8
(C)
The GharacteristiG equation is,
j ’ + s ’ + 1 0 ( * , + 2 ) i + 1 0 * ,= 0 .
Apply Routh’s array criteria.
1
I O ( * ,+ 2 )
I
lOit,
I0 (* ,+ 2 -* ,)
m ,
For stability.
10jlr,> 0
k ,> 0
And,
10( * , + 2 - * ,) > 0
* , + 2 -* ,> 0
* ,> * ,-2
The open loop transfer with respect to reference tracking is.
” ''>■18
10
+ S + 2 0 )
( V + t,) ( 1 0 )
Therefore, the system is Type 1 with respect to reference tracking.
Step 6 of 8
Calculate the emor constant.
ATy s lim 4G (4)
r (v + * ,) ( io ) '
s lim j
^ » ( s * + i + 20)
i^
i^ +
+s
s ++ :2 0
•-••I
J
( t,(0 )f* ,) (1 0 )
(0 )^ + 0 + 2 0
lOA;
20
k,
2
Therefore, the velocity constant,
with respect to reference tracking is
I
Step 7 of 8
(d )
The open loop transfer with respect to disturbance rejection is.
s
Therefore, the system is Type 1 with respect to disturbance rejection.
Step 8 of 8
Calculate the en^or constant.
= Um sG (5)
=lim(V+*;)
=* , ( 0 ) + t ,
-k ,
Therefore, the velocity constant, AT, with respect to disturbance rejection is 0
Problem 4.31 PP
Consider the second-order plant with transfer function
(f+IX5f+I)’
and in a unity feedback structure.
(a) Determine the system type and error constant with respect to tracking polynomiai reference
inputs of the system for P [Dc = kP\, PD [Dc (s) = kP + kD s], and [Z>c(s) =
+ ^
+
controllers. Letk P = ^ Q ,k l= 0.5, and
(b) Determine the system type and error constant of the system with respect to disturbance
inputs for each of the three regulators in part(a) with respect to rejecting polynomial
disturbanceswffj at the input to the plant.
ueiermine me system type ana error constani or me system witn respect to aisturoance
inputs for each of the three regulators in part(a) with respect to rejecting polynomial
disturbanceswffj at the input to the plant.
(c)
Is this system better at tracking references or rejecting disturbances? Explain your response
briefly.
(d)
Verify your results for parts(a) and(b) using Matlab by plotting unit-step and -ramp responses
for both tracking and disturbance rejection.
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 4.32PP
The DC motor speed control shown in Fig. is described by the difTerential equation
y+ 60y= 600va-1500w,
where y is the motor speed, i^a is the armature voltage, andw is the load torque. Assume the
armature voltage is computed using the PI control law
Va = - ^*j>e + t / j T
e d t^ ,
where e = r - y .
(a) Compute the transfer function from W toYas a function of kP and kl.
(b) Compute values for kP and k l so that the characteristic equation of the closed-loop system
(b) Compute values for kP and k l so that the characteristic equation of the closed-loop system
will have roots at -60 ± 60/.
Figure DC Motor speed-control block diagram
Step-by-step solution
step 1 of 4
(a)
Consider the differential equation of the DC motor.
+60;>=600v, - l.SOOw
Apply Laplace transform on both sides.
jK(j)+60y(s) =600K.(5)- 1.500FK(5)
Consider the armature voltage value in PI control.
Apply Laplace transform on both sides.
Step 2 of 4
Substitute
for
in the Laplace transform of the differential
equation.
(i +60)r(i)-600^-*,,£(j)-^£(j)j-l,500»'(s)
{s + 60)Y(s) = - 6 0 0 ^ ^ * , j £ ( j ) j - l ,5 0 0 » '( i )
( j + 6 0 ) r ( j ) + 6 0 0 ^ ^ * , + ^ j £ ( j ) j = - l , 5 0 0 » '( j )
^ s + 60 + 6 0 0 t , + 6 0 0 ^ j £ ( s ) = - l, 5 0 0 » '( » )
E (s)
-1 ,5 0 0
+ 6 0 + 600*, + 6 0 0 ^ j
- l,5 0 0 i
” ( i “ + 6 0 j + 6 0 0 * ,* + 60 0 * ,)
Consider the input R as zero. The error function is,
£(s)=-r(s)
y (j)
________ l,5 0 to __________
W{s)
« '+ 6 0 ( l + I O * ,)s + 6 0 0 * ,
1,5005
Thus, the transfer function from 1/Yto Y is
*’ +60(1+10*,)i +600*,
Step 3 of 4 ^
The roots of characteristics equation are - 6 0 + 6 0 y ,-6 0 -6 0 y
Write the characteristic equation from the roots.
( s + 6 0 + 6 0 y ){ i+ 6 0 - 6 0 y ) = s’ + 120*+7200
The characteristic equation of the system is,
, * + 6 0 ( l + 1 0 * , ) i + 6 0 0 * ,= 0
Compare the two characteristic equations.
600Jt,»7200
60(U10Jfcp)»120
Calculate integral constant kf ■
_ 7200
600
-12
Thus, the value of kf is m
Step 4 of 4
Calculate proportional constant kp ■
60(1+ 1 0 *,) = 120
1+ 10*, = 2
* ,= 0 .1
Thus, the value of kf, is I P
Problem 4.33PP
For the system in Fig., compute the following steady-state errors:
(a) to a unit-step reference input;
(b) to a unit-ramp reference input;
(c)
to a unit-step disturbance input;
(d)
for a unit-ramp disturbance input.
(e)
Verify your answers to (a) and (d) using Matlab. Note that a ramp response can be generated
(e)
Verify your answers to (a) and (d) using Matlab. Note that a ramp response can be generated
as a step response of a system modified by an added integrator at the reference input.
Figure DC Motor speed-control block diagram
Step-by-step solution
Step 1 of 16
Refer to Figure 4.44 in the text book for the block diagram of a DC motor speed-control.
The controller transfer function is,
D ,(5 ) = * , + i
Consider the differential equation of the DC motor.
+ 60;> = 600v, - l.SOOw
Apply Laplace transform on both sides.
» r(i)+ 6 0 1 '(s ) = 6 0 0 F .(i)- l,500»'(j)
Consider the armature voltage value in PI control.
Apply Laplace transform on both sides.
Step 2 of 16
Calculate the transfer function of the PI controller.
( i + 6 0 ) r ( i ) - 6 0 0 ^ - * , , £ ( j ) - ^ £ ( j ) j - l , 5 0 0 » '( s )
{s + 6 0 )Y (s ) =
6 0 0 ^ ^ * , j £ ( j ) j-1 ,5 0 0 » '(* )
( j + 6 0 ) r ( j ) + 6 0 0 ^ ^ * , + ^ j £ ( j ) j = - l,5 0 0 » '( j)
^s + 60 + 6 0 0 t , + 6 0 0 ^ j £ ( s ) = -l,5 0 0 » '(» )
-1,500
^ i+ 6 0 + 600*, + 6 0 0 ^ j
-1.500s
( j “ +60s + 600*,j + 600*,)
Consider the value of R as zero and write the value of error detector.
£ ( » ) — l'( s )
1,500s
s '+ 6 0 (l + 101:,)s+6001:,
»'(s)
Step 3 of 16
The roots of characteristics equation is - ^ + 6 0 J y - 6 0 - 6 0 J ■
Write the characteristic equation from the roots.
(s+60+60y){s+60 -6 0 y ) = s’ + 120s+7200
Write the general characteristic equation.
5*+2C o ^ + o),* « 0
Calculate natural frequency
from equations (7) and (8).
« ),= V 7 i2 0 0
< t t= i0 j2
Step 4 of 16
Calculate damping ratio, ^ .
120
2 ( 6 0 j2 )
5 = 0.707
Calculate integral constant k, ■
600k, -(6 0 > ^ )“
k,^12
Step 5 of 16
Calculate proportional constant k ^ .
60(1+1 Ot, ) = 2(0.707)(60V2)
60
2 (0 .7 07 )(6 0^ )
lO/t,
^
60
* , = 0.1
Step 6 of 16
(a)
To determine the transfer function with reference input, take disturbance input, » '( s )
as zero.
Determine the transfer function from the output to the reference input.
1+600
6 0 0 ( V + it ; )
s(s + 60) ♦ 600(^^ + k,)
Step 7 of 16
Determine the steady state error.
£ (s ) = « ( s ) - l '( s )
I
y (» )
£ (s )
£ (s)
I
600(*,s + * ,)
s(s+60) + 6 0 0 (*,s + *,)
s(s + 60)
s (s + 60)+600( t ,s + * ,)
j ( j -f 60)
£ (s ) =
s(s + 60) + 600(k^ + k,)
g(s)
Step 8 of 16
The unit-step reference input is,
I
£ (s ) = l
Determine the steady-state emor to unit-step reference input.
= lin i5 £ (^ )
s lim j
+ 60)
1
5 (j + 6 0 ) + 6 0 0 ( it^ + /t,) 5
Thus, the steady state error to unit step reference input is
.
Step 9 of 16
(b)
The unit-ramp reference input is.
Determine the steady-state emor to unit-step reference input.
g |im j£ ( f )
g (j+ 6 0 )
s lim ;
1
g ( g + 6 0 ) + 6 0 0 ( it^ + * ,) 4 ^
(j+ 6 0 )
= lim -
• 5 (g + 60) + 6 0 0 ( + Jfc,)
60
600A;,
I
" lo t ,
The value of k, is 12.
1
10(12)
1
*120
Thus, the steady state error to unit ramp reference input is
Step 10 of 16
(c)
The transfer function from the output to the disturbance input is,
£ ( i)
B '( i)
-1,500»
« '+ 6 0 (l + 10*,)s+600*,
Substitute 12 for k, and 0.1 for k^-
-l,5 0 te
j ’ +6 0 (l + 10x0.1)f + 600(12)
£ (£ )= .
^ (s )
j’
-l,5 0 0 i
+ 120j + 7200
The error function is.
E (s ) = ~i-------------------fY(s)
g’ +120g + 7200
The unit-step disturbance input is.
Step 11 of 16
Determine the steady-state emor to unit-step reference input.
e „^ \m sW {s)
-hSOOs
1
slim gg* + 120g + 7200g
=0
Thus, the steady state error to unit step disturbance input is
.
Step 12 of 16
(d)
The unit-ramp disturbance input is.
Determine the steady-state emor to unit-step reference input.
e„ = \m sE {s)
-USOQg
1
s lim g - r
<->a g* + 120g + 7 2 0 0 g *
..
-1,5 00
^
.................
« - » i' + 120f + 7200
1500
= -0.208
Thus, the steady state error to unit ramp disturbance input is |-0.2081-
Step 13 of 16
(e)
The error function is.
g(s + 60)
E (s) =
s (s + 6 O )+ 6 O 0 {k p S + k ,
j( g + 60)
■ , ’ + (6 0 + 6 0 0 t , ) j + 600*,
Write the MATLAB code to verify the steady state emor for unit step reference input.
ki=12;
kp=0.1;
n=[1 60 0];
d=[1 60+600*kp 600*ki];
sys=tf(n,d);
step(sys)
Step 14 of 16
Get the MATLAB output for the step response.
Step 15 of 16
The transfer function to disturbance input is.
y ( j)
» '( i)
-l,50to
+120s+ 7200
Note that the ramp response is obtained by added an integrator tenn to the transfer function.
y (s )
-1.500s
B '( s ) " s ( s ’ + 120$ + 7200)
Write the MATLAB code to verify the steady state emor for unit ramp disturbance input.
n=[-1500 0];
d=[1 120 7200 0];
sys=tf(n,d);
step(sys)
Step 16 of 16
Get the MATLAB output for the ramp response.
Hence, the result is verified.
Problem 4.34PP
Consider the satellite-attitude control problem shown in Fig. where the normalized parameters
are
J = 10 spacecraft inertia, N m sec2/rad
dr= reference satellite attitude, rad.
6 = actual satellite attitude, rad.
Hy= ^ sensor scale, factor V/rad.
Hr = 1 reference sensor scale factor, V/rad.
w = disturbance torque. N m.
(a) Use proportional control, P. with Dc (s) = kP. and give the range of values for kP for which the
system will be stable.
use piupuiuuiieii ia
^ (o/ —A/~, eiiiu ^ive me teiii^e l veiues i«
system will be stable.
(b) Use PD control, let Dc(s) = {kP + kDs), and determine the system type and emor constant
with respect to reference inputs.
(c)
Use PD control, let Dc(s) ={kP + kDs), and determine the system type and error constant with
respect to disturbance inputs.
(d)
Use IP control, let D c(s) = {kp + ^ ) ,
determine the system type and error constant
with respect to reference inputs.
(e)
Use IP control, let Dc(s) = {kp + y ) ,
determine the system type and emor constant
with respect to disturbance inputs.
(f)
Use IPD control, let Dc{s) = {kp-\- ^
^nd determine the system type and error
constant with respect to reference inputs.
(g)
Use IPD control, let Dc{s) = {kp + ^ + kps) >^^d determine the system type and e
constant with respect to disturbance inputs.
Figure Satellite attitude control
-O 0
S te p -b y -s te p s o lu tio n
step 1 of 9
J=10, H y=l, H j= l, ‘W=Disturbance torque
Step 2 of 9
(a)
From the block dis^gram of the sjrstem,
Js’0(s)= ((6t.e)D + W )
>e(s)=
'■ '
D+Js" ^
D+Js"
E(s)=%(s)e(s)
D+Js^
D+Js^
Step 3 of 9
Characteristic equation is D+Js^ = 0
i.e., 10s^-HCp=0 (Proportional control)
By routh's criterion for stability, [kp > 0| and
system is marginally stable with roots on jo) axis.
Step 4 of 9
(b)
D (s)^p-H cpS
(PD control)
c .= lim sE (s)
lOs^
1
=lim s k ,4t„s+ 10s"^ 7s12.
1
But,as
e,f= —
(k^=Accelaration const.)
^"T o
step 5 of 9
(c)
D(s)=fcp4fcps
E (0 _ -Y (0
W (s)
W (s)
= i(!l
w (0
kfl+k^s+lOs
>S3Tstemis of |Type-0| [Bror const: 4Cp j
Step 6 of 9
D (s) ^ p + — (PI control)
s
lOs'
E (0 =
^ (0
kp+ — +10s'
s
10s’
T ><r(0
kj-HtpS+lOs'
e_=lim sE(s)
* S-kO
s lim s
^ '
1
10s-
10
k|-HcpS+10s^
System is of [Type -3] and error const | ^ .
Step 7 of 9
W
D ( s ) = k ,+ ^
E ( 0 _ _____ L
k ,+ ^ + 1 0 s ’
k|-HcpS+10s
[Type -Ij "With error const = |
Step 8 of 9
(£)
D ( s ) ^ p + — "Hc^s
s
—
(PID control)
lOs^
k ---------------7
rW
kp+—
lOs'
sE (s)----------------- >-------w ^ ( s )
^ ^kpS+4ki-HcpS^+10s^ ^
‘
[Type -31 witti error const = —
Step 9 of 9
(g )
D (s )^ p + -^ -K tp S
ECO ________
^C O
[Type ^
k ,+ ii-H c„s+ 10s’
s
with error const. =
Problem 4.35PP
Automatic ship steering is particularly useful in heavy seas when it is important to maintain the
ship along an accurate path. Such a control system for a large tanker is shown in Fig., with the
plant transfer function relating heading changes to rudder deflection in radians.
(a) Write the differential equation that reiates the heading angle to rudder angle for the ship
without feedback.
(b) This control system uses simple proportional feedback with the gain of unity. Is the closedloop system stable as shown?( Hint: use Routh’s criterion.)
(c) Is it possible to stabilize this system by changing the proportional gain from unity to a lower
value?
(c) Is it possible to stabilize this system by changing the proportional gain from unity to a lower
value?
(d) Use Matlab to design a dynamic controller of the form Deis) = K
^
closed-loop system is stable and in response to a step heading command it has zero steadystate error and less than 10% overshoot. Are these reasonable values for a large tanker?
Figure Ship-steering control system
—N.
Rudder
angle
r
L
- 0 .I 6 4 ( j -f 0.2K j - 0 .3 2 )
H e a d in g ^
*’ (* + 0 .2 5 X * - 0.009)
*
_
Step-by-step solution
step 1 of 8
(a)
Refer figure 4.46 in the textbook and write the differential equation without consider the feedback
path.
, ■
- 0 .1 6 4 ( t + 0 . 2 ) ( j - 0 . 3 2 )
(1)
» * ( * + 0 .2 5 ) ( j- 0 .0 0 9 )
Multiply out the denominator and numerator terms in equation (1).
- 0 .I 6 4 ( j ’ - 0 .1 2 j - 0 .0 6 4 )
(2)
G (* ) =
i V 0 . 2 4 1 s ’ -0.0025»*
Write the differential equation that heading angle to ruder angle for the ship without consider the
feedback path by apply Laplace transform.
^ + 0 . 2 4 1 ^ - 0 . 0 0 2 5 ^ = -0 .1 6 4 f^ - 0 .1 2 ^ - 0 .0 6 4 tfl
<**
dr’
dr’
dr’
)
dr
(3)
Hence, the differential equation of the system without feedback is
^ + 0 J 4 I ^ - 0 . 0 0 2 5 ^ ~ - 0 . I 6 4 ( ^ - 0 . 1 2 — -0 .0 6 4 s ]
dt*
d t^
d t^
dt
'
Step 2 of 8
(b)
Write the general characteristic equation of the system.
l + G ( s ) / f ( s ) = 0 .......(4)
Substitute equation (2) in equation (4).
- O .I6 4 ( s ’ - 0 .1 2 * - 0 .0 6 4 )
*’ + 0 .2 4 U ’ -0 .0 0 2 5 * ’
,
^ (4 -®
-0 .1 6 4 * ’ + 0.01968* + 0 .0 1 0 4 9 6 „ , .
.
1+........ ............... ......... ,------ ff(t) = 0
-(5)
*‘ + 0.241*’ -0 .0 0 2 5 * ’
Substitute 1 for f f ( s ) in equation (5.
.
-0 .1 6 4 * ’ + 0.0 1 9 6 8 * + 0 .0 1 0 4 9 6
*‘ + 0.241*’ -0 .0 0 2 5 * ’
( 1 ).0
*‘ +0.241** -0 .0 0 2 5 * ’ - 0 .1 6 4 * ’ + 0.0 1 9 6 g * + 0 .0 1 0 4 9 6 = 0
-(B)
Step 3 of 8
The system is stable if the equation satisfies the following condition.
• All the terms in the first column of the Routh’s array should be positive sign.
• The first column of Routh’s array should not possess any sign change.
Apply Routh-Hurwitz criteria in equation (6).
Figure 1
Step 4 of 8
From the Figure (1), all values in first column of Routh’s array are not positive. So the system is
lunstablel
step 5 of 8
(c)
Write the closed loop characteristics polynomial equation.
i
+ a: , g ( * ) = o .......(7)
1+ JC.
'
-0 .1 6 4 * ’ + 0 .0 1 9 6 8 * + 0 .0 1 0 4 %
„
*‘ +0.241*’ -0 .0 0 2 5 * ’
*‘ +0.241** -0 .0 0 2 5 * ’ -0 .1 6 4 A :,* ’ + 0 .0 1 9 6 8 A :^ + 0 .0 1 0 4 9 6 i:, = 0
['*‘ + 0.241*’ + ( - 0 .0 0 2 5 - 0 .1 6 4 8 :,) * ’ ')
^
sO
(8)
[+ 0 .0 1 9 6 8 8 :,* + 0 .0 1 0 4 9 6 8 :,
From equation (8), all coefficients of characteristics equation are not same sign.
Step 6 of 8
Therefore only the proportional constant is
|not sufRcientI
bring the system from unstable to
stable.
Step 7 of 8
(d)
Consider the value of 0 ^ ( 4 ) •
Consider the value of a.
a = 0.002 ...... (10)
Consider the value of b.
*=2 . 8 ... (11)
Consider the value of K.
K ^ 2 S ...... (12)
Write the MATLAB program for design the dynamic controller of the form from equations (2), {10),
(11) and (12).
numP=-0.164*(s+0.2)*(s-0.32);
denP=s''2*(s+0.25)*(s-0.009);
numDc=25*(s+0.002)''2;
denDc=(s+2.8)''2;
sys=tf(numP*numDc/(denP*denDc));
sysCL=feedback(sys, 1);
step(sysCL);
The output of MATLAB program is given below.
Figure 2
Step 8 of 8
From figure (2), the closed loop is step response and its rise time is very short.
Hence these values are
[nnfealistic for the large tanker].
Problem 4.36PP
The unit-step response of a paper machine is shown in Fig.(a) where the input into the system is
stock flow onto the wire and the output is basis weight (thickness). The time delay and slope of
the transient response may be determined from the figure.
(a) Find the proportional-, PI-, and PID-controller parameters using the Ziegler-Nichols transientresponse method.
(b)
Using proportional feedback control, control designers have obtained a closed-loop system
with the unit impulse response shown in Fig.(b). When the gain Ku = 8.556, the system is on the
verge of instability. Determine the proportional-, PI-, and PID-controller parameters according to
the Ziegler-Nichols ultimate sensitivity method.
Figure Paper-machine response data
—i^ioi luio ului IIdle oci loiuviiy 111011100.
II ic
Figure Paper-machine response data
m e (mc )
H m efee
0>)
(I)
Step-by-step solution
step 1 of 8
(a)
From the step response vs. time graph,
We get,
6
L sl
Step 2 of 8
For proportional control,
k p = ^ ^ = r6 ]
Step 3 of 8
For PI control,
k„= —
= ls!^
T i= — ^
‘ 0.3
Step 4 of 8
1.2
For PID control, k p * —^ « 7 .2
' RL
1 ^ 5L=0.5|
Step 5 of 8
(b)
K,=8.556
Pq = 2.25 (From unit impulse response g r ^ h )
Step 6 of 8
For proportional control, RpsO.5 , K^=4.278
Step 7 of 8
For PI control,
Rp=0.45, K^=3.8502
T,= .^ = 1 .8 7 5
1.2
Step 8 of 8
For PID control,
Rp=0.6,K,=5.1336
T ,= ^ = U 2 5
T „ = is. =0.28125
“
8
The control is given by
D (s)=Rp 1+— +Tps
V.
)
Problem 4.37PP
A paper machine has the transfer function
<Hs) =
3 s+ l'
where the input is stock flow onto the wire and the output is basis weight or thickness.
(a) Find the PID-controller parameters using the Ziegler-Nichols tuning rules.
(b) The system becomes marginally stable for a proportional gain of Ku = 3.044 as shown by the
unit impulse response in Fig. Find the optimal PID-controller parameters according to the
Ziegler-Nichols tuning rules.
Figure Unit impulse response
0.0201--------- 1--------- 1-------
1— r
n
Figure Unit impulse response
Step-by-step solution
Step 1 of 3
A = l,
r=
t= 3 , z <
3s=2
A=1
z
3
L=z<i^2
Step 2 of 3
(a)
For PHD control,
T ,= 2 L = 0
T„=0.5I^|i]
Step 3 of 3
(b)
K„=3.044
For PID control, R b=0.6K„H1-82641
T,=S.=lT75l
4
Tb = | = I
Problem 4.38PP
Consider the DC motor speed-control system shown in Fig. with proportional control, (a) Add
feedfonward control to eliminate the steady-state tracking error for a step reference input, (b) Also
add feedforward control to eliminate the effect of a constant output disturbance signal, w, on the
output of the system.
Figure Block diagram
Step-by-step solution
step 1 of 7
(a)
Consider the gain of the plant.
C ( .) =
59.292
i"+6.978s + 15.123
(1)
Consider the gain of the controller.
= 3 ......(2)
Consider the gain of sensor.
H{$) = \ ..... (3)
Determine the DC gain of the plant.
15.123
G " ( 0 ) = i59.292
C -'(0 ) = 0.2551 ..... (4)
Write the output equation for error detector.
£ ( j ) = j t ( j ) - y ( 5 ) ...... (5)
step 2 of 7
Refer from Figure 4.49 in the textbook and write the transfer function equation with respect to
zero Wvalue.
y (« ) = G (» )[*,£ (»)+G -'(0 )J ? (» )]
(6)
Substitute equation (5) in equation (6).
y ( » ) = G W [ * , ( « M - r W ) + G - '( o ) J iW ]
i'( 4 ) = G ( 4 ) [ M ( 4 ) - t , i'( 4 ) ) + c r - ( o ) « ( * ) ]
r ( j ) = G (f ) * , « ( j ) - G ( i ) t , l ' ( j ) + G ( j ) G - ' {0)R(s)
( l + G ( * ) * , ) y ( , ) = « W [ G W * , + G W G - '( 0 ) ]
y(»)
f t ,-fG -'(0 )lG (» )
« ( i)
l+ * ,G ( i)
(7)
Step 3 of 7
Write the MATLAB program for tracking response with feedfonward from equations (1), and (7).
clc;
s=tf('s');
G=59.292/(s''2+6.978*s+15.123);
kp=3;
dcgain1=dcgain(G);
T1 =G*{1/dcgain1 +kp)/(1 +kp*G);
t=0:.01:5;
y1=step(T1,t):
plot(t,y1);
xlabel{'Time (sec)');
ylabel{'$y(t)$'.'interpreter'.'latex'):
The output for MATLAB program is given below.
Figure 1
Step 4 of 7
The figure 1 clarifies the influence of feedfonward control in eliminating the steady-state tracking
error.
Hence, the addition of feedfonward control results in zero steady-state tracking error for a step
reference input and the dc gain of the closed loop system is Iunity!.
Step 5 of 7
(b)
Refer from Figure 4.49 in the textbook and write the transfer function equation with Wvalue.
r ( j) = iy ( i) + G ( s ) [ * , £ ( j) + G - '( o ) » '( j) ]
(8)
Write the output equation for error detector.
f(4 ) = J?(4)-y(5) ..... (9)
Consider the value of J?(4 )= 0 ...... (10)
Substitute equation (10) in equation (9).
£ ( 4 ) = - r ( 4 ) ...... (11)
Substitute equation (11) in equation (8).
y ( , ) = » '( * ) + G W [ * , ( - y M ) - G - '( o ) » '( s ) ]
y ( s ) . ( y ( i ) - t , G ( j ) r ( s ) - G - '( o ) G ( s ) ) r ( j )
[ N . t , G W ] y ( s ) = [ i- G - ( o ) G ( * ) ] iy W
r(4
[ i- G - '( o ) G M ]
W (s)~
[l+ t,G (j)]
(12)
Step 6 of 7
Write the MATLAB program for disturbance rejection response with feedfonvard systems from
equations (1), and (12).
s=tf('s');
G=59.292/(s''2+6.978*s+15.123);
kp=3;
dcgain1=dcgain(G);
t=0:.01:5;
Tw1=(1-1/dcgain1*G)/(1+kp*G);
yw1 =step(Tw1 ,t);
figureO
plot(t,yw1);
xlabel('Time (sec)');
ylabel('$y(t)$'.'interpreter'.'latex');
nicegrid
The output for MATLAB program is given below.
Figure 2
Step 7 of 7
The figure 2 clarifies the influence of feedfonward control in eliminating the steady-state tracking
error for a step output disturbance.
Hence, the dc gain of the closed loop system is |zero|.
Problem 4.39PP
Compute the discrete equivalents for the following possible controllers using the trapezoid mle of
Eq. in Appendix W4.5. Let Ts = 0.05 sec in each case.
(a) £>c1 {s) =(s + 2)/2,
(b) D d (s ) = 2 ^ ,
(=) D c s W —
fd) D ^ ( s ^ = 5 (j +2)(5-44).1)
[O)
— J(,+|0)(s+ 0.01)*
/.n
rv
e
( 5 + 2 ) ( 5 + 0 .1 )
(d) O c4 W - 5 ( ,^ .1 0 ) ( ,+ 0 .0 |) Eq.
l + GDci = 0,
Step-by-step solution
step 1 of 4
(a)
Consider the sample period.
? ;« 0 .0 5 s e c ...... (1)
Write the value of Z ),,^j)o r
.
Write the formula for discrete operator.
t; z
+1
...... (3)
Substitute equation (1) in equation (3).
2 z -\
* “ 0 .0 5 z + l
z = 4 0 i 4 .......('*>
Z+ 1
U (z )
Calculate — “
£ (z)
t/( z )
from equation (2) and (4).
z -f2 |
2
4 0 z-4 0 + 2 z+ 2
=____ thJ____
2
4 2 Z -3 8
-
z tl
2
£ (z)
...... (5)
z+ 1
Hence, the value of discrete equivalent for given transfer function is
2 IZ -1 9
z+l
Step 2 of 4
(b)
£ (z )
’
£ (z)
■(6)
z+4
l/(z )
Calculate — “ from equation (4) and (6).
£ (z)
£ l £ i = 2 £ ± l|
£ (z )
z+4|,_ffl«-i
+2
+4
+ 2- 2z z+ +2 2" \j
^ 4 0 z-4 0 H
=2
4 0 z -4 0 + 4 z + 4 j
z+l
=2
42Z-38
4 4 Z -3 6
I /(z )
4 2 ( z - 0 .9 )
£ (z)”
4 4 ( z - 0 .8 )
I /( z )
1.909Z-1.72
£ (z)
z - 0 .8
■(7)
Hence, the value of discrete equivalent for given transfer function is
1.909Z-1.72
z - 0 .8
Step 3 of 4
(c)
Write the value
■(8)
a - M l#
l/(z )
Calculate — “ from equation (4) and (8).
£ (z)
£ (£ l = 5 ± tij
£ (z)
z+10L,„>+
+2
'F i l
+10
l '4 0 z - 4 0 + 2 z + 2 'l
_____ £±1_____ L
r 4 0 z -4400++l<
10z + 10j
I
=5
I /( z )
z+l
4 2 Z -3 8
SO z-30
4 2 (z-0 .9 0 S )
£ ( z ) " ’ 5 0 ( z - 0 .6 )
4.2 ( z - 0.905)
( z - 0 .6 )
4.2Z-3.801
I /( z )
£ (z)°
z - 0 .6
(9)
Hence, the value of discrete equivalent for given transfer function is
4.2Z-3.801
z - 0 .6
Step 4 of 4
(d)
m .
£ (z )
’
E {s )
(z + 10)(j+0.01)
.(10)
l/(z )
Calculate — “ from equation (4) and (10).
£ (z)
U (z )
^ (z + 2 )(z + 0 .1 ) I
£ (z)
( i+ 1 0 ) ( i+ 0 .0 1 ) L _ ^
^ 4 0 z - 4 0 + 2 z + 2 j ^ 4 0 z - 4 0 + 0 .1 z + 0 .1 j
° ^ ^ 4 0 z - 4 0 + 1 0 z + 10j ^ 4 0 z - 4 0 + 0 .0 1 z + 0 .0 1 j
z+l
,
( 4 2 z - 3 8 ) ( 4 0 .1 z - 3 9 .9 )
° ’ ( 5 0 z - 3 0 ) ( 4 0 .0 1 z - 3 9 .9 9 )
t/( z )
l,6 8 4 .2 z ^ -3 ,1 9 9 .6 z + l,516.2
£ (z)
’ 2,0 0 0 .5 z’ -3 ,1 9 9 .8 z + l,1 9 9 .7
^ l,6 8 4 .2 (z ’ - 1 .8 9 8 z + 0 .9 )
2 ,0 0 0 .5 (z ‘ - 1 .6 z + 0 .5 9 9 7 )
4 .2 (z ’ - 1 .8 9 8 z + 0 .9 )
z ’ - 1 .6 z + 0 .5 9 9 7
I/(z )
4 .2 z * - 7 .9 7 z + 3 .7 8
£ (z)
z ’ - 1 .6 z + 0 .5 9 9 7
(11)
Hence, the value of discrete equivalent for given transfer function is
4.2z* -7 .9 7z+ 3.78
z ^ - l .6 z + 0 .5 9 9 7
Problem 4.40PP
Give the difference equations corresponding to the discrete controliers found in Problem,
respectively.
(a) Part 1.
(b) Part 2.
(c) Part 3.
(d) Part 4
(d) Part 4
Compute the discrete equivalents for the following possible controllers using the trapezoid mle of
Eq. in Appendix W4.5. Let Ts = 0.05 sec in each case.
(a) £>c1 {s) =(s + 2)/2,
(b) D t i( s ) = 2 ^ ,
(=) D c3( s) — 5 - ^ j ,
Eq
l+ G D c / = 0,
Step-by-step solution
step 1 of 4
(a)
Refer part (a) in the solution of 4.39P and write the discrete equivalent equation for the controller.
H £ i= ll£ z li
(1)
E (z )
' '
z+ 1
Modify equation (1).
I /( z )
E {z ) ~
z ( 2 l- 1 9 z - ')
z ( l + r - ')
2 1 -I9 Z -'
1+Z-'
( l+ r - ') l /( z ) = ( 2 1 - 1 9 z - ') ( £ ( z ) )
(2)
Convert z to /f form:
Consider the general form of
z -'U (z ) = u { k - l )
•
(3)
Consider the general form of U {z ).
l/ ( z ) = « l ( t )
(4)
Substitute equations (3) and (4) in equation (2).
» ( * ) + u (t-l)= 2 1 « (* )-1 9 e (* -l)
ii(* ) = K ( * - l ) + 2 1 « ( * ) - 1 9 e ( * - l )
Hence, the difference equations to the discrete controller is
|ir(it) = -ii(lfc-l)+ 2 1 e(lfc)-1 9 e(it-l)|
Step 2 of 4
(b)
Refer part (b) in the solution of 4.39P and write the discrete equivalent equation for the controller.
I / ( r ) _ 1.909z-1.72
£ (z)
z - 0 .8
' '
Modify equation (5).
U {z )
z(l.9 0 9 -1 .7 2 z-‘)
E (z )
z ( l - 0 .8 z - ')
U ( z ) ( l- 0 .8 z - ') = ( l.9 0 9 - 1 .7 2 z - ') £ ( z ) ...... (6)
Convert z to /f form from equation (6).
1/ (z ) - 0 .8 z - 't/ (z ) = 1.909£(z) - 1 .7 2 z-'£ (z)
» ( * ) - 0 . 8 a ( * - l ) = l.9 0 9 e (* )-1 .7 2 « (t-l)
a (* ) = 0.8« (* - 1 ) + 1 .9 0 9 e(t) - 1.72«(* -1 )
Hence, the difference equations to the discrete controller are
|a(ifc) = 0 .8 a ( it-l)+ l.9 0 9 z (* )-1 .7 2 z (* -'i)|.
Step 3 of 4
(c)
Refer part (c) in the solution of 4.39P and write the discrete equivalent equation for the controller.
U {z )
£ (z )~
4.2Z-3.801
z - 0 .6
(7)
Modify equation (7).
l/( z )
z ( 4 .2 - 3 .8 0 1 z - ')
£ (z)
z ( l - 0 . 6 z ”')
t/( z )
( 4 .2 -3 .8 0 1 Z -')
£ (z)~
( l - 0 . 6 z - ')
U ( z ) ( l - 0 . 6 z - ') = ( 4.2 - 3 .8 0 I z - ') £ ( z ) ....... (8)
Convert the equation from z to /f form.
l / ( z ) - 0 . 6 z - '£ / ( z ) = 4 . 2 £ ( z ) - 3 . 8 0 1 z - '£ ( z )
a ( it) - 0 .6a ( £ - 1) = 4 .2 z (£ ) - 3.801e(lt - 1 )
a ( i t ) = 0 . 6 a ( i t - l ) 4 4 .2 e ( i t ) - 3 . 8 0 1 z ( i t - l )
Hence, the difference equations to the discrete controller is
|a (it) = 0 .6 a (it-l)+ 4 .2 z (it)-3 .8 0 1 e (it-l)|
Step 4 of 4
(d)
Refer part (d) in the solution of 4.39P; write the discrete equivalent equation for the controller.
U {z )
£ (z)
4.2z* -7 .9 7 z+ 3.78
z '-1 .6 z + 0 .5 9 9 7
■(9)
Modify equation (9).
U {z )
z’ (4 .2 -7 .9 7z-'+ 3.78z-^)
£ ( r ) “ z* (l-1 .6 z-'+ 0.5997z-^ )
U {z )
(4 .2 -7 .9 7 z-'+ 3 .7 8 z -’ )
£ (z)
(l- 1 .6 z -' + 0.5997z-=)
(l- 1 .6 z -' + 0.5997z-=)£/(z) = (4 .2 -7 .9 7 z -'+ 3 .7 8 z ^ )£ (z )
t/(z )-1 .6 z -'£ /(z )+ 0 .5 9 9 7 z -'U (z ) = 4 .2 £ (z )-7 .9 7 z -'£ (z )+ 3 .7 8 z -'£ (z )
U ( z ) ( l- 0 .6 z - ') = ( 4 .2 - 3 .8 0 1 z - ') £ ( z ) ...... (8)
Convert z io k form from equation (8).
a ( * ) - 1 .6 B ( il- l) + 0 .5 9 9 7 a (il-2 ) = 4 .2 e (it)-7 .9 7 e (il-l)+ 3 .7 8 e (it-2 )
a ( * ) = 1 .6 B (* -l)-0 .5 9 9 7 a (it-2 )+ 4 .2 e (it)-7 .9 7 e (il-l)+ 3 .7 8 e (it-2 )
Hence, the difference equations to the discrete controller is
|a(ifc) = l.6 a ( it- l) -0 .5 9 9 7 a (it-2 )-t-4 .2 z (it)-7 .9 7 e (it-l)+ 3 .7 8 z (t-2 )]
Problem 5.01 PP
Set up the listed characteristic equations in the form suited to Evans’s rootlocus method. Give
L(s), a(s), and b(s) and the parameter K in terms of the original parameters in each case. Be sure
to select K so that a(s) and b(s) are monic in each case and the degree of b(s) is not greater than
that of a(s).
(a) s + {^/T) = 0 versus parameter r
(b)
s2 + cs + c + 1 = 0 versus parameter c
(c)
(s + c;3+iA (T s+1) = 0
(i)
versus parameter/A.
|N) versus j)arameter T,
(i) versus parameter/A.
(ii) versus parameter T,
(iii) versus the parameter c. if possible. Say why you can or cannot. Can a plot of the roots be
drawn versus c for given constant values of A and T by any means at all?
(d) 1l + [| ^^ + **,/W
( , ) + ^ | G] (gs(s)) = 0. Assume that G {s) =
w^i®re c(s) and d(s)
are monic polynomials with the degree of d(s) greater than that of c(s).
(i) versus kp
(ii) versus kl
(iii) versus kD
(iv) versus
t
Step-by-step solution
step 1 of 9
(a)
Write the general formula for the characteristics equation.
a + * * = 0 ...... (1)
The characteristic equation is,
(2)
^+1
Compare Equation (1) and (2).
a^s
r
Therefore, the required parameters are.
a^s
* = i
Step 2 of 9
(b)
The characteristic equation is,
5 * + c f+ c + l* 0
j ' + l + c ( i + l ) = 0 ...... (3)
Compare Equation (1) and (3).
k -e
Therefore, the required parameters are.
a = s’ + l
k -c
b -s + l
Step 3 of 9
(C)
(i)
The characteristic equation is.
( j + c )’ + y<(7i + l ) = 0
( j + c ) ’ + .4 7 - ^ s + ^ j = 0 ...... (4)
Compare Equation (1) and (4).
a *(^ + c )’
k = AT
b = s+ —
T
Therefore, the required parameters are.
<i = ( j + c )
k^A T
1
_____ T
Step 4 of 9
(ii)
ai doici isuo CLiuduui i
( s + c ) ’ + . 4 ( 7 i+ l) = 0
( 5 + d ) + A T s + A —0
[ ( j + c ) ’ + 4 l]+ > l7 i = 0 ...... (5)
Compare Equation (1) and (5).
a = ( 4 + c ) ’ + 4(
k^A T
b=s
Therefore, the required parameters are.
a = (5 + c )* + i<
k = AT
bss
Step 5 of 9
(iii)
The parameter c enters in a nonlinear equation, so the standard root locus does not apply.
Therefore, using a polynomial solver, the roots are plotted versus c.
Step 6 of 9
(d)
(i)
The characteristic equation is.
(6)
Substitute
'A ( s )
U
A
in Equation (6).
d (s )
s d (s )
( ts + 1)A( s )
* ^ ^ , A ^ ^ .
'd ( s )
s d (s )
+—
Akf
)
+ —j
c
(
=0
j
)
+
— s^Ac{s
[ 4 + ; ] < /( *
^5
)+
Ak,
+ —^ 0( 5) + ^
jH
(7)
=0
Compare Equation (1) and (7).
= ^4^5 + —j< /(4 ) + Ak, ^5 + —^ 0 ( 4) + ^ 4 * ^ c ( 4 ) j
k ^=kk^A
^A
b = * ( 4 + 7 : ] cW
Therefore, the required parameters are.
Step 7 of 9
(ii)
Rearrange Equation (7).
= 0 ...... (8)
*Ak,
Compare Equation (1) and (8).
= 4^4 + —j r f { 4) + ^ 4 * ^ c ( 4 ) + Ak^ ^4 + —^ c (4 )
k^A k,
= ( * + 7 ] '( ® )
Therefore, the required parameters are.
Step 8 of 9
(iii)
Rearrange Equation (7).
4^4 + i j < / ( 4 ) + ^ * , ^ 4 + i j c ( 4 )
sO ■
(9)
^4 + - jc ( 4 ) + - ^ ^ 4*0 ( 4 )
Compare Equation (1) and (9).
k-
Ak„
T
b ^ s ^ c [s )
Therefore, the required parameters are.
Step 9 of 9
(iv)
Rearrange Equation (7).
/,
[4</ ( 4 ) + ^ jfc,4c( 4 ) + A k , c { s ) +
A k^s^c ( 4 ) ] —
*0
+ [ 4^ ( 4 ) + i< *^ * c (4 ) + i< *,4c(4)]
Compare Equation (1) and (10).
a = s^d{s)+ A k ^ ^ c [s )+ Ak,sc[s)
b ^ s d ( s ) + A k ^ ( s ) + A k ,c { s ) + A k p S ^ c { s )
Therefore, the required parameters are.
a * 4 V (4 ) + d<it^4*c(4)+ ^iky4c(4)
* =r
b = s d {s)+ ^ ik ^ c (4 )+ A k,c(s)+ v4ifcfl4*c(4)
(10)
Problem 5.02PP
Roughly sketch the root loci for the pole-zero maps as shown in Fig. without the aid of a
computer. Show your estimates of the center and angles of the asymptotes, a rough evaluation of
arrival and departure angles for complex poles and zeros, and the loci for positive values of the
parameter K. Each pole-zero map is from a characteristic equation of the form
I + i C ^ = 0.
a ( j)
where the roots of the numerator b(s) are shown as small circles o and the roots of the
denominator a(s) are shown as ’<’s on the s-plane. Note that in Fig.{c) there are two poles at the
origin.
Figure Pole-zero maps
Step-by-step solution
step 1 of 6
We had to make
some numbers to do it on MATLAB, so the results partly depend on
what was dre amed up, but the idea here is just get the basic rules right.
w
a(ff) =
+ s And
b (s)= s+ l
Break-in(ff) is -3.43
Breakawayfis) is -0.586
Sketch the root locus for the given pole zero plot
Real Axis
Thus, the root locus for the given pole zero plot is sketched.
Step 2 of 6
(b)
fl(s)=s'+0.2f+l
i(s ) = s+ l
Angle o f departure is 135.7
Break-in(s) is -4.97
Sketch the root locus for the given pole zero plot
Thus, the root locus for the given pole zero plot is sketched.
Step 3 of 6
w
a (s) = s^ b (ff) = (ff+ 1)
b(s) = ( s + l)
Break-in(s’) is - 2
Sketch the root locus for the given pole zero plot
Real Aids
Thus, the root locus for the given pole zero plot is sketched.
Step 4 of 6
(d)
a (s) =
+5e+ 6
i( s ) =
Break-in(ff) is -2.37
Breakawa3r(s) is-0.634
Sketch the root locus for the given pole zero plot
Thus, the root locus for the given pole zero plot is sketched.
Step 5 of 6
(e)
a ( s ) = s ^ + 3 s ^ + 4 s -8
Center o f asymptotes is -1
Angles o f asjm^totes are ±60 and 180
Angle o f departure is -56.3
Sketch the root locus for the given pole zero plot
Thus, the root locus for the given pole zero plot is sketched.
Step 6 of 6
(f)
a (s) =
+ 3 e ^ - 5
i>(s) = s+1
Center o f asymptotes is -0.667
Angles o f asymptotes are ±60 and -1 8 0
Angle o f departure is -90
Break-in(s') is -2.06
Breakaways) is 0:503
Sketch the root locus for the given pole zero plot
Thus, the root locus for the given pole zero plot is sketched.
Problem 5.03PP
For the characteristic equation
1 + ^ - ^ _____ = 0 .
l2 ( i+ l) ( j+ 5 )
(a) Draw the real-axis segments of the corresponding root locus.
(b) Sketch the asymptotes of the locus for K -
(c)
Sketch the locus
(d)
Verify your sketch with a Matlab plot.
(d)
Verify your sketch with a Matlab plot.
Step-by-step solution
(a)
Consider the characteristics equation.
1+-
■(1)
The roots of the general form of an equation by the root locus method is.
(2)
The roots of
= 0 are called the zeros
The roots of D(s) = 0 are the poles
Step 1:
Compare Equation (1) and Equation (2),
To find zero put numerator jv W = o .
Thus, there is no zero in the transfer function.
To find poles put denominator D (» ) = 0
j “ (** + l) ( i+ 5 ) = 0 ...... (3)
The roots of the equation are 0.0, -1 , and -5.
Thus, the real axis is
.
(b)
Step 2:
Consider the formula for the number of paths.
Numberof path « Numberof poles
»4
Step 3:
Write the expression for the angle of asymptotes.
_rl8(P
0^
a
n —m
...... (4)
Here,
Number of poles is n
Number of zero is m
r = ± I,± 3 ,± 5 ...
Substitute
for r, 4 for n and 0 for m in the equation.
{±1)1802
^
4 -0
= ±45*
Substitute ±3 for r, 4 for n and 0 for m in the equation.
*
4 -0
= ±135«
Thus, the angle of asymptote, 6, is |±4S^| and
is |±13S^|.
Step 4:
Calculate the value of centroid (<r^)-
_
(sumoffinitepoles)-(sumoffimtezeros)
* (numberof fioitepoles)-(nuiiU>eroffiiutezeros)
{“ ) - ( 0 )
“ 4
• -I.5
Thus, the value of centroid
is b U
(C)
step 5:
Consider the breakaway points.
Recall Equation (1).
K
1-f
j'+ 6» ’ +5i'
K
i* + 6 s ’ + 5 j*
=0
=-l
X = - ( j ‘ +6»’ +5s=)
X = - ( j * + 6 j ’ + 5 j = ) ...... (5)
Differentiate the equation with respect to s and equate to zero.
^ = 0
ds
— ( j ^6 j *5 s* ) = 0
4 j ’ + 185*+ IO j = 0
s (4 s ’
+ I 8 j + I0)=*0.. (6)
The roots of the equation are 0, -0.65 and -3.85
Substitute the value -0.65 for s in Equation (5).
a: = -(-0.65)* -6(-0.65)’ -5(-0.65)’
--0.1785+1.64775 - 2.1I25
= -0.64325
Substitute the value -3.85 for s in Equation (5).
K = -(-0.65)" -6(-0.65)’ -5(-0.65)“
-48.5807
The value of K is positive. So, the breakaway points are 0 and -3.85.
Thus, the breakaway points are
and
|»3.85l-
Step 8 of 11 ^
Step 6:
Procedure to draw root locus plot for the uncompensated system:
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the centroid on the real axis, and draw the asymptotes from centroid at an angle of 45 *
and 135*.
• Locate the breakaway point on the real axis.
• Draw the root locus.
Thus, the root locus diagram for the given system is verified
Step 10 of 11
(d)
MATLAB program for the root locus with the characteristic equation:
» num=[1];
» den=[1 6 5 0 0];
» sys=tf(num,den):
» rlocus(sys)
Step 11 of 11
The root locus plot is shown in Figure 2.
F^nre2
Thus, the root locus diagram for the given system is verified by MATLAB.
Problem 5.04PP
Real poles and zeros. Sketch the root locus with respect to K for the equation 1+ KL(s) = 0 and
the listed choices for L(s). Be sure to give the asymptotes, and the arrival and departure angles
at any complex zero or pole. After completing each hand sketch, venty your results using Matlab.
Turn in your hand sketches and the Matlab results on the same scales.
(a) Us) -
(b)
Us) =
(<-■) Us) = ,<,+*f)'J2SjS.ii))
Step-by-step solution
step 1 of 26
(a)
Consider the general form of characteristics equation.
1+ A 2,(i) = 0 ......(1)
Substitute
a(a+l)(s+5)(a+IO)
2
l+ J t-
a(5+I)(j+5)(a+IO)
for L ( i) in Equation (1).
= 0 ... (2)
Consider the roots of the general form of an equation by the root locus method.
l* K
^
(3)
= 0.
D(s)
Where.
The roots of JV(j) = 0are called the zeros of the problem.
The roots of £>(a) = 0 are the poles.
Consider the number of poles and zeros from the characteristics equation.
Compare the Equation (2) and the Equation (3).
To find zeros put numerator N(s) = 0
Thus, there is no zero in the transfer function.
To find poles put denominator D(s) = 0 .
j (« + 1)(4+5)( j + I0)
= 0 ..... (4)
The roots of the equation (4) are 0, - 1, - 5 and - 10.
Thus, the four poles are 0. - 1. - 5 and - 10.
Step 2 of 26 ''V
Consider the formula for the asymptotes.
n -m
(sumoffinitepoles)-(sumoffinitezeros)
(numberoffinitepoies)-(nuinberoffinitezeros)
(0-l-5-l0)-(0)
(4)-(0)
=-4
Thus, the asymptotes is
E 3
Consider the formula for the angle of asymptotes.
^ l80°-t-360»(/-l)
n -m
Where.
Number of poles is n
Number of zeros is m
/ =i,2,..ji-m
Step 3 of 26
Substitute 1 for /, 4 for n and 0 for m in equation (5).
I8 0 °+ 3 6 0 °(I-I)
4 -0
= 45*
Substitute 2 for /, 4 for n and 0 for m in equation (5).
f8 0 ° -t-3 6 0 ° ( 2 - l)
4 -0
= 135»
Substitute 3 for /, 4 for n and 0 for m in equation (5).
,
i8 0 » + 3 6 0 » ( 3 - l)
4 -0
«225«
= -45»
Substitute 4 for /, 4 for n and 0 for m in equation (5).
l8 0 * + 3 6 0 » ( 4 - l)
4 -0
-3 1 5 «
= -l3 5 »
Thus, the angle of asymptotes are |4 S«»|, [ 135**!, |-4 5 ° |and | - 135**|.
There is no complex zero or pole in the given function.
Hence, there are no arrival and departure angles in the transfer function.
Step 4 of 26 s\-
Procedure to draw root locus plot;
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes from centroid at an angle of
4r
• Draw the root locus.
Step 5 of 26
The root locus plot is shown in Figure 1.
Figure 1
Hence, the root locus is plotted for the given transfer function and it is shown in Figuret.
Step 6 of 26
Consider the following given function.
£ (*) = -
z(z+l)(s+5)(s+10)
Wnte the MATLAB program to obtain root locus.
s=tf('s');
sysL=2/{s*(s+1 )*(s+5)*(s+10));
rlocus(sysL)
The root locus plot is shown in Figure 2;
Step 7 of 26
Thus, the root locus is verified from MATLAB output.
Step 8 of 26
(b)
(4 ^ 3 )
i ( z + l)(x + 5 )(z + I0)
for L ( j) ir i Equation (1).
(6)
1+ J f - : -------------------------- r = 0
z(z+l)(j+5)(z+I0)
Consider the number of poles and zeros from the characteristics equation.
Compare the Equation (6 ) and the Equation (3).
To find zeros put numerator N(s) =
0
Thus, the one zero is - 3.
The roots of the equation (4) are 0, - 1, - 5 and - 10.
Thus, the four poles are 0. - 1. - 5 and - 10.
Step 9 of 26
Consider the formula for the asymptotes.
n -m
(sumoffinitepoles)-(sumoffinitezeros)
(numberoffinitepoies)-(numberoffinitezeros)
(0-l-5-l0)-(-3)
(4)-(l)
=-4.33
Thus, the asymptotes is E n a
Substitute 1 for /, 4 for n and 1 for m in equation (5).
f80°+360o(l-l)
4-1
=60»
Step 10 of 26
Substitute 2 for /, 4 for n and 1 for m in equation (5).
180°+360°(2-l)
4-1
=180»
Substitute 3 for /, 4 for n and 1 for m in equation (5).
, 180°+360°(3-l)
4-1
=300“
=-60»
Thus, the angle of asymptotes are |6()v|. | | 8Qe|and | - 60^1 ■
There is no complex zero or pole in the given function.
Hence, there are no arrival and departure angles in the transfer function.
Step 11 of 26 -rv
Procedure to draw root locus plot;
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes from centroid at an angle of
6cr
• Draw the root locus.
The root locus plot is shown in Figure 3.
IUMLocm
30
K-poles
o - zeros
a - Ceam of the «yni|]Mes
yy
y/
..
. . A
V
-
V " * ’’
a — 4J3
VV
Vv
■Ov
•30
■*45
-2S
-20
-IS
-to
-5
XnlAiit(ftc<aO>4)
S
to
IS
Figures
Hence, the root locus is plotted for the given transfer function and it is shown in Figure3.
Step 12 of 26
Consider the following given function.
i ( j ) = -------------------------------
' ' i(i+l)(s+5)(s+10)
Write the MATLAB program to obtain root locus.
s=tf('s');
sysL=(s+3)/(s*(s+1 )*(s+5)*(s+10));
rlocus(sysL)
The root locus plot is shown in Figure 4.
Figure 4
Step 13 of 26
Thus, the root locus is verified from MATLAB output.
Step 14 of 26
(c)
(s+2)(j+4)
u a
:
4(z+1)(j+5)(j+10)
=0.
(7)
Consider the number of poles and zeros from the characteristics equation.
Compare the Equation (6 ) and the Equation (3).
To find zeros put numerator JV(j) = 0
Step 15 of 26 A
Thus, the two zeros are - 2 and - 4.
The roots of the equation (4) are 0, - 1, - 5 and - 10.
Thus, the four poles are 0. - 1. - 5 and - 10.
Consider the formula for the asymptotes.
n -m
(sumoffinitepoles)-(sumoffinitezeros)
(numberoffinitepoles)-(numberoffinitezeros)
(0-l-S-10)-(-2-4)
(4)-{2)
=-5
Thus, the asymptotes is E H
step 16 of 26 A
Substitute 1 for /, 4 for n and 2 for m in equation (5).
180°+360«(1-1)
4-2
= W>
Substitute 2 for /, 4 for n and 2 for m in equation (5).
180»+360°(2-l)
4-2
-270“
— 90“
Thus, the angle of asymptotes are |90<>|and | - 9(y>|.
There is no complex zero or pole in the given function.
Hence, there are no arrival and departure angles in the transfer function.
Step 17 of 26 ^
Procedure to draw root locus plot;
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes from centroid at an angle of
• Draw the root locus.
The root locus plot is shown in Figure 5.
Root Locos
Figures
Hence, the root locus is plotted for the given transfer function and it is shown in Figures.
Step 18 of 26
Consider the following given function.
i ( j ) = ------(4 + 2 ) ( 4 -i- 4 ) -------
' '
i( i + l ) ( s + 5 ) ( s + 1 0 )
Write the MATLAB program to obtain root locus.
s=tf('s');
sysL=(s+2)*(s+4)/{s*(s+1)*(s+5)*(s+10));
rlocus(sysL)
Step 19 of 26
The root locus plot is shown in Figure 6 .
Step 20 of 26
Thus, the root locus is verified from MATLAB output.
Step 21 of 26
(b)
Substitute
(s+2)(i+6)
, V
r
, for A (j)in Equation (1).
i ( z + l) ( s + 5 ) ( z + 10)
ua :
0.
z (z + l ) ( j+ 5 ) ( j+ 1 0 )
(7)
Consider the number of poles and zeros from the characteristics equation.
Compare the Equation (6 ) and the Equation (3).
To find zeros put numerator JV(j) = 0
Thus, the two zeros are - 2 and - 6 .
The roots of the equation (4) are 0, - 1, - 5 and - 10.
Thus, the four poles are 0. - 1. - 5 and - 10.
Consider the formula for the asymptotes.
n -m
(sumoffinitepoles)-(sumoffinitezeros)
(numberoffinitepoles)-(numberoffinitezeros)
(0-l-5-10)-(-2-6)
(4)-(2)
= -4
Thus, the asymptotes is E 3
step 22 of 26 ^
Substitute 1 for /, 4 for n and 2 for m in equation (5).
180“+360“(1-1)
4-2
=90“
Substitute 2 for /, 4 for n and 2 for m in equation (5).
, 180“+360“(2-l)
4-2
=270“
—90“
Thus, the angle of asymptotes are |9(y>|and | - 9(y*|.
There is no complex zero or pole in the given function.
Hence, there are no arrival and departure angles in the transfer function.
Step 23 of 26 ^
Procedure to draw root locus plot;
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes from centroid at an angle of
9 tt
• Draw the root locus.
Step 24 of 26
The root locus plot is shown in Figure 7.
Figure?
Hence, the root locus is plotted for the given transfer function and it is shown in Figure7.
Step 25 of 26
Consider the following given function.
(s*2){s+ 6)
' '
i(i+l)(s+5)(s+10)
Write the MATLAB program to obtain root locus.
s=tf('s');
sysL=(s+2)*(s+6)/{s*(s+1)*(s+5)*(s+10));
rlocus(sysL)
The root locus plot is shown in Figure 8 .
Step 26 of 26
Thus, the root locus is verified from MATLAB output.
P r o b le m
iltiple poles at the origin. Sketch the root locus with respect to K for the equation 1 + KL(s)
d the listed choices for L(s). Be sure to give the asymptotes and the arrival and departure
gles at any complex zero or pole. After completing each hand sketch, verify your results u
itlab. Turn in your hand sketches and the Matlab results on the same scales.
t ( j) =
r
'5+S)
(1+1)^
- ? c rn «
Step-by-step solution
Step 1 of 23
U s ):
^ (s + 8 )
Step 2 of 23
K = 0 points: 0, 0, - 8
K = apoints: a, a, a
Asymptotes: 60®, 180®, 300®
Centroid = — = - 2 .6 6
3
Intersection o f root loci on the Imaginary axis:
Characteristic equation:
s^+S s^+K = 0
Step 3 of 23
1
0
s’
8
K
s’
-K
0
s“
8
E
Step 4 of 23
As the necessary condition o f Routh’s criterion itself is not satisfied, we can say
that the system is unstable.
As the roots are not complex conjugate, angles o f departure o r arrival need not be
calculated.
Step 5 of 23
I
£ (S ) :
ff^ (s + 8 )
Step 6 of 23
K = 0 points: 0, 0, 0, - 8
K = apoints: a, ot, a, a
Asymptotes: 45®, 135®, 225®, 315®
Centroid = — = - 2
4
Characteristic equation:
s*+ S ^+ K = 0
As all the coefficients o f the characteristic equation are not present, we can say
that the system is not stable.
Step 7 of 23
s*(s+ 8 )
Step 8 of 23
K = 0 points: 0, 0, 0 ,0 , - 8
K = apoints: a, ot, a,
a
Asymptotes: 36®, 108®, 180®, 252®, 324®
Centroid = — = - 1 .6 .
5
Step 9 of 23
s+2
U s ):
^ (s + 8 )
Step 10 of 23
K = 0 points: 0, 0, - 8
K = apoints: ot» ot* - 3
Asymptotes: 90®, 270®
^
-8 -(-3 )
Centroid = ------ i— i = -2.5
2
Characteristic equation:
s ^ ( s + 8 ) + i: ( s + 3 ) = 0
s^+S s^+K s+3K = 0
Step 11 of 23 A.
s^
1
0
s’
8
3K
s’
8 E ’- 3 ^
0
s“
8
3K
Step 12 of 23
For system stability,
K >0
—
> 0
i.e .,
> 0
8
Therefore, in order to determine, the point at which, the root loci cross the
imaginary axis.
W e have to substitute K = 0 into the equation,
8 s“+3JT = 0
8s“ = 0
s = 0
Therefore, we can say that, the ro o t loci start &om the origin and they do not
intersect the imaginary axis anywhere.
Step 13 of 23 A
s+3
L (S ):
^ {s + A )
Step 14 of 23
K = 0 points: 0, 0, 0, - 4
K = apoints: a, a, a, - 3
Asymptotes: 60®, 180®, 300®
Centroid =
= z l = -0 .3 3
3
3
Characteristic equation:
s*+ A s^+ K {s+ 3 ) = 0
s*+ A ^+ K s+ 3K = 0
As the coefficient o f s^ term is absent, we can say that the system is unstable.
Step 15 of 23
L (S ):
s ^ (s + 4 )
Step 16 of 23
K = 0 points: 0, 0, 0, - 4
K = apoints: -1 , - 1 , a, a
Asymptotes: 90®, 270®
Centroid = >
-(-1 -1 ).
z2 = _ ,
2
Characteristic equation:
s ^ ( s + 4 ) + A r ( s + l) ^ = 0
f f * + 4 s ^ + J ^ ( s ^ + 2 s + l) = 0
Step 17 of 23 ^
1
K
s’
4
2K
s’
AK-2K
K
s*
K
4
s’
0
E ’-4 E
XI2
K
s“
Step 18 of 23
> 0
> AK
( 2)
K >4
The condition which satisfies both the equations (1) and (2) is
K >A
Now, in order to get point at w hich the root loci intersect the imaginary axis, let
us substitute K = 4 into the equatioa
— s’ + i r = 0
2s’+ 4 = 0
s’ = - 2
s= ±J^ .
Step 19 of 23
s’ (s-l-lO)
Step 20 of 23
K = 0 points: 0, 0, 0, -1 0 , -1 0
K = apoints: -1 , - 1 , a, a, a
Asymptotes: 60®, 180®, 300®
(-lO -lO )-(-l-l)
Centroid = '
= -6
3
Characteristic equation:
s’ ( s + i o ) V j i : ( s + i ) “ = 0
s’ (s’ -M 0 0 + 2 0 s )+ i:(s ’ +l-l-2s) = 0
s’ + io o s ’ + 2 0 s *-i-J i:s ’ - i- 2 i! : s + ji: = o
s’ + 2 0 s * - i- io o s ’ -i-Ji:s’ - i - 2 i : s + j i : = o
step 21 of 23
s’
1
100
2K
s*
20
K
K
2000K-K
20
200K-K^-ia0K
2000-K
XY-WZ
r
AOK-K
20
s’
s’
K
0
K
s"
step 22 of 23
■Where fT =
2000-K
20
X :
40K-K
20
r : 2000K-X^-7S0K
2000-K
Z
Step 23 of 23
From the above array,
K >0
39K^{\220-K)-K{2000-K)
> 0
3 9 a : ’ ( 1 2 2 0 - j!:) > A : ( 2 0 o o - i: ) ^
4 7 5 8 0 ^ -3 9 1 !:’ > 4xl0 * + J!:’ -4 0 0 0
4 0 i : ’ -5 1 5 8 0 ii:+ 4 x l0 ‘ > 0 .
5 .0 7 P P
<
5 .0 7 P P
Mixedrealandcomplexpoles.SketchtherootlocuswithrespecttoKfortheequation1+KL{s)
=0andthelistedchoicesforL(s). Besuretogivetheasymptotesandthearrivalahddeparture
anglesatanycomplexzeroorpole.Aftercompletingeachhandsketch,verifyyourresultsusing
Matlab.TurninyourhandsketchesandtheMatlabresultsonthesamescales.
... r... _
(a) —
j(n.io(W)
K**.+2f+2)
(c) u n = ,Jo+iS!S+iSi+ai
....
(*+3M.^44l+6a)
Step-by-step selution
(a)
Considerthegeneralformofcharacteristicsequation.
l-hiX(s).0...(1)
...
d+3
j(s+I0)(s+l-ky)(5-f1-y)for inEquation(1).
s-l-3
l- k if= » .....(2)
5(s-l- 10)(j +1 -hy)(5 +1 - y )
Considertherootsofthegeneralformofanequationbytherootlocusmethod.
.(3)
D(s)
W
here,
Therootsof jy($)=Oarecalledthezerosoftheproblem.
Therootsof Dis)=0arethepoles.
Determinethenumberofpolesandzerosfromthecharacteristicsequation.
ComparetheEquation(2)andtheEquation(3).
Tofindzeros, putnumeratorJV(a)=0
Thus,thezerois-3.
TofindpolesputdenominatorDis)=0
s(a+10)(s-i-l+y)(s+I-j)=0...(4)
Therootsoftheequation(4)are0,-10, —
1—
j and—
l-i-j.
Thus,thefourpolesare0,-10, —
\—
j and—
1-t-y.
Considertheformulafortheasymptotes.
n-m
(stunoffinitepoles)-(sumoffinitezeros)
(ntimberoffinitepoies)-(nuinberoffinilezeros)
-10-l-y-Hj-(-3)
4-1
*3
Thus,theasymptotesis E3
Considertheformulafortheangleofasymptotes.
l80°-i-360°(f-l)
n-m
W
here,
Numberofpolesisrr
Numberofzerosisrrr
/ • l,2 ,..ji- m
Substitute1fori,4forrrandIforrrrinequation(5).
, 180+360(1-1)
3
3
.60“
Substitute2fori,4forrrandIforrrrinequation(5).
, 180+360(2-1)
3
3
-180“
Substitute3fori,4forrrandIforrrrinequation(5).
, 180+360(3-1)
3
_300
3
--60“
Thus,theangleofasymptotesare1^^, jlgg^and
Considerthefollowingequation.
61=180“-tnn-'^ij
.135“
^-90“
,%= B n - [ i]
-26.56“
« .-n n .- [i]
Considerthedepartureanglesinthetransferfunction.
sumofangleofvectorto
thecomplexpoleA
angleofdepature1 180“- fio
otherpoles
fiomarxnnplexAJ Jsiimofanm
gleofvernoiatothel
[complexpoleAfiomzeros J
6[,-180“-[6!+6)+6!,]+6)
Substitute 135“for61, 90“for6(, 6.34*for 6( and 26.56“for 6( intheequation.
61,=180“-[l35“+90”+6.34“]+26.56“
—24.78“
Thus,thedepartureangleinthetransferfunctionis 24TO^ats=1+y.
Step5of37 -o
Proceduretodrawrootlocusplot:
•TakerealandimaginarylinesonXaxisandYaxisrespectively.
•M
arkthepolesontherealaxis.
•Locatetheasymptotesontherealaxis,anddrawtheasymptotesfromcentroidatanangleof
60•Drawtherootlocus.
TherootlocusplotisshowninFigure1.
Root Locus
Real Axis (seconds'*)
Figure1
Thus,therootlocusisplottedforthegiventransferfunctionanditisshowninFigurel.
ConsiderthefollowinggivenfuntXion.
. ___ s+3___
'*'“s(j+10)(s’+2s+2)
M
ATLABprogramtoobtainrootlocus:
s=tf('s');
sysL=(s+3)/(s*(s+10)‘(s''2+(2's)+2));
riocus(sysL)
TherootlocusplotisshowninFigure2.
Thus,therootlocusisverifiedfromM
ATLABoutput.
Substitute —
zc-- (j+3)
77—
;----- ofor7(«1inE^quation(1).
s‘(s+10)(s’+6t+25)
s*(s+10)(s“+6s+25)
. _________ s j j __________
■T=0.....(6)
s*(s+10)(s+3+4y)(s+3-4y)
step9of37 XV
Comparetheequation(6)andequation(3).
TofindzerosputnumeratorJV(s)—
0
Thus,thezerois-3.
TofindpolesputdenominatorD(s)=0
s’(s+I0)(s+3+4/)(s+3-4y)=0...(7)
Therootsoftheequation(7)are0,0,-10, —
3—
4yand—
3+4y
Thus,thepolesare0,0,-10, —
3—
4yand—
3+4y.
Considertheformulafortheasymptotes.
a . Z ic lfL
n-m
(sumoffinitepoles)-(sumoffinitezeros)
(ntimberoffinitepoies)-(numberoffinilezeros)
-10-3+4y-3-4y-(-3)
4
-13
4
Thus,theasymptotesis l-3.25i
Substitute1fori,5forrrand1forminequation(5).
, 180“+360”(1-1)
5-1
=45“
Substitute2fori,5forrrand1forminequation(5).
, 180“+360“(2-l)
* “ «_i
Substitute3fori,5forrrand1forminequation(5).
+360“(3-l)
-225“
—45“
Substitute4fori,5forrrand1forminequation(5).
180“+360“(4-l)
5-1
=315“
=-135“
Thus,theangleofasymptotesare , jl3^, |—
45“|andj_^y3^.
Considerthefollowingequation.
61=180“-lan-'^ij
-126.93“
6(=180“-lan-'^|j
-126.93“
61-90“
61-90“
9, = B n - 'i
-29.74“
Considerthedepartureanglesinthetransferfunction.
Stunofangleofvectorto
thecomplexpoleA
angleofdepature1 180“- fio
otherpoles
fiomacomplexAJ Jsiimofanm
gleofverXoistothe)
[complexpoleAfiomzeros j
61„=180“-[61+6l,+6!,]+61
Substituteall ffvaluesinequation.
61„=180“-[126.93“+126.93“+90“+29.74“]+90“
Thus,thedepartureangleinthetransferfunctionis I0L5^ats=—
3+4y
step12of37 xv
Proceduretodrawrootlocusplot:
•TakerealandimaginarylinesonXaxisandYaxisrespectively.
•M
arkthepolesontherealaxis.
•Locatetheasymptotesontherealaxis,anddrawtheasymptotesfromcentroidatanangleof
45•Drawtherootlocus.
TherootlocusplotisshowninFigure3.
Real Axis (seconds'*)
Figure3
Hence,therootlocusisplottedforthegiventransferfunctionanditisshowninFigures.
ConsiderthefollowinggivenfuntXion.
i(,).
(44-^)
' ’ s’(s+10)(s’+6s+25)
M
ATLABprogramtoobtainrootlocus:
s=tf('s');
sysL=(s+3)/(s''2‘(s+10)'(s''2+(6*s)+25));
riocus(sysL)
TherootlocusplotisshowninFigure4.
Thus,therootlocusisverifiedfromM
ATLABoutput.
(0)
Substitute
j ’ (i
(443)’
+I0)( s’ +6!+25)
forL(s)inEquation(1).
(443)’
l+ K -
s'(s+10)(s“+6s+25)
,+jC_--- j £ ± 3 i i l 3 L
.(8)
5*(s+10)(s+3+4y)(s+3-4y)
Comparetheequation(8)andequation(3).
TofindzerosputnumeratorJV(s)=0
Thus,thezeroare-3and.3.
TofindpolesputdenominatorDis)=0
s’(s+I0)(s+3+4/)(s+3-4y)=0...(9)
Therootsoftheequation(9)are0,0,-10, —
3—
4yand—
3+4y
Thus,thepolesare0,0,-10, —
3—
4yand—
3+4y.
Considertheformulafortheasymptotes.
n-m
(sumoffinitepoles)-(sumoffinitezeros)
(ntimberoffinitepoies)-(mimberoffinilezeros)
-10-3+4y-3-4y-[-3-3]
3
-10
3
Thus,theasymptotesis j^^33j.
Substitute1fori,5fornand2forrrrinequation(5).
, 180“+360”(1-1)
^' k_o
=60“
Substitute2fori,5fornand2forrrrinequation(5).
, 180“+360“(2-l)
5-2
540“
3
-180“
Substitute3fori,5fornand2forrrrinequation(5).
, 180“+360“(3-l)
rr_o
eangleofasymptotesarejgQ^, jlgg^and
Considerthefollowingequation.
61 =180“- la n - '^ i j
= 126.93“
6(=180“- la n - '^ l j
-126.93“
61-90“
9 .- h m - [l]
-29.74“
61-90“
61-90“
Considerthedepartureanglesinthetransferfunction.
sumofangleofvectorto
thecomplexpoleA
angleofdepature[ 180“- fio
otherpoles
fiomacomplexAJ Jsiimofanm
gleofvectoistothe]
[complexpoleAfiomzeros j
61 = 180“-[61+61+61+61]+61+61
Substituteall ffvaluesinequation.
61 = 180“-[126.93“+126.93“+90“+29.74“]+90“+90“
Thus,thedepartureangleinthetransferfunctionis
13^^atx=—
3+4y
step19of37 XV
Proceduretodrawrootlocusplot:
•TakerealandimaginarylinesonXaxisandYaxisrespectively.
•M
arkthepolesontherealaxis.
•Locatetheasymptotesontherealaxis,anddrawtheasymptotesfromcentroidatanangleof
60•Drawtherootlocus.
TherootlocusplotisshowninFigure5.
RootLocus
RealAxis(seconds *)
step20of37 xv
Hence,therootlocusisplottedforthegiventransferfunctionanditisshowninFigures.
ConsiderthefollowinggivenfunrXion.
' '
(443)‘
r ’ (j+ 1 0 )(j’ +61+25)
M
ATLABprogramtoobtainrootlocus:
s=tf('s');
sysL=(s+3)“2/(s“2‘(s+10}*(s“2+(6‘s}+25));
riocus(sysL)
TherootlocusplotisshowninFigure6.
Step22of37 .
Thus,therootlocusisverifiedfromM
ATLABoutput.
(x+3)(x*
+4s+68)-forDs)inEquation(1).
Substitute ---1—
-----L
j’(s+10)(s’+4i+85)
= (4+3)(x*+4x+68)
s'(i+10)(s“+4!+85)
(x+3)(l+2+8y)(x+2-8y)
j’(x+10)(s+2+9y)(s+2-9y) ...' '
Comparetheequation(10)andequation(3).
TofindzerosputnumeratorJV(x)=0
Thus,thezeroare-3, —
2+8yand—
2—
8y.
TofindpolesputdenominatorD(s)=0
s’(s+IO)(j+2+9/)(i+2-9y)=0 ...(11)
Therootsoftheequation(11)are0,0,-10, -2-9J and-2+9JThus,thepolesare0,0,-10, -2-9J and—
2+9J.
Considertheformulafortheasymptotes.
n-m
(sumoffinitepoles)-(stimoffinitezeros)
(ntimberoffinitepoies)-(mimberoffinitezeros)
-IO-2+9J-2-9y-[-3-2+8y-2+8y1
3
“ 2
Thus,theasymptotesis m
Substitute1for),5fornand3forrrrinequation(5).
, 180“+360”(1-1)
5-3
=90“
Substitute2for),5fornand3forrrrinequation(5).
, 180”+360“(2-l)
ft “
X X
5-3
540“
2
=270“
—90“
Thus,theangleofasymptotesare |9Q^and .
Considerthefollowingequation.
61=180“-lan-'^|j
-102.52“
l%=180“-tan-'[|j
-102.52“
61-90“
61-90“
61-90“
-83.65“
-48.36“
Considerthedepartureanglesinthetransferfunction.
sumofangleofvectorto
- thecomplexpoleA
angleofdepature[
motherpoles
fiomacomplexAJ Jsiimoffio
angleofverXoistotheJ
[complexpoleAfiomzeros j
61=180“-[«,+61+«,+6l]+[«,+61+61]
Substituteall ffvaluesinequation.
61=180“-[102.52“+102.52“+90“+48.36“]+[90“+90“+83.65“]
Thus,thedepartureangleinthetransferfunctionis|^^I002^ats=—
2+9y.
Considerthefollowingequation.
61 =180“- ta n - '^ i j
-104.03“
61=180“- t a n - '^ |j
= 104.03“
61-90“
61-90“
61-90“
f.l-tan-[i]
-82.87“
-45“
Considerthearrivalangleinthetransferfunction.
sumofangleofvectorto
- thecomplexzeroA
angleofarrival [
motherpoleszeros
fiomacomplexAJ Jsumoffio
angleofvetXoistothe[
[complexzeroAfiompoles J
61=180-J61+«,]+[61+61+61+61+«,]
Substituteall ffvaluesinequation.
61=180“-[90“+82.86“]+[l04.03“+104.03“+90“+45“+(-90“)]
=260.2“
Thus,thearrivalangleinthetransferfunctionis
ats=—
2+87
step27of37 xv
Proceduretodrawrootlocusplot:
•TakerealandimaginarylinesonXaxisandYaxisrespectively.
•M
arkthepolesontherealaxis.
•Locatetheasymptotesonthelealaxis,anddrawtheasymptotesfromcentroidatanangleof
9V
•Drawtherootlocus.
TherootlocusplotisshowninFigure7.
RootLocus
RealAxis(seconds *)
Figure7
Hence,therootlocusisplottedforthegiventransferfunctionanditisshowninFigure7.
ConsiderthefollowinggivenfunrXion.
(j +3)( x4+4x +68)
i( x ) .
r’(j+10)(j’+4j+85)
M
ATLABprogramtoobtainrootlocus:
s=tf('s');
sysL=((s+3}*(s“2+(4‘s)+68})/(s''2‘(s+10)‘(s''2+(4*s)+85));
riocus(sysL)
TherootlocusplotisshowninFigure8.
30of37
Step
Thus,therootlocusisverifiedfromM
ATLABoutput.
Substitute [(j+
L' Ox’ + ilJ forlis)inEquation(1).
x*(x+2)(x+3)
[(4 +l)‘ +l]
‘ ■^^x’V + 2 )(x + 3 )
“
, ^ ^ (x +1+7K x + 1 - / ) , ( ,
z*(z+2)(x+3)
...,,,,
Comparetheequation(12)andequation(3).
TofindzerosputnumeratorJV(x)=0
Thus,thezeroare —
1+yand—
1—
j.
TofindpolesputdenominatorDis)=0
Thus,thepolesare0,0,-2and—
3.
Considertheformulafortheasymptotes.
n-m
(sumoffinitepoles)-(stimoffinitezeros)
(mimberoffinitepoies)-(mimberoffinilezeros)
-2-3-1-1+7-1-Jl
2
-5+2
2
Thus,theasymptotesis EEI!
Substitute1for),4fornand2forrrrinequation(5).
, 180“+360“(1-1)
4-2
=90“
Substitute2for),4fornand2forrrrinequation(5).
, 180“+360“(2-l)
4-2
540“
2
=270“
=-90“
Thus,theangleofasymptotesarejg^andj^go^.
Considerthefollowingequation.
6l=180“- t t n - '^ l j
-135“
61=180“- tt n - '^ ij
-135“
61-90“
6,-ten-[l]
-4 5 “
6,-ten-[i]
-26.56“
Considerthearrivalangleinthetransferfunction.
sumofangleofvectorto
- thecomplexzeroA
angleofarrival [
motherpoleszeros
fiomacomplexAJ Jsumoffio
angleofvectorstothe[
[complexzeroAfiompoles J
61 =180“ -[61]+[61+61+6,+«,]
Substituteall ffvaluesinequation.
61=180“-[90“]+[l35+135+45+26.56]
-431.56“
Thus,thearrivalangleinthetransferfunctionis|^^7L5^atJ=—
1+y.
Step34of37 XV
Proceduretodrawrootlocusplot:
•TakerealandimaginarylinesonXaxisandYaxisrespectively.
•M
arkthepolesontherealaxis.
•Locatetheasymptotesonthelealaxis,anddrawtheasymptotesfromcentroidatanangleof
9<r
•Drawtherootlocus.
TherootlocusplotisshowninFigure9.
Root Locus
Real Axis (seconds'*)
Step 35 o
f37 XV
Hence,therootlocusisplottedforthegiventransferfunctionanditisshowninFigures.
ConsiderthefollowinggivenfunrXion.
r(44*)’ 4 i]
s‘(s+2)(z+3)
M
ATLABprogramtoobtainrootlocus:
s=tf('s');
sysL=((s+1}"2+1}/(s“2'(s+2}'(s+3»;
riocus(sysL)
TherootlocusplotisshowninFigure10.
Step37of37
Thus,therootlocusisverifiedfromM
ATLABoutput.
RHP and zeros. Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the
listed choices for L(s). Be sure to give the asymptotes and the arrival and departure angles at
any complex zero or pole. After completing each hand sketch, verify your results using Matlab.
Turn in your hand sketches and the Matlab results on the same scales.
(a)
L(s) =
(b)
L(s) =
model for a case of magnetic levitation with lead compensation.
magnetic levitation system with integral control and lead
compensation.
(c)
L(f) = i
(d)
L {s) = —j—
. What is the largest value that can be obtained for the damping
ratio of the stable complex roots on this locus?
-
(i-l)((i+ 2 F + 3 ]
S te p -b y - s te p s o lu tio n
step 1 of 17
(a)
Write the expression for £ ( i) r/ \
^+ 2
1
There are 3 poles and one zero.
Calculate the number of asymptotes, P.
P=« -«
Here, n represents number of open loop poles.
m represents number of open loop zeros.
P =3 -l
s2
Step 2 of 17 ^
Determine the angle of asymptotes.
, = ,,2 .3 .....-™
n -m
360°(1-1)
2
= 0°
36 0 °(2 -l)
2
= 180”
Therefore, the asymptotes are |g()® apart from each other.
Determine the intersection point of asymptotes on the real a)s
n -m
3 .1 6 -3 .l6 - 1 0 -(-2 )
2
Step 3 of 17
Write the MATLAB code to obtain root locus for i ( j ) •
num=[1 2];
den=[1 10-1 -10];
sys=tf(num,den):
rlocus(sys);
Root Locus
Step 4 of 17
(b)
Write the expression for £ (s)K \-_ £ ± 2
_ I
There are 4 poles and one zero.
Calculate the number of asymptotes, P.
P = 4 -l
»3
Step 5 of 17 ^
Determine the angle of asymptotes.
, = , .2 .3 .....- „
n -m
,
360”(1-1)
3
= 0”
,
360”( 2 - l)
3
120”
=
Therefore, the asymptotes are 120* apart from each other.
Determine the intersection point of asymptotes on the real axis.
n -m
0 - 1 0 + l-l-(-2 )
2
Write the MATLAB code to obtain root locus for L { s y
Step 6 of 17
num=[1 2];
den=[1 10-1 -10 0];
sys=tf(num,den):
rlocus(sys);
Step 7 of 17
(c)
Write the expression for £ (s)-
There are 2 poles and one zero.
Calculate the number of asymptotes, P.
P= 2 -I
si
Step 8 of 17 ^
Determine the angle of asymptotes.
, = 1,2.3.....-™
n -m
,
360®(I-I)
3
= 0®
Determine the intersection point of asymptotes on the real axis.
Step 9 of 17 ^
Write the MATLAB code to obtain root locus for
Z.(j)•
num=[1 -1]:
den=[1]:
sys=tf(num,den):
rlocus(sys);
Root Locus
0.2
0 .1 5 *
0.1
•
h .05 ‘
I
^
.^.0 5 •
- 0.1 •
-0.15 •
-0.5
0
Real Axis (seconds**)
(d)
Step 10 of 17 ^
Write the expression for £ (s)L( 1
+2s+l
'* ^ ” i ( i + 2 0 ) ' ( j ’ - 2 s + 2 )
There are 5 poles and two zeros.
Calculate the number of asymptotes, P.
P = 5 -2
s3
Step 11 of 17 ^
Determine the angle of asymptotes.
=
, = 1,2,3.....-™
n —m
,
360”(1-1)
3
= 0”
360”( 2 - l )
3
Therefore, the asymptotes are 120* apart from each other.
* ”
120*
=
Determine the intersection point of asymptotes on the real axis.
a .IS lId L
n -m
0 -2 0 - 2 0 + \+ l- { -l-\)
3
= -I2
Determine the angle of departure from the pole i+ y .
A = = i8 0 ” - 2 ; ^ , + ! ; » '.
= l80*-140.45*-h 53.13*
=92.68*
The angle of departure from the conjugate pole l —J.
^^ = -(9 2 .6 8 * )
= -92.68*
Step 12 of 17
Write the MATLAB code to obtain root locus for i ( j ) •
num=[1 2 1];
den=[1 38 322 -720 800 0];
sys=tf(num,den):
rlocus(sys);
Root Locus
Step 13 of 17 ^
(e)
Write the expression for £ (s)-
' '
j(i-l)(i+ 6 )’
There are 4 poles and one zero.
Calculate the number of asymptotes, P.
P = 4 -l
-3
Determine the angle of asymptotes.
* = 5 5 M
, = 1 ,2.3 .....-™
n -m
, 360”(1-1)
"■
3
= 0”
,
360”( 2 - l )
"■
=
3
120”
Therefore, the asymptotes are 120* apart from each other.
Write the MATLAB code for £ ( j ) •
Step 14 of 17 ^
num=[1 2];
den=[1 11 24 -36 0];
sys=tf(num,den):
rlocus(sys);
Step 15 of 17
Write the expression for £ (s)I
L {s ) =
( i- l) [ ( » + 2 ) - + 3 ]
There are 3 poles and no zeros.
Calculate the number of asymptotes, P.
P = 3 -0
-3
Determine the angle of asymptotes.
* = 5 5 M
n -m
,
, = 1 .2.3 .....-™
360”(1-1)
3
= 0”
,
360”( 2 - l )
3
=
120”
Therefore, the asymptotes are 120* apart from each other.
Step 16 of 17 ^
Determine the angle of departure from the pole —2+1.73J ■
= 1 8 0 * - ^ t a n - '^ ^ j + 9 0 * j
= 180 *-I20*
= 60*
The angle of departure from the conjugate pole —2 —1.73y.
= -60*
Step 17 of 17
Write the MATLAB code for £ ( j ) •
num=[1);
den=[1 3 3 -7);
sys=tf(num,den):
rlocus(sys);
Put the characteristic equation of the system shown in Fig. in root-locus fonn with respect to the
parameter a, and identify the con'esponding L(s), a(s), and b(s). Sketch the root locus with
respect to the parameter a, estimate the closed-loop pole locations, and sketch the
corresponding step responses when a = 0, 0.5, and 2. Use Matlab to check the accuracy of your
approximate step responses.
Figure Control system
-O K
s(s+2)
S te p -b y - s te p s o lu tio n
step 1 of 23
Refer to control system in Figure 5.52 in the textbook.
/f(i) = l + Oi
The characteristic equation is,
1 + G ( s ) f f { i) = 0
f ( « + 2 )+ 5 (1-I-as) s 0
s*+ 2 s+ 5 + 5 as = 0
(1)
s* + 2 s+ 5 '
The above equation is in the form.
1+ J C ^ = 0 ...... (2)
a{s)
Step 2 Of 23
Equate equation (1) and (2).
a ( s ) = s* + 2s + 5
A(s) = 5s
The open loop transfer function is.
s* + 2s + 5
Therefore, the values of £ ( s ) . a ( s ) and ^^s) are,
a ( s ) » s * + 2s + 5
A (s )> Ss
Step 3 of 23
MATLAB code to plot the root locus:
num=[5 0];
den=[1 2 5];
sys=tf(num,den):
rlocus(sys)
Step 4 of 23
The root locus plot is shown in Figure 1.
R e a l A xis (secoods*^)
Figure 1: Root locus
Step 5 of 23 ^
Calculate the closed loop transfer function.
s (s + 2 )
rW -
5
s(s+ 2 )+ 5 + 5 a s
5
s*+ 2 s+ 5 + 5 as
5
s * + (2 + 5 a )s-f5
Step 6 of 23 ^
Calculate the closed loop pole locations when a —0s*
+ ( 2 - h 5 o ) s -i-5
s* + [2 + 5 (0 )]s+ 5
s* + 2 s -1-5-0
S i^ = - l± J 2
Therefore, the closed loop pole locations when a —0
ET±7g
Step 7 of 23
The closed loop transfer function is,
n s )_
s
s*+ 2s+ 5
R{s)
Substitute — for X (s) in the above equation,
s
Calculate the step response when a —0 -
s ^ s*+ 2 s+ s)
1
s+ 2
( s + I ) " + 2*
1
s+ 1
1
s (s + l) * + 2^ ( s + l)* + 2 *
Apply inverse Laplace transform.
^ (f)= « (/)-e"'co s(2 /)tt(f)-0 .5 e"'sin (2 r)tf(/)
step 8 of 23
MATLAB code to plot the step response:
t=0:0.01:5;
y=1-exp{-t).*cos(2.*t)-0.5.*exp(-t).*sin{2.*t);
plotfty)
title('step response when a=0');
Step 9 of 23 -A.
The step response is shown in Figure 2.
s te p resp o B se w faea a= 0
Step 10 of 23
MATLAB code to plot step response when a —0 a=0 ;
num=[5];
den=[1 2+5*a 5];
sys=tf(num,den):
step(sys);
title('step response when a=0');
Step 11 of 23
The step response when ^ s 0
shown in Figure 3.
The step responses in Figure 2 and Figure 3 are same.
Step 12 of 23 ^
Calculate the closed loop pole locations when g s 0 ,5 .
j* + (2 -» -5 a )s + 5
s ’ + [ 2 + 5 ( 0 .5 ) ] j + 5
l ’ + 4 . 5 i+ 5 « 0
j ,j= - 2 ,- 2 .5
Therefore, the closed loop pole locations when f f = 0.SPtP |- 2 , —2.5|-
Step 13 ot 23
The closed loop transfer function is.
5
j* + 4 .5 s + 5
J l( j)
! — for X (s) in the above equation.
I 'M Calculai the step response when a —0 Calculate
5
I 'M -
( 5 + 2 ) ( j + 2.5)
j
_ I ^ ___5
^
s s+2
4
5+2.5
Apply inverse Laplace transform.
) = « ( /) - Se~^u ( f ) + A e '^ u (/)
Step 14 of 23
MATLAB code to plot the step response:
t=0:0.01:5;
y=1-5.*exp(-2.*t)+4.*exp(-2.5.*t):
plot(t,y)
title('step response when a=0.5');
Step 15 of 23
The step response is shown in Figure 4.
stq > r e s p o n s e w k e n a= 0 ,5
Step 16 of 23
MATLAB code to plot step response when a —0.5 a=0.5;
num=[5];
den=[1 2+5*a 5];
sys=tf(num,den):
step(sys);
title('step response when a=0.5');
Step 17 of 23 ^
The step response when g s 0,5 is shown in Figure 5.
s te p re s p o n s e w h e n ^ 0 , 5
Figure 5
The step responses in Figure 4 and Figure 5 are same.
Step 18 of 23 ^
Calculate the closed loop pole locations when a —2 + (2 + 5 a )5 + 5
ff^ + [2 + 5 (2 )]s + 5
j *+12 j
+5 -0
s ,j= -0 .4 3 2 .-II.5 5 6 7
Therefore, the closed loop pole locations when a —2
|-Q.432>-11.5S67|.
Step 19 of 23
The closed loop transfer function is,
5
I 'M
« ( i)
3’ +12 s+5
Substitute i. for R{s) in the above equation.
s
5
r(3 )=
j( i^ + 1 2 s + s )
Calculate the step response when a —0 ■
K (5 )-
5 ( j + 0 .4 3 2 )( s +11.5567)
.0015
1.0404
0.039
s
5+0.432
5+11.5567
Apply inverse Laplace transform.
,t-(f) = 1 .0 0 15ii(/)-1.0404«-*""ii(t)+ 0.039e‘ " '“ ’'i((()
= [ l . 0 0 1 5 - l .0 4 0 4 « - ^ +0.039« - '‘^ ' ] h (»)
step 20 of 23 ^
MATLAB code to plot the step response:
t=0:0.01:5;
y=1.0015-1.0404.*exp(-0.432.*t)+0.039.*exp(-11.5567.*t):
Plotfty)
title('step response when a=2');
Step 21 of 23 ^
The step response is shown in Figure 6.
s t ^ r e s p o n s e w h e n a= 2
Step 22 of 23
MATLAB code to plot step response when
a —2 -
a=2 ;
num=[5];
den=[1 2+5*a 5];
sys=tf(num,den):
step(sys);
title('step response when a=2');
Step 23 of 23
The step response when a —2
shown in Figure 7.
s te p re s p o n s e w h e n a= 2
Figure 7
The step responses in Figure 6 and Figure 7 are not same.
Problem 5.1 OPP
Use the Matlab function ritool to study the behavior of the root locus of 1 + KL(s) for
U s) =
(i + o)
j (j
+ 1)(j 2 + 8j + 52)
as the parameter a is varied from 0 to 10, paying particular attention to the region between 2.5
and 3.5. Verily that a multiple root occurs at a complex value of s for some value of a in this
range.
Step-by-step solution
step 1 of 4
Step 1 of 4
The loop transfer function is.
w/ \ ________ Js+a
T « ______
' * ^ " i ( j + l ) ( i ’ + 8 s+ 5 2 )
Determine the characteristic equation of the system.
I + X Z ,(i) = 0
I+ if j ( i + l ) ( j * + 8 j + 52)
Write the MATLAB code to draw the root locus when the parameter a is varied from 0 to 10 using
ritool function.
a=0:0.5:10;
num=[1 a];
den=conv([1 1 0],[1 8 52]);
sys=tf(num,den);
rltool(sys)
Step 2 of 4 ^
Execute the code at the MATLAB command window and study the behavior of the root locus.
Step 3 of 4
Go to Analysis Plots. Select Pole/Zero and Response as Closed Loop r to y. Click on Show
Analysis Plot.
CenlanlsafPWs---------------------------------------------------------------nod
Response
1 2 3 4 s 6 AM
B r: r r n r n OoMdloMrtev
n r' n r r n n OotidLoMrtOM
Q r r. r r n n OotidleeedutoV
c r n r r □ n OotidUid^iev
n r n r r n n OotidImp ntoir
n r: n r' r' n n Oodiloopl
n r r r r n n ComptAtMotC
nn n r r n n
n r n r r n n RIMO
r n r r r n n SmotH
Step 4 of 4
Get the MATLAB output for the pole-zero plot.
Observe from the pole/zero plot that there is a multiple complex root in the region of a between
2.5
and 3.5.
Problem 5.11 PP
Use Routh’s criterion to find the range of the gain K for which the systems in Fig. are stable, and
use the root locus to confirm your calculations.
Figure Feedback systems
Step-by-step solution
step 1 of 9
Step 1 of 9
(a)
Refer to Figure 5.54 (a) in the text book for the block diagram of a system.
The open loop transfer function is.
The characteristic equation of the system is,
1 + G (j)ff(* ) = 0
j’ + j+2
r=0
+ 5 ) (5 + 6 ) (5 * + 2 j + 1)
1 + iT -
j (j
j ( j + 5 ) ( j + 6 ) ( j = + 2 i + 1 )+ JC (s’ + j + 2 ) = 0
I * + 1 3 j ‘ + 5J s ’ + ( 7 1 + ^ :) *’ + ( 3 0 + a: ) j + 2 ^ : = 0
Step 2 of 9
Construct the Routh table.
30-fr
2t
**: 1
»
»xS3-(7i.»r) < ii-r
13
—
iX 30*x)-2r 390*nr
13
13
’
13
^- g iT T --------------------—
13
( • r* *«M:*3<K>»Y3w*iig'|
.1
tfU-X
A u
u } ~iiP*9t72P~>«7ttOr*lS13S120
161t
step 3 of 9
For the system to be stable, all elements in the first column of the Routh table should be greater
than zero.
2K>0
K>0
.........( 1 )
-13A:’ + %72Ar*-67860Ar + 1513SI20
(6 1 8 -A :)’
(K + 739. I)[(A: + 2.46)’ + 39.6’ ] > 0
The conditions are,
A: + 7 3 9 .I5 < 0
( a: + 2.46)’ +39.6’ < 0
K < -m .\5
(2)
-AT’ +404A: + 38808
6 18-A :
^
(A :-484.16)(A :+ 80.16)> 0
A :<484.16
a:
(3)
< -8 0 .1 6
J i:< 6 i8
...... (4)
The range of
satisfying all the four conditions. (1). (2). (3) and (4) is,
0 < ^ < 4 8 4 .! 6
Thus, the range of AT for stability is |Q< JC<484.16l-
Step 4 of 9
The characteristic equation is,
, . „
s’ + f + 2
r=0
s (s + 5 ) ( i + 6 ) ( i ’ + 2 j + 1 )
Draw the root locus using MATLAB to find the range of the gain JC for stability.
num=[1 1 2]:
den=conv([1 11 30 0],[1 2 1]);
sys=tf(num,den);
rlocus(sys)
Step 5 of 9
Get the MATLAB output for the root locus.
Observe from the root locus that the gain should be less than 482 for the system to be stable.
Thus, the root locus result agrees with the Routh’s result.
Step 6 of 9
(b)
Refer to Figure 5.54 (b) in the text book for the block diagram of a system.
The open loop transfer function is.
The characteristic equation of the system is,
l + G (s )H (j) = 0
1+^-;—
— t :t = 0
s(s-2)(s’+2s+t0)'
i ( j - 2 ) ( i ’ + 2 i + 10)+AT(s + 2) ■ 0
* ' + 6 s’ + ( A : - 2 0 ) i + 2A: = 0
Step 7 of 9
Construct the Routh table.
/
1
6
s ’
0
AT-20
s’
2g
—
0
As the first column element is infinity, the system is unstable for all values of
.
Step 8 of 9
The characteristic equation is,
i+ ic ——
----------, = 0
5 ( j - 2 ) ( j * + 2 j + 10)
Draw the root locus using MATLAB to find the range of the gain
for stability.
num=[1 2];
den=conv([1 -2 0],[1 2 10]);
sys=tf(num,den);
rlocus(sys)
Step 9 of 9
Get the MATLAB output for the root locus.
Observe from the root locus that the locus does not touch the zero-axis. Hence, the system is
unstable for all values of ICThus, the root locus result agrees with the Routh’s result.
Problem 5 .1 2PP
Sketch the root locus for the characteristic equation of the system for which
and determine the value of the root-locus gain for which the complex conjugate poles have the
maximum damping ratio. What is the approximate value of the damping?
Step-by-step solution
step 1 of 5
Consider the General form of characteristics eouation.
Step 1 of 5 -
Consider the general form of characteristics equation.
I + AX(i) = 0
Substitute
1 + a:
(1)
( s + 2)
a’ ( ^ + 5 )
for I / j ) in Equation (1).
(2)
a* ( a + 5 )
Consider the roots of the general form of an equation by the root locus method.
...... (3)
D is )
Where,
The roots of A f(» )= 0 are called the zeros of the problem.
The roots of D ( j) = 0 are the poles.
Consider the number of poles and zeros from the characteristics equation.
Compare the Equation (2) and the Equation (3).
To find zeros put numerator J V (j)= 0
Thus, the one zero is - 2.
To find poles put denominator D (s) = 0 •
»“ ( i + 5 ) = 0 ...... (4)
The roots of the equation (4) are 0,0, and - 5.
Thus, the three poles are 0.0, and - 5.
Step 2 of 5
Consider the formula for the asymptotes.
n -m
(sum o f finite p oles) - (sum o f finite zeros)
(n u m b e ro f finite poles )~ (n u n d > e ro f finite zeros)
(°-S )-(-2 )
■
(3)-(l)
= -1.5
Thus, the asymptotes is E O Consider the formula for the angle of asymptotes.
^
l80°+ 3 6 (y * (/-l)
n —m
Where.
Number of poles is n
Number of zeros is m
/ = l,2 ,..Ji—m
Substitute 1 for /, 3 for n and 1 for m in equation (5).
180°+360»(1-1)
3 -1
= 90“
Substitute 2 for /, 3 for n and 1 for m in equation (5).
180” + 3 6 0 ° ( 2 - l)
3 -1
= 270»
Thus, the ahgle of asymptotes are |9(y>|ahd |27Q**|.
Step 3 of 5 ^
Procedure to draw root locus plot:
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes from centroid at an angle of
gffana 270"
• Draw the root locus.
Consider the following given function.
' ’
* ’ (s + 5)
Write the MATLAB program to obtain root locus.
s=tf('s');
sysL=(s+2V(s''2*(s+5)):
rlocus(sysL)
[K]=rlocfind{sysL)
Step 4 of 5
The MATLAB output for the value K.
Select a point in the graphics window
selecteu_point —
-0.663 + 2.32i
K=
10.7
The root locus plot is shown in Figure 1.
Figure 1
Step 5 of 5
From Figure 1, the maximum damping ^ is 0.275 when the value of gain K is approximately
10.7.
Thus, the maximum damping ^ is |Q.275l when the value of gain K is [|Q,7 | and the root locus
is plotted for the given transfer function and it is shown in Figurel.
Problem 5.13PP
For the system in Fig.,
(a)
Find the locus of closed-loop roots with respect to K.
(b)
Is there a value of K that will cause all roots to have a damping ratio greater than 0.5?
(c)
Find the values of K that yield closed-loop poles with the damping ratio ^ = 0.707.
(d)
Use Matlab to plot the response of the resulting design to a reference step.
Figure Feedback system
Ul
I . or
_
_
Figure Feedback system
j*+«l
100>
____________
S te p -b y -s te p s o lu tio n
step 1 of 10
(a)
Obtain the characteristic equation for feedback system shown in Figure 5.54 to find the locus of
closed -loop roots with respect to x ■
Since system contains cascaded blocks,
Closed loop transfer function of the system is,
G(»)
l + C ? (i)//(j)
r ( j) =
V
j- fl
J*+81
U + 1 3 j [ i ' ( f ‘ + 100)
\s + \3 )
The characteristic equation of the system is.
■(1)
Equation (1) is of the fonn 1+ K L (s) = 0 . Where
( j + l ) ( i ’ + 8 l)
i( j) =
i ’ ( j + l 3 ) ( j ’ +100)
Loop gain, £( 5) has both poles and zeros near the imaginary axis and should expect to find the
departure angles of particular importance.
Step 2 of 10
Procedure to draw root locus:
RULE 1 : There are five branches to the locus, three of which approach to finite zeros and two of
which approach to Infinity.
RULE 2 : The real-axis segment defined by —13 ^ ^
] is part of the locus.
Draw pole-zero plot for the transfer function.
RULE 3 : calculate the centre of asymptotes.
5 -3
_ -12
2
s -6
The angles of asymptotes are at
.
RULE 4 : Compute the departure angle from the pole at ^
+ylO-The angle at this pole we will
define to be ^ . The other angles are marked in Figure 2.
Figure 2: Measuring the angle o f departure
Step 3 of 10
The root locus condition is,
* = v , + '5 ', + H « , - ( ^ + A + ^ i ) - i 8 c r .
= l a n - '^ Y j + W +90‘ -|^ 9 0 '+ 9 (r + 90‘ + tan-'
= 8 4 .2 8 '- 9 0 '- 3 7 .5 6 '- 1 8 ( r
= -22 3 .2 T
Step 4 of 10
RULE 5: Find the breakaway points.
j* + 8 l
\-¥K\-
sO
U + 1 3 j [ j ’ ( i ' + 100)J
j
* ( s + 1 3 ) ( s * + 1 0 0 )+ A :( j * + 8 1 ) ( j + 1) = 0
Solve for X f * ( j + 1 3 )(j* + 100)
( j ’ + 8 l) ( s + l)
j* + i 3 s * + i o a ! ’ + i3 o a ! '
j* + j* + 8 1 j+ 8 l
Differentiate the equation with respect to s .
(i’
+ 8 U + 8 1 )(5 j * + 52i’ +300 j ’ +2600 i )-
( i ’ +13 « V 100^*+130fts’)(3*’ + 2 i +81)
ds
+ s * + 8 U + 8 IJ
dK
Equate -----to zero.
ds
y
+ » * + 8 1 i+ 8 l) ( 5 / + 52s" +300s’ + 2 6 0 0 s)-
( i ’ +13 i‘ +1 OOj ’ +1 300j " ) (3 * ’ + 2 i + 8 1)
(s’ + i'+ 8 1 s + 8 l)
(j* +»’ +81i+8l)(5j*+52s’ +300J*+2600j ) =0
( j ’ +13j * + 100j ’ + 1300j ')(3 j *+2 i +81)
5i’ +57 s‘ +757 j ’ + 7517J* +31112j’ +234900*' + 210600*'
=0
-(3*’ +4I*‘ + 4 0 7 / + 5153** + 10700*’ +105300*')
2 * '+ 1 6*‘ +350*’ +2364** + 20412*’ +129600*'+ 210600* = 0
Step 5 of 10
Use MATLAB to solve the equation.
»solve('x'‘7■^16*x^6+350*x''5■^2364*xM+20412*x''3■^129600*x''2■^210600*x');
ans = 0
-5.6417302144823378589757341357966-2.3112175660279159755982613661971
All the three roots are valid break-away points since the real-axis segment defined by
- 1 3 ^ j S - 1 is part of the locus.
Remaining four roots are complex roots which are not valid for breakaway points.
The locus of closed loop poles with respect to x
shown in Figure 3.
Therefore, root locus has been implemented.
Step 6 of 10
(b)
Draw the root locus for i ( * ) .
( * + ! ) ( * ' + 8 l)
i( * ) =
’ * '( * + 1 3 ) ( * '+ 1 0 0 )
Enter the following code in MATLAB to draw the root locus and asymptotes at ^ s 0.5 •
» num=conv([1 1],[1 0 81]);
» den=conv([1 13 0 0],[1 0 100]);
» sys=tf(num,den):
» rltool(sys)
Step 7 of 10
Observe the MATLAB output on the Figure window.
Root Locus Edior for Op«n Loop 1(OL1)
From the root locus in Figure 4, it is observed that the asymptote at ^ = 0.5 cuts the root locus
at a point of the form,
In under damped system, roots are of form -^6)^ ± Jo>^ ( complex conjugates), where ^ value
is in the range 0 < ^ < 1 •
From the value of ^ it is clear that system is under damped.
There exists a value o f X ^bat causes roots to have damping ratio greater than 0.5.
Step 8 of 10
(c)
To find the value of x ^bat yield closed loop poles with damping ratio ^ s 0.707.
Consider the magnitude condition
1
ii( * ) i ■
* ' ( * ’ + 100 )(*+13)
K =
( * + ! ) ( * ' + 8 l)
(3)
To find value of x that yields closed loop poles with
0.707, d value is to be found.
As Rvalue is known, calculate the angle 0 ,
cos$=^
9 ■ cos"' ^
= cos"'(0,707)
= 45'
Draw a line at an angle ^ —45*00 the root locus plot and mark the intersecting point as shown in
Figure (4).
Figure 5; Graph to find
5
value
Step 9 of 10
The intersecting point in Figure 4 is the value of s .
Substitute value of
5
In equation (3).
( - 5 .6 + 6.5 i f ( ( - 5 .6 + 6.5y)’ + 1 0 0 )(-5 .6 + 6.5y+13)
K =
( - 5 .6 + 6.5y+ 1)((-5.6+ 6 . 57 ) ' + 8 I)
[V(-5.6)'+(6.5)' ] ’ [V(89.1 D' +(72!iF][>/(7.4)' +(6.5)'
[ V ( ^ . 6 ) ' + (6 .5 ) '][ V ( 7 0 .1 D ' + ( 7 2 .8 7 ] ”
= 103.685
Thevalueof jfth a t yields closed loop poles with the damping ratio ^ = 0.707 is |lQ3.68Sl-
Step 10 of 10
(d)
Enter the following code in MATLAB to plot the response of the closed loop system to a reference
input.
» num=conv([1 1],[1 0 81]);
» den=conv([1 13 0 0],[1 0 100]);
» sys=tf(num,den);
» sys1=feedback(sys,1);
» step(sysl)
The MATLAB output for the step response is shown in Figure 5.
Thus, the response of the closed loop system using MATLAB is plotted.
Problem 5.14PP
For the feedback system shown in Fig., find the value of the gain K that results in dominant
closed-loop poles with a damping ratio ^ = 0.5.
Figure Feedback system
Step-by-step solution
Step-by-step solution
step 1 of 6
Refer to Figure 5.55 in the textbook.
Simplify the feedback system {Figure 5.55) as shown in Figure 1.
Figure 1
Step 2 of 6
Calculate the transfer function of the feedback system.
yw
R(s)
GW
1+G(i)ff(j)
1+
K M
The characteristic equation for the feedback system is.
U ( j+ i) J L i- i& J
luation (1) is of the fonn
Equation
1+KL(s) - 0. Where
10
( 5 + l) ( l- A : r )
L ( j) has two poles, one on real axis and other pole value depends on value
.
To make system stable consider the ^ value as -1.
Then,
10
i( s ) .
(s+l)(i +l)
step 3 of 6
Procedure to draw root locus:
RULE 1 : There are two branches to the locus, both of which approach asymptotes.
RULE 2 : There is no part of root locus.
RULE 3 : Calculate the centre of asymptotes.
1-1
2-0
-
< r*-
-2
The angles of asymptotes are at ^ 9q* .
Step 4 of 6
To find the value of fc that yield closed loop poles with damping ratio ^ =
0.5•
Consider the magnitude condition.
^
I
| i( * ) | •
K =
(j +l)(i+l)
(1)
10
To find value of ^ that yields closed loop poles with ^ = 0.5. rvalue is to be found.
As Rvalue is known, calculate the angle $ ,
cos$=^
6
■cos"'C
cos"'(0.5)
*60*
-
Draw the root locus using sisotool function in MATLAB and add design parameter as 0.5 damping
ratio to get asymptotes that cut the root locus as shown in Figure 2.
» num=10;
» den=[1 2 1];
» sys=tf(num,den):
» sisotool(sys)
Step 5 of 6
MATLAB output:
Root Locus EdHorfor Open Loop 1(OL1)
Figure 2
Step 6 of 6
From Figure 3, observe that the 0.5 damping ratio asymptote cuts the root locus at
5=-l±yT.732.
Substitute value of s in equation (1).
(-1+yi.732+l)(-l +;i.732+l)|
10
-2.991
10 I
=0.299
Therefore, the value of
that yields closed loop poles with the damping ratio ^ = 0.S is
Problem 5.15PP
A simplified model of the longitudinal motion of a certain helicopter near hover has the transfer
function
<H*)- («+0.66)(«2-a24c + 0.IS>
and the characteristic equation 1 + Dc(s)G(s) = 0. Let Dc(s) = kp at first.
(a)
Compute the departure and arrival angles at the complex poles and zeros.
(b)
Sketch the root locus for this system for parameter K = 9.6kp. Use axes -4 < x < 4; -3 < y <
3.
(c)
Verify your answer using Matlab. Use the command axis([-4 4 -3 3]) to get the right scales.
(c) Verify your answer using Matlab. Use the command axis([-4 4 -3 3]) to get the right scales.
(d)
Suggest a practical (at least as many poles as zeros) alternative compensation Dc(s) that wil
at least result in a stable system.
S te p -b y -s te p s o lu tio n
step 1 of 6
9.8(s’ -0 .5 s+ 6 .3 )
a {s ):
(s+ 0 .6 6 )(s^-0.24s+ 0.15)
Step 2 of 6
w
Step 3 of 6
Angles o f departure:
= 1 8 0 ° -« t>
Where
= 2 < b > -Z 'fe
2 4 ^ = 90®+25.25'>
= 115.25"
2<tir = 106.38"+253.61"
= 360"
^
= 180"+244.75"
= 424.75"
= 64.75"
And,
^
« -64.75"
Step 4 of 6
Angles o f arrival:
^ = 180"+<t»
■Where if = S f e - Z ' f e
= 70°+86.5°+87.4”
= 244°
Z f e = 90°
If, = 334°
And,
<bi, = -3 3 4 " .
Step 5 of 6
(b)
, ,
A r(5^-0.5ff+6.3)
a f s ) = --------- V r .; ------------i ^
where K = 9 .8 ^ ,
(s+0.66)(ff^-0.245+0.15)
'
K = 0 points: -0 .6 6 . 0.12±j0.368
iT s a points: 0.25±j2.5
As
^
<ti, = 64.75"
= -64.75"
= 334"
And,
^ = -3 3 4 "
We can plot the root locus o f the given transfer fimction as shown
below.
Step 6 of 6
Problem 5.16PP
(a)
For the system given in Fig., plot the root locus of the characteristic equation as the
parameter K^ is varied from 0 t o w i t h A = 2. Give the corresponding L(s), a(s), and b(s).
(b)
Repeat part (a) with /\ = 5. Is there anything special about this value?
(c)
Repeat part (a) for fixed K^ = 2, with the parameter K = A varying from 0 to
Figure Control system
Step-by-step solution
step 1 of 15
Refer to Figure 5.56 in the textbook.
Draw the modified block diagram.
Figure 1
Step 2 of 15
of a block as shown in Figure 2.
Move the summing point
R
Figure 2
Step 3 of 15
Draw the reduced block diagram of Figure 2.
Figure 3
Step 4 of 15 ^
Move the summing point ahead of a block as shown in Figure 4.
Figure 4
Step 5 of 15
Draw the reduced block diagram.
Step 6 of 15 ^
(a)
Calculate the characteristic equation.
I + G (s )ff(j) = 0
f
lOiC,
Y 0.U(^-fA)-fg,(0.2»-Mn .
K,
)-
j ( j + IO )(j + A)
j (j + 10)(5 + A)
step 7 of 15
Simplify further.
i ’ + ( 1 0 + A ) i’ + 10.1i
irgf
1-0
(1)
Substitute 2 f o r A in equation (1).
‘ ■ ^ ^ { j ’ + { l l + 2 ) j ‘ + l l ( 2) j ] “ “
From the characteristic equation, the loop transfer function is.
r(
2 £ + I2 _
+ 22 s
o ( i ) = »* + lJ s ’ + 2 2 i
j ’ +13 j ‘
And,
i ( j ) = 2 i+ 1 0
Step 8 of 15
Write the MATLAB code to draw the root locus.
» num=[2 10]:
» d e n = [1 13 22 Oj;
» sys=tf(num,den)
sys =
2 S + 10
s ^ 3+ 13s^2 + 22s
Continuous-time transfer function.
» rlocus(sys)
Step 9 of 15
Draw the root locus plot using MATLAB.
Root Locn$
Real Axis (seconds'*)
Figure 6
Step 10 of 15
(b)
Substitute 5 for >l in equation (1).
2j + 10
1 + Jf,l
^ j’ + ( l l + 5 ) i " + l l( 5 ) ;
2*+10
'I
1 + Jf,( '
k^j’ + 1 6 f* + 5 5 5 j
0
1+ JC,|
^ j( j+ 5 ) ( j + l l ) j
l+ if ,l'
*
1
0
From the characteristic equation, the loop transfer function is.
And,
b (s )^ 2
Step 11 of 15
Write the MATLAB code to draw the root locus.
» num=2;
» d e n = [1 11 Oj:
» sys=tf(num,den)
sys =
2
s^2 + 11 s
Continuous-time transfer function.
» rlocus(sys)
Step 12 of 15
Draw the root locus plot using MATLAB.
Root Locus
^ 2 ■n
1
1-
I®
fr
e -1
^ -2 h
-12
-10
Real Axis (seconds'*)
Figure 7
step 13 of 15 A
(c)
From the block diagram, the characteristic equation is,
1+ G ( s ) f f ( s ) = 0
,.r
lOiC,
[i(j+10)(i+A)Jl,
„
K,
)
s{s+ X )+ K ,{2 s*\0)
j ( i + 1 0 )(j + 2 )
Substitute 2 fo ritr,
3 (3 ^ 2 )> 2 (2 3 ^ I0 )
i ( s + 1 0 )(s + 2 )
j (j
j
1-2)+43H-20
(3 + 1 0 ) ( * + 2 )
j( j- F ll)
1 + .1 j ’ + lls * + 4 r + 2 0
From the characteristic equation, the loop transfer function is.
' ^
* ’ + lb * + 4 ff + 2 0
fl( 5 )= 5 * + I I j* + 4 s + 2 0
And.
i(j)=j(s+ll)
Step 14 of 15
Write the MATLAB code to draw the root locus.
» num=[1 11 0];
» d e n = [1 11 4 20];
» sys=tf(num,den)
sys =
s^2 + 11 s
s^3+ 11 s'‘2 + 4 s + 20
Continuous-time transfer function.
» rlocus(sys)
Step 15 of 15
Draw the root locus plot using MATLAB.
Root Locus
Problem 5 .1 7PP
For the system shown in Fig., determine the characteristic equation and sketch the root locus of it
with respect to positive values of the parameter c. Give L(s), a(s), and b(s), and be sure to show
with arrows the direction in which c increases on the locus.
Figure Control system
Step-by-step solution
step 1 of 4
Step 1 of 4
Refer to Figure 5.57 in the textbook.
From the block diagram, the process transfer function is.
/'C + I 65 V 9 ''
G ( ,)
V C+S
Calculate the characteristic equation.
1+ G (» )H (» ) = 0
( c + j) i* + 9 c + I 4 4 5
(c + j)j*
cy* + 5* + 9 c + 144s * 0
c (s* + 9) + s ( s * + 144) = 0
l+ c i- = 0
j(j" + 1 4 4 )
(1)
(s*+9)
s(s* +144)
Therefore, the characteristic equation is
Step 2 of 4
From the equation {1), the loop transfer function is.
s (s ^ + 1 4 4 )
From the characteristic equation.
a(s)=s*+144s
And.
6(s)=s*+9
Therefore, the loop transfer function, L { s ) i:
p(s*+144)
are |a(j)sj^+1 44 s and ^(s) = s*
Step 3 of 4
Write a MATLAB code to plot the root locus.
» num=[1 0 9];
» den=[1 0 144 0]:
» sys=tf(num,den)
sys =
s^2 + 9
s^3 + 144 s
Continuous-time transfer function.
» rlocus(sys)
Step 4 of 4
Draw the root locus for all positive values of c .
( ) andi(s)
e values of 0 5
Problem 5.18PP
Suppose you are given a system with the transfer function
d+z)
where z and p are reai and z > p. Show that the root iocus for 1 + KL(s) = 0 with respect to K \ss
circle centered at z with radius given by
r= {z -p ).
Hint Assume s + z = rej<p and show that L(s) is real and negative for real <punder this
assumption.
Step-by-step solution
Step-by-step solution
step 1 of 2
L (s )= -^
let s=a+jflo
T / \l
_
(cT+p)^V+2jcD(<T+p)
Phaseof L ( s ) ^ 1 8 0 “
Step 2 of 2
Imaginary part of L (s)
=0
- 2 g j ( o + z ) ( o + p )-I- co^(<H- p ) ^ - o j ^ J = 0
(D^
+CD^+2oz+2pz-p^ ] =0
(or) a^+© ^2pz+z^^^+z^-2oz
or o^+©^+2oz+z^^^+z^-2pz
(o+z)* +©^=(z-p)^
IWhich is a circle with radius; z-pl
Problem 5.19PP
The loop transmission of a system has two poles at s = -1 and a zero at s = -2. There is a third
real-axis pole p located somewhere to the /e/tof the zero. Several different root loci are possible,
depending on the exact location of the third pole. The extreme cases occur when the pole is
located at infinity or when it is located at s = -2. Give values for p and sketch the three distinct
types of loci.
Step-by-step solution
step 1 of 4
T .M s -
s+2
Step 1 of 4
L ( s ) = ----- ---------( = + l) C = ^ )
Step 2 of 4
(i)
Let p = -2
Step 3 of 4
(ii)
p= ^
Step 4 of 4
(iii)
p = -oo
For the feedback configuration of Fig., use asymptofes, center of asymptofes, angies of
deparfure and arrivai, and fhe Routh array to sketch root ioci for the characteristic equations of
the iisted feedback controi systems versus the parameter K. Use Mattab to verity your resuits.
(a) C(i) = ;p T n ^ r F I= 57'
(») C ( j) = 5 ,
(c )
=
m
= '+ 3 *
Figure Feedback system
m
Figure Feedback system
S tep-by-step s o lu tio n
S le p t of 30
(a)
Consider the foiiowing equation.
« , ) = (.
a (j+ l+ 3 y )( s + l- 3 y )
i( a ) =
s+2
a (a + I+ 3 y ) (a + I- 3 y ) ( s + 8 )
Consider the generai torm of characteristics equation.
l + * i( a ) = 0 ......( 1)
fyfor ^ 5)in Equation (1).
i( a + I 0 ) ( a + I+ y ') ( a + I - y )
a+2
1+ i r = 0 ...... (2)
a ( a + l+ 3 y ) ( a + l- 3 y ) ( i+ 8 )
Consider the roots of the generai form of an equation by the root iocus method.
l +iC - ^ - O
ZKs)
......(3)
Where.
The roots of ^ ( 5) = 0 are catted the zeros of the probiem.
The roots of O(a) = 0 are the potes.
Consider the number of potes and zeros from the characteristics equation.
Compare the Equation (2) and the Equation (3).
To find zeros put numerator fV(s)=0
Thus, the zero is -2.
To find potes put denominator D(_s) = 0.
j ( j + l+ 3 y ) ( i+ l- 3 y ) ( » + 8 ) = 0 ......(4)
The roots of the equation (4) are 0 .-8 . —1—3y and —l+ 3 y .
Thus, the tour poies are 0. - 8. —I - 3 y and - l+ 3 y .
Step 2 of 30 ^
Consider the formuta for the asymptotes.
n -m
(sum o f finite poles)-(sum o f finite zeros)
(numberof finite poles )-(n i]in b e ro f finite zeros)
- l- 3 y - l+ 3 y - 8 - ( - 2 )
3
3
Thus, the asymptotes is |-2.67|.
Consider the formuta for the angte of asymptotes.
180<-f360°(/-l)......,5,
n -m
Where.
Number of potes is n
Number of zeros is m
/ = I,2 ,..ji- m
Substitute f for /. 4 for n and 1 for m in equation (5).
,
180+3«0(l-l)
" “
a
180
“
3
= 60°
Substitute 2 for /. 4 for n and 1 for m in equation (5).
,
180+360(2-1)
ft “
e
540
3
= I80»
Substitute 3 for /. 4 for n and 1 for m in equation (5).
,
180 + 360(3-1)
ft”
+
300
3
= -60«
Thus, the angte of asymptotes are |6Q**| . |]go«*|and | - 60**l-
Step 3 of 30
Consider the foiiowing equation.
6 (-l8 0 °-ta n[3
|^1
= 108.43“
6^=90“
[3
6( -tanj^Y
= 71.56“
= 23.19“
Consider the departure angies in the transfer function.
sum o f angleof vector to'
180“ -
angleof depature )
the complex pole A
from otherpoles
from a complex A j
sum o f angleof vectors to the'l
{
complex pole A from zeros J
0 j = lS « ° -[0 ,+ e ,+ e ,]+ 0 ,
Substitute ait
vatues in equation.
61, = 180“-[108.43“+90“+23.l9“]+71.56“
Thus. the departure angie in the transfer tunction is |6i^ =-29.94“ | at s = —l+ 3 y .
Step 4 of 30
Consider the foiiowing transfer function.
C (s )W (s )
l+ G ( s ) // ( s )
s ( s + l+ 3 y ) ( s + l- 3 ./)
(
_______ ^
(_s(s+ I+3y)(s
+ 3 y )(s + l-3 y )J (. s + 8 /
Consider denominator poiynomiai to find the K vaiue.
s(s + l+ 3 y ) ( s + l- 3 y )
~ s (s + l+ 3 y )(s + l-3 y )(s + 8 )+ A :(s+ 2 )
s ( s + l + 3 y ) ( s + I - 3 y ‘) (s + 8 )
_______________ 8T (s+8)______________
“ s (j+ l+ 3 y )(i+ l-3 y )(j+ 8 )+ 6 r(i+ 2 )
Consider the denominator poiynomiai is equai to zero.
s [s ’ + s -3 y s + s + l- 3 y + 3 y s + 3 y -9 y * ](s + 8 )+ 8 r( s + 2 ) = 0
[s ’ + s * -3 y s “ + s *+ s -3 y s + 3 y s *+ 3 y s -9 y *s j(s + 8 )+ H r(s + 2 ) = 0
(s’ + 2s’ + s - 9 y ’s ) (s + 8) + I f ( s + 2 ) = 0
s *+ 2s’ + s’ - 9 y V + 8s’ +16s’ + 8s - 72y’ s + a: (s + 2) = 0
[s ’ + 10s’ + s ’ + 9s’ + 8s’ + 16s’ + 8 s+ 72s]+ K ( s + 2 ) - 0
s‘ +10s’ +26s’ + 8 0 s + K (s + 2 ) - 0
s“+10s’ + 26s’ +(80+AT)s +2A: = 0
......(6)
From equation (6). the characteristics equation is given beiow.
A ( s ) = s‘ + 1Os’ + 26s’ + (8 0 + K ) s +28:
s’ + l0 s ’ + 26s’ + ( 8 0 + i: ) s + 2 8 : = 0 .......(7)
By appiying Routh-Huiwitz criteria to equation (7).
1
26
10
80 + / :
2K
2K
1 8 - - ^
10
lOOAT+14400
- I f l -
1 8 0 - /:
2K
Figure 1
step 5 of 30 ^
The sysfem is stabie if fhe equation satisfies the foitowing condition.
• Aii the terms in the tirst coiumn of the Routh’s array must be positive sign.
From Figure f. the stabie condition is 0 < g f <gQ.
Thus, the range of/Cis 0< 8 :< 8 0und it is satistied the stabiiity condition.
Substitute 80 for K in equation (7).
s’ +10s’ +26s’ +(80 + 80)s+160=0
s“+I0 s’ +26s’ + 160s +160 = 0
W
The roots of the equation (8) are ±4 y. —S—y J lS ^ ^ ^ lS —SThus, the imaginary axis crossings are
and
step 6 of 30 ^
Procedure to draw root iocus piot:
• Take reai and imaginary iines on X axis and Y axis respectiveiy.
• Mark the potes on the reai axis.
• Locate the asymptotes on the reai axis, and draw the asymptotes from centroid at an angte of
W
• Draw the root iocus.
The root iocus piot is shown in Figure 2.
Hence, the root iocus is ptotted for the given transfer function and it is shown in Figure2.
Step 7 of 30 ^
Consider the foiiowing given function.
j+ 3
x(x + 10)(x+l+ y )(« + l-y )
L {s )-
Write the MATLAB program to obtain root iocus.
s=tf('s');
sysL=(s+3)/(s*(s+10)*(s+1+1j)*(s+1-lj));
riocus(sysL)
The root iocus piot is shown in Figure 3;
Step 8 of 30 ^
Thus, the root iocus is verified from MATLAB output.
Step 9 of 30
(b)
Consider the foiiowing equation.
>-m]
i( j) =
■ j ’ (j+ 3)
Step 10 of 30
Substitute
)
* for £ (j)in Equation (1).
i ( j+ 3 )
.(9)
i ’ (i+3)
Compare the Equation (9) and the Equation (3).
To find zeros put numerator N {s^= 0
Thus, the zero is -1.
To find poies put denominator D(_s) = 0.
The roots of the equation (4) are 0.0 and -3.
Thus, the tour poies are 0. 0 and -3.
Step 11 of 30 -rv
Consider the formuta for the asymptotes.
n -m
(sum o f finite poles)-(sum o f finite zeros)
(numberof finite poles )-(m u n b e ro f finite zeros)
2
~ 2
Thus, the centre of asymptotes is
.
Substitute 1 for /. 3 for n and 1 for m in equation (5).
180+ 360(1-1)
ft “
+
180
”
2
=90“
Substitute 2 for /. 3 for n and 1 for m in equation (5).
180+ 360(2-1)
540
2
= 270“
= -9 0 “
Thus, the angle of asymptotes are |90“ |and
Step 12 of 30
Consider the following transfer tunction.
T -r
0 (s )ff(s )
l+ G (s )W (s )
X (» + 0
” s’ (s+ 3 )
l+ X
s’ (s+ 3 )
Consider denominator polynomial to find the K value.
s’ +3s’ + J&+A:
Consider the denominator polynomial is equal to zero.
s’ +3s’ +&+A: = 0
(10)
From equation (10). the characteristics equation is given below.
A ( s ) = s ’ +3s’ + A i + j :
s’ +3s’ +&+A: = 0
(11)
By applying Routh-Huiwitz criteria to equation (11).
a
:
a:
2a :
K
Figure 4
Step 13 of 30 ^
The sysfem is stable if the equation satisties the following condition.
• All the terms in the tirst column of the Routh’s array must be positive sign.
From Figure 4. the stable condition is JK > Q.
Thus, the value of K is JK = Qand it is satistied the stability condition.
Substitute 0 for X in equation (11).
*’ +3»’ + k » + a: = o
j =0
Thus, the imaginary axis crossing is
.
Step 14 of 30
Procedure to draw root locus plot:
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes from centroid at an angle of
9(r
• Draw the root locus.
The root locus plot is shown in Figure 5.
RootLocus
Hence, the root locus is plotted for the given transfer function and it is shown in Figure 5.
Step 15 of 30
Consider the following given function.
(j+ 1 )
Write the MATLAB program to obtain root locus.
s=tf('s');
sysL=(s+1)/(s“ 2*(s+3));
riocus(sysL)
The root locus plot is shown in Figure 6:
Step 16 of 30
Thus, the root locus is verified from MATLAB output.
Step 17 of 30
(0
Consider the following equation.
Substitute
Ifo r L (j)in Equation (1).
U + lA s + 3 j
■•'(Srlef)-".. '■»
step 18 of 30
Compare the Equation (12) and the Equation (3).
To find zeros put numerator N {s^= 0
Thus, the zero is -5 and -7.
To find poles put denominator D(s) = 0.
The roots are -1 and -3.
Thus, the tour poles are -1 and -3.
Step 19 of 30 ^
Consider the formula for the asymptotes.
Asymptodes=(numberofUnitepoles)-(numberoffinltezeros)
= 2-2
=0
Thus, the asymptotes is
.
s 0 and simplity equation is given below.
Consider
ds
8s’ +64s+I04 = 0
From the above equation, the break in and break away points is -2.27 and -5.73.
Thus, the break in and break away points is -2.27 and -5.73.
Step 20 of 30
Procedure to draw root locus plot:
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes from centroid.
• Draw the root locus.
The root locus plot is shown in Figure 7.
Root Locus
Hence, the root locus is plotted for the given transfer function and it is shown in Figure 7.
Step 21 of 30
Consider the following given function.
i( i) =
M S I)
MATLAB program to obtain root locus:
s=tf('s');
sysL=((s+5)*(s+7))/((s+1)*(s+3)):
riocus(sysL)
The root locus plot is shown in Figure 8.
Figure 8
Step 22 of 30
Thus, the root locus is verified from MATLAB output.
Step 23 of 30
(d )
Consider the following equation.
w A - (< ^ 3 s )( s + 3 + 4 y )( s + 3 - 4 ;)
j( s + l+ 2y ) ( j + l - 2y)
Substitute ( '• ’• ^ ) ( 4 ’* '3 + y ) ( s + 3 - 4 y ) ^^^
i ( j + l + 2y)(s + l - 2y)
Equation (1).
*4 f
(l+ 3s)(3+3+ 4y)(s+3- 4y)
i ( i + l + 2y ) ( j + l - 2y)
u
(13)
Compare the Equation (13) and the Equation (3).
To find zeros put numerator ^ ( s ) = 0
Thus, the zero is -0.33. —3 -4 ya n d —3 + 4 y.
To find poles put denominator /}(s) = 0 ■
The roots are -1 and -3.
Thus, the poles are 0. —l-2 y a n d —l+ 2 y .
Step 24 of 30 ^
Consider the formula for the asymptotes.
A sym ptodes=(num ber o f Unite poles )-(n u m b e r o f finltezeros)
= 3 -3
=0
Thus, the asymptotes is
.
Step 25 of 30
Consider the following equation.
6 (-I8 0 “-tanj^Y]
= 116.56“
= 108.26“
6!, = 90“
6)4 = tan
[I]
=71.56“
[2 1
6( = tan |
= -45“
Consider the departure angles in the transfer function.
sum o f angleof vector to'
180“-
the complex pole A
ang leof depature 1
from otherpoles
from a complex A j
sum o f angleof vectors to the'l
{
complex pole A from zeros J
6>, = 180“-[61 +6l,]+[tf, +6I4 + 6I5]
Substitute all 0 values in equation.
6>, = 180“-[1 16.56“+90“]+[108.26“+71.56“-45“]
Thus, the departure angle in the transfer tunction is \0^ =108.26“| at s = —1+2/.
Step 26 of 30
Consider the following equation.
d l- lO O - ta n j^ jj
= 126.86“
6 ^ .1 8 0 - ta n [Y ^ ]
= 123.62“
6( =lS0-tan|^Y
= 135“
6)4 = 180-tan
= 108.43“
6 1 ,-9 0“
Consider the arrival angle in the transfer function.
ang leof arrival
1
from a complex A J
sum o f angleof vector to'
= 180“ — the complex zero A
from other poles zeros
sum o f ang leo f vectors to the]
{
complex zero A from poles J
6>. = 180“ - [ 6i + 6) ] + [ 6l + 6) + ^ 4]
step 27 of 30
Substitute all 0 values in equation.
6!.-180“-[l23.62“+90“]+[l26.86“+135“+108.43“]
= 336.67“
Thus, the arrival angle in the transfer tunction is |6t^ = -22.33“| at s = -3 + 4 /.
Step 28 of 30 ^
Procedure to draw root locus plot:
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes from centroid.
• Draw the root locus.
The root locus plot is shown in Figure 9.
R oot Locus
Hence, the root locus is plotted for the given transfer function and it is shown in Figure 9.
Step 29 of 30
Consider the following given function.
w A - 0 ^ 3 s )(s+ 3 + 4 y )(s+ 3 -4 ;)
s(s + l+ 2 /) ( s + l- 2 /)
MATLAB program to obtain root locus:
s=tf('s');
sysL=((1+(3*s))*(s+3+4i)*(s+3^j)V(s*(s+1+2i)*(s+1-2j));
riocus(sysL)
The root locus plot is shown in Figure 10.
Root Loon
Step 30 of 30
Thus, the root locus is verified from MATLAB output.
Problem 5.21 PP
Consider the system in Fig..
(a) Using Routh’s stability criterion, determine all values of K for which the system is stable.
(b) Use Matlab to draw the root locus versus K and find the values of K at the Imaginary-axis
crossings.
Figure Feedback system
ELh-
S te p -b y -s te p s o lu tio n
step 1 of 6
Refer to Figure 5.59 in the textbook.
Calculate the characteristic equation.
1+ G ( j ) f f ( j ) = 0
1+
'
K (s + 3 )
m -’
i(j+ l)(s ^ + 4 i+ 5 )+ A :(j+ 3 )
j ( j + l) ( i^ + 4 i + s j
+ j){ » ^ + 4 » + 5 )+ K s + 3 X = 0
t S s * + 9 j ^ + 5 j + * s +3A: = 0
j*+5j’
+ 9 j ^ + ( 5 + A :) s + 3 X = 0
( 1)
Step 2 of 6 ^
Calculate the range of
using Routh’s stability criteria.
I
9
5
4 S -5 -K
3K
5^K
3AT
4 0 -K
5
3AT
From the Routh’s stability criteria, first column elements should be greater than zero.
Therefore,
S
4 0 -K > 0
K<A0
And.
3^>0
^>0
Therefore, the values of
for which the system is unstable is |0 ^ AT ^
Step 3 of 6
(b)
Calculate the loop transfer function.
L (s ) = G (» )W (i)
^ ^(^-^3)
Y
1 j
[ i( i^ + 4 s + 5 ) J li+ l
y (» + 3 )
j(j+ l)(j* + 4 s + S )
A T (j+ 3)
+ & ’ + 9 i* + 5 i
Consider that,
=
Therefore,
(^ ^ 3 )
i( s ) .
s * + J j’ + 9»"+ 5s
Step 4 of 6 ^
Write a MATLAB program to find the root locus of the system.
» num=[1 3]:
» den=[1 5 9 5 0];
» sys=tf(num,den)
sys =
s+3
sM + 5 s'^3 + 9 s'^2 + 5 s
Continuous-time transfer function.
» riocus(sys)
Step 5 of 6
Draw the root locus.
Figure 1
Step 6 of 6
The root locus is crossing the imaginary axis at ^ = ± J { ,38.
Therefore, the value of gain,
is I S U
Problem 5.22PP
C (j) =
Using root-locus techniques, find values for the parameters a, b, and K of the compensation
Dc(s) that will produce closed-loop poles at s = -1±yforthe system shown in Fig.
Figure Unity feedback system
Step-by-step solution
Step-by-step solution
step 1 of 2
Refer to Figure 5.53 in the textbook.
From the Figure 5.53, the loop transfer function is,
£ (* ) = G (« )i)(5 )
I
i(,)=
(5 + 2 )( j +3)
______ A '( j+ o )
( j + 2 ) ( j + 3 )^ 5 + 6 )
Since, the closed loop poles are at S = - l ± J
Thus, the characteristic equation of the 2nd order transfer function is,
(j+i+ j ) { s + \ - y)»o
( j + l)^ + l = 0
s^+2s+2 = 0
Since, the closed loop poles are poles than the poles of the transfer function G (^ ) •
Thus consider that the zero of D ( s ) is at 3 to cancel the pole at 3.
That is, a s - 3 .
Step 2 of 2
Calculate the closed loop transfer function.
r(,)
G (^ )g M
R (s)
i+ G (s )D {s )H {s )
\+
(^ + 2 )(^ + 3 )
K
( i+ 2 ) ( s + i)
“
K
(s * 2 )(s + b )
_________ K ________
* i'+ ( 2 + f t ) s + 2 * + ^ :
From the transfer function, the characteristic equation is,
j* + ( 2 + i) j+ 2 4 + K = 0
(2)
Compare equation (1) with equation (2).
2+i=2
i =0
And.
24 + A : = 2
2 (0 )+ ^ = 2
K =2
Therefore, the parameters Oy b and K of the compensation V ( s ) to produce closed loop
poiesat » = - l ± y are |a = - 3 , t = 0 a n < |g = 2|
Problem 5.23PP
Suppose that in Fig.
=
,,
!.— ^ and
5)
Dc(s) =
j (j2 + 2 s +
s
+2
Without using Matlab, sketch the root locus with respect to K of the characteristic equation for the
ciosed-loop system, paying particuiar attention to points that generate multiple roots. Find the
value of K at that point, state what the location of the mulitple roots is, and how many multiple
roots there are.
Figure Unity feedback system
- • a»)
Step-by-step solution
step 1 of 8
Consider the general form of characteristics equation
i+ o , w c w = o
(1)
I
7 1 — Z—
(■s + 2 j + 5 j
Substitute
K
G( j ) and -------for D^(ff)in Equation (1).
j+ 2
1
1 + ii: -
(2 )
+ 2 j+ 5 j
Consider the roots of the general form of an equation by the root locus method.
(3)
Where,
The roots of J V (j)= 0 are called the zeros of the problem.
The roots of D ( j) = 0 are the poles.
Consider the number of poles and zeros from the characteristics equation.
Compare the Equation (2) and the Equation (3).
To find zeros put numerator JV(j) = 0
Thus, there is no zero in the transfer function.
To find poles put denominator D {s) = 0 •
j ( j + 2 )( j '+ 2 j + 5 ) = 0 ...... (4)
The roots of the equation (4) are 0 , - 2 , —\ + 2 J and —l - 2 y .
Thus, the four poles are 0, - 2, —l + 2y'and —1—2y-
Step 2 of 8
Consider the formula for the centre of asymptotes.
(sum of finite poIes)-(sum of finite zeros)
(numberof finite poles )-(nunU>er of finite zeros)
( 0 - 2 - l + 2 y -l-2 y )-(0 )
(4 )-(0 )
=-I
Thus, the centre of asymptotes is O
Step 3 of 8
Consider the formula for the angle of asymptotes.
,
l8 0 °+ 3 6 0 « (/-l)
(5)
n —m
Where,
Number of poles is n
Number of zeros is m
/ = l,2 ,..ji-m
Substitute 1 for /, 4 for n and 0 for m in equation (5).
I80°+360°(1-1)
^ ”
>
41- 0A
= 45*
Substitute 2 for /, 4 for n and 0 for m in equation (5).
l8 0 °+ 3 6 0 °(2 -l)
4 -0
= 135»
Substitute 3 for /, 4 for n and 0 for m in equation (5).
,
180»+360»(3-l)
4 -0
«225«
= -45»
Substitute 4 for /, 4 for n and 0 for m in equation (5).
180**+ 3 6 0 » (4 -l)
4 -0
-315«
= -135»
Thus, the angie of asymptotes are [45 ^ . | 135<>|, |-45 °|and | - 135<>|.
Step 4 of 8
Consider the following formula for the departure angle ^ ^ fro m the pole at - l+ 2 y .
j
= - t a n - ' - ta n -'^ I j - 90°+180°
— 116.6°-63.4°-90°+180°
= -90°
Thus, the departure angle ^ ^ fro m the pole at —l°h2y is |-9Q^|.
Step 5 of 8
From equation (2), the characteristics equation is given below.
A(a)=a(a+2)(j’ +2a+5)+X
A(a)=a‘ +4a’+9a“+10a+X... (6)
Apply Routh-Hurwitz criteria to equation (6 ).
1
9
^
4'
^10
6.5
10
tS - l
K
-
13
K
50
Figure 1
Step 6 of 8
The system is stable if the equation satisfies the following condition.
• All the terms in the first column of the Routh’s array must be positive sign.
From Figure 1, the stable condition is 0 < AT < 16.25 •
Thus, the range of /CIs 0 <
AT< 16.25^nd it is satisfied the stability condition.
Substitute 16.25 for K in equation (6 ).
a*+ 4 a ’ +9a’ +10a + 16.25 = 0
P)
The roots of the equation (7) are —2±1.58114y and ±1.58114y •
Thus, the imaginary axis crossings are |l.58114y| and |-l-5 8 1 1 4 y |.
step 7 of 8
Take differentiate to equation (6) with respect to £
*45’ +12 j * + 18«+10
ds
4 j ’ + 12«*+ I8 j + 1 0 - 0
(8)
The roots of the equation (8 ) are - 1 and —l±1.225y •
Substitute - 1 for s in equation (6).
( - | ) % 4 ( - I ) ’ + 9 (-I)*+ 1 0 (-1 )+ A : = 0
a: =
4
Substitute —l+ 1 .2 2 5 y fo rs in equation (6 ).
(-1 -I-1.225j Y + 4 ( - l +1.225j f + 9 ( - l -t-1.225y) ' +
I0(-I-H.225y)-FA:
a:
=
0
= 16.25
Thus, the location of multiple roots are |_ \\ and | - 1 ± 1 . 2 2 ^ .
Hence, the corresponding value of/C is 4 and 16.25.
Step 8 of 8
Procedure to draw root locus plot:
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes from centroid at an angle of
45T
• Draw the root locus.
The root locus plot is shown in Figure 2.
Figure2
Hence, the root locus is plotted for the given transfer function and it is shown in Figure2.
Problem 5.24PP
Suppose the unity feedback system of Fig. has an open-loop plant given by G(s) = 1/s2. Design e
lead compensation Dc(s) =
to be added in set
of the closed-loop system are located at s = - 2 ± 2j.
Figure Unity feedback system
- • a»)
Step-by-step solution
_
Step 1 of 2
The loop transfer function of the system with lead compensation is.
G W £ > ,W - A r ( ^ ) ( l)
The characteristic equation of the system is,
l + G ( j) D c W = 0
„ j f f i ± £ ¥ 'l = o
5* ( 5 + p ) + ( 5 + z ) » 0
s **p s **K s *K z ^ 0
Step 2 of 2 ^
The dominant poles are at s = - 2 ± J 2 .
Assume a real pole at j s —2The characteristic equation of the system with dominant poles at s = —2 ± j 2 and a real pole at
5 = - 2 is.
( i + 2 ) [ ( i + 2 ) ' + 2’ ] = 0
( j + 2)( j *+45 + 8) = 0
j ’ + 6 s ‘ + I f a + I6 = 0
Compare the characteristic equation of the system with the desired characteristic equation.
p =6
K ^16
Kz = 16
So.
Z»1
s designed lead compensator
Problem 5.25PP
Assume that the unity feedback system of Fig. has the open-loop plant
^ J(l + 3)(s + 6)'
Design a lag compensation to meet the following specifications;
• The step response settling time is to be less than 5 sec.
• The step response overshoot is to be less than 17%.
• The steady-state emor to a unit-ramp input must not exceed 10%.
Figure Unity feedback system
-T
.
Step-by-step solution
step 1 of 4
Refer to Figure 5.60 in the text book for the unity feedback system.
The open loop plant transfer function is,
j ( s + 3 ) ( i+ 6 )
The lag compensator transfer function is,
D { s ) - - ^ —\ 2 > P
5+ P
The loop transfer function is.
5+ Z
^(^ + />)(s + 3 )(s + 6 )
The error equation is,
I
E ^1+GZ>
«
1
1+
-
s { s+ p ){ s * 3 ) ( j + 6 )
The input is unit ramp.
E=
1+
^ (^ + p )(^ + 3 )(^ + 6 )
Step 2 of 4
Determine the steady state error to unit ramp input.
e . = I im j£ ( 4
I
s t im j
s-tO
+ p ) ( 5 + 3 )(5 + 6 )
s lim j
«-»o .y(5 + /> )( i + 3 )(^ + 6 ) + ( « + «)
« l im
*- *
1 1
(» + p )(n -3 )(j+ 6 )
1
+ ;>)(^ + 3)(5+6) + (j + r) J
_ l Bp
The steady state error to unit ramp input is less than 10%, that is. 0.1.
Z
z> \Z 0 p
Select z = 2 0 0 p
Step 3 of 4 ^
Select p and z using MATLAB by trial and error method.
» p=0.0001;
» z=200*p:
» num=25*[1 z];
» den=conv([1 p 0],[1 9 18]);
» sys=tf(num,den):
» sys_^feedback(sys,1);
» step(sys_f)
Step 4 of 4
Get the MATLAB output for the step response.
Observe from the step response that the settling time is 3.42 s (less than 5 sec) and the
overshoot is 13.9% { less than 17%).
Hence, the design conditions are met with the selected values.
Thus, the designed lag compensation is.
Problem 5.26PP
A numerically controlled machine tool positioning servomechanism has a normalized and scaled
transfer function given by
C ( j) =
I
■ J(5+ !)■
Performance specifications of the system in the unity feedback configuration of Fig. are satisfied
if the closed-loop poles are located at :s = —l ± j ^ .
(a) Show that this specification cannot be achieved by choosing proportional control alone, Dc(s)
= kp.
(b) Design a lead compensator
Dc(s) =
that will meet the specification.
Figure Unity feedback system
Figure Unity feedback system
* °
—■ Gt*)
I
o
Step-by-step solution
step 1 of 3
(a)
Assume
is the proportional control.
Write the loop transfer function.
r ( « ) = f l. W G W
1
Substitute • ] ±
for s ,
r(«)=*.
(-\± jS )(-i± jS + i)
' (-i± jS ){± j^/3 )
( ± jS ± i)
Closed loop poles lie on root locus and m ust satisfy the angle criterion.
Take positive sign and check the angle criteria Z r ( 5 ) s - l g 0 " .
^ T (s ,
— if ]
— 30«
Take negative sign and check the angle criteria.
^ T [s )
— 30®
Therefore, if D ^(s) is replaced by proportional control alone then the obtained closed loop
system is not satisfied the angle criteria. So, required specifications cannot be achieved using
proportional control alone.
Step 2 of 3
(b)
is the lead compensator.
A (*)= « —
' '
s+p
Substitute Ja> for s .
Determine the phase angle of lead compensator £ .
D (jeo)
=tan''| —|-tan''| —I
(1)
Differentiate equation (1) with respect to
d*.
1
r n
^2)
I
(1
( a»Y l i
'n f j
dw
p
p ‘ *<^
2^+0^
Equate
equal to zero and find & .
r
-
^
=0
r/7* + 2d ^ =
Since the closed loop pole is located at
.
pz=Z
Step 3 of 3
Determine the phase angle at ^ s > 1 ± y'-Ts •
r.R\
-
Substitute
120®
for
A = -> /3
i+ i-
J ^ ( £ Z £ l= - V 3
Substitute 3 for pz.
1
3 (2 )
3
Substitute — for z ■
P
B
3 (2 )
) ..,
;? * + 6 p -3 * 0
p — 6.46
Therefore, calculate the value of z .
3
Z-—
P
3
- 6 .4 6
— 0.46
Hence, the lead compensator transfer function is fC
5 + 0 .4 6
5 + 6 .4 6
Therefore, to meet the required specifications
,5 + 0 .4 6
D ,{ s ) = K
s + 6 .4 6
Problem 5.27PP
A servomechanism position control has the plant transfer function
G (j) = -
10
j(j+ l)(j+ IO )
You are to design a series compensation transfer function Dc(s) in the unity feedback
configuration to meet the following closed-loop specifications;
• The response to a reference step input is to have no more than 16% overshoot.
• The response to a reference step input is to have a rise time of no more than 0.4 sec.
• The steady-state emor to a unit ramp at the reference input must be less than 0.05.
(a)
Design a lead compensation that will cause the system to meet the dynamic response
specifications, ignoring the error requirement.
specifications, ignoring the emor requirement.
(b)
What is the velocity constant Kv for your design? Does it meet the emor specification?
(c)
Design a lag compensation to be used in series with the lead you have designed to cause the
system to meet the steady-state error specification.
(d)
Give the Matlab plot of the root locus of your final design.
(e)
Give the Matlab response of your final design to a reference step.
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
5.28PP
Problem 5.28PP
Assume that the closed-loop system of Fig. has a feed forward transfer function
Design a lag compensation so that the dominant poles of the closed-loop system are located at s
= -1 ± yand the steady-state error to a unit-ramp input is less than 0.2.
Figure Unity feedback system
Step-by-step solution
Step-by-step solution
step 1 of 3
Consider the following feedfonvard transfer function.
1
<?(*)
Calculate the velocity error constant for uncompensated system.
= Urnj G(5)
= lim ^l —r ^ — r 1
s lim
(5 + 1)
>1
Assume the lag compensator having transfer function,
Where the values of z andp are small.
Step 2 of 3
The required steady state error to a unit ramp input Is less than 0.2.
Write the formula for steady state error to a ramp input for type 1 system.
— < 0 .2
K,
K ,> —
0.2
’
K ,> 5
The factor by which steady state gain is to be increased is,
\p j
---------------K y o f uncom pensated
0.2
-S
Place the zero and pole of the lag compensator very close to the origin.
Let zero of compensator at a s - 0 .1 That is,
z = 0.1
Calculate the value of pole using the following factor.
i= 5
P
M .5
P
p = 0M
Therefore, the transfer function of the lag compensator is.
d
( , ) = a:
.
' '
'( * + 0 . 0 2 )
( 1)
Therefore, the transfer function of the compensated system is.
(a + 0 .1 )
' s(j +1)(5+0.02)
Step 3 of 3
The dominant poles of the closed loop system is, j s -1 ± y .
Apply magnitude condition at this location of dominant closed loop pole.
( * + 0.1)
=1
'* ( * + l) ( * + 0 .0 2 )
( -i+ y + 0 1 )
' ( - l + ; ) ( _ l + y + l) ( - l+ y + 0 .0 2 )
i-O-9+j)
'( - i* J ) { J ) ( - 0 - 9 S * j)
1.345K.
=1
(1.414)(1)(1.4)
X =1.47
1.47 for
D(*) =1.47
in equation (1).
(*+ 0 .1 )
(*+ 0 .0 2 )
Therefore, the transfer function of the compensated system is
D(*) =
1.47(*+0.1)
(*+ 0 .0 2 )
Problem 5.29PP
An elementary magnetic suspension scheme is depicted in Fig. For smail motions near the
reference position, the voltage e on the photo detector is related to the ball displacement x (in
meters) by e = 10Ox. The upward force (in newtons) on the ball caused by the current / (in
amperes) may be approximated by f= 0.5/+ 20x. The mass of the ball is 20 g and the
gravitational force is 9.8 N/kg. The power amplifier is a voltage-to-current device with an output
(in amperes) of i = u + VO.
(a)
Write the equations of motion for this set up.
(b)
Give the value of the bias VO that results in the ball being in equilibrium at x = 0.
(c)
What is the transfer function from u to e?
(c) What is the transfer function from u to e?
(d)
Suppose that the control input u is given by u = -Ke. Sketch the root locus of the closed-loop
system as a function of K.
(e)
Assume that a lead compensation is available in the form
of K. z, and p that yield improved performance over the one proposed in part (d).
Figure Elementary magnetic suspension
Step-by-step solution
step 1 of 5
=0.5i+20rH3flg
Step 2 of 5
b.
For Equilibriiun at X = 0, 0.5Vg^ng
V„=2mg=0.392V
Step 3 of 5
c.
sm X (s)= 0 .5 V (s)+ 20X (s)
0.5V (s)= X (s)[0.02s’ -20]
E (s )_ r
V (s)
50
0.02s^-20
Step 4 of 5
K50
2500K
0.02s"-20
s"-1000
Step 5 of 5
D (s)= ^ fc !^ l
^^
s4p
p=50
for ^ . 7 ( H .
let 1=31.6
K=0.665
D (s)= 0.665^ ^ ^ ' ^ ^
'■ '
(s+50)
values
Problem 5.30PP
A certain plant with the nonminimum phase transfer function
is in a unity positive feedback system with the controiler transfer function Dc(s).
(a)
Use Matlab to determine a (negative) value for Dc(s) = K so that the closed-loop system with
negative feedback has a damping ratio ^ = 0.707.
(b)
Use Matlab to plot the system’s response to a reference step.
Step-by-step solution
Step-by-step solution
step 1 of 4
4 -2 s
= -y
s^ + s + 9
Is in a unity positive feedback system with the controller transfer function D(s).
A certain plant with the nonmininmm phase transfer function
'With all the negatives, the problem statement might be confusing. W ith the 0 (^ )
as given, MATLAB needs to plot the negative locus, which is the regular positive
locus for -G . The locus is plotted below.
Thus, [the value of gain for closed loop roots at dangling o f 0.7 is ^ = -1.04|.
Step 2 of 4 ^
(b)
The MATLAB program to find the root locus of the given S3rstem is
n u m = [0 2 - 4 ] ;
d e n = [l 1 9 ];
k = 0 :0 .0 0 5 :1 0 ;
r l o c u s (nuihr den^ k) ;
s g rid
%plot ( r , ’ x ’ ) ;
t i t l e ( ' R o o t - l o c u s ’ );
x l a b e K ’ r e n l p a r t ’ );
y l a b e l ( ’ imag p a r t ’ );
figure
s t e p (niunr d e n ) ;
Step 3 of 4 ^
The root locus obtained on execution o f the above program is.
Step 4 of 4
The step response obtained is
RwipnnM*
04
■
A ......................
■. 1
<
. A
.........
-0 4
m
■• ■ o ' - .
m
. '2 •
: : . 4 .■
•6 ■ . . 8
Tim* {wei • " ■
ic
12
The final value o f the step response plotted below is -0.887 .To get a positive
ou ^u t we would use a positive gain in positive feedback.
Thus, the required step response is plotted.
Problem 5.31 PP
Consider the rocket-positioning system shown in Fig.
(a) Show that if the sensor that measures x has a unity transfer function, the lead compensator
stabilizes the system.
(b) Assume that the sensor transfer function is modeled by a single pole with a 0.1 sec time
constant and unityDCgain. Using the root-locus procedure, find a value for the gain K that will
provide the maximum damping ratio.
Figure Block diagram for rocket-positioning control system
Figure Block diagram for rocket-positioning control system
X
Step-by-step solution
step 1 of 2
Root locus shows that the system is a stable system
Step 2 of 2
b.
H ,( s ) *
O .ls+ l
At K=5, Damping ratio i s maximum
Problem 5.32PP
For the system in Fig.,
(a) Find the locus of closed-loop roots with respect to K.
(b) Find the maximum value of K for which the system is stable. Assume /C= 2 for the remaining
parts of this problem.
(c)
What is the steady-state error {e = r - y) for a step change in r?
(d) What is the steady-state error in y fo r a constant disturbance w1?
(d)
What is the steady-state error in y fo r a constant disturbance w1?
(e)
What is the steady-state error in y for a constant disturbance w2?
(f) If you wished to have more damping, what changes would you make to the system?
Figure Control system
S te p -b y -s te p s o lu tio n
Step 1 of 8
(a)
iooii:(s+i)
D O H {s ) =
s ’ (s + 6 + 2 j) ( s + 6 0 - 2 j)
K = 0 points: 0, 0 , -6 ± J 2
K = a points: - 1 , ot, a a
Asyn^totcs: 60®, 180®, 300®
Centroid = --------- ^— - = -3 .6 6
3
Break away point: s = -5 .8
Characteristic equation:
s'(s'+ 1 2 a + 4 0 )+ 1 0 0 J!:(s+ l) = 0
s‘ +12s’ + 4 0 s^ + 1 0 0 rs+ 1 0 0 r = 0
Step 2 of 8
7
7
7
1
40
12
100 K
480-lOOJi:
100 K
100 K
12
7
I0 0 ji:(4 8 0 -i0 0 ji:)-i4 4 0 0 i:
7
480-100^:
100 K
Step 3 of 8
From the above array,
K >0
iooi!:(480-iooj!:) > 14400^
4 8 0 0 0 ^ -1 0 0 0 0 i:^ -1 4 4 0 0 £ ’ > 0
- 1 0 ,0 0 0 ^ “ +33600^: > 0
10,000A: > 33600
K > 3.36
M 8 0 -1 0 0 ^> 1 a
^
— - — js ^ + m x = 0
At K = 3.36,
M 8 0 - 336^ a
I
12
+336 = 0
r
12s^+336 = 0
s = ±J5.29
The root loci intersect the imaginary axis at 5 s ± ^5.29.
Step 4 of 8
Angles o f departure:
= 180®-<t>
Where
= 2 < b > -Z 'fe
2 < t^ = 161.56®+161.56®+90®
= 413.12®
2<t!r = 158.2®
^ = 254.9®
= -74.92®
And,
^
= 74.92®
Step 6 of 8
100 jg'
r(s)
s ^ (s“+ 12s+ 40)
(c)
W )
1+
(1+ 5)100^
s’ (s^+ 12s+ 40)
^
lOQg_________
s ^ (s“+ 1 2 s+ 4 0 )+ (H -s)1 0 0 i:
^
lOQg__________
7 + i2 7 + 4 o 7 + T o o F s + Io o Z
^ s*+125^+405^+ i o o j : s + i o o g - i o o j :
s ' +12s" + 40s" + 1 0 0 r s + 1 0 0 r
s (s ’ + 1 2 s^ + 4 0 s+ 1 0 0 r)
R{s)
^
” s * + 1 2 ^ + 4 0 s ’ + 1 0 0 ^ s + 1 0 0 i:
t „ = I m s S [s)
■When R(s) = -
Step 7 of 8
100
s’ + 12s+ 40
m
. ____ ____
100
7+127+40
________ lOOs^___________
7 ( 7 + 1 2 s + 4 0 )+ ir(l+ s )(1 0 0 )
l'( s )
ir,( s )
1007
7 + T 2 7 + 4 o 7 + io o ^ 7 fio o Z
B (s) = » I ( s ) - r ( s )
s*+ 1 2 7 + 4 0 7 + 1 0 0 iT s+lO O ^-lO O s’
W,(s)
^ s * + 1 2 7 - 6 0 7 + 1 0 0 i: s + 1 0 0 g ^ , .
s* + 1 2 7 + 4 o7
+
iooa : s + iooji:
‘
«„ = Im s £ (s )
When
= constant
I «„ = 0 I .
Step 8 of 8
100
(e)
l (
s^(s^+ 12s+ 40)
i .
ir ,( s )
l + k ( ' + l ) Ts “(s“+
7 T 12s+ 40)
100
7 + 1 2 7 + 4 0 7 + 1 0 0 ^ ( s +1)
J’(s) _
100
(r,(s) “ 7 + 1 2 7 + 4 0 7 + 1 0 0 i:(s + l)
E {s ) = W r,[s )-r(s )
,,
7 + 1 2 7 + 4 0 7 + io o i:(s + i)-io o
“ 7 + i2 7 + 4 o 7 + io o i:s + io o ^
Substituting K = 2,
,
^^
s*
+ 12s* + 4 0 7 + 200s + 1 0 0 „ , .
sV l2s^+ 40s^+2005+200
When Ifa(^) ~ constant
*„ = Im s fl(s )
I e. = 0 I .
^
, ,
Problem 5.33PP
Consider the plant transfer function
bs + k
C(i) =
' fiimMfi + (Af +m)te + (M + m)k]
to be put in the unity feedback loop of Fig. This is the transfer function relating the input force u(t)
and the position y(t) of mass M in the noncoilocated sensor and actuator problem. In this
problem, we will use root-locus techniques to design a controller Dc(s) so that the closed-loop
step response has a rise time of less than 0.1 sec and an overshoot of less than 10%. You may
use Matlab for any of the following questions;
(a) Approximate G(s) by assuming that m ^ 0, and let M = 1. k = 1. b = 0.1, and Dc(s) = K. Can K
be chosen to satisfy the performance specifications? Why or why not?
(b) Repeat part (a) assuming that Dc(s) = K(s + z), and show that K and z can be chosen to meet
(b) Repeat part (a) assuming that Dc(s) = K(s + z). and show that K and z can be chosen to meet
the specifications.
(c)
Repeat part (b), but with a practical controller given by the transfer function
Dc{s)=K*-
s+ p
Pick p so that the values for K and z computed in part (b) remain more or less valid.
(d)
Now suppose that the small mass m is not negligible, but is given by m = MA 0. Check to see
if the controller you designed in part (c) still meets the given specifications. If not, adjust the
controller parameters so that the specifications are met.
Figure Unity feedback system
O—
— » G (i)
I
O
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
A SK A N EX PERT
Problem 5.34PP
Consider the Type 1 system drawn in Fig. We would like to design the compensation Dc(s) to
meet the following requirements: (1) The steady-state value of y due to a constant unit
disturbance
wshould be less than V s, and
(2) the damping ratio ^ = 0.7. Using root-locus techniques,
(a)
Show that proportional control alone is not adequate.
(b)
Show that proportional-derivative control will work.
(c)
Find values of the gains kp and kD for Dc(s) = kp + kDs that meet the design specifications
with at least a 10% margin.
ic f i-ina values of me gams
Kpana kuror uc{sj = Kp+ kusmat meet me aesign specincations
with at least a 10% margin.
Figure Control system
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 5.35PP
Using a sample rate of 10 Hz, find the Dc(z) that is the discrete equivalent to your Dc(s) from
Problem using the trapezoid rule. Evaluate the time response using Simulink, and determine
whether the damping ratio requirement is met with the digital implementation. {Note: The material
to do this problem is covered in the Appendix W 4.5at www.FPE7e.com or in Chapters.)
Problem
Mixed real and complex poles. Sketch the root locus with respect to K for the equation 1 + KL(s)
= 0 and the listed choices for L(s). Be sure to give the asymptotes and the arrival and departure
angles at any complex zero or pole. After completing each hand sketch, verify your results using
Matlab. Turn in your hand sketches and the Matlab results on the same scales.
(a)
Us) =
(b)
^(*) — fZ(r4iO)(yi6s+2S)
(c) UA) —
iA\ ,
10)^1^4.25)
(*+3)(J^+4»4«)
- F(i+ iW ?+ 4l+ 5)
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 5.36PP
Consider the positioning servomechanism system shown in Fig., where
ei = Kodi, eo = Kpotdo, /Co = 10 V/rad,
T = motor torque = Kt ia,
km = /Cf= torque constant = 0.1 N m/A,
Ke = back emf constant = 0.1 Vsec
Ra = armature resistance = 10Q,
Gear ratio = 1:1,
JL + Jm = total inertia = 10-3 kg m2,
va = KA(ei - ef).
(a)
What is the range of the amplifier gain KA for which the system is stabie? Estimate the upper
limit nranhicallv u<;inn a rnnt-lncii<; nint
va = KA(ei - ef).
(a)
What is the range of the amplifier gain KA for which the system is stabie? Estimate the upper
limit graphicaily using a root-locus plot.
(b)
Choose a gain KA that gives roots at
0.7. Where are all three closedloop root locations fo
this value of KA7
Figure Positioning servomechanism
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
A SK A N EX PERT
Problem 5.37PP
We wish to design a velocity controi for a tape-drive servomechanism. The transfer function from
current l(s) to tape velocity QfsJ (in miiiimeters per miilisecond per ampere) is
Q (l)
I3 (l2 + 0.91 + 0.8)
7 w “ (* + 1)(J^+ ! . ! » + 1 ) '
We wish to design a Type 1 feedback system so that the response to a reference step satisfies
tr< 4 msec, ts < 15 msec, Mp < 0.05.
(a)
Use the integral compensator k l /s to achieve Type 1 behavior, and sketch the root locus with
respect to k l . Show on the same plot the region of acceptable pole locations corresponding to
the specifications.
(b)
Assume a proportional-integral compensator of the form kp(s + a)/s, and select the best
(b)
Assume a proportional-integral compensator of the form kp(s + a)/s, and select the best
possible values of kp and a you can find. Sketch the root-locus plot of your design, giving values
for kp and a, and the velocity constant Kv your design achieves. On your plot, indicate the
closed-loop poles with a dot (-) and include the boundary of the region of acceptable root
locations.
Step-by-step solution
step 1 of 3
a.
G (s)D (s)
_ 1 5 k ,(s “+0.9s+0.S)
s(s+ l)(s“+ U s + l)
M ,= e ^
t = ----- ^ 15ms
,, k0.381rad/msec
ii-cos’V § ) ^ .
t-= ------= = ^ ^ 4 m s
>|a^>0.840rad/m8ec|
Step 2 of 3
Step 3 of 3
d
(0 = k . ( ! 1 ^
S4*-0.58&+:j 0.588
for a.=1.2 and Kp=0.05
= 0.815 = 0.75x0.8 = 0.6
Problem 5.38PP
The normalized, scaled equations of a cart as drawn in Fig. of mass me holding an inverted
uniform pendulum of mass mp and length / with no friction are
Eq.1
0-9 =-v,
y + p0=v,
where fi =
is a mass ratio bounded by 0 < /3 < 0.75. Time is measured in terms of r =
toot where
motion y is measured in units of pendulum length as
^
C(4«c-Hllp)
y = | | . and the input is force normalized by the system weight v = . These
equations can be used to compute the transfer functions
Eq. 2-3
e
1
Eq. 2-3
e
1
r
V
J2CJ2-1)'
In this problem you are to design a control for the system by first closing a loop around the
pendulum, Eq. (2), and then, with this loop closed, closing a second loop around the cart plus
pendulum, Eq. (3). For this problem, let the mass ratio be me = 5mp.
(a) Draw a block diagram for the system with V input and both Y and 6 as outputs.
(b) Design a lead compensation Dc(.s) =
for the d loop to cancel the pole at s = -1 and
place the two remaining poles at -4 ±;4. The new control is U(s), where the force is
V(5)
=
U(s) + De(s)&(s).- Draw the root locus of the angle loop.
(c)
Compute the transfer function of the new plant from U to Y with De(s) in place.
(d)
Design a controller De(s) for the cart position with the pendulum loop closed. Draw the root
locus with respect to the gain of De(s).
(e)
Use Matlab to plot the control, cart position, and pendulum position for a unit step change in
cart position.
Figure Figure of cart pendulum
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 5.39PP
Consider the 270-ft U.S. Coast Guard cutter Tampa (902) shown in Fig. Parameter identification
based on sea-trials data (Trankle, 1987) was used to estimate the hydrodynamic coefficients in
the equations of motion. The result is that the response of the heading angle of the ship ip to
rudder angle 5 and wind changes w can be described by the second-order transfer functions
f(5)
-0.0184(j-I-0.0068)
0.2647)(5+ 0.0063) *
S is )
G^(s) =
s is +
fis )
■wW
~ s is +
0.0000064
0.2647)(s+ 0.0063) *
where
(fj = heading angle, rad,
if)r= reference heading angle, rad,
r = yaw rate, rad/sec,
if)r= reference heading angle, rad,
r = yaw rate, rad/sec,
6 = rudder angle, rad,
w = wind speed, m/sec.
(a) Determine the open-loop settling time of rfo r a step change in &
(b) In order to regulate the heading angle ifj. design a compensator that uses ifj and the
measurement provided by a yaw-rate gyroscope (that is, by ^ = /■;). The settling time of
to a
step change in ipr is specified to be less than 50 sec, and for a 5° change in heading, the
maximum allowable rudder angle deflection is specified to be less than 10°.
(c)
Check the response of the closed-loop system you designed in part (b) to a wind gust
disturbance of 10 m/sec. (Model the disturbance as a step input.) If the steady-state value of the
heading due to this wind gust is more than 0.5°, modify your design so that it meets this
specification as well.
Figure USCG cutter Tampa (902)
If Vf J f Xf M/ M
Step-by-step solution
step 1 of 10
Refer to Figure 5.67 in the textbook for the 270 ft U.S. Coast Guard cutter Tampa.
The second-order transfer function that shows the response of the heading angle is.
-0.01g4(t+0.0068)
i(s+0.2647)(s+0.0063)
The second-order transfer function that shows the wind changes is.
w ( i)
0.0000064
j(s+0.2647)(j+0.0063)
Here, ^ is the heading angle, g is the rudder angle, and w is the wind speed.
Step 2 of 10
(a)
Consider a step change in rudder angle to calculate the settling time of r .
Here, r isthe yaw rate in rad/s.
Find the transfer function of r(s)-
r
-0.018 4 (t+ 0.0068)
|_s(j+0.2647)(i+0.0063)
j(j+ 0 .2 i
-0.0184(1+0.0068)
"(j+0.2647)(i+0.0063)
Draw the step response of the function and locate the settling time for a single step change.
Use the following MATLAB code to plot the step response of the function:
num = -0.0184*[1 0.0068];
den = conv([1 0.2647],[1 0.0063]);
Gd = tf(num,den);
step(Gd)
Step 3 of 10 ^
The open-loop step response of the function r ( ^ ) is.
Figure 1
Thus, the open-loop settling time of r is |21Q si •
Step 4 of 10
(b)
Design a compensator to regulate the heading angle.
Consider that the maximum allowable rudder angle deflection is less than |0« for a 5^ change
in heading angle.
Note that the maximum deflection in rudder is possible at the initial instant. Therefore,
Substitute 10 for ^ (O ) and $ for
io = /:5
5
Consider that the settling time of ^ to a step change of
is less than 50 s.
The general expression for the step response is.
Here,
is the negative real part of the pole.
Substitute 50 for 71 to find the value of <r.
4.6
50
*0.092
Step 5 of 10
Use the following MATLAB code to find the gain and roots on the root locus;
num = -0.0184*[1 0.0068];
den = conv ([1 0],conv([1 0.2647],[1 0.0063]));
Gd = tf(num,den);
sysci = feedback(Gd,1);
rlocus(syscl)
[K,roots] = rlocfind(syscl)
The result of the code is the root locus. Select a value on the root locus to get the values for gain
and roots. Note that the value selected for gain should be less than 2.
The output of the code is.
Select a point in the graphics window
selectedjx)int =
0.1111
-
0.00021
K=
1.2605
roots =
-0.3754
0.1111
-0.0068
Step 6 of 10
The value of fc is less than the required gain value. Observe the roots to note that -0.0068 is
present to eliminate the zero in the system and -0.3754 is present to eliminate the pole in the
system. Take the pole -0.1111 for the compensator.
Hence, the transfer function of the compensator is,
G (5 ) = I.2605(5-0.1III)
Use the following MATLAB code to sketch the step response.
% r(s) is taken in the open-loop to make the overall system equal to second order othenvise take
Gd(s) with additional pole (that is 0.3754) in the compensator.
num = -0.0184*[1 0.0068];
den = conv([1 0.2647],[1 0.0063]);
sys = tf(num,den);
numi = 1.2605*[1 -0.1111];
deni =1;
sysl = tf(num1,den1);
sysT=series(sys,sys1);
sys2 = feedback(sysT,1);
step(sys2)
Step 7 of 10
The step response of the closed loop system is.
Figure 2
Observe that the settling time is less than 50 s.
Thus, the required compensator is I j - Q . m ! ] Note that the design of compensators can be done in many ways.
Step 8 of 10
(c)
10
Consider that the disturbance, W ( j ) = — •
'
Calculate the response
s
with disturbance.
Response for disturbance,
is calculated when
= 0 >so the controller <7^
j acts
as a feedback.
y (4
G A s)
» 'W
l + G .W G c ( ^ )
Use the following MATLAB code to find the response of the system for 10 m/s disturbance:
num = -0.0184*[1 0.0068];
den = conv([1 0.2647],[1 0.0063]);
Gd = tf(num,den);
numi = 1.2605*[1 -0.1111];
deni =1;
Gc = tf(num1,den1);
sysw =10;
sysci = sysw*feedback (Gd,Gc);
step(syscl)
Step 9 of 10 ^
The step response of the system for 10 m/s disturbance is.
Step 10 of 10
Use the following command to find the steady-state error of the system:
Kv= 10*1.2605*-0.0184*0.0068*-0.1111/0.2647/0.0063/0.0068;
ess = 1/Kv
The output of the code is,
ess =
0.0647
Therefore, the steady state error is very much less than 0.5. Therefore, there is no change
required in the design.
Problem 5.40PP
Golden Nugget Airlines has opened a free bar in the tail of their airplanes in an attempt to lure
customers. In order to automatically adjust for the sudden weight shift due to passengers rushing
to the bar when it first opens, the airline is mechanizing a pitch-attitude autopilot. Figure shows
the block diagram of the proposed arrangement.We will model the passenger moment as a step
disturbance Mp(s) = MO/s, with a maximum expected value for MO of 0.6.
(a) What value of K is required to keep the steady-state error in 6 to less than 0.02 rad
1“)?
(Assume the system is stable.)
(b) Draw a root locus with respect to K.
(c)
Based on your root locus, what is the value of K when the system becomes unstable?
(c)
Based on your root locus, what is the value of K when the system becomes unstable?
(d)
Suppose the value of K required for acceptable steady-state behavior is 600. Show that this
value yields an unstable system with roots at
s = -2.9,-13.5,+1.2±6.6).
(e)
You are given a black box with rate gyro written on the side and told that, when installed, it
provides a perfect measure of 0, with output K T ' 6. Assume that K = 600 as in part (d) and draw
a block diagram indicating how you would incorporate the rate gyro into the autopilot. (Include
transfer functions in boxes.)
(f)
For the rate gyro in part (e), sketch a root locus with respect to K T .
(g)
What is the maximum damping factor of the complex roots obtainable with the configuration
part (e)?
(h)
(i)
What is the value of KT for part (g)?
Suppose you are not satisfied with the steady-state errors and damping ratio of the system
with a rate gyro in parts (e) through (h). Discuss the advantages and disadvantages of adding an
integral term and extra lead networks in the control law. Support your comments using Matlab or
with rough root-locus sketches.
Figure Golden Nugget Airlines autopilot
Step-by-step solution
step 1 of 10
(a)
Refer to the block diagram in Figure 5.68 in the text book.
The plant transfer function of the system without disturbance is.
V
AT(5 + 3)
<7(j) » — ------- ' '----------- 7
' ' * ( s + 10)( j ’ + 4 j + 5)
Determine the steady state error of the system.
£ s l im 5 G ( f )
A:( j + 3 )
s lim j
i ( i + 1 0 ) ( j* + 4 5 + 5 )
^ ( j+ 3 )
s lim
-»“ ( j + I0 )(5 “ + 45 + 5)
50
Equate the steady-state error to 0.02.
* 0 .3 3
Thus, the value of K is
Step 2 of 10
(b)
The transfer function of the system without disturbance is.
K {s + y )
j ( 5 + 1 0 )(i* +
+ 5)
5 ( j + I0 )( j ^ + 4j + 5)
The characteristic equation of the system is.
1+^i ( j + 10)
r» 0
+ 4 s + 5)
Enter the following code in MATLAB to draw the root locus.
» num=[1 3]:
» den=conv([1 10 0],[1 4 5]);
» sys=tf(num,den):
» rlocus(sys)
The following is the MATLAB output:
Step 3 of 10
(c)
From Figure 1, observe that the value of A for which the system roots move to right-half is 150.
Thus, the value of A is h sol-
Step 4 of 10
(d )
The value of K is 600.
The transfer function of the system is.
y ( j + 3)
T (s)
j ( i + 1 0 ) ( j ^ + 4 i + 5 ) + 3 r ( i + 3)
Substitute 600 for K
■pi^\ _________
6 00(^+ 3)__________
' ^ " s ( j + 1 0 ) ( s ’ + 4 j + 5) + 60 0 (i + 3)
600(^ + 3)
'
s (j
+ 1 0 )( s * + 4 j + 5) + 600( j + 3)
________ 600(^4-3)________
** + 14j ’ + 45 j ’ +650 s + 1800
Enter the following code in MATLAB to find the roots of the system.
» C E = [1 14 45 650 1800];
» roots(CE)
ans =
-13.5014+ 0.0000i
1.2183+ 6.6284i
1.2183-6.6284i
-2.9352 + O.OOOOi
The roots of the system are at |5 = -2 .9 ,-1 3 .5 ,+ l.2 ± 6 .6 y | •
As the complex roots lie to the right-half of the s-plane, the system is unstable.
Step 5 of 10
(e)
The value of K is 600.
Connect the Rate gyro as feedback at the input.
Determine the loop transfer function.
K (s + 3 )
(s + 1 0 ) ( V + 4 i+ 5 )
<?w =
e)
1,
(* + 1 0 )(j* + 4 j + 5)
Substitute 600 for K
600(»4-3)
(s + 1 0 ) ( s '+ 4 i+ 5 )
a { s )-
0
N ._ M £ ± 3 )
(* + 1 0 )(j* + 4 j + 5)
Draw the block diagram of the system by connecting the block.
ft
Figure 2
Step 6 of 10
(f)
The loop transfer function is.
600(»4-3)
(s + 1 0 ) ( s '+ 4 i+ 5 )
<?w =
0
N ._ M £ ± 3 )
(* + 1 0 )(j* + 4 j + 5)
The characteristic equation is,
(. +3)
=
< + ^ r7
’^( j + 1 0 )(5 *+ 4 j + 5)
0
Enter the following code in MATLAB to draw root locus.
» num=[1 3]:
»den=conv([1 10],[1 4 5]);
» sys=tf(num,den):
» rlocus(sys)
Step 7 of 10
The following is the MATLAB output for root locus with respect to K j. '■
Root Locus
[System:sy$
Gain: 0
Pole:-2+ li
Dancing: 0.894
Overshoot (%): 0.187
(rad/s):2.24
10
.
V.
-
•
•10
-15.
.12
*10
-6
.4
.2
R eal A xis (seccmds*^)
Figure 3
Thus, the root locus is shown in Figure 3.
Step 8 of 10
(g)
Observe from Figure 3 that the maximum damping factor of the complex roots is
|Q.894|•
Step 9 of 10
(h )
Observe from Figure 3 that for maximum damping factor, the value of
is [ ^ .
Step 10 of 10
(i)
Adding lead compensation to the design lowers damping.
Consider the following lead network;
„
J+2
^ 1 + 20
The plant loop transfer function is,
A:(h -2)(» + 3)
<?w =
’ i ( i + 10)(s + 2 0 )(j’ + 4 i + 5)
Enter the following code in MATLAB to draw root locus.
» num=conv([1 2],[1 3]);
» den=conv([1 30 200],[1 4 5 0]);
» sys=tf(num,den):
» rlocus(sys)
The following is the MATLAB output:
Observe from Figure 4 that, the range of K for stability is reduced with the introduction of extra
lead network.
Problem 5.41 PP
Consider the instrument servomechanism with the parameters given in Fig. For each of the
foliowing cases, draw a root locus with respect to the parameter K, and indicate the location of
the roots corresponding to your final design:
(a)
Lead network: Let
s+z
De(.s)=Ks+ p
=
£=6.
Select z and K so that the roots nearest the origin (the dominant roots) yield
^2
f> O A
(b)
Output-velocity (tachometer) feedback: Let
H(s) = 1 + /Crs and Dc(s) = K.
(b) Output-velocity (tachometer) feedback: Let
H(s) = 1 + /Crs and Dc(s) = K.
Select K T and K so that the dominant roots are in the same location as those of part (a).
Compute Kv. If you can, give a physical reason explaining the reduction in Kv when output
derivative feedback is used.
(c)
Lag network: Let
and
J+1
Deis) = Ks+ p
Using proportional control, is it possible to obtain a Kv = 12 at
0.4? Select K and p so that the
dominant roots correspond to the proportionalcontrol case but with Kv = 100 rather than Kv = 12.
Figure Control system
Step-by-step solution
g
W =t t
s(s“+51s+550)
Step 2 of 4
a.
let
z=7
p=42 ,
K=55000
D ( s ) = f - ^ lx 5 5 0 0 0
|8a=-11.5+j26.5|
Step 3 of 4
b.
H (s)=l+KjS
for |Kt=0.04|
|K=18000 I, roots are at the samepositioo.
Step 4 of 4
c.
^^
s4p
-= ^ = 1 0 0 ^
550p
let p=2.
K=55000p
Then, |K=110000|
Problem 5.42PP
Plot the loci for the 0* locus or negative K for each of the following:
(a) The examples given in Problem 1
(b) The examples given in Problem 2
(c) The examples given in Problem 3
(d) The examples given in Problem 4
(e) The examples given in Problem 5
(e) The examples given in Problem 5
(f) The examples given in Problem 6
Problem 1
For the characteristic equation
1+
-
j 2(j
K
■ = 0,
+ 1 )( j + 5 ) ■
(a) Draw the real-axis segments of the corresponding root locus.
(b) Sketch the asymptotes of the locus for K -
(c) Sketch the locus
(d) Verify your sketch with a Matlab plot.
Problem 2
Real poles and zeros. Sketch the root locus with respect to K for the equation 1+ KL(s) = 0 and
the listed choices for L(s). Be sure to give the asymptotes, and the arrival and departure angles
at any complex zero or pole. After completing each hand sketch, verify your results using Matlab.
Turn in your hand sketches and the Matlab results on the same scales.
(a) Us) - j(,+i)(,+5)(,+io)
(b) K s) =
(c) Us) —
fd) K s) =
f+ray
Problem 3
Complex poles and zeros. Sketch the root locus with respect to K for the equation 1 + KL(s) = 0
and the listed choices for L(s). Be sure to give the asymptotes and the arrival and departure
angles at any complex zero or pole. After completing each hand sketch, verify your results using
Matlab. Turn in your hand sketches and the Matlab results on the same scales.
W «*> = ?T 5 T i5
(b) t ( j ) _
Problem 4
Multiple poles at the origin. Sketch the root locus with respect to K for the equation 1 + KL(s) = 0
and the listed choices for L(s). Be sure to give the asymptotes and the arrival and departure
angles at any complex zero or pole. After completing each hand sketch, verify your results using
Matlab. Turn in your hand sketches and the Matlab results on the same scales.
(b) W
=
(<=)
=
(t)
= ^
Problem 5
Mixed real and complex poles. Sketch the root locus with respect to K for the equation 1 + KL(s)
= 0 and the listed choices for L(s). Be sure to give the asymptotes and the arrival and departure
angles at any complex zero or pole. After completing each hand sketch, verify your results using
Matlab. Turn in your hand sketches and the Matlab results on the same scales.
(a) L(J) =
(b) Ufl) — j2(,+io)(y^+25)
(c) Us) —
10)^1^4.25)
(* + 3 )(J^+ 4 » 4 « )
- F (i+ iw ? + 4 l+ 5 )
Problem 6
RHP and zeros. Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the
listed choices for L(s). Be sure to give the asymptotes and the arrival and departure angles at
any complex zero or pole. After completing each hand sketch, verify your results using Matlab.
Turn in your hand sketches and the Matlab results on the same scales.
(a) Us) = ^ io ^ | . | ;the model for a case of magnetic levitation with lead compensation.
^ 2^ ; t h e magnetic levitation system with integral control and lead
(b) Us) =
compensation.
(c) Us) = ^
(d) l,( s ) _____ What Is the largest value that can be obtained for the damping
5(j-f-20)^ —2i+2)
ratio of the stable complex roots on this locus?
(*>
-
( j- l) t( jW + 3 ]
S te p -b y -s te p s o lu tio n
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
5.43PP
Problem 5.43PP
Suppose you are given the plant
^ ^
a®+ (I +
+ (I + o)*
where a is a system parameter that is subject to variations. Use both positive and negative rootlocus methods to determine what variations in a can be tolerated before instability occurs.
S te p -b y -s te p s o lu tio n
step 1 of 7
Step 1 of 7
Consider the following given function.
+ (1 + or) j + ( I + t f )
1
^ * + j + l+ o rj+ o r
1
# * + j+ l+ o r ( j+ l)
Consider the characteristics equation for the given function.
5+ 1
1 + o rj + i+ i
(1)
To find zeros put numerator
Thus, the one zero is - 1 .
To find poles put denominator D(s) = 0 •
j '+ J + l- O
(2)
The roots of the equation (2) are -0 .5 + 0.866y'and -0 .5 -0 .8 6 6 y Thus, the two poles are -0 .5 + 0.866y’ and -0 .5 -0 .8 6 6 y
Step 2 of 7 ^
Consider the formula for the asymptotes.
n -m
(sumoffinitepoles)-(sumoffinitezeros)
(numberoffinitepoles)-(nund>eroffinitezeros)
(-0.5+0.866y-0.5- 0.866y)-(-I)
(2 )-0 )
=0
Thus, the asymptote is
.
Consider the formula for the angle of asymptotes.
^ l80°+36(y*(/-l)
n —m
Where,
Number of poles is n
Number of zeros is m
Substitute 1 for /, 2 for n and 1 for m in equation (3).
180°+360o(l-l)
2-1
=180»
Substitute 2 for /, 3 for n and 1 for m in equation (3).
_ 180^+360^(2-1)
2-1
*540*
Thus, the angle of asymptotes are | i 3Qo|and |54Q^|.
Step 3 of 7
Procedure to draw the positive root locus plot:
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes from centroid at an angle of
I go* and 540«.
• Draw the root locus.
The root positive locus plot is shown in Figure 1.
RootLocus
Figure 1
Hence, the positive root locus is plotted for the given transfer function and it is shown in Figure 1.
Step 4 of 7 ^
Consider the characteristics equation for the given function.
1
5+ 1
_
l + a -3----- -= 0
Multiply by - 1 in the above transfer function and rearrange the equation for negative root locus
form.
i+ „ (^ K i± l)= o
m
S -t-i + l
To find zeros put numerator N {s ) = 0
Thus, the one zero is - 1 .
To find poles put denominator D{s) = 0 •
»*+S + I = 0 ...... (5)
The roots of the equation (5) are -0 .5 + 0.866y'and -0 .5 -0 .8 6 6 y Thus, the two poles are -0 .5 + 0.866yand -0 .5 -0 .8 6 6 ^ -
Step 5 of 7
Consider the formula for the asymptotes.
n -m
(sumoffinitepoIe$)-(sumoffinitezeros)
(numberoffinitepoles)-(nunU>eroffinitezeros)
(-0.5+0.866j - 0.5- 0.866y)-(-!)
(2)-0)
=0
Thus, the asymptote is
.
Consider the formula for the angle of asymptotes.
* =
«
: ! ) ...... (6)
n -m
Where,
Number of poles is n
Number of zeros is m
/ = l, 2 , . . ji - m
Substitute 1 for /, 2 for n and 1 for m in equation (3).
360“(1-1)
2-1
= 0»
Substitute 2 for /, 3 for n and 1 for m in equation (3).
360®(2-l)
2-1
=360®
Thus, the angle of asymptotes are [ ^ a n d |36Q^|.
Step 6 of 7
Procedure to draw the negative root locus plot:
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes from centroid at an angle of
0»and 360®
• Draw the root locus.
The negative root locus plot is shown in Figure 2.
RootLocus
step 7 of 7
Thus, the negative root locus is plotted for the given transfer function and it is shown in Figure 2.
From Figure 1 and Figure 2, the given system is stable for all ^
Hence, the system is stable for all l a > - l l
.
Problem 5.44PP
Consider the system in Fig.
(a) Use Routh’s criterion to determine the regions in the /C1, /<2 plane for which the system is
(b) Use ritool to verify your answer to part (a).
Figure Feedback system
. ( . + IX .+ IX 5 )
Step-by-step solution
step 1 of 1
SA: 760
SR: 3976
Sketch the given figure.
1
4 ( 4 + l ) ( j + 0 .5 )
We know that
= ir^and k^ =
and the characteristic equation is
s* + 1.5s’ + 0.5s^ + i , s + * ,= 0.
The Routh table is
/
1
0.5
0
1.5
1.5x0.5-ifc_
^
1.5
'
0
-\.5 k ,
s'
For the Routh table, we get
Jt^>0
1.5x0.5-yfc,
—
1.5x0.5-A ^ > 0
k^ <0.75
So,
\K, <0.75
Again we have,
(l.5 x 0 .5 - * ,) t,- 2 .2 5 i,> 0
k ‘, + 2.25k,-0.15k, <0
4 * J - 3 i , + 9 * ,< 0
|4 i!:i^ - 3 i!:i+ 9 j:ig , < o|
This represents aparabola in ^k^, ib,Jplane.
w
The region of stability is the are a under the parabola and above the k^ axis as
shown in the figure below.
Thus, the regions in the (iT^,
determined.
plane for which the system is stable is
Problem 5.45PP
The block diagram of a positioning servomechanism is shown in Fig.
(a) Sketch the root locus with respect to K when no tachometer feedback is present KT = 0.
(b) Indicate the root locations corresponding to /< = 16 on the locus of part (a). For these
locations, estimate the transient-response parameters tr. Mp. and ts. Compare your estimates to
measurements obtained using the step command in Matlab.
(c)
For K =
draw the root locus with respect to K T .
(d)
For K = 16 and with KT set so that Mp = 0.05 = 0.707). estimate fr and fs. Compare your
tn th p
(d)
actual v p Iii p p nf tranrl /.<; nhtpinprl ii.<;inn Mntlph
For K = 16 and with KT set so that Mp = 0.05 = 0.707). estimate fr and fs. Compare your
estimates to the actual values of tr and ts obtained using Matiab.
(e)
For the values of K and KT in part (d). what is the velocity constant Kv of this system?
Figure Control system
Step-by-step solution
S te p i of 15
Given block diagram o f a positioning servo mechanism.
Step 2 of 15
The root locus with respect to . ^ v ^ n no tachometer feedback is present is
Root Loois
Step 3 of 15
(b)
The root locations corresponding to .^ = 16 on the locus o f part (a) is
Step 4 of 15
'With iT s 16 p&om the root locus, the transient response parameters are
The overshoot is
Af, =44.4% .
The dancing ratio is
^ = 0 .2 5 .
The frequency is
at^=A rad/s.
The rise time is
2 .16^+ 0.6
ir = -
2.16x0.25+0.6
= 0.285 8
C = 0.285 s|
The settling time is
4.6
=-
4.6
■0.25x4
. = 4.6 s
Step 5 of 15
The MATLAB code to plot the step response of the system
»
»
»
num = 16;
d e n = [ l 2 16] ;
s y s = t f (mun, d e n )
Step 6 of 15
T ra n s fe r fu n c tio n :
16
3 ^2 + 2 s + 16
»
»
»
»
»
»
t= D : 0. 0 0 0 1 :5 ;
y = s te p ( s y s , t ) ;
p lo t ( t,y , ’b * )
g rid
x la b e l t
y la b e l y
Step 7 of 15 •
The step response ofthe system using MATLAB is
Step 8 of 15
From the step response.
The overshoot is
Af^ = (1 .4 4 -l)x l0 0
i / =44.4%
The rise time is
/^ = 0.4322-0.1173
= 0.3149s
The settling time is
It. =4.783 s I
Hence the parameters almost agree with that obtained from root locus.
Step 9 of 15
(c)
For
»
»
»
^ = 1 6 th e root locus with respect to ^j<is
num = 16;
d e n = [ l 3 0] ;
s y s = t f (niuttf d e n )
T ra n s fe r fu n c tio n :
16
»
»
rlo c u s (s y s )
g rid
Step 10 of 15
The transfer function with iT = 16 and with tacho feedback is
r{ \
^ ® ^ ° s ^ + (2+JC-..)s+16
Given,
ilf, = 0.05
C = 0.707
Step 11 of 15
So, we have,
a, = ^
= 4 rad/s ec
Also,
2Cat^ = 2 + K j.
2x0.707x4 = 2+£'j.
K j. = 3 .6 5 6
K j. «
3.7
Now, the transfer function is
16
T {s) =
s" + (2 + 3 .7 )s+ 1 6
16
s" + 5.7s+16
Step 12 of 15
The rise time is
_ 2 .1 6 ^+ 0 .6
2.16x0.707+0.6
= 0.532 s
J = 0.532 s|
The settling time is
4.6
C<On
4.6
'0 .7 0 7 x 4
= 1.63 s|
Step 13 of 15
The step response using MATLAB is
Step 14 of 15
From the above response.
The rise time is
i , = 0.6677-0.1265
L = 0.5412 s
The settling time is
h =1.683 s|
Hence the results agree with that obtained &om the estimatioa
Step 15 of 15
w
The velo city constant o f the system is
= lim s
(__
l_ s ( s + 2 + i: r ) J
K
~2-\-K j.
16
“ 3.7 + 2
= 2.81
Thus, we get
|J:, = 2.81|
Problem 5.46PP
Consider themechanical systemshown in Fig., where g and aO are gains. The feedback path
containing gs controls the amount of rate feedback. For a fixed value of aO, adjusting g
corresponds to varying the location of a zero in the s-plane.
(a) With g = 0 and r = 1. find a value for aO such that the poles are complex.
(b) Fix aO at this value, and construct a root locus that demonstrates the effect of varying g.
Figure Control system
Step-by-step solution
Step 1 of 4
L (s) = =>11( g ^ ^ )
s ( ts+1)
Step 2 of 4
a.
■ts^+s+a^s+a,=0
s^+s+a^=0
for
g=0 and x=l.
For getting complex conjuguate roots,
l< 4a^
““4
Step 3 of 4
4s^+4s+l+gs=0
H - ^ = 0
( 2 ^
Step 4 of 4
Vu
Root Locus
X
-
1/2
>
-0 -
lu iu iu u e ii.
5 .4 7 P P
Step-by-step solution
step 1 of 1
Step-by-step solution
step 1 of
1
G (s)— 5s
For the above transfer function, as3miptoti c angles are 6 0 '. 180', 300',
Two roots will always be in right half plane if pole is not cancelled
Hence,plant G (s )= -^ cannot be made unconditionally stable if pole
s
cancellation is forbidden
Problem 5.48PP
Prove that the plant G(s) = 1/s3 cannot be made unconditionally stable if pole cancellation is
forbidden.
luiuiuueii.
Step-by-step solution
ste p 1 of 1
G (s)— 5s
For the above transfer function, asymptoti c angles are 6 0 '. 180', 300',
Two roots will always be in right half plane if pole is not cancelled
Hence,plant G (s )= -^ cannot be made unconditionally stable if pole
s
cancellation is forbidden
Problem 5.49PP
For the equation 1 + KG(s). where
J (J + P )[(I+ 1 )2 + 4 ]’
use Matlab to examine the root locus as a function of/<for p in the range from p = 1 to p = 10,
making sure to include the point p = 2.
Step-by-step solution
step 1 of 2
Step 1 of 2
G (s)= -
A t p=2. Two roots are asymptotisc to imaginary axis
Step 2 of 2
Problem 6.01 PP
(a) Show that oO in Eq. (4), with A = Uo and (jjo = oj, is
«o = [ g W —
= - V o G ( - jo ) ) ^ ,
and
(b) By assuming the output can be written as
y ( 0 = O b*->“ '
Hariwo Pnc
=
derive Eqs. (1)-(3)Eqs. 1
y(») = AM cos(aib> + ^),
Eqs. 2
M = |G (7 < h ,)| = |C(j ) | ^
= y (Re[GC/<o„)]}2 + {Im[G(yii^)])2,
Eqs. 3
Eqs. 4
S -p \
M -p i
M -P a
S + J te t
S -J to ,
S te p -b y -s te p s o lu tio n
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 6.02PP
(a) Calculate the magnitude and phase of
1
G is) = s+ lO
by hand foroj = 1, 2, 5.10. 20, 50. and 100 rad/sec.
(b) Sketch the asymptotes forGfsj according to the Bode plot rules, and compare these with your
computed results from part (a).
S te p -b y -s te p s o lu tio n
S te p -b y -s te p s o lu tio n
step 1 of 2
(a).
G (s)= —
^ ^ s+10
M i^ itu d e = |G(jo[))|=
Vffi^+ioO
Phas e = Z.G (joo) =-tan'^
s.no
j
ai(rad/s)
Magnitude
Phase
1
1
0.0995
-5.7*
2
2
0.0980
-11.3"
-26.6"
3
5
0.0894
A
10
0.07071
-45"
5
20
0.0447
-63.4"
6
50
3.846x10-^
-78.7"
7
100
9.95X10-’
-84.3"
Step 2 of 2
Problem 6.03PP
Sketch the asymptotes of the Bode plot magnitude and phase for each of the following open-loop
transfer functions. After completing the hand sketches, verily your result using Matlab. Turn in
your hand sketches and the Matlab results on the same scales.
(a) L(s) =
2000
j(s+ 20 0)
(b) L(s) = l(O .I> +100
l)(O J s + l)
1
(c) L ( i ) = 5 (s + T 7 ^0
7 E r+ T y
I
tdl L(s) =
'"
(d) L(s) =
■
1
(T P T p tT P W
(e) L (I) =
m +4)
i( J + l) ( J + l( I0 )
(f) L(s) =
I000(<-1-0.1)
i ( j + l) ( i+ 8 ) ^
(g) L(s) = i ((J-|-5)(J-|-10)
j+ l) ( j+ l0 0 )
10)
(h) L(s) = (s + 4I0j (5-|0 )(j+ 3 0 0 )
(i)
L(s) =
5
S te p -b y -s te p s o lu tio n
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 6.04PP
Real poles and zeros. Sketch the asymptotes of the Bode plot magnitude and phase for each of
the listed open-loop transfer functions. After completing the hand sketches, verify your result
using Matlab. Turn in your hand sketches and the Matiab results on the same scales.
(a)
= iii+ T T § T W + T D 5
"
(d)
- KJ-l-l)(J+S(a+10)
L (s ) = _tJ + 2 )(a + 4 )
Step-by-step solution
step 1 of 25
L (a )-
a(j +lXa-l-SXa+10)
t ( /<») -------------------- i ----------------’ y-fflOaH-lX/ai+SX/ffl+lO)
0.02
>
y < » ( / < » + l) [ ^ + l) g + l)
Breakorcomerfiequencies: ai| = Irad/sec
0^=5 rad/sec
ai^=10lad/sec
Step 2 of 25
The magnitude of L(ja>) in dB is,
k>g|£.(yV»)| = 2 0 Io g |0 .0 2 |-2 0 Io g |/< ti|-2 0 Io g |y < » + l|-2 0 lo g r ^ + I - 2 0 l o g ^ + I
=20log(0.02)-20log(<»)-20log(Vl+<»’]-201og|^l+^yj
2 0 > o g [f(^ ]
step 3 of 25
Follow the steps to draw the magnitude plot.
(i) The constant term 0.02 causes an increase in magnitude of -33.97 dB.
-20dB/decadeat=1 lad/sec■ ^Idpe changes from .20dPfdfradfto-40dB/decadedue to the
presence of (ya>+l) in the denominator.
(N) The Initial low frequency slope due to pole at the ohgin is
(ill) At
(iv)
At a r = S rad/sec >tde slope changes from -40 dB/decade t o -60 dB /decade due to the
presence of (y'ffl+5) in the denominator.
(v )
At dt=10rad/sec.dteslopechangesfrom -60 dB/decade t o -80 dB /decade due to the
presence of
(yat+lO) in the denominator.
Step 4 of 25
Consider the phase values.
oXrad/sec)
*
0.1
-97.428“
1
-152“
-240.26°
5
10
-282.T
100
-350.8°
IK
-359.13
Step 5 of 25
Draw the magnitude and phase plots as shown in figure 1.
Bodeouopom
Step 6 of 25
Execute the following MATtAB code;
num=1;
den=[1 16 65 50 0];
sys=tf(num,den);
bode(sys)
Step 7 of 25
Obtain the magnitude and phase plots as shown In figure 2.
Step 8 of 25
(b )
i(s) =
i(j+1X4+5X4+10)
O'ar+2)
L(jd,) -
yaX/aH-l)0'a>+S)(/ar+10)
0 .0 4 ^ ;^ + lj
i(ya»)=
^‘^'^'>(f+')(f+')
B reak o r c o m e r fiequencies:
cs, = I rad/sec
a>2 -2 rad/sec
ai,= S rad/sec
a i,= 1 0 rad/sec
step 9 of 25 ^
Follow the steps to draw the magnitude plot.
(i) The constant term 0.04 causes an increase in magnitude of -28 dB.
(ii) The Initial low frequency slope due to pole at the ohgin is -20dB/dccade (Hi) At u > = 1 rad/sec >Ibe slope changes from -20 dB/decade to -40 dB /decade due to the
presence of ( y ® + i) in the denominator.
(iv)
At ^ = 2 rad/sec >lbe slope changes from -40 dB/decade t o -20 dB /decade due to the
presence of (y<u+2) in the numerator.
(v )
At u r = 5 rad/sec >Ibe slope changes from -20 dB/decade t o -40 dB /decade due to the
presence of (ya>+ 5) in the denominator.
(vi)
At ^ = 1 0 rad/sec >lbe slope changes from -40 dB/decade t o -60 dB /decade due to the
presence of (yVu+lO) in the denominator.
Step 10 of 25
Consider the phase values.
oXrad/sec)
*
0.1
-94.56°
1
-125.44°
5
-172°
10
-204°
100
-261.94°
IK
-269.24
Step 11 of 25
Draw the magnitude and phase plots as shown in figure 3.
Bodeefoonn
Step 12 of 25
Execute the following MATtAB code;
num=[1 2];
den=[1 16 65 50 0];
sys=tf(num,den);
bode(sys)
Step 13 of 25 A
Obtain the magnitude and phase plots as shown In figure 4.
Step 14 of 25
(c )
L U ) _____
' ’
4(4 + 1X4 + 5X4 + 10)
(/ar+2X/flH6)
L(ja>) -
yaX/oH-1)(/ru+5)(/ru+10)
B reak o r c o m e r fiequencies:
to, * 1 rad/sec
a >,=2 rad/sec
<o,=5 rad/sec
0)4 -6 rad/sec
0 , - 1 0 rad/sec
step 15 of 25
Follow the steps to draw the magnitude plot.
(i) The constant term 0.24 causes an increase in magnitude of -12.4 dB.
(ii) The Initial low frequency slope due to pole at the ohgin is
intersects the 0 dB line at
(Hi) At
-20dB/decade- / ^ d this
o)=l rad/sec-
u)=1rad/sec>Ibe slope changes from -20dB/decadeto-40dB/decadedue
(yV»+l) in the denominator.
at=2 rad/sec, the slope changes from .40 dB/dfcadfto-20dB/decadedue
presence of
(iv) At
presence of (y ru + 2 ) in the numerator.
(v )
At <u= 5 rad/sec. tbe slope changes from -20 dB/decade t o -40 dB /decade due
presence of
(vi)
(ya)+5) in the denominator.
At 4u = 5 lad/seC'lbe slope changes from -40 dB/decade t o -20 dB /decade due
presence of ^yu )+ 6 ) in the numerator.
rad/sec. the slope changes from -20dB/decadeto-40dB/decadedi
(yVu+lO) in the denominator.
(vii) At a i= l0
presence of
Step 16 of 25
Consider the phase values.
<o(rad/sec)
*
0.1
-93.6°
1
-115.97°
5
-132.2°
10
-145°
100
-175.37°
IK
-179.58°
Step 17 of 25
Draw the magnitude and phase plots as shown in figure 5.
B oeetsogram
Step 18 of 25
Execute the following MATtAB code;
num=[1 8 12];
den=[1 16 65 50 0];
sys=tf(num.den);
bode(sys)
Step 19 of 25
Obtain the magnitude and phase plots as shown In figure 6.
Figure6
step 20 of 25
(d)
i ( j ) = -----(4 -f 2 )(4 + 4)----' ’ 4(4 + 1X4 + 5X4 + 10)
(/<ir+2)()ar+4)
L (y ® ) =
,/a)(/a)+1)(/ru+5)(/a)+10)
B reak o r c o m e r fiequencies:
ra, = 1 rad/sec
atj=2 rad/sec
CD,= 4 rad/sec
0)4=5 rad/sec
0 , - 1 0 rad/sec
step 21 of 25 A
Follow the steps to draw the magnitude plot.
(i) The constant term 0.24 causes an increase in magnitude of -12.4 dB.
(ii) The Initial low frequency slope due to pole at the ohgin is -20dB/decade - /
intersects the 0 dB line at o ) = l ra d /s e c (ill) At 4u = I ra d /s e c . Ibe slope changes from .2 0 dB/decade to -40 dB/deca
presence of (yV »+l) in the denominator.
(iv) At 0) = 2 rad /se c , the slope changes from .40 dB/decade to -20 dB/deca
presence of (y u )+ 2 ) in the numerator.
(v) At <u= 4 rad /se c . tbe slope changes from -20 dB/decade to 0 dB/decade
presence of (y ru + 4 ) in the numerator.
(vi)
At n , = 5 lad /se c. the slope changes from 0 dB/decade t o -20 dB/decad<
presence of (y a )+ 5 ) in the denominator.
(vii) At rp s 10 rad/sec. the slope changes from -20 dB/decade to -40 dB/de<
presence of (yru+ 10) in the denominator.
Step 22 of 25
Consider the phase values.
<o(rad/sec)
0.1
1
5
*
-92.16°
-111.4°
-120.6°
10
-135.8°
100
-174.23°
IK
-179.46°
Step 23 of 25
Draw the magnitude and phase plots as shown in figure 7.
Bodsooponi
Step 24 of 25
Execute the following MATtAB code;
num=[1 10 24];
den=[1 16 65 50 0];
sys=tf(num.den);
bode(sys)
Step 25 of 25 ^
Obtain the magnitude and phase plots as shown In figure 8.
B o d e D ia g ra m
Problem 6.05PP
Complex poles and zeros. Sketch the asymptotes of the Bode plot magnitude and phase for each
of the listed open-loop transfer functions, and approximate the transition at the second-order
break point, based on the value of the damping ratio. After completing the hand sketches, verify
your result using Matlab. Turn in your hand sketches and the Matlab results on the same scales.
i (j 2 + L + 1 0 )
(e)
L(S) =
+ 1)
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 6.06PP
Multiple poles at the origin. Sketch the asymptotes of the Bode plot magnitude and phase for
each of the listed open-loop transfer functions. After completing the hand sketches, verily your
result with Matiab. Turn in your hand sketches and the Matlab results on the same scales.
(a)
U s) -
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 6.07PP
Mixed real and complex poles. Sketch the asymptotes of the Bode plot magnitude and phase for
each of the listed open-loop transfer functions. Embellish the asymptote plots with a rough
estimate of the transitions for each break point. After completing the hand sketches, verify your
result with Matlab. Turn in your hand sketches and the Matlab results on the same scales..
i (j
+ 1 oW + 2 j + 2 )
^ ( j + 1 0 )(J + 6 i+ 2 5 )
(c) r ( x \ =
______
^ ( j+ 1 0 )(j^ + 6 s + 2 5 )
(c) r ( x \ =
____ f’ + p ^ ______
^ ( j+ 1 0 )(j^ + 6 s + 2 5 )
' '
j ^( j + I0 ) ( s^ + 4 j +85 )
(e) U s ) =
+
ji( l+ 2 M l+ 3 )
S te p -b y -s te p s o lu tio n
Step 1 of 20
(a).
(s+2)
L (s) =
s(r+10)(s’+2s-t2)
(Jm+2)
L ( J cd) =
J(D(Jo&t-l 0) (-(D^-t-2J(Brf2)
U
L(Jm) =
J
■ fe)’
+ Jo + l
= ->J2 rad/sec
Break or comer frequenci es:
gq>
2=2 rad/sec
is^~10 rad/sec
Step 2 of 20
Magnitude plot:
(a) The constant term '0.1' causes an increase in magnitude of 20 log 0.1—-20 dB.
(b) The initial low frequency slope due to pole at tiie origin is -20dB/decade,
andttiis slope intersects the OdB line at 0) —Irad/sec
(c) A t o = J2 rad/sec, the slope changes from -20dB /decade to -60dB /decade
due to presence of
+Ja5rl-1 in die denominator.
■teJ
Since 2^gd^ = 2 and
cd^ = - ^ ,
I.-. ; = 0.707]
(d)
A t ixi= 2 rad/sec, the slope changesfrom -60dB/decadeto -40dB /decade
(Jot
\
due to presence of I “^■*'11
erator.
(e) At iX)= 10 rad/sec, the slope changes from -40dB /decade to -60dB /decade
(J<x> ^
due to the presence of I — +11 in the denominator.
Step 3 of 20
Phase plot:
0(rad/sec)
0.1
-93.42°
1
-132.58°
2
7.13°
5
-24.86°
10
44.77°
100
-84.29°
Step 4 of 20 ^
BooeoKQion
Step 5 of 20
W-
(s+2)
L (0 =
s^(s+10)(s^+6s+25)
(Jarf2)
L(Jco) =
-<D^(Jco+1 0 ) ( - o ^ + 6 J cd+ 2 5 )
0 .0 0 8 ^ ^ |+ lj
L(Jcj) =
24JfiH-l
Break or comer frequencies:
0 ^ = 2 rad/sec
0)2=5 rad/sec
a>3=10 rad/sec
Step 6 of 20
Magnitude plot:
(a) The constant term *0.008' causes an increase in m s ^ t u d e of 20 log 0.008= -42 dB.
(b) The initial low frequency slope due to the presence o f two poles at the origin is -40
dB/ decade. And this asyrnptote intersects the OdB line at ® = Irad/sec
(c) At CD= 2 rad/sec, the slope changes from -40dB /decade to -20dB /decade
{Ja, \
due to presence of I — +11 in the numerator.
At Q0= 5 rad/sec, the slope changes from -20dB /decade to -60dB /decade
{
"I y j +0.24JCO+1 in the denominator.
due to presence of
Since 2 1 ^ = 6 and cc^=5 ,
M = 0 .6 |
(e)
At CD= 10 rad/sec, the slope changes from -60dB /decade to -80dB /decade
due to the presence of I — +11 in the denominator.
Step 7 of 20
Phase plot:
0(rad/sec)
0.1
-179°
1
-173.15°
5
-228.36°
10
-107.65°
100
-172°
Step 8 of 20
Booe otagnm
step 9 of 20
(s+2)^
L (0 =
(c).
s^ (s+10) (s^+6s+25)
(-<D^+4JayH)
L(Jm) =
-<D^(Jco+1 0)(-co^+6Jcd+25)
0.016
-(f)
L(Jco) = •
■H).24J<»I-1
Break or comer frequencies:
® j= 2 rad/sec
0)2=5 rad/sec
0)3=10 rad/sec
Step 10 of 20
M s^ itu d e plot:
(a) The constant term *0.016* causes an increase in magnitude of 20 log 0.016 =
-3 6 d B
(b) The initial low frequency slope due to the presence of two poles at the origin
is -40dB/decade, and this asymptote intersects the OdB line at CD= Irad/sec
(c) A t CD= 2 rad/s e c , the slope change s from - 40dB/decade to OdB/decade due to
in the numerator.
presence o f I -I ^ I + Jfl^l
'[-(fj-
Since 2^cc^ = 4
and cc^=2
(<Q At ®= 5 rad/sec, the slope changes from OdB /decade to -40dB /decade
due to presence of
Since 2 ^
in ttie denominator.
+ 0 .2 4 JC D + 1
-(!)■
= 6 and cc^=5 ,
. 4 = 0.6
(e)
At CD= 10 rad/sec, the slope changes from -60dB /decade to -80dB /decade
n®
due to the presence o f -----1*1 in &e denominator.
UO )
Step 11 of 20
Phase plot:
0(rad/sec)
0.1
-180.52°
1
-146.58°
2
-131°
5
-340.16°
10
-208.97°
100
-263.15°
Step 12 of 20 ^
BoMOogum
Step 13 of 20
T
_
(!+2)( s“ -M s+68)
- ^ ( s+ 1 0 )( = M s+85)
jf j
\ _
CT(d+^(-< d^44J0^-68)
flea
^
r ^ mV
lio
J
[
1
L ( J cd) = ■
-1 —
1 +0.05J(i>fl
Break or comer frequencies:
J
0 ^ = 2 rad/sec
0)2=8-24 rad/sec
0)3=9.22 rad/sec
0)4=10 rad/sec
Step 14 of 20 ^
Magnitude plot:
(a) The constant term *0.16'causes an increase in magnitude of 20 log 0.16=-16 dB.
(b) The initial low frequency slope due to tiie presence of two poles at the origin is
-40dB/decade,and this ass3nnptote intersects the OdB line at o = Irad/sec
(c) At CD= 2 rad/sec, the slope changes from -40dB /decade to -20dB /decade
/■j® ^
due to presence of I — +11 in the numerator.
At 0 = 8.24 rad/sec, the slope changes from -20dB /decade to +20dB /decade
due to presence of
Since 24o^ = 4
+0.06J1S+1
-f—T
U . 24 J
in the numerator.
and 13^=8.24 ,
I.-. 4 = 0.2431
(e)
At 0D= 9.22 rad/sec, the slope changes from +20dB /decade to -20dB /decade
due to presence of
05JO+1 in dte denominator.
Since 2413^=4 and o^=9.22 ,
I.-. 4 = 0.2171
(e)
At 0D= 10 rad/sec, the slope changes from -20dB /decade to -40dB /decade
{J a
due to tile presence of I — +11 in the denominator.
Step 15 of 20
Phase plot:
0(rad/sec)
0.1
-177.64°
1
-158.46°
8.24
-111.06°
9.22
-301.55°
100
-175.44°
Step 16 of 20
BodedKQKm
Step 17 of 20
w.
L (s ) =
'■ '
L ( J cd)
s^s+2)(s+3)
_
(-a)*+2JGi+2)
~ -m’ (J(ii+3)(Jai+2)
0.33
r ^V
1
L(Jm) = ■
comer frequencies:
02=2 rad/sec
03=3 rad/sec
Step 18 of 20
Magnitude plot:
(a) The constant term *0.33'causes an increase in magnitude of 20 log 0.33=-9.63 dB.
(b) The initial low frequency slope due to tiie prasence of two poles at the origin is
•40dB/decade, and tiii s slope intersects tiie OdB line at 0 = Irad/sec
(c)
At 0D= -72 rad/sec, the slope changes from -40dB /decade to OdB /decade
due to presence of
■ te)'
+J©+1 in the numerator.
Since 24co^ = 2 and <Og='j2,
I.-. 4 = 0.70^
At 0 = 2 rad/sec, the slope changes from 0 dB/decade to -20 dB/decade
fJot
due to tiie presence of I — +11 in the denominator.
(e)
At 0D= 3 rad/sec, the slope changes from -20 dB/decade to -40 dB/decade
(Jot
due to tiie presence of I — +11 in the denominator.
Step 19 of 20
Phase plot:
0(rad/sec)
0.1
0
-179°
1
-161.56°
>12
-150.5°
5
-24.86°
10
44.77°
100
-84.29°
Step 20 of 20
Problem 6.08PP
Right half-plane poles and zeros. Sketch the asymptotes of the Bode plot magnitude and phase
for each of the listed open-loop transfer functions. Make sure that the phase asymptotes properly
take the RHP singularity into account by sketching the complex plane to see how thexLfsJ
changes as s goes from 0 to +
After completing the hand sketches, verity your result with
Matlab. Turn in your hand sketches and the Matlab results on the same scales.
=. 1 ± 2
(a)
1
(The model for a case of magnetic levitation with lead compensation)
C-* U s ) =
^ (The magnetic levitation system with integral control and lead
compensation)
(c) L(s) =
(d) L (f) _
(f) L(s) - ( , _ |) [ ( , + 2)X+ 3]
Step-by-step solution
s te p ! of27
(a)
Write the expression for
a-f2
L(a)
.
I
( i + I O ) ( j" - 4 )
Substitute Jio for s .
( y « )+ io )((y v » )* _ 4 )
2(i+o.sy«i)
10(l+0.1yV») 4(o.25(yV»)’ - l )
0.05(l+0.5y<g)
” (-0.25<»*-l)(l+0.1yV»)
0.05 x ,/(P - i-(0.5 i»)M
|l O'<u) | = ------------ iu —
(-0.25<»’ -l)^(l*+ (0.l< »)')
ZL^Ja)= -180®+tan"' (O.Sm) - tan"' (0. I<»)
Step 2 of 27
Determine the breakpoint frequencies.
__1_
*■' " 0.5
-2
—
sec
_1_
°0.1
= 102^
sec
Step 3 of 27
0.05
start with the low frequency asymptote
-0.25<»’ - r
This is a second-order pole with ^ = 0.
The magnitude plot of this temr has a slope of -4 0 dB per decade.
0.05
Calculate the magnitude of
indB at n) = iO "'— .
-0.25® =-1
0.05
A , - 2 0 log
0.25(y(0.1))’ - l
0.05
= 20log
0.25(y(0.1)) -1
° -28.028 dB
Step 4 of 27
Consider the second low frequency asymptote (l+0.5y® )
The magnitude plot of this temr has a slope of 20 dB per decade.
At the break point frequency of the zero, the slope shifts to -2 0 dB per decade.
I
(1+O.ly®)
Consider the third low frequency asymptote
The magnitude plot of this temr has a slope of -2 0 dB per decade.
At the break point frequency of the pole, the slope shifts to -4 0 dB per decade -
Step 5 of 27
The phase curve starts at ^ = -180® corresponding to the low frequency asymptote
i= -4
Determine the phase Z i ( / ® ) at n t - 2 ^ ^ sec
ZL (J2) - (-180®+tan"' (0 .5 (2 ))- tan"' (0.1(2)))
— 146.3®
. rad
Therefore, the phase curve rises to —146.3® at at - 2 ---- sec
ste p 6 of 27
Determine the phase Z 6(y® ) at a t - 6 ^ ^ sec
Z £ ( y 2 ) - (-180®+tan"' (0 .5 (6 ))-ta n "' (0.1(6)))
= -140®
rad
Therefore, the phase curve rises to —140® at ar - 6 ---sec
Determine the phase Z £.(/® ) at a t = 1 0 ^ ^ sec
Z £ (y i0 ) -(-1 8 0 ® + tan"' (0 .5 (1 0 ))-tan "' (0.1(10)))
= -1463®
- tad
The phase curve again falls to —146.3® at a r= 1 0 ---- sec
rad
Determine the phase ZL (y® ) at at = 500---Z £ (y 5 0 0 ) = (-1 8 0® + tan"'(0.5(500))-tan"'(0.1(500)))
^ rad
The phase curve again falls to —lg0« at at = 500---- sec
Combine all the asymptotes to draw a composite curve.
Therefore, sketch the asymptotes of the bode plot magnitude and phase for the open loop
transfer function L{s).
B o d e d iig iM i
Step 7 of 27
The transfer function is,
i( s ) =
s+2
1
(s + 1 0 )(s = -4 )
j+ 2
s’ + 1 0 s = -4 s -4 0
Draw the MATtAB code to plot the response of £ ( s ) .
num=[1 2] den=[1 10-4 -40] sys=tf(num,den) bode(sys)
Step 8 of 27
The Bode plot for the function L (s ) by using Matlab is shown in Figure 2.
Step 9 of 27
(b)
Write the expression for L{s) s+2
1
£ ( ,) = s ( s + 1 0 ) ( s = -l)
Substitute j a for s .
y® (y ® + 1 0)((ya))= -l)
2(l+ 0.5y® )
■ioy®(i+o.iy®)((ya>)’ -i)
0.2(l+0.5y® )
y®((y®)’ -i)(i+o.iy®)
The magnitude and phase of the function are.
^(l=+(0.5®)=)
|£ (y ® ) |= -
® (-< t^ -l)^ (l= + (0 .1 ® )')
Z L { ja ) = -90® -180® + tan"'(0.5® )-lan"'(0.1® )
Determine the breakpoint frequencies.
_ l rad
sec
at = “
1
_ l rad
sec
0.5
= 2—
sec
_1_
°0.1
= 10
rad
Step 10 of 27
Start with the low frequency asymptote
0.2
(J< s)
The magnitude plot of this temr has a slope of -2 0 dB per decade.
Calculate the magnitude of . ^ . in dB at at = 10"'
(y®)
sec
0.2
A, = 2 0 log
(-/(«•>))
= 20 log A
= 6.02dB
Step 11 of 27
1
Start with the second low frequency asymptote
This is a second-order pole with ^ = 0.
The magnitude plot of this temr has a slope of -4 0 dB per decade.
At the break point frequency of the pole, the slope shifts to -6 0 dB per decade.
Consider the third low frequency asymptote (l+0.5y® )
The magnitude plot of this temr has a slope of 20 dB per decade At the break point frequency of the zero, the slope shifts to -4 0 dB per decade 1
Consider the fourth low frequency asymptote
(1+o.iy®)
The magnitude plot of this terni has a slope of -2 0 dB per decade.
At the break point frequency of the pole, the slope shifts to -6 0 dB per decade -
Step 12 of 27
The phase curve starts at ^ = -270® corresponding to the low frequency asymptotes
0.2
,(a = -l)
Determine the phase Z £ (/® ) at ^ s 2 ^ ^ sec
Z £ (y 2 ) - (-270®+tan"' (0 .5 (2 ))- tan"' (0.1(2)))
= -236.3®
. rad
Therefore, the phase curve rises to —236.3® at n> = 2 ---- sec
ste p 13 of 27
Determine the phase Z £(y® ) at
sec
Z £ ( y 2 ) - (-270®+tan"' (0 .5 (6 ))-ta n "' (0.1(6)))
= -230®
rad
Therefore, the phase curve rises to —230® at n» = 6 ---Determine the phase Z £(y® ) at nr=10
rad
Z£.(yi0) -(-2 7 0 ® + tan"' (0.5(10))-tan"' (0.1(10)))
— 2363®
. rad
The phase curve again falls to —236.3® at ® = 1 0 ---- sec
rad
Determine the phase Z £(y® ) at nr = 500----
Z£.(y500) = (-2 7 0® + tan"'(0.5(500))-tan"'(0.1(500)))
= -270®
^rad
sec
The phase curve again falls to —270® at nr = 500---- Combine all the asymptotes to draw a composite curve.
Therefore, sketch the asymptotes of the bode plot magnitude and phase for the open loop
transfer function £ ( s ).
Step 14 of 27
Draw the MATtAB code to plot the response of £ ( s ) .
num=[1 2] den=[1 10-1 -10 0] sys=tf(num.den) bode(sys)
The Bode plot for the function £ ( s ) by using Matlab is shown in Figure 4.
Step 15 of 27
(c)
Write the expression for £ ^ s ) .
Substitute y® for s .
(js>)
|£(y“^^(y®)=[(^i+tan-'(-®)
ste p 16 of 27 ^
Tabulate the terms of L{jfp) in the increasing order of their frequencies.
Term slope changeinslope(<l%^)
-40
( i) '
-1+y® 20
-40+20=-20
Step 17 of 27
rad
Calculate the magnitude of - — ^ in dB at nr= 10'
( »
sec
A, = 2 0 log
(7(0.01))=
= 201og!10®|
= 80 dB
Determine the phase at various values of ® .
®
2X(y®)
10-=
0
10“
-4 5
10
-84.2
10=
-9 0
Combine all the asymptotes to draw a composite curve.
Therefore, sketch the asymptotes of the bode plot magnitude and phase for the open loop
transfer function £ ( s ).
rad
Frequency —
sec
F ig u re s
ste p 18 of 27
Draw the MATtAB code to plot the response of £ ( s ) .
num=[1 -1] den=]1 0 0] sys=tf(num.den) bode(sys)
The Bode plot for the function £ ( s ) by using Matlab is shown in Figure 6.
Step 19 of 27
(d)
Write the expression for £ ^ j ) .
s= +
£ (s) = -
2j + 1
i( i+ 2 0 ) = ( s = - 2 s + 2 )
Substitute y® for s .
(y®)‘ + 2 (y®)+i
£ (y ® )= —
y®(y®+ 20 )= ((y®)= - 2 (y® )+ 2 )
■irS)
ste p 20 of 27 ^
Tabulate the terms of £ (y ® ) in the increasing order of their frequencies.
Teim
slope
_1_
change in s l o p e ( < l % ^ J
-2 0
i<o
(y ® )= + 2 ( y ® )+ i
1
40
- 2 0 + 40 = 20
-4 0
20 - 40 = - 2 0
-4 0
- 2 0 - 4 0 — 60
( y ® + 20)=
1
( ( y ® ) = - 2( y ® ) + 2)
The break point frequencies are pp s l i ^ . e r
“
= 1 ,4 1 4 ! ^ . et = 2 0 ^ ^ .
sec
sec
sec
Since the break point frequencies are very closer, cannot observe the accurate change in slope
on the scale of logarithm.
So. the approximate hand sketch of magnitude of asymptotes is as shown in Figure 7.
Step 21 of 27
Combine all the asymptotes to draw a composite curve.
Therefore, sketch the asymptotes of the bode plot magnitude and phase for the open loop
transfer function £ ( s ).
F igure?
ste p 22 of 27
Draw the MATtAB code to plot the response of £ ( s ) .
num=[1 2 1] den=]1 38 322 -720 800 0 0] sys=tf(num,den) bode(sys)
The Bode plot for the function £ ( s ) by using Matlab is shown in Figure 8.
Step 23 of 27
(e)
Write the expression for £^ 5 ).
4+ 2
£ ( 4) = -
4(4-1){4+6)=
Substitute y® for 4 .
L i j a , ) -------------------------------y ® ( y ® - i) ( y ® + 6 )
2(i+yo.5®)
y m ( y ® -i) (( y ® ) = + i2y ® + 6=)
^ { j a ) = -9 0 ® + tan"' (0.5®) - tan"' ( - ® ) - tan"'
j
Tabulate the terms of L{jip) in the increasing order of their frequencies.
Term
slope
change in s l o p e ( « l % ^ J
-2 0
y®
y ® -i
-2 0
- 2 0 - 20 = -4 0
(i+ y o .5 ® )
20
-4 0 + 2 0 = - 2 0
(y ® )= + i2y ® + 6=
-4 0
-2 0 - 4 0 = -6 0
Determine the phase at various values of ® .
®
2 X (y ® )
i0"=
-270
10“
-217.43
10'
-193
10=
-270
Step 24 of 27
Therefore, sketch the asymptotes of the bode plot magnitude and phase for the open loop
transfer function £ ( 4 ).
Figixc9
Step 25 of 27
Write the MATtAB code to plot the response of £^ 4 ).
num=[1 2] den=[1 11 24 -36 0] sys=tf(num,den) bode(sys)
The Bode plot for the function £ ( 4 ) by using Matlab is shown in Figure 10.
Step 26 of 27
(f)
Write the expression for £^ 4 ).
£ ( 4) = ^ 4 -1 )((4 + 2 )= + 3 )
Substitute y® for 4 .
^ ^
( y ® -l) (( 7 ® + 2 )= + 3 )
Z £ (y ® ) = -90® -tan"'
Tabulate the terms of £ (y ® ) in the increasing order of their frequencies.
Term
slope
fe )
-2 0
1
-4 0
change in s l o p e ( < l % ^ J
-2 0 -4 0 = -6 0
(y ® + 2 )= + 3
Determine the phase at various values of ® .
®
z £ (y ® )
i0"=
-180°
10“
123.7
10=
-180®
Therefore, sketch the asymptotes of the bode plot magnitude and phase for the open loop
transfer function £ ( 4 ).
Step 27 of 27
Write the MATtAB code to plot the response of £^ 4 ).
num=[1] den=]1 3 3-7] sys=tf(num.den) bode(sys)
The Bode plot for the function £ ( 4 ) by using Matlab is shown in Figure 12.
Problem 6.09PP
A certain system is represented by the asymptotic Bode diagram shown in Fig. Find and sketch
the response of this system to a unit-step input (assuming zero initial conditions).
Figure Magnitude portion of Bode plot
Step-by-step solution
step 1 of 5
Refer to asymptotic Bode diagram in Figure 6.85 in the textbook.
The initial slope represents the term —.
s
10 , the slope has been changed to OdB/dec, it represents the presence of zero.
And as
That is 11 4 — 1 in the numerator.
I loj
From the asymptotic bode plot,
201ogX'-201ogl0=l
20IogAT-20=l
.
21
■ o g A T --
k>gA: = 1.05
a: =10'“
= 10
Step 2 of 5
The transfer function <?{*) is,
C ( ,)
(s + lO )
(1)
s
Calculate the response to unit step input.
y ( j) = c ( j) c /( i)
J410
Apply partial transforms to simplify the equation.
a + 1 0 _ i<
^
s
s+ \0 = As + B
(2)
Equate s coefficients.
^ =1
Equate constant terms.
£ = 10
Step 3 of 5
The response to unit step input is,
Apply Laplace transform.
> A (f)s tf(/)+ 1 0 fu (/)
=(l+10f)i/(/)
Therefore, the response of the system,
to a unit step input is
Step 4 of 5
MATLAB code to plot output response to unit step input.
» t=0:0.1:10;
» y=1+10.*t;
» plot(t,y)
Step 5 of 5
The response to unit step input is shown in Figure 1.
Figure 1
Therefore, the response to unit step input is shown in Figure 1.
[(l+ 1 0 /) i/(f)|.
Problem 6.1 OPP
Prove that a magnitude slope of -1 in a Bode plot corresponds to -20 db per decade or -6 db
per octave.
Step-by-step solution
step 1 of 1
Step 1 of 1
The m ^ i t u d e of G(jo)) in dB is obtained by multiplying
the logarithm to the base 10 of |G (-j0)| by 20
|G(jcD)|dB=2Caog|G(j(!))|
If |G ( - j..) |= t
201og]QK=201ogK,ffi
|G (jco) IdB=201 ogu,E-201ogjp(D
Which has solpe of -20dB per decade
Also if
= 2c0i then itis called as Octave
since -201og2 = -6dB , the slope could also be expressed as-6dB/octave
Problem 6.11 PP
A normalized second-order system with a damping ratio ^ = 0.5 and an additional zero is given by
G(s) =
s /a + \
j^ + j+ r
Use Matlab to compare the Mpfrom the step response of the system fo ra = 0.01,0.1,1.10, and
100 with the Mrfrom the frequency response of each case. Is there a correlation between M r and
Mp?
Step-by-step solution
step 1 of 15
Step 1 of 15
Peak overshoot K
)
is the maximum peak value of the transient response.
Peak overshoot K
)
is generally related to the damping of the system expressed as.
Step 2 of 15
Consider the seconder order system for a s 0.01 •
G(s)>
7+ 1
0.01
5 *+ J+ l
1005+1
5*+S + l
Write the MATLAB code for the step response of the transfer function.
» num=[100 1]:
» d e n = [1 1 1]:
» sys=tf(num,den):
» step(sys);
» hold on
Step 3 of 15
Consider the seconder order system for a s 0.1 •
5*+S + l
105 + 1
5*+5 + l
Write the MATLAB code for the step response of the transfer function.
» num=[10 1]:
» d e n = [1 1 1]:
» sys=tf(num,den):
» step(sys)
Step 4 of 15
Consider the seconder order system for a s 1.
i+ 1
G(j). _L
5*+5 + l
5+ 1
5*+5 + l
Write the MATLAB code for the step response of the transfer function.
» num=[1 1]:
» d e n = [1 1 1]:
» sys=tf(num,den):
» step(sys)
Step 5 of 15
Consider the seconder order system for a s 10 •
5*+5 + l
0.15 + 1
5*+5 + l
Write the MATLAB code for the step response of the transfer function.
» num=[0.1 1];
» d e n = [1 1 1]:
» sys=tf(num,den):
» step(sys);
Step 6 of 15
Consider the seconder order system for a s 100 •
u ^ 5 ;«
*7 + 1
inn
5*+5 + l
0.015 + 1
5*+5 + l
Write the MATLAB code for the step response of the transfer function.
» num=[0.01 1];
» d e n = [1 1 1]:
» sys=tf(num,den):
» step(sys);
Step 7 of 15
Step response for various values of a is,
Therefore, peak overshoots for various values of a has been determined.
Step 8 of 15
The maximum value of the frequency-response magnitude is referred to as resonant peak
Frequency response resonant peak
Resonant-peak magnitude
.
is related to the closed loop frequency response.
is generally related to the damping of the system expressed as.
For ^ = 0.5. the resonant peak magnitude
is,
________1
J14, = 2 ( 0 .5 ) ^ l - ( 0 .5 ) *
= 1.1547
Step 9 of 15
Consider the seconder order system for q s 0.01 •
*7 + 1
G {s).
0.01
5*+5 + l
1005 + 1
5*+5 + l
Write the MATLAB code for the frequency response of the transfer function.
» num=[100 1]:
» d e n = [1 1 1]:
» sys=tf(num,den):
» bode(sys)
» hold on
Step 10 of 15
Consider the seconder order system for j s 0.1 ■
5*+5 + l
105 + 1
5*+5 + l
Write the MATLAB code for the frequency response of the transfer function.
» num=[10 1]:
» d e n = [1 1 1]:
» sys=tf(num,den):
» bode(sys):
Step 11 of 15
Consider the seconder order system for a s 1.
i+ 1
G(j). _L
5*+5 + l
5+ 1
5*+5 + l
Write the MATLAB code for the frequency response of the transfer function.
» num=[1 1]:
» d e n = [1 1 1]:
» sys=tf(num,den):
» bode(sys):
Step 12 of 15
Consider the seconder order system for a s 10 •
0 {s)-
To*'
5*+5 + l
0.15 + 1
5*+5 + l
Write the MATLAB code for the frequency response of the transfer function.
» num=[0.1 1];
» d e n = [1 1 1]:
» sys=tf(num,den):
» bode(sys):
Step 13 of 15
Consider the seconder order system for a s 100 •
G { S ):
*7 + 1
100
5*+5 + l
0.015 + 1
5*+5 + l
Write the MATLAB code for the frequency response of the transfer function.
» num=[0.01 1];
» d e n = [1 1 1]:
» sys=tf(num,den):
» bode(sys):
Step 14 of 15
Bode plot response of for various values of a is.
Therefore, resonant peak for various values of a has been determined.
Step 15 of 15
Tabulate the peak overshoot and resonant peak value for various values of a .
Tab le 1
a
0.01
0.1
1
10
100
Mr
98.8
9.93
1.46
1.16
1.15
"r
54.1
4.94
0.30
0.16
0.16
As a is Increased, the resonant peak
in frequency response Is decreases. This leads to
expect extra peak overshoot M p in transient response.
From the Table 1, No Significant change in the frequency response is observed from a = 10
onwards. Similarly significant change In the transient response is observed at a = 0.01,0.1 and
a -\
Therefore, the resonant peak M ,. in frequency response and peak overshoot response M p in
transient response are correlated at a = 0.01,0.1 and a = l-
6 .1 2PP
Problem 6 .1 2PP
A normalized second-order system with ^ = 0.5 and an additional pole is given by
^ -------------- .
[(■s/p) + 1 ](*^ + 1 + 1)
Draw Bode plots with p = 0.01, 0.1,1,10. and 100. What conclusions can you draw about the
effect of an extra pole on the bandwidth compared with the bandwidth for the second-order
system with no extra pole?
Step-by-step solution
step 1 of 6
Step 1 of 6
Forp=0.01,
Step 2 of 6
Forp=0.1,
Step 3 of 6
For p = l.
Step 4 of 6
For p=10.
Step 5 of 6
For p=100,
Step 6 of 6 ^
Conclusions:
(1)
.The bandwidth is increased due to the introduction of extra pole.
(2)
.The bandwidth value goes on increasing, as the p value goes on increasing.
Problem 6.13PP
For the closed-loop transfer function
j'/ g \ — ________5______
s^ + 2((OnS + < i^ ’
derive the following expression for the bandwidth ojB W of T(s) in tenns of u)n and ^ :
= a > n y jl\ - 2 ( '^ + ^ 2 + 4 t * ^ ^ ^ .
Assuming that oj/7 = 1, plot ojSlVfor 0 < ^ < 1.
Step-by-step solution
Step-by-step solution
step 1 of 2
T ( s ) = t ------ --------------J.
8 + 25oa^s+cc^
The Bandwidfti of the standard second - order system
can be found by Bnding that frequency for which M = - |=
u^ere M s|T(ja>)|
(’■ M ) " '’M
Step 2 of 2
solving for -^5”
^
get
^ = l- 2 4 = ± j4 ? 4 ? t i
since
^5SL>Q,
Oil
<d^ = o>.^(1-2?= )+ V 45'^?“ +2
Hence proved.
Problem 6.14PP
Consider the system whose transfer function is
i4o<wo^
G (s )^
+ <3
This is a model of a tuned circuit with quality factor Q.
(a)
Compute the magnitude and phase of the transfer function analytically, and plot them for Q =
0.5,1.2, and 5 as a function of the normalized frequency oj/wO.
(b)
Define the bandwidth as the distance between the frequencies on either side of cuO where th
magnitude drops to 3 db below its vaiue at uX), and show that the bandwidth is given by
magnitude drops to 3 db below its vaiue at ciX), and show that the bandwidth is given by
BW
(c)
What is the relation between Q and ^
Step-by-step solution
step 1 of 4
(a)
Consider the magnitude and phase of the transfer function.
(1)
Q s ^ + a ^ + o ii’ Q
Substitute Ja>tor s in Equation (1).
Gijosi)—
Q (ja ) +a,(ja)+<a^Q
A„aja>
-o ^ Q + a J a + C D ^ Q
S]
A
G (j0 )-
(2)
j0
Take modulus for Equation (2).
1
^
V
,3,
...... "
^ = la n - |— - :2 - |
1.^1 «>)
(4)
Consider the following formula for the normalized frequency.
-
V*
Substitute 0.5,1 and 2 for Q, in equation (3). We get phase with normalized frequency —
varying from 0.1 to 10 for magnitude.
The bode plot is shown in Figure 1.
if r - i
io «
10'
Figure 1
Similarly, substitute Q value in equation (4), we get phase with normalized frequency — varying
from 0.1 to 10 for phase margin.
Step 2 of 4
(b).
Bandwidth at 3db point from centre frequency
from bode plot, there is even symmetry from
centre frequency 0 ^.
From bode plot. Magnitude at frequency 0 ^ s
magnitudeatfrequeocyoiji
Where a)[<d)^and 0 ^ > 0 ^.
Hence
|G(M)|-|C(M)|
From bode plot, the frequency relationship is given below.
And equa magnitude from symmetry, we have
From the above equation, the centre frequency is 0 ^ > 0 ^0 ^.
Step 3 of 4
Where magnitude at 0 ^ and 0 ^ are equation which may be frequency which also include e
frequency at which magnitude drops to 3db, from mean value
Betweenat3dbpoints uppercutoff-lowercutoff
2x
^ 0 j- 0 t
2x
=aL.
Substitute the
value in equation (4),
BW
-(5)
At 3db point which corresponding to magnitude is given below,
^
■
3
1
I 2
"F F F
Take square on both sides
1
,+ e » f^ _ 3 .V
W
OH)
W
"I2
o^J
o>^)
0 ),J
=l
(6)
Compare the equation (5) and equation (6).
BW ^
Thus, the bandwidth is BW :
1 ( e '''
Step 4 of 4
(c).
Compare quality factor Q and damping factor ^
Consider the given transfer function.
C (* ) =
0 s * + < V + « ( ,’ g
A ^
G (4 -
(7)
Consider the standard second order transfer function.
s^+ 2C a ^+ ef
(8)
Compare the equation (7) and equation (8).
» ' + + «H,’ = » '+
Consider
s scentreFrequency
Thus, the quality factor is
Problem 6.15PP
A DC voltmeter schematic is shown in Fig. The pointer is damped so that its maximum overshoot
to a step input is 10%.
(a)
What is the undamped natural frequency of the system?
(b)
What is the damped natural frequency of the system?
(c)
Plot the frequency response using Matlab to determine what input frequency will produce the
largest magnitude output?
(d)
Suppose this meter is now used to measure a 1-V AC input with a frequency of 2 rad/sec.
pftpr in itip l trp n p ip n tp h p v p HIp H n u t ? W h p t i.<; th p nhp<;p \an
W h flt a m n litiiH p w ill th p m p t p r in riin p tp
(d)
Suppose this meter is now used to measure a 1-V AC input with a frequency of 2 rad/sec.
What amplitude will the meter indicate after initial transients have died out? What is the phase lag
of the output with respect to the input? Use a Bode plot analysis to answer these questions. Use
the Isim command in Matlab to verily your answer in part (d).
Figure Voltmeter schematic
/= 4 0 X I( r« k g m 2
J t= 4 x l( r « k g iii2 /s e c 2
T = input torque = Kg^v
v = input voltage
j r - = 4 x H T ® N i ii / V
Step-by-step solution
Step 1 of 7
(a)
The DC voltmeter is shown in Figure 1.
Step 2 of 7
Consider the given data.
/ = 4 0 x l0 - ‘ k g .m '
* = 4 0 x l 0 '‘ k g .m V se '
r = 4 0 x l 0 ‘‘ kg.m ’
K , = 4 0 x l 0 '‘ N .m /v
Consider the equation of motion.
dt
...... (1)
dt
From equation (1). the transfers function is given c
e (s )
(2)
Consider the standard second order transfer function.
(3)
J _
Compare the equation (2) and equation (3).
... (4)
" I
(5)
Substitute the given values in equation (4) and equation (5).
s 0.316 rad/sec.
Thus, the natural frequency is 0.316 rad/sec •
Step 3 of 7
(b)
Consider the following formula for the peak overshoot.
......
Substitute 0.1 for A/^in equation (6).
0.1 » e
Take logarithm on both sides.
In (0 .1 )= I
f = 0.59
Consider the formula for the damping frequency.
......
Substitute 0.316 for ajj^and 0.59 for ^ in equation (7).
ffl.= 0 .3 lW l-0 .5 9 ’
=0.25 rad/sec
Thus, the damping frequency is 0.25 rad/sec ■
Step 4 of 7
(c)
Consider the Magnitude and phase response of the given function.
F (4 )
K JI
g (» )
K(s) 4’ +2^<»,s+<»,’
(8)
Consider the value of s is
»(J<»)
IC J l
V (M )
{ j( u ) ‘ + 2 i a , ( j a ) + a ’
(9)
From equation (9), the Magnitude equation is given as.
^
---------------
(10)
+ {2 C a ,a > y J
Take differentiate with respect to a>.
deo
.(11)
*
When d \ n j < 4
0 , rearrange the Magnitude equation.
d&
<»= 0.549<».......(12)
Substitute 0.316 for <»^in equation (12).
<9= 0.549 x 0.316
=0.173 rad/sec
Step 5 of 7
Substitute0.316for <9^, 0.173for <», 4x10"*^°'^
4 0 x 1 0 ' * ^ ° ' ^ ^ ® * ^ * ^ ^inequation
(9)8(j<o )
(4 xl0 -‘ /4 0 x l0 -‘ )
V(J<»)
(y®)^+2(0.59)(0.316)(y<»)+(0.316)’
0.1
( > ) ' + 0.37288y®+0.0998
Write the Matlab program and draw the bode plot.
bode(tf([0.1],[1 0.37288 0.0998]))
The output of the Matlab program is given in Figure 2.
BodeDiagnm
Step 6 of 7
5 peak frequency is [Q.173rad/se^
Step 7 of 7
(d)
With 2rad/secfrom Figure 2, the amplitude and phase angle are 0.0256radand —1<
Thus, the amplitude and phase angle are |0.0256rad|and |-169°I
Problem 6.16PP
Determine the range of K for which the ciosed-ioop systems (see Fig) are stabie for each of the
cases beiow by making a Bode piot for K = 1 and imagining the magnitude plot sliding up or
down until instability results. Verity your answers by using a very rough sketch of a root-locus
plot.
(a)
js:g ( s)
= ' ^
(b) K G (s) -
v r * t ^ \ _ ^(a-F IO)(j-F 1)
* G ( * ) - ( , + , oo)(,+ 5 )1
(0
Fiaure Block diaoram for YfsVRfst = KGfst/tl + KGtstl
v r> t-\
A T fa-F I 0 ) ( J - F I )
« G (** ) - ( . + , oo,(,+ 5 )3
r#-r
(0
Figure Block diagram for Y{s)/R(s) = KG{s)/[1 + KG(s)]
—■ act*)
I or
Step-by-step solution
step 1 of 31
(a)
Consider the given transfers function.
'
...... (1)
a+30
’
Rewrite equation (1).
a ( j* 3 i
K [ ? * ']
Substitute 1 for K in Equation (1).
' ’
■•fe-')
ii:<;(a) . f e * " - ' )
■(2)
■ ( & • ')
Step 2 of 31
Plot the straight Bode diagram for the transfer function K G {s ) ■
Consider the standard form of transfer function.
g ( lF a M )
■(3)
(1 + a M )
Compare Equations (2), and (3).
Number of zeros present in the system is 1
Number of poles present in the system is 1
=30aodo)^ =30-
The corner frequencies are
Substitute j a for s in Equation (2).
( f ^ ')
K G (jo> ) = 0 . \ ^ ----- 4
( f * ')
a
v
, r
. , -
-(4)
y
Step 3 of 31 'v
Find the magnitude of K G [jiii) in dB using Equation (4) as listed in table (1).
Table 1
Corner Frequency Slope
Change in slope
Term
© ©
(S)
(5-)
®.,=30
20
-
ax,, =30
-20
0
Calculate gain A using table 1.
Assign lower frequency
Calculate gain A,,
'
= 1 ^ ^ and higher frequency
sec
=1000^^-
sec
at
A „,=|G (y<i.)|
= -2 0 d b
Calculate gain
at At = a>^ ■
A ^ , =|G(y<»)|
= -9.S42db
Calculate gain (A) at At=At^2 -
A=-^=|G(y<»)|
= 0db
Calculate gain (A) at
A „ 2 = |C (H
= 0db
Step 4 of 31
Find the phase plot for equation (4).
*
de
©
g
1
16.5
10
54.8
30
39.3
100
15.1
1000
1.58
The Bode diagram is shown in Figure 1.
Step 5 of 31
Thus, the bode diagram is plotted for the given transfer function and it is shown in Figure 1.
From Figure 1, the gain is increased or decreasing on the bode diagram and the phase angle is
not less than -18CP
Step 6 of 31
Hence, the system is stable for |jc > Q |.
Step 7 of 31 'v
Consider the number of poles and zeros from Equation (1).
To find zeros put numerator JV(j) = 0
Thus, the zero is -3.
To find poles put denominator D (s) = 0 ■
Thus, the pole is - 30.
Step 8 of 31 'v
Consider the formula for the asymptotes.
n -m
(sumoffinitepoles)-(sumoffinitezeros)
(numberoffinitepoles)-(niunberoffinitezeros)
-3 0 -(-3 )
1-1
= co
Thus, the asymptotes is
.
Step 9 of 31
Procedure to draw root locus plot;
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes.
• Draw the root locus.
The root locus plot is shown in Figure 2.
Step 10 of 31 A
Thus, the root locus is plotted for the given transfer function and it is shown in Figure2.
From Figure 2, the system is stable for |J(f >Q|.
Step 11 of 31
Hence, the results are verified from root locus plot.
Step 12 of 31
(b)
Consider the transfer function.
K
K G (s ) = -^
■(5)
( j + 10)(s + l)
Rewrite equation
equ
(5).
K
JCG(4)=7
( j + 1 0 )(j+ l)(s + l)
Substitute 1 for K in above equation.
Substitute
_1_^____ 1
ATG(j) =
lo r ^ T T v - ^ . v
<6>
Plot the straight Bode diagram for the transfer function KG{^s) ■
Consider the standard form of transfer function up to third order system.
3r(l-F4/r».)
G (z)
-(7)
0 + 4 M )0 + 4 M )0 + V < » > )
Compare Equations (6), and (7).
Number of zeros present in the system is 0
Number of poles present in the system is 3.
The corner frequencies are
=1,<»^2
andfiVa -1 0 -
Substitute j a for s in Equation (6).
7 ------------
-(8)
Find the magnitude of K G [ ja ) In dB using Equation (8) as listed in table (2).
Table 2
Corner Frequency Slope
Change in slope
Term
(y-AX+l)
(S)
m
"'■=T
-20
-
m
-1
(y-AX+l)
AX„=1
-20
■40
(if*')
AX„=10
-20
-60
Calculate gain A using table 2.
Assign lower frequency ax = 0 . 1 ^ ^ and higher frequency ax = 1 0 0 ^ ^ -
sec
sec
Calculate gain (A) at a —a , -
34= |G ( » |
= -20db
Calculate gain A
at ax = a ^ ■
A„^, =|G(yAx)|
= -20db
Calculate gain (A) at ax= ax,- A,,
wIslopefiomAX^, toAx^jxlcxg^T. +A „ ^
I
= - 4 0 1 o g ^ Y j- 2 0 d b
= -« 0 d b
Calculate gain (A) at ax= A^.
A
^ s^sk x p c fitx m A x ^jto A jjiX lo g -^ j+ A ^
= -6 0 1 o g ^ ^ ^ j-6 0 d b
= - 1 2 0 dB
Find the phase plot for equation (8).
*
de
g
0.1
-12.1
1
-95.9
10
-213
100
-263
Step 13 of 31
The Bode diagram is shown in Figure 3.
Step 14 of 31 A
Thus, the Bode diagram is plotted for the given transfer function and it is shown in Figure 3.
Step 15 of 31
From Figure 3, the gain is equal to the 242 when phase angle crosses —IgCP.
Step 16 of 31
Hence, the system is stable for |j(f < 242 |ond the system unstable for |j(f > 2421 ■
Step 17 of 31 ^
Consider the number of poles and zeros from Equation (5).
To find zeros put numerator JV(j) = 0
Thus, there is no zero.
To find poles put denominator D (s) =
0■
Thus, the pole is - 10, - 1 and - 1.
Consider the formula for the asymptotes.
n -m
(sumoffinitepoles)-(sumoffinitezeros)
(numberoffinitepoles)-(numberoffinitezeros)
.z lizM
3-0
=-4
Thus, the asymptotes is a
step 18 of 31
Consider the formula for the angle of asymptotes.
2 180°+360«(/-l)
w—
-(9)
n —m
Where.
Number of poles is n
Number of zeros is m
/ =l,2,..ji-m
Substitute 1 for /, 3 for rx and 0 for m in equation (9).
180+360(1-1)
3
180
s ----3
-60»
Substitute 2 for /, 3 for rx and 0 for m in equation (9).
180+360(2-1)
3
540
3
=180*
Substitute 3 for /, 3 fOr rx and 0 for m in equation (9).
, 180+360(3-1)
3
300
3
—60*
Thus, the angle of asymptotes are |60v|, | 180°|and | - 60**l-
Step19of31
Procedure to draw root locus plot;
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes.
• Draw the root locus.
Step 20 of 31
The root locus plot is shown in Figure 4.
Step 21 of 31 A
Thus, the root locus is plotted for the given transfer function and it is shown in Figure 4.
Step 22 of 31 A
From Figure 4, the system is stable for |j(f < 242 |und the system unstable for |jc > 2 4 2 l ■
Step 23 of 31 ^
Hence, the results are verified from root locus plot.
Step 24 of 31 ^
(c)
Consider the transfer function.
^ , ^ ( 4 + 10X4 + 11
' ^
.(10)
(4 + 1 0 0 )(4 + 5 )
Rewrite equation (10).
K
(4 + 1 0 )(4 + l) ( 4 + l)
Substitute 1 for K in above equation.
x e |.).-L
.(11)
Plot the straight Bode diagram for the transfer function KG{^s) ■
Substitute J a fOr s in Equation (11).
.(12)
K G {J a )=
Find the magnitude of K G [ ja ) In dB using Equation (12) as listed In table (3).
Table 3
Corner Frequency Slope
Change in slope
Term
(y-AX+l)
AX„=1
-20
-
(f-)
ax„ =
-20
■40
(if-)
AX„=10
-20
-60
(£*■)
AX„ = 100
-20
-80
5
Calculate gain A using table 3.
Assign lower frequency ax = 0 . 1 ^ ^ ond higher frequency ax = 1 0 0 0 ^ ^ -
sec
sec
Calculate gain (A) at a —a , ■
=-62db
Calculate gain A
at ax = a ^ ■
A^=-59db
Calculate gain (A) at ax= ax^, .
A,^=-S6db
Calculate gain (A) at ax = AX., -
Step 25 of 31
Calculate gain (A) at ax= ax^, .
A ^ = -8 3 d B
Calculate gain (A) at a —a,,-
A ^ = -1 2 0 d B
Find the phase plot for equation (4).
The Bode diagram is shown in Figure 5.
Step 26 of 31
Thus, the Bode diagram is plotted for the given transfer function and it is shown in Figure 5.
From Figure 5, the phase angle is never crosses —180®.
Step 27 of 31
Hence, the system is stable for |jc > Q |.
Step 28 of 31 ^
Consider the number of poles and zeros from Equation (1).
To find zeros put numerator JV(4) =
0
Thus, the zeros are -10 and -1.
To find poles put denominator ^ ( 4) =
0-
Thus, the pole is - 100, - 5, - 5 and - 5.
Consider the formula for the asymptotes.
n -m
(sumoffinitepoles)-(sumoffinitezeros)
(numberoffinitepoles)-(numberoffinitezeros)
. z iiH im
4 -2
=-52
Thus, the asymptotes is □ u
Step 29 of 31
Substitute 1 for /, 4 fOr rx and 2 for m in equation (9).
180+360(1-1)
"■
^ 180
2
2
=90®
Substitute 2 for /, 4 fOr rx and 2 for m in equation (9).
180+360(2-1)
2
540
B----2
=270®
Thus, the angle of asymptotes are |9Q®|and |270®l-
Step 30 of 31
Procedure to draw root locus plot;
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes.
• Draw the root locus.
The root locus plot is shown in Figure 6.
Root Locus
Step 31 of 31
Thus, the root locus is plotted for the given transfer function and it is shown in Figure 6.
From Figure 6, the system is stable for |jc > Q |.
Hence, the results are verified from root locus plot.
Problem 6 .1 7PP
Determine the range of K for which each of the iisterJ systems is stabie by making a Bode piot for
K = 1 and imagining the magnitude piot siiding up or down untii instabiiity resuits. Verity your
answers by using a very rough sketch of a root-iocus plot.
(c) K G {s) -
( , ^ 2 K j2 + 9 )
tr4 'i)t» ? H
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 6.18PP
(a)
Sketch the Nyquist plot for an open-loop system with transfer function 1/s2; that is. sketch
s= C x
where C1 is a contour enclosing the entire RHP. as shown in Fig. 1. {Hint Assume Cl takes s
small detour around the poles at s = 0. as shown in Fig. 2)
(b)
Repeat part (a) for an open-loop system whose transfer function is
Figure 1 An s-piane plot of a contourCI that encircles the entire RHP
Figure 1 An s-piane plot of a contourCI that encircles the entire RHP
c,
/
V
Figure 2C1 contour enclosing the RHP for the system
Step-by-step solution
step 1 of 3
M o p in g imaginadty axis ( 0 < ( 9 < od)
idiich lies on the negative real axis
Step 2 of 3
Step 3 of 3
Problem 6.19PP
Sketch the Nyquist plot based on the Bode plots for each of the following systems, and then
compare your result with that obtained by using the Matlab command nyquist; Don’t be
concerned with the details of exactly where the curve goes, but do make sure it crosses the real
axis at the right spot, has the correct number o f -1 encirclements and goes off to infinity in the
correct direction.
(a) X G ( S ) =
(b) K C ? ( s y =
(C )
if '\
(C )
(d)
K C ^ i^ s y
K C ^ i^ s y
—
—
i s -+- 1 0 )C ^ + 2> 2
+ 1 0 ) ( j + 1)
+ 1 0 ) ( j + 1)
( ,_ j_ io o ) ( ;5 +
2 )a
Using your plots, estimate the range of K for which each system is stable, and qualitatively
verify your result by using a rough sketch of a root-locus plot.
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 6.20PP
Draw a Nyquist plot for
1)
KG(s) = Ks{s(s+
+ 3) *
choosing the contour to be to the right of the singularity on the yw-axis. Next, using the Nyquist
criterion, determine the range of K for which the system is stable. Then redo the Nyquist plot, this
time choosing the contour to be to the left of the singularity on the imaginary axis. Again, using
the Nyquist criterion, check the range of K for which the system is stable. Are the answers the
same? Should they be?
Step-by-step solution
Step-by-step solution
step 1 of 10
Consider the following open loop transfer function of the system:
' '
(1)
j(j+ 3 )
Nyquist plot represents frequency response of the system.
Magnitude of ATG(y<9) ■'
■(2)
Phase of JirG(y(9)is,
j
=ta n -' - 9<r- ta n -'
j
(3)
Step 2 of 10
Magnitude of ATG(y<9) at ^ s O is.
Phase of JSrG(yfl») at
=
\KG{jo>) = tan"' f y l
” tan"' f j
* -9 0 *
Magnitude of ATG(y<9) at n>sco i
Step 3 of 10
Take ^ c o m m o n from the above equation.
n
if n . J - V + 1
l^ c o < » )L = M f—
=0
Phase of ATG(y(9) at <n>oois
|^G(ya>)| =tan-'
[ j]
= 9 0 --9 tr-9 (r
= -9 0 ‘
Follow the above procedure and calculate magnitude and phase of KG[^jG>) tor different values
of I
The Nyquist plot will always be symmetric with respect to the real axis.
Step 4 of 10
To choose contour to the right of the singularity on the
axis, the value of
in the equation
(1) must be positive.
Enter the following code in MATLAB to draw the Nyquist plot with contour to be to the right of the
singularity of y<naxis;
» numerator=[1,1];
» denominator=conv([1,0],[1,3]);
» system=tf(numerator,denominator);
» nyquist(system)
Step 5 of 10
The Nyquist plot for the feedback system is shown in Figure 1.
Step 6 of 10
Choosing the contour to the right of the singularity on the imaginary axis, the plot is as shown in
Figure 2.
FtgDre2
Step 7 of 10
The numbers of clockwise encirclements of -1 {denoted by
). by observing the Nyquist plot of
Figure 2 are zero.
Number of unstable (RHP) poles of KG[^jG>). denoted by p are zero.
Numbers of unstable closed loop roots Z are,
Z = JV+P
=
0+0
=0
Hence, there are no characteristic equation roots in the RHP.
So the system is stable for positive values of k ■
Thus, the range of
for which system is stable when contour is to the right of the singularity on
the ja> axis is |j^ > Q |.
Step 8 of 10
To choose the contour to the left of the singularity on the imaginary axis, value of x. in equation
(1) must be chosen negative.
The Nyquist plot for the feedback system, choosing contour to the left of the singularity on the
imaginary axis is as shown in Figure 3.
Step 9 of 10
The numbers of counter-clockwise encirclements of _ j by the contour in the Nyquist plot of
Figure 4 is.
iV = - l
The number of unstable (RHP) poles of ATG(yfl>). denoted by p a re ,
P =\
Numbers of unstable closed loop roots Z are,
Z = A ^+I
= -
1+1
=0
Hence, there is no characteristic equation root in the RHP.
So the system is stable for negative values of
.
Thus, the range of x for which system is stable when contour is to the left of the singularity on
the y<»axisis |jj^ > Q |.
Step 10 of 10
Hence, it is clear that the range of x for the system to be stable when the contour is to the right
of singularity on the imaginary axis is same as that of range of x for the system to be stable
when the contour is to the left of singularity on the imaginary axis.
Thus, the way of choosing the contour around singularity on the j & axis does not affect the
system stability criterion. The results should be same in either way.
Problem 6.21 PP
Draw the Nyquist plot for the system in Fig. Using the Nyquist stability criterion, determine the
range of K for which the system is stable. Consider both positive and negative values of K.
Figure Control system
Step-by-step solution
Step-by-step solution
step 1 of 12
Consider the feedback system shown in Figure 1.
Step 2 of 12
The closed loop transfer function for the above feedback system is,
y (t)
KG{s)
R {s )° l + KG(s)H{s)
^
I
( i ’ + 2 i + 2)
I
l+Ky
(s ' + 2 i+ 2 ) ( i+ l)
From the closed loop transfer function, the open loop transfer function
G (s)H (s) =
is obtained
(s“ + 2 j + 2 ) ( i + l)
Step 3 of 12
Nyquist plot represents frequency response of the system. Hence,
K
1
(2 -ffl’ + J2to) (y'ffl+l)
Step 4 of 12
Magnitude of G { ja ) H ( ja ) is.
^ ( 2 - e ^ f + A e ^ y ll + a ^
Phase of G {ja > )H {ja) is,
ZG{j(o)H(ja>) = 0 - a a -‘ [ 3 ^
] -
[f]
Step 5 of 12
Magnitude of G ( ja ) H { ja ) at (psQ is.
^ ( 2 - 0 ) ’ + 4 ( 0 ) ’ V i+ 0
£
' 2
Phase of G {ja > )H {ja) at (psQ is,
ZG(ya))W(yffl)=0 - tan-'
j-
(0)
= (T
Step 6 of 12
Magnitude of G { j a ) H ( j a ) at d>»oo is,
\G (ja \H {jo ii
■
^ ( 2 - 00)* + 4 ( 00)^ V l +«>“
=0
Phase of G (j< o )H (ja) at <2>sco is,
Z .G (ja)H ( J a ) = O - to n - '
j"
[7 ]
= - 9 0 '- W
= -180'
Follow the above procedure and calculate magnitude and phase of G { j o \ H { j a ) for different
values of to.
The Nyquist plot will always be symmetric with respect to the real axis.
The plot is normally created by the NYQUIST MATLAB m-file.
Step 7 of 12
Enter the following code in MATLAB to draw the Nyquist plot;
» num=1;
»den=conv([1 2 2],[1 1]);
» sys=tf(num,den):
» nyquist(sys)
Step 8 of 12
The Nyquist plot for the feedback system is as shown in Figure 2.
FigDre2
Step 9 of 12
Determining stability of closed loop system based on frequency response of system’s open loop
transfer function is Nyquist stability criterion.
Consider positive value of k ■
V
The maximum gain of G (7 ® )« (y < » ) isobtainedat ^ s Q ^ n d the value is — .
The numbers of clockwise encirclements of -1 {denoted by U ), by observing the Nyquist plot of
Figure 2 are zero.
Number of unstable (RHP) poles of G (7 ® )« (y < » ). denoted by /» are zero.
Step 10 of 12 ^
The numbers of unstable closed loop roots is,
Z = JV+P
=
0+0
=0
For positive value of
, Z = 0 : that is there are no characteristic equation roots in the RHP.
So the system is stable for positive values of j^ .
Step 11 of 12
Consider
value as negative.
The number of clockwise encirclements have to be now determined depending on value of
When j f > _ 2 , Nyquist plot doesn’t encircle _ j.
Number of clockwise encirclements of -1 . A ^=0
Number of unstable (RHP) poles of G { i a ) H ( j a \ denoted by /> are zero.
Z = JV+P
=
0+0
=0
For
> —2 . Z = 0 : that is there are no characteristic equation roots in the RHP.
So the system is stable for
> —2 •
Step 12 of 12 ^
When j f < _ 2 , Nyquist plot encircles _ j.
Number of clockwise encirclements of -1 . JV = 1
Number of unstable (RHP) poles of G { i a ) H ( j a \ denoted by /» are zero.
Z = J\7+P
= 1+0
=1
For fC < ~ 2 ’ Z = l^ that is there are characteristicequation roots in the RHP.
So the system is unstable for
< —2 •
Thus, it is clear from the obtained results that the range of
[Z H 5 I
for which system is stable is
.
Problem 6.22PP
(a)
For w = 0.1 to 100 rad/sec, sketch the phase of the minimum-phase system
and the nonminimum-phase system
G(s) = --y - 1
I
noting that A{joj - 1) decreases with oj rather than increasing.
(b)
Does an RHP zero affect the relationship between the -1 encirciements on a poiar piot and
the number of unstable closed-loop roots in Eq. (1)?
(b)
Does an RHP zero affect the relationship between the -1 encirciements on a poiar piot and
the number of unstable closed-loop roots in Eq. (1)?
Eq. (1)
Z ~ N + P.
(c)
Sketch the phase of the following unstable system for w = 0.1 to 100 rad/sec:
G{s) =
(d)
J+ 1I
j- lO ,
Check the stability of the systems in (a) and (c) using the Nyquist criterion on KG(s).
Determine the range of K for which the closed-loop system is stable, and check your results
qualitatively by using a rough root-locus sketch.
Step-by-step solution
step 1 of 4
G(S): ■ - M )
G (= )= —
'■ '
S+10
s+10
Step 2 of 4 ^
b.
No, it does not have any eSect on £Q (6 .2 8 )
Step 3 of 4
Step 4 of 4
d.
s+1
F o r G f s ) —-----^ ^ s+10
For G (S) =
s+10
s+1
F o r G ( s ) = - ;— r
-(s-l)
is stable for all values of K
0 < K < 1 for stabili^
0 < K < 1 forstability
Problem 6.23PP
Nyquist plots and the classical plane cun/es; Determine the Nyquist plot, using Matlab, for the
systems given below, with K =
and verity that the beginning point and end point for the jw > 0
portion have the correct magnitude and phase;
(a) The classical cunre called Cayley’s Sextic, discovered by Maclaurin in 1718;
1
KG(s) = K-
(» + !)’ ■
(b) The classical cunre called the Cissoid, meaning Ivy-shaped;
JTG(a) = K
1
i ( i + 1)
(c) The classical cunre called the Folium of Kepler, studied by Kepler in 1609;
CG(a) = K
1
( a - l) ( s - H ) 2 '
(d) The classical cunre called the Folium (not Kepler's);
KG(s) = K( j - l ) ( j1- F 2 )
(e) The classical cunre called the Nephroid, meaning kidney-shaped;
KG{s) = K -
(1 -1 )3
(f) The classical cunre called Nephroid of Freeth, named after the English mathematician T J.
Freeth;
( i- F l) ( j3 - F 3 )
KG(s) = K
4 ( r - l) 3
(g) A shitted Nephroid of Freeth;
_____
Step-by-step solution
step 1 of 14
(a)
Consider the open loop transfer function of the classical cunre called Cayley’s Sextic.
1
KG{s) = K -
Substitute j a for s.
K G (ja > ) = K — L ^
[ ja + l)
Determinethemagnitudeat ^ s Q a n d Jtf = l
1
K G (jta)= -
■ ( i/T iv ) ’
X G (0 ) = I
Determine the phase at ^ s Q a n d Jtf = l
z a :g ( o) = o°
Enter the following code in MATtAB to plot the Nyquist plot for
=i
» numG=1;
» denG=conv(conv([1 1],[1 1]),[1 1]);
» sysG=tf(numG,denG);
» nyquist(sysG)
Step 2 of 14
The Nyquist plot for the cunre is shown in Figure 1.
The magnitude of the beginning point is 1 and phase is Q* and hence has the correct magnitude
and phase.
Step 3 of 14
(b)
Consider the open loop transfer function of the classical cunre called Cissoid, meaning ivy­
shaped.
Substitute jo> for s.
Jta(ja>+\)
1
= AT
-a
+ ja
step 4 of 14
Determine themagnitudeat ^ s Q a n d Jtf = l
1
KG{Ja>)
■<lar+a
X G (0 )= « >
Determine the phase at ^ s Q a n d Jtf = l
Z K G O v») = - t a i T ' ^ - ^ j
- ( i )
ZATG(0) = 0°
Step 5 of 14
Enter the following code in MATtAB to plot the Nyquist plot for
=i
» numG=1;
» denG=[1 1 0];
» sysG=tf(numG,denG);
» nyquist(sysG)
The Nyquist plot for the cunre is shown in Figure 2.
The magnitude of the beginning point is oo and phase is 0® and hence has the correct
magnitude and phase.
Step 6 of 14
(c)
Consider the open loop transfer function of the classical cunre called Folium of Kepler;
1
KG (s) = K -
(4 -l)(4 + l)'
Substitute j a for s.
K G (ja )-K —
(y V » -l)(y « > + l)’
^ ( l + a i‘ ) . ^ ( l - a ‘ )^ + 4 a ‘
step 7 of 14
Determine themagnitudeat ^ s Q ^ n d Jtf = l
1
K G {ja) = -
^ { l+ a ’ ) ^ j{ l- a ? f+ 4 a ‘
K G {0 )-\
Determine the phase at ^ s Q ^ n d Jtf = l
Z K G (ja )= -lS iy‘
Enter the following code in MATtAB to plot the Nyquist plot for
=i
» numG=1;
» denG=conv([1 -1],[1 2 1]);
» sysG=tf(numG,denG);
» nyquist(sysG)
Step 8 of 14
The Nyquist plot for the cunre is shown in Figure 3
Hence the magnitude and phase of the beginning point are verified.
Step 9 of 14
(b)
Consider the open loop transfer function of the classical cunre called Folium;
1
KG(s) = K j
(» -1 )(4 + 2 )
Substitute J a for s.
1
K G (ja ) = K -
{ J a - l) ( J a * 2 )
1
=K -
Determine themagnitudeat ^ s Q ^ n d Jtf = l
K G ( ja ) = -
'
'4 + a
JCG(0) = i = 0.S
Determine the phase at ^ s Q ^ n d Jtf = l
■ - tan " '(r» ) - t a n " ' j
Enter the following code in MATtAB to plot the Nyquist plot for Jtf = i
» numG=1;
» denG=conv([1 -1],[1 2]);
» sysG=tf(numG,denG);
» nyquist(sysG)
Step 10 of 14
The Nyquist plot for the cunre is shown in Figure 4.
Hence the magnitude and phase of the beginning point are verified.
Step 11 of 14
(e)
Consider the open loop transfer function of the classical cunre called Nephroid, meaning kidney
shaped;
KG (s) = K
2 ( j + 1)(4’ - 4 j + i )
(4 -1 )’
Ja
Substitute
for s.
( y o - l)
KG(0) = 2
Determine the phase at at = 0 b n d Jtf = l
^ G [ja )
■tan"' (nr)+tan"' ^|-^^j-3tan"' {- a )
= 0»
Enter the following code in MATtAB to plot the Nyquist plot for
s |
» numG=2*conv([1 1],[1 -4 1]);
» denG=conv(conv([1 -1],[1 -1]),[1 -1]);
» sysG=tf(numG,denG);
» nyquist(sysG)
Step 12 of 14
The Nyquist plot for the cunre is shown in Figure 5.
Nyaebl Dtoaraw
Hence the magnitude and phase of the beginning point are verified.
(f)
Consider the open loop transfer function of the classical cunre called Nephroid of Freeth.
4 ( j- l)
Ja
Substitute
for s.
4 (y < » -l)
^ 2 7 ( l4 ‘t»’ ) ( 3 - « . ’ )
4 ( V IT ^ )’
Determine themagnitudeat nt = 0 b n d Jtf = l
4 ( v T iV ]
XG(0) = j = 1.5
Determine the phase at a> = 0 b n d Jtf = l
Z K G (ja ) ■tan"' (ar)-3tan"' (-®)
=0
Enter the following code in MATtAB to plot the Nyquist plot for Jtf = i
> numG=2*conv([1 1],[1 0 3]);
» denG=4*conv{conv([1 -1],[1 -1]),[1 -1]);
» sysG=tf(numG,denG);
» nyquist(sysG)
Step 13 of 14
The Nyquist plot for the cunre is shown in Figure 6.
Hence the magnitude and phase of the beginning point are verified.
Step 14 of 14
(g)
Consider the open loop transfer function of the classical cunre called shifted Nephroid of Freeth.
r V
(^’ + 0
^’
4 (4 -1 )’
KG (s) = K-^ ------ V
Ja
Substitute
for s.
(y»-l)
■
(v /n v )’
Determine themagnitudeat a> = 0 b n d Jtf = l
( V T iV )
a:g ( o) = i
Determine the phase at a> = 0 b n d Jtf = l
ZKG(ja) ■ -3tan"' (-<»)
=0
Enter the following code in MATtAB to plot the Nyquist plot for
s |
» numG=[1 0 1];
» denG=conv(conv([1 -1],[1 -1]),[1 -1]);
» sysG=tf(numG,denG);
» nyquist(sysG)
The Nyquist plot for the cunre is shown in Figure 7.
Hence the magnitude and phase of the beginning point are verified.
Problem 6.24PP
The Nyquist plot for some actual control systems resembles the one shown in Fig. What are the
gain and phase margin(s) for the system of Fig., given that a = 0.4, /5 = 1.3. and 0 = 40*.
Describe what happens to the stability of the system as the gain goes from zero to a very large
value. Sketch what the corresponding root locus must look like for such a system. Also, sketch
what the corresponding Bode plots would look like for the system.
Figure Nyquist plot
Im [C (j)]
^
/
/
/
I
I
1
\
/
/
/
-1
V
U -a..
1
/n
1
1
1
X 1
1
1
1
t /W tf/
ll
Re[G(s)]
S te p -b y -s te p s o lu tio n
Step 1 of 2
P M = q ) = Aiy
G a fl = - = — = 2.5
2 0.4
a n f 2 = - = 0.77
If gain lies between 0.77 and 2.5 system is stable
If gain > 2.5 system becomes unstable
Root locus will look like
Step 2 of 2
Re(s)
Problem 6.25PP
The Bode plot for
1 0 0 [(i/1 0 ) + l ]
G (J) =
4 W D - 1 ] [ ( V 1 0 0 ) + 1]
is shown in Fig.
(a)
Why does the phase start at -270° at the low frequencies?
(b)
Sketch the Nyquist piot for G(s).
(c)
Is the closed-loop system for the Bode plot shown in Fig. stable?
(c)
Is the closed-loop system for the Bode plot shown in Fig. stable?
(d)
Will the system be stable if the gain is lowered by a factor of 100? Make a rough sketch of a
root locus for the system, and qualitatively confirm your answer.
Figure Bode plot
S t e p - b y - s t e p s o lu t io n
step 1 of 8
Consider the transfer function of the system.
looj^
G (,)
(1)
j[i-l]
(a)
Calculate the phase at low frequencies.
Refer to the Figure 6.91 from the text book.
It is clear from the Figure that the phase of —27CP is obtained at ^ s 0.01 •
Calculate phase of the system at <ps0.01 for the system shown in equation (1).
= 0 ® + 0 .0 5 7 2 ® -9 0 ® -(l8 0 * -0 .5 7 2 ® )-0 .0 0 5 7 9 *
— 269.37®
»270®
Step 2 of 8
It is clear that the pole at 5 s l in equation (1) contributes to a phase of —igo*
d> = 0.01<
pole at 5 s 0 io equation (1) contributes to a phase of _90* for ^ s 0.01 >pole at 5 s -10 0 io
equation ( 1) contributes to a phase of 0* for ^ s 0.01 ^od zero at 5 = -1 0 contributes to a phase
of O' fo r® = 0.01
The phase at low frequencies is obtained by calculating the total phase,
total phase=-180®-90®+0®-0®
= -270®
Thus, it is clear from the total phase obtained that the phase at low frequencies is —270®. so the
phase graph in Figure 6.91 started from the angle of —270®-
Step 3 of 8
(b)
Nyquisf plof represents frequency response of the system.
fe-')
100
100
G (j< o )H (ja )= -
UOO
)
Magnitude of G { ja ) H ( ja ) Is,
I/--/
M
ur/
+1
Phase of G{ja>)H {ja) Is,
| £ ( y ® ) £ ^ = 0 » + t a n - '[ ^ j - 9 0 » - [ ^ 1 8 0 » - t a n - ' [ ^ j j - t a n - ' ( ^ j
Step 4 of 8
Magnitude of G ( ja ) H { ja ) at (2> a 0 is
100
7(0)’ +1
|g ( H w ( / < » ) L =
= 00
Phase of G{ja>)H {ja) at a^sOis
| G ( ^ f f l ) £ ^ = 0 '> + t a i r '^ ^ j- 9 0 ‘> - ^ 1 8 0 » - t > i r ' ^ ^ j j - t a i r ' ^ ^ j
= -2 7 0 «
Magnitude of G { ja ) H ( ja ) at a> = oo I
"vU .
100
+0
100
=0
Step 5 of 8
Phase of
at a> = ools
|G ( ja > ) g ( y a > ) = 0 ° - f a n - ' ^ ^ j - 9 0 » - ^ 1 8 0 ° - t a n - ' [ y ] ] - ! ” ' ' [ ^ ]
= 9 0 ° - 9 0 ° - ( 1 8 0 ° - 9 0 '> ) - 9 0 »
= -1 8 0 »
Follow the above procedure and calculate magnitude and phase of G{ja>)H {ja) for different
values of a>.
The Nyquist plot will always be symmetric with respect to the real axis.
The plot is normally created by the NYQUIST MATLAB m-file.
Enter the following code in MATLAB to draw the Nyquist plot:
» n u m =[10 100];»den=conv ([1 -1 0],[0.01 1]);
»sys=tf(num,den);
»nyquist{sys)
Step 6 of 8
The Nyquist plot for the system of equation (1) is as shown in Figure 1.
Figure 1
Step 7 of 8
(c)
It is clear from the Figure that there is a counter clock-wise encirclement, so jy = 1.
It is clear from the equation (1) that there is one right hand side pole at $ s 1.
Numbers of unstable closed loop roots Z are,
Z = JV+/*
= 1-1
=0
For any value of
Z = 0 : that is there are no characteristic equation roots in the RHP.
So the system is stable for any value of f c .
Thus, for all values of ^ the system is stable.
Step 8 of 8
(d)
It is clear from the equation (1) that the gain of the system is 100. So the system becomes
unstable if the gain condition is
< 1 , that is gain becomes 1. It is clear the gain of the system
equals 1 if the gain is decreased by a factor of 100 .
Thus, the system becomes unstable when the gain decreases by a fector of 100.
Problem 6.26PP
Suppose that in Fig.,
2 5 (s+ l)
G{s) = s(s + 2 ) ( j 2 + 2 r + 1 6 )
Use Matlab’s margin to calculate the PM and GM for G(s) and, on the basis of the Bode plots,
conclude which margin would provide more useful information to the control designer for this
system.
Figure Control system
-O Y
G(s)
Step-by-step solution
step 1 of 4
The loop transfer function is,
C(s)=
2 S (j-t-l)
s (5+ 2)( s^+2 s + 6)
Write the MATLAB code to calculate the gain margin and phase margin.
num=[25 2 5 ];» den=conv{[1 2 0],[1 2 1 6 ]);» sys=tf{num,den)sys =
25 s + 25 —
------------ sM + 4 s^'3 + 20 s^'2 + 32 s Continuous-time transfer function.» margin(sys)
Step 2 of 4
Draw the bode plot.
Bode Diagram
Frequency (rad/s)
Figure 1
Step 3 of 4
From Figure 1,
The gain margin is, <7A /s3.91dB
The phase margin is,
s 101^
Therefore, the gain margin, G M
|3.9! dBl
phase margin, /> A /Is | | q | o|.
Step 4 of 4
The phase margin is more commonly used to specify control system performance it is most
closely related to the damping ratio of the system.
Problem 6.27PP
Consider the system given in Fig.
(a)
Use Matlab to obtain Bode plots for K = 1, and use the plots to estimate the range of K for
which the system will be stable.
(b)
Verify the stable range of K by using margin to determine PM for selected values of K.
(c)
Use hocus to determine the values of K at the stability boundaries.
(d)
Sketch the Nyquist plot of the system, and use it to venfy the number of unstable roots for the
unstable ranges of K.
(d)
Sketch the Nyquist plot of the system, and use it to venfy the number of unstable roots for the
unstable ranges of K.
(e)
Using Routh’s criterion, determine the ranges of K for closed-loop stability of this system.
Figure Control system
Step-by-step solution
step 1 of 15
Refer to Figure 6.93 in the textbook.
From the block diagram, the loop transfer function is.
L ( » ) = A rC ( * ) W ( j)
•)+ l
( i - l ) ( i ’ + 2 j+ 2 )
Step 2 of 15
(a)
Substitute i for
in loop transfer function.
iW = 7
( j - 1 ) ( s * + 2j + 2 )
( j - l ) ( s ’ + 2 i+ 2 )
Write the MATLAB code to plot the bode diagram.
» num=[1 2];den=conv([1 -1],[1 2 2]);sys=tf{num,den)bode(sys)sys =
s-2
s^2 - 2 Continuous-time transfer function.
Draw the bode diagram.
Bode D iig ram
Freqaeaqr (rtd/s)
Figure 1
Step 3 of 15
From the bode plot, the closed loop system is unstable for ATs 1. But this can make the closed
system stable with positive gain margin by increasing the gain K up to the crossover frequency
reaches at ^ s 1.42rad/s. where phase plot crosses the —]80^ line.
Therefore, the range of x for the stable system is ||< jj^< 2| •
Step 4 of 15
(b)
= 1.5 •
Calculate the phase margin for
Write the MATLAB code to plot the bode diagram.
» num=[1.5 3 ] ;» den=conv([1 -1],[1 2 2 ]);» sys=tf(num,den)sys =
1.5 s + 3
+ s^2 - 2 Continuous-time transfer function. » margin{sys)
Step 5 of 15
Draw the bode plot.
Bode Diagram
Freqieacy (rad/s)
Figure 2
Step 6 of 15
From the plot, the phase margin is, PM = 6.66^^^ Al =1.5Therefore, the phase margin at A T s l.5 is
Step 7 of 15
(c)
Calculate the characteristic equation.
l + J C G (j)W ( i) = 0
s*2
\ + K\
= 0 ...... (1)
( * - i) ( ( * + i) ^ + i)
Step 8 of 15
Write the MATLAB code to plot the root locus.
» num=[1 2 ] : » den=conv([1 -1],[1 2 2 ]);» sys=tf{num,den)sys =
s+2 •
- 2 Continuous-time transfer function.» rlocus(sys)
Step 9 of 15
Draw the root locus plot.
Figure 3
Step 10 of 15
From the characteristic equation,
( j - l ) ( j ' + 2 i+ j)+ /:( j + 2)«0
j ’ + « ^ - 2 + / r ( i+ 2 ) = 0
»’ + * ’ + K t-2 + 2 A : = 0
(2)
Step 11 of 15
Apply Routh-Hurwitz criterion to find the intersection of the RL with the imaginary axis.
s’
1
s’
1
s' K -2 + 2 K
s'
-2 + 2 K
K
-2+2iC
0
0
For stable system, all first column elements should be positive.
Thus,
K -2 -2 K > 0
K<2
And,
-2 + 2 K > 0
K>\
Therefore, the range of x fo'’ fbe stable system is |} < j^ < 2 | •
Step 12 of 15
(d)
Write the MATLAB code to plot the Nyquist plot.
» num=[1 2 ] : » den=conv([1 -1],[1 2 2 ]);» sys=tf{num,den)sys =
s+2 -
- 2 Continuous-time transfer function.» w=logspace(-2,2);» nyquist(sys,w)
Step 13 of 15
Draw the Nyquist plot.
P ^ o is t Diagram
Step 14 of 15
(i)
For 0 < K < \ .
The number of encirclements are,
Pole, p = l .
Therefore,
Z = N+P
= 0+1
=1
The system has one unstable closed loop root.
(ii)
For 1 < a: < 2 ,
The number of encirclements are. JV = -1
Pole. p = \ .
Therefore.
Z = N+P
= - 1+1
=0
Thus the system is stabie.
(Ill)
For 2 < K .
The number of encirclements are. JV *1
Pole, P a l .
Therefore,
Z = N+P
= 1+ 1
=2
Thus, the system has two unstable closed loop roots.
Therefore, the range of x for the stable system is | | < jj^ < 2 | •
Step 15 of 15
(e)
Recall the characteristic equation.
»’ + i ’ + K t - 2 + 2 A : = 0
Apply Routh-Hurwitz criterion to find the intersection of the RL with the imaginary axis.
1
1
K
-2 + 2 K
K -2 + 2 K
-2 + 2 K
0
0
For stable system, all first column elements should be positive.
Thus,
K -2 -2 K > 0
K<2
And,
-2 + 2 J ^ :> 0
^>1
Therefore, the range of x for the stable system is | | < jj^ < 2 | •
Problem 6.28PP
Suppose that in Fig.
3 .2 (1 + 1 )
<Hs) = -
j (i
+ 2)( i 2 + 0.2 s + 1 6 ) '
Use Matlab’s margin to calculate the PM and GM for GfsJ, and comment on whether you think
this system will have well-damped closed-loop roots.
Figure Control system
|cw|—p o r
Sten-hv-<;ten ^ nliitinn
Step-by-step solution
step 1 of 4
Refer to the control system in Figure 6.90 in the textbook.
The open loop transfer function is.
,
3 .2 ( i + l)
g (5 )= ■
^
V
' ' j ( i + 2 ) ( s ’ + 0 .2 j+ 1 6 )
3.2(i + l)
/ + 2 . 2 i ’ +16.4i*+32*
Step 2 of 4
The MATLAB code to obtain gain and phase margins is,
num=[3.2 3.2];
den=[1 2.2 16.4 32 0];
sys=tf(num,den);
bodeplot(sys)
margin(sys)
grid
Step 3 of 4
The Bode plot is shown in Figure 1.
Bode Diagram
Freqaeaqr (rad/s)
Figure 1
step 4 of 4
From Figure 1, it is clear that the gain margin is.
G M = l d B a t ^ = 4 . 0 2 rad/s
The phase margin is,
PM = 9 2 . 8 'at ffl = 0 .1 n u j/s
Since p |^ * 92.8* ■
damping will be 1. That is the roots of the system are real. However from
the Figure, it is clear that a very small change gain would cause instability to believe that the
damping of roots is very small.
Problem 6.29PP
For a given system, show that the ultimate period Pu and the corresponding ultimate gain Ku for
the Ziegler-Nichois method can be found by using the following:
(a)
Nyquist diagram
(b)
Bode plot
(c)
Root locus
Step-by-step solution
Step-by-step solution
ste p 1 of 4
K g and Tg are related to gain
andintegtal dehvative
times Tr and Tn
Step 2 of 4 ^
a.
In Nvquist diagram the point at
plot crosses real axis
2fr
is the desired point. At that point Pg = —
Tiriiere a is phase
cros sover frequency and gain is
Step 3 of 4 ^
In Bode plot the point at vidiich plot crosses 0 axis is the
desired point to find K g andPg
Step 4 of 4
In root locus plot the point at which plot crosses imaginary
axis is the desired point to find R g and Pg
Problem 6.30PP
If a system has the open-loop transfer function
G (s) = -
s(s + 2(o>n)
with unity feedback, then the closed-loop transfer function is given by
Q)i
T (s) =
-
^ " --------y.
+ 2f0)nS +
Verify the values of the PM shown in Fig. f o r^ = 0.1,0.4, and 0.7.
Figure Damping ratio versus PM
Step-by-step solution
step 1 of 2
G (0 = -
s(s+2?<ii.)
A nd closed loop transfer function is given by
T ( s) = t -----^
------ 5-
s +25oa^s+cc^
Step 2 of 2
PM as given by equation 6.31 is
PM = tan"
2^
Jbr 4 = 0.\ , P i f = tan
2x0.1
= 110.42*1
V > f t + 2 x 0 .f - 2 x - 0 .f
Jbr ^ = 0.4 . PM=taa~^
2x0.4
=1
^
.V V l+2x0.4" -2 X -0 .4 V
Sor ^ = 0.7 , PM
2x0.7
_> /V n-2x0.7*-2x-0.7^
All these values matche closely with fig: 6.36
= 170.33*1
Problem 6.31 PP
Consider the unity-feedback system with the open-ioop transfer function
K
GW = ■ s(s + l) [ ( P / 2 5 ) + 0 .4 ( i/S ) + 1] ■
(a)
Use Matlab to draw the Bode plots for G( jo)). assuming that K =
(b)
What gain K is required for a PM of 45°? What is the GM for this value of K7
(c)
What is Kv when the gain K is set for PM = 45“?
(d)
Create a root locus with respect to K, and indicate the roots for a PM of 45*.
(d)
Create a root locus with respect to K, and indicate the roots for a PM of 45*.
Step-by-step solution
step 1 of 10
The open-ioop transfer function of the system is,
G (4 =
K iR iH
+1
” f [Q.CMi"+0.08s’ + i+ 0 .0 4 j’ +0.08*+ l]
____________ K ___________
“ j (0 .0 4 j ’ + 0 . 1 + 1 ,08 j + 1)
^ __________ K __________
0.04j* + 0 .1 2 i’ + 1.08i’ + j
G ijoA —
^ i
.
*
o.04(y<») + o . i 2 ( y « ) ) + i . 08(y<») + ;«>
Step 2 of 10
(a)
Write the MATLAB code to draw the Bode plot for
with J ^ s l.
n u m G = [ 1 ] ;
d e n G = [ 0 . 0 4
0 . 1 2
1 . 0 8
1
0 ] ;
s y s G = t f ( n u m G j d e n G ) ;
[ m a g , p h a s e , w ] = b o d e ( s y s G ) ;
l o g l o g ( w , s q u e e z e ( m a g ) )
s e m i l o g x ( w , s q u e e z e ( p h a s e ) )
Get the MATLAB output for the Bode Magnitude plot.
Step 3 of 10
Get the MATLAB output for the Bode phase plot.
Step 4 of 10
(b)
Write the MATLAB code to get the gain K for a phase margin of 45®.
n u m G = [ 1 ] ;
d e n G = [ 0 . 0 4
0 . 1 2
1 . 0 8
1
0 ] ;
s y s G = t f ( n u m G , d e n G ) ;
m a r g i n ( s y s G )
Step 5 of 10
iviai i\ u ic pun II ui I u ic pi laoc piui wi icic ii ic pi lasc lo
" luiuaico a pi laoc iiiaigii i '
45» Mark the corresponding point on the magnitude plot to get the gain in dB.
Bode Diagnzn
Step 6 of 10
The magnitude in dB for a phase margin of 45 ® is,
2 0 lo g |C ( > ) | = -0.996
|G(7<»)| = 0.892
Determine the gain K for a phase margin of 45® using the condition that the magnitude is 1 at
the phase margin.
|( ^ L « .) ( |G ( y < » L - r ) |= '
___ 1 _
* 0.892
=
1.12
Thus, the gain K for a phase margin of 45® is m
step 7 of 10 ^
Write the MATLAB code to determine the gain margin for fC = 1.12 •
n u m G = [ 1 . 1 2 ] ;
d e n G = [ 0 . 0 4
0 . 1 2
1 . 0 8
1
0 ] ;
s y s G = t f ( n u m G , d e n G ) ;
m
a r g i n ( s y s G )
Step 8 of 10
Get the MATLAB output for the Bode plot.
BodeD U grw
Thus, the gain margin for K - \ . \ 2 is 14.9dB .
Step 9 of 10 ^
(c)
The open loop transfer function of the system for Jir = 1.12 is,
gf
_
1.12_________
^ ' " j (0.04 s ’ + 0.12 j ’ + I.08 j + 1)
Determine the velocity constant,
.
K, = Lt sG (s)
1.12
-U s
j
(0.045*+0.12 j * + 1.08j + 1)
=U
"
(0.04j * + 0.\2s^ +\.0K s + 1)
=
1.12
Thus, the velocity constant, K , for phase margin equal to 45® is m
step 10 of 10
(d)
Draw the root locus of the system using MATLAB and indicate the roots for a PM of 45®, that is,
when the gain K \s^ .12.
R o o tL o c u
Observe from the plot that the roots are -0.468± y0.971.
Thus, the roots for a PM of 45® are |-Q .468± y 0 .9 7 l|.
Problem 6.32PP
For the system depicted in Fig{a), the transfer-function blocks are defined by
1
1
G (s ) =
(5 -1 -2 )2 (5 -I-4 )
'H ( s ) =
5-1-1
(a) Using hocus and riocfind, determine the value of K at the stability boundary.
(b) Using hocus and riocfind, determine the value of K that will produce roots with damping
corresponding to
(c)
0.707.
What is the GM of the system if the gain is set to the value determined in part (b)? Answer
this question without using any frequency-response methods.
(c)
What is the GM of the system if the gain is set to the value determined in part (b)? Answer
this question without using any frequency-response methods.
(d)
Create the Bode plots for the system, and determine the GM that results for PM = 65®. What
damping ratio would you expect for this PM?
(e)
Sketch a root locus for the system shown in Fig (b). How does it differ from the one in part
(a)?
(f)
For the systems in Figs, (a) and (b). how does the transfer function V2(s)/R(s) differ from
Y1 {s)^(s)7 Would you expect the step response to r(t) to be different for the two cases?
Figure Block diagram; (a) unity feedback; (b) H(s) in feedback
S te p -b y -s te p s o lu tio n
Step 1 of 25
(=0
Given
o ( s ) = ------------2------------
^ '
( s + 2 ) \s + A )
H (s ) — ! -
^ '
s+1
( s + 2 ) ^ ( s + 4 ) ( s + l)
Step 2 of 25
MATLAB program for root locus
» b = [1 4 ];
»
»
»
»
c=[l 1];
<^onv(a,b);
den=conv(c,<5;
num=[l];
» ^£(n um ,den )
Step 3 of 25
Transfer fimction:
1
8^4 + 9 s^3 + 28 s^2 + 36 s + 16
» rlocus(x)
» rloc£ind(x)
Select a point in the grs^hics window
selected_point =
0.0179 + 2.00821
ans = 81.9598
Step 4 of 25
From the matlab code for .AT= 81 root locus crosses imaginary a
For ^ > 8 1 system is unstable
Step 5 of 25
b)
^F=[14 4];
» b = [1 4 ];
»
»
»
»
»
c=[l 1];
^conv(a,b);
dei^o n v (c,d );
num=[l];
^£(num ,den)
Step 6 of 25
Transfer function:
1
sM + 9s^3 + 28 s^2 + 36 s + 16
» rlocus(x)
» v=[.6 0 -3 3];
» sgrid([0.7a7],[])
» gtext('\2eta*0.707')
» axis(v);
» rlocfmd(x)
Select a point in the graphics window
Step 7 of 25
selected_point =
-0.9092 + 0.90861
ans = 5.9288
Step 8 of 25
Root locus is
K = 5.9288 Produces roots with damping ratio corresponds to
0.707
Step 9 of 25
c)
Gain margin is
g
= -? L
5.928
G = 13.66
Step 10 of 25
<0
Given
0 ( s ) « ( s ) = -------- 3—^^-----------^ ' ( s + 2 ) '( s + 4 ) ( s + l )
Bode plot for open loop transfisr for iT = 1
a ^ l 4 4];
» b = [1 4 ];
» c=[l 1];
» ^conv(a,b);
» den=conv(c,<5;
» num=[l];
» }^f(num,den)
Transfer function:
1
sM + 9 s^3 + 28 s^2 + 36 s + 1 6
» bode(x)
» grid
Step 11 of 25
Bode plot is
Step 12 of 25 ^
Phase margin
^ = 1 8 0 ° + ^ where
is the pahse o f G(j(zr) of a =
v4ien y = 65®
65=180®+^^
d>, = -115®
Step 13 of 25 ^
With
1 the db gain at
135® is -2 9 .6 d b . This gain shouldbe made zero to have
to PM of 65®. Hence to every point o f magnitude plot a db gain of 26.6db shouldbe
added. The corrected magnitude plot is obtained by shifting the plot with R=1 by 29.6db
iqswards. The magnitude correction is independent of frequency. Hence the magnitude of
29.6db is contributed by the term K. The value is calculated by equating 2 0 1 o ^ to
29.6db
201og£'=29.6
294
A T =10»
^ = 30.12
Step 14 of 25
With K = 30.12 open loop transfer function is
a { s ) H [ s ) ----------------------------^ ' ( s + 2 )’ ( s + 4 ) ( s + l)
Now the bode plot for above system is
» a = [ 1 4 4];
» b = [1 4 ];
»
»
»
»
»
c=[l 1];
^conv(a,b);
den*conv(c,^;
num=[30.12];
^tf(num ,den)
Transfer function:
30.12
sM + 9 s^3 + 28 s^2 + 36 s + 1 6
» bode(x)
» grid
Step 15 of 25
Step 16 of 25
From the above bode plot
Gain margin = 8.56db
Gain margin in linear scale
201ogK=8.56
8.S6
K = -[0 ^
K = 2.6B
So gain margin is 2.68
Step 17 of 25
Approximate danq^ing ratio
PM
100
65
c=-
100
^ '* 0 .6 5
Step 18 of 25
Fig (a)
Step 19 of 25
Fig(b)
Step 20 of 25
Root locus is drawn for 0(^b)H(^ s)
For figures
0 { s ) H { s ) = K O { s )H (s )
So Root locus for both the figures is same
And root locus is
Step 21 of 25
£)
For figure (a) Transfer function is
T,(s)
K a (s )H {s )
R [s )~ l+ K O {s )H {s )
1
yi{s)
(s + 2 ) ^ ( s + 4 )
(s+ 2 ) (s+ 4 )
J^(s)
. . 1
s+ 1
s+1
1
( s + 2 ) ^ ( s + 4 ) ( s + l)
(ff+2) (ff+ 4)(ff+ l)
W
__________ f
R{s) J5 T + (s+ 2 )^(s+ 4 )(s+ l)
Step 22 of 25
For figure (b) Transfer function is
r,( s ) _
K a js )
R (s )~ \+ K O [s )H [s )
jv
r^ (s)
^ W " l+ -
1
(s + 2 )^ (a + 4 )
K
1
(s+ 2 ) (s+ 4 )
jv
^
s+1
]
(s + 2 )^ (s + 4 )
^
-----------( s + 2 ) ( s + 4 ) ( s + l)
r ,( s )
i:(s + i)
5 (s)
J i:+ (s + 2 )’ ( s + 4 ) ( s + l)
step 23 of 25
M AILAB program for step response
» a = [ 1 4 4];
» b = [1 4 ];
» c=[l l]i
» ^conv(a,b);
» dei^o n v (c,d );
» n u m = [ l 1];
» ^f(n u m ,d en )
Step 24 of 25
Step 25 of 25
Problem 6.33PP
For the system shown in Fig., use Bode and root-locus plots to determine the gain and frequency
at which instability occurs. What gain (or gains) gives a PM of 20*? What is the GM when PM =
20*7
Figure Control system
Step-by-step solution
Step-by-step solution
step 1 of 1
Open loop transfer function of the unity feedback
system is given by
T (s) =
s^ (s+3) [s^+2s+25j
From Madab we get &at at ^d= 4.96 instability occurs
A t \tD— 4.96 and Gain —S 2 |, instability occurs
A !soalffl= 034rad/sec , |K =4.17 to get PM = 20°|
Also
G M = 2 2 .4 d B
at
PM = 20“
Problem 6.34PP
A magnetic tape-drive speed-control system is shown in Fig. The speed sensor is slow enough
that its dynamics must be included. The speedmeasurement time constant is rm = 0.5 sec; the
reel time constant \sTr= J/b = A sec. where b = the output shaft damping constant = 1 N m- sec;
and the motor time constant is r1 = 1 sec.
(a)
Determine the gain K required to keep the steady-state speed error to less than 7% of the
reference-speed setting.
(b) Determine the gain and phase margins of the system. Is this a good system design?
Figure Magnetic tape-drive speed control
I Amplifier
p and motor
[ 7 ™ ,^
I ^ drive
Step-by-step solution
Step 1 of 10
Refer to the Figure 6.94 from the text book.
It is clear that the value of speed measurement time constant
is 0.5 seconds, the value of
reel time constant r^is 4 seconds, the value of output shaft damping constants is 1 N.nLsecand
the value of motor time constant r . i s l second.
Obtain the transfer function of the system.
Thus, the transfer function of the system for the Figure 6.94 is as follows.
f— If—1
T {s ).
1 -1-
( 1)
( s r , + ! ) ( « , + l)( y s + i) + iT
Step 2 of 10
(a)
It is clear from equation (1) that the system is of Type 1.
It is clear that for Type 1 system, there exists positional error constant
.
The steady state error of the system is.
1
1+*.
(2 )
It is clear that it^in equation (2) represents a positional error constant.
Calculate the steady state error.
sR{s)
l + G{s)H{s)
= lim1+
n
i+ ( i) ( jr )
Step 3 of 10
It is clear that the value of b is 1 N.m.sec
The steady state error of the system is.
I
Substitute 1 N.m.sec
^ in steady state emor.
1
Equate this expression to equation (2).
1
_
1
1 + /:= !+ * ,
Thus, it is clear that the value of K is equal to k^ .
Step 4 of 10
Calculate the value of gain such that the steady state error is less than
.
e^£7%
1
7 ^ 0 .0 7
1 + ^ :"
1
z^\+K
0 .0 7 "
I4 .2 8 S 1 + /:
1 4 .2 8 -!£ a:
13.28£JC
Hence, approximately the value of K is.
Thus, for the emor less than 7% of the reference, the gain must be
.
Step 5 of 10
(b)
Recall equation (1).
r(j)=
A r(tr, + 1)
{ « ’, + l ) ( i r , + l ) ( J s + 4 ) + ^ T
i ( s T , + l)(*r,+
+
1j + X
Substitute 13 for/C, 1 N.m.sec for b. 0.5 secfor
. and 1 sec for r,-
13(0.5 j + 1)
'■ '
( 0 .5 i + l ) ( j + l ) ( 4 j + l)+ 1 3
1 3 (0 .5 i + l)
0.5*’ + 3 .5 * '+ 5 .5 * + 1 4
Step 6 of 10
The transfer function of the fonvard loop is.
<?(*)
1
( j + l)(4 i + 1 )
Simplif
Simplify the expression further.
C (*) =
1
(4i* + j + 4 j + l)
I
+ J5a> + 1
I
( l + 4<P*) + JS49
step 7 of 10
Find the magnitude of G(s)
| c ; ( * ) U - , ------------!--------------^ ( l + 4® ’ ) ' + (5® )’
Find |G (* )|a tZ G (* ) = -180'
The phase shift is.
Step 8 of 10
Hence.
I G (*) I = 0.79 at ZG(s) = -180’ .
The GM can be calculated as the inverse of the magnitude of G(s) at Z G (4 ) » -180*Find the GM.
G M «t
I
1
0.79
-1 .2 7
Thus, the gain margin GM is ||,27l-
Step 9 of 10
Find the Z G (4 ) when | G ( j ) | - 1 •
Thus, we get Z G ( f ) = -I73®The phase margin PM is obtained by adding Z G ( 4 ) when | G(iS)| - 1 with 180^.
Find the PM.
PM = [ ^ C ( * ) | | ^ , , | . , ] + 180»
■ -I73 ® - f 180®
-7 ®
Thus, the phase margin PM is 0
Step 10 of 10
Since GM is low, the system is very close to instability and since PM is low, the damping ratio is
low and the overshoot is high.
For the system to be more stable, the gain should be low. If the gain is less, the steady state
error e„ will be high.
Thus, this is not a good design and needs a compensator.
P ro b le m 6 .3 5 P P
For the system in Fig., determine the Nyquist plot and apply the Nyquist criterion
(a)
to determine the range of values of K (positive and negative) for which the system will be
(b)
to determine the number of roots in the RHP for those values of K for which the system is
unstable. Check your answer by using a rough rootlocus sketch.
Figure Control system
S te p -b y -s te p s o lu tio n
step 1 of 14
(a)
Refer to the feedback system in Figure 6.95 in the textbook.
The closed loop transfer function for the above feedback system is:
R
I+ G (s)N (s)
3
j( j+ I) ( j+ 3 )
1 + a:
3
|i
« ( j+ l) ( « + 3 ) J
3R
5 ( j + 1) (j + 3 ) + 3 ^
From the above transfer < 7 ( ^ )//( j) c a n be obtained as shown below.
1(5 + l)(*+3)]
Step 2 of 14 ^
Nyquist plot represents frequency response of the system.
The magnitude of
is,
| G ( y - » ) « ( H ------ 1
(2)
+
V 9+ < ir
The phase of
is,
(y®) = 0 -90* - I a n " ' t a n " '
Step 3 of 14
Magnitude of G ( j a ) f / ( j a ) at (psQ is
*00
Phase of G ( j ( » ) f / ( j a ) at ® a Q is
|G (y '® )tf(y ® ) = 0 -9 0 * - tan"'
j “
^j
a -9 0 *
Step 4 of 14
Magnitude of G ( j a ) f f ( j a ) at ® a 00 Is
=0
Phase of G ( M ) f / ( j a ) at ® so o ls
|G(y®),ff(y®)=0-90* - tan"' j - tan"'^yj
= -90*-90*-90*
= -270*
Follow the above procedure and calculate magnitude and phase of G ( j o ) f / ( j a ) for different
values of ® for drawing the Nyquist plot.
Step 5 of 14
The Nyquist plot will always be symmetric with respect to the real axis.
The plot is normally created by the NYQUIST MATLAB m-file.
G (s)ff(s)^/C
5(« + I)(5 + 3 )
i( i * + 4 s + 3 )
+3s]
Enter the following code in MATLAB to draw the Nyquist plot;
» num=[3];
» den=[1 4 3 0];
» sys=tf(num,den):
» nyquist(sys)
The Nyquist plot for the feedback system is shown in Figure 2.
Step 6 of 14 ^
Determining stability of closed loop system based on frequency response of system’s open loop
transfer function Is Nyquist stability criterion.
To determine the range of
for the stability, gain of the system when the Nyquist plot touches
the real axis is to be known.
Procedure to determine gain of the system when the Nyquist plot touches the real axis is
explained as follows;
Find ® value by equating phase of the system to —igQ*0 - 9 0 - - t»d-‘
j - t a « - ' = -18<r
.(M l.,
t a n '- i- : — ^ = 9 < r
-T
4ffl
3 -a ^ b O
e )= J i
Step 7 of 14
Substitute the above obtained value of ® in equation (2).
ZK
° 4
Consider positive value of k ■
3 gain of G { ja ) H ( ja ) at
=
is y .
The numbers of clockwise encirclements of -1 (denoted by ^ ), by observing the nyquist plot of
Figure 1 are zero.
Number of unstable (RHP) poles of G { ja ) H ( ja ) . denoted by p are zero.
Numbers of unstable closed loop roots Z are,
Z^N +P
=
=
0+0
0
For positive value of K - Z - 2 ’ that is there are no characteristic equation roots in the RHP.
Thus, the system is stable fo r positive values o f
.
Step 8 of 14
Consider k value as negative.
The number of clockwise encirclements have to be now determined depending on value of
When
.
Nyquist plot doesn’t encircle » j.
Number of clockwise encirclements of - 1. A^ = 0
Number of unstable (RHP) poles of G ( ja ) H ( ja ) . denoted by p are zero.
Z = JV+/*
=
=
0+0
0
For AT > - 4 , Z = 0 : that is there are no characteristic equation roots in the RHP.
So the system is stable for
>-4.
When j ^ < - 4 , Nyquist plot encircles
Number of clockwise encirclements of - 1. JV = 1
Step 9 of 14
denoted by p are zero.
Number of unstable (RHP) poles of
Z = N -¥ P
=
1+0
=1
For K < - A ’ Z = I ’ that is, there are characteristic equation roots in the RHP.
So, the system is unstable for K < - A Thus, it Is clear from the obtained results that the range of x. tor which system is stable is
S > 3
Step 10 of 14 ^
(b)
The number of roots in the RHP for those values of x tor which system is unstable can be
determined from the rough root locus sketch.
Draw the root locus for equation (1).
Procedure to draw root locus:
RULE 1 : Number of poles of the feedback system shown in Figure 1 are three. So, there are
three branches to the locus, all three poles approach asymptotes.
RULE 2 : The real-axis segment defined bv j< - 3 a n d —1 <
< 0 is part of the locus.
RULE 3 : Calculate the centre of asymptotes.
0-1-3
3-0
- l l
“ 3
= -1.33
The angles of asymptotes are at 60*,180*,300*-
Step 11 of 14
RULE 4: calculate break away point.
^
-1
_
j ( j + l)( j+ 3 )
Perform — = 0-
ds
j(j+ l)(j+ 3 )
3
Simplify the above equation.
- (3 » * + 8 s + 3 ) '
3
=
0
3s* + 8 j + 3 = 0
The roots obtained for the above equation are
5, = - 0 .4 5
j, « - 2 . 2 1
Step 12 of 14 ^
The root —2.21 doesn’t lie in the range j < - 3 of root locus or —l < ^ < 0 > s o ^ d o e s n ’t
represent break away point.
The root - 0.45 lies in the range —] < 5 < 0 of root locus, so 5, ''^prosents break away point.
Enter the following code in MATLAB to draw the root locus:
» num=[3];
» den=[1 4 3 0];
» sys=tf(num,den):
» rlocus(sys)
Step 13 of 14 ^
The locus of closed loop poles with respect to x
shown in Figure 2:
F ^o re 2
Step 14 of 14
Thus, it is clear from Figure 2 that the roots in RHP are zero.
Consider transfer function shown in equation (1).
[ j ( j + l)(*+3)J
The number of roots in RHP is zero.
Thus, it is clear from the root locus plot shown in Figure 2 that the number of roots in RHP for
those values of x for which system is unstable is zero.
Problem 6.36PP
For the system shown in Fig., determine the Nyquist plot and apply the Nyquist criterion
(a)
to determine the range of values of K (positive and negative) for which the system will be
(b) to determine the number of roots in the RHP for those values of K for which the system is
unstable. Check your answer by using a rough rootlocus sketch.
Figure Control system
Step-by-step solution
step 1 of 3
K G
( 0 = ^
Step 2 of 3
a.
OM = 2
K<2
Also for negative K ,
G M =1, R > 1
Hence 1-1 < K < ^
Step 3 of 3
b.
In both regions of instability, two roots lies on right half plane
Problem 6.37PP
For the system shown in Fig., determine the Nyquist plot and apply the Nyquist criterion
(a) to determine the range of values of K (positive and negative) for which the system will be
(b)
to determine the number of roots in the RHP for those values of K for which the system is
unstable. Check your answer by using a rough rootlocus sketch.
Figure Control system
Step-by-step solution
step 1 of 9
Consider the feedback system shown in Figure 1.
Figure 1
Step 2 of 9
(a)
Write the closed loop transfer function for the feedback system shown in Figure 1.
r
CM
R
1 + G (r )ff(*)
s-\
4 -1
\+ K
.1
[ ( « + !)■
From the transfer function, the expression for G ( r ) f f ( r ) is,
( 1)
(s + \)
Nyquist plot represents frequency response of the system.
■ (^+ 1)’
ic O v » -i
C { ja ) H { ja )
K ( j a - ')
Magnitude of G ( j a ) H { j a ) is,
K ^\ +0 ‘
|g ( » « ( H = +Aa?
K y l \ + e i‘
K
Phase of G (ja > )H {ja ) is,
Z G ( ja i) H {ja f) = 0 + tan"'
= 1 8 0 » - t a n " '( < » ) - t a n - '^ j ^ j
Step 3 of 9
Magnitude of G { j a ) H ( j a ) at ^ s O ■
Phase of G (y«l))H 0V ») at ® = 0 is
^ G ( J a ) H (J a ) = 1 8 0 ° - la n - ' ( 0 ) - tan"'
= 180“
Magnitude of G ( j a ) H ( j a ) at
is
step 4 of 9
Phase of G ( j< o ) H ( ja ) at n>soois
Z C O 'o ) H (j& i) = 180®- tan"' («>) - tan"' I
1
U -H
J
«l80® -9 0 ® -9 0 ®
=
0®
Follow the above procedure and calculate magnitude and phase of G ( j a ) H ( j a ) for different
values of of.
The Nyquist plot will always be symmetric with respect to the real axis.
The plot is normally created by the NYQUIST MATLAB m-file.
Enter the following code in MATLAB to draw the Nyquist plot;
» num=[1,-1]:
» den=[1 2 . 1]:
» sys=tf(num,den):
» nyquist(sys)
The Nyquist plot for the feedback system of Figure 1 is as shown in Figure 2.
Step 5 of 9
Determining stability of closed loop system based on frequency response of system’s open loop
transfer function is Nyquist stability criterion.
Consider positive value of
.
The maximum gain of G ( j t t , ) H ( ja ) isobtainedat ^ s Q a n d the value is jff.
The numbers of clockwise encirclements of -1 (denoted by f j ). by observing the nyquist plot of
Figure 2 are one.
Number of unstable (RHP) poles of
denoted by p are zero.
Numbers of unstable closed loop roots Z are,
Z = N -¥ P
=
1+0
=1
For positive value of K - Z —V that is there are characteristic equation roots in the RHP.
So the system is unstable for positive values of k ■
Step 6 of 9
Consider x value negative.
The number of clockwise encirclements have to be now determined depending on value of x ■
When
Nyquist plot doesn’t encircle _ j.
Number of clockwise encirclements of -1 . AT=0
Number of unstable (RHP) poles of G ( j a ) H ( j a ) . denoted by p are zero
Z = JV+/*
= 0+0
=
0
For AT > - 1 , Z = 0 : that is there are no characteristic equation roots in the RHP.
So the system is stable for AT > - I •
Step 7 of 9
When at < -1 >Nyquist plot encircles _ ].
Number of clockwise encirclements of -1 . N ^ \
Number of unstable (RHP) poles of G ( j t t , ) H ( ja ) . denoted by p are zero.
Z = N+P
= 1+0
= 1
For A T <-1 . Z = I ; that is there are characteristic equation roots in the RHP.
So the system is unstable for AT < -1 •
It is clear from the above results that there is no range of x tor which system is stable.
Step 8 of 9
(b)
Number of roots in the RHP should be determined from Nyquist plot first and then verified using
root locus.
Draw the root locus for equation (1).
Procedure to draw root locus:
RULE 1 : There are two branches to the locus, one of which approach finite zero and other of
which approach asymptotes.
RULE 2 : The real-axis segment defined bv 5 < —l a n d —1 < ^ < 1 is part of the locus.
RULE 3 : Calculate the center of asymptotes.
H z IH !)
2-1
-3
S ----
1
= -3
The angles of asymptotes are at 180®.
RULE 4: calculate break in point.
K =-
(» -•)
Perform — = 0
ds
Simplify the above equation.
2( ^ - l ) ( j + l ) - ( i + l)‘
( * - ') ’
2j ' - 2 - i ’ - l - 2 i
=
0
(* + •)’
j^ -2 j
-3 = 0
The roots obtained for the above equation are
s ,= 3
5j«-l
The root 3 doesn’t lie in the range —1 < 5 < 1of root locus, so ^ doesn’t represent break ii
point.
The root -1 lies in the range $ < —lo f root locus, so 5. represents break in point.
Step 9 of 9
Enter the following code in MATLAB to draw the root locus:
» num=[1 - 1]:
» den=[1 2 1]
» sys=tf(num,den):
» rlocus(sys)
The locus of closed loop poles with respect to x
shown in Figure 3:
R oo tL o o u
0.4
0J •
0.2
•
0.1
■
Ifr »
.§ .0 .1 ■
•
•
■“ • I i
-5
-4
J
-2
-I
0
1
2
R e d AzbCsccoMb'*)
Figures
Thus, it is clear that root at $ s 1 is in the RHP which results in system to become unstable.
Thus, the numbers of roots in the RHP are checked from the root locus as shown in Figure 3.
Problem 6.38PP
The Nyquist diagrams for two stable, open-loop systems are sketched in Fig. The proposed
operating gain is indicated as KO. and arrows indicate increasing frequency. In each case give s
rough estimate of the foilowing quantities for the closed-loop (unity feedback) system;
(a) Phase margin;
(b) Damping ratio;
(c) Range of gain for stability (if any);
(d) System type (0, 1. or 2).
(d) System type (0, 1. or 2).
Figure Nyquist plots
\
1
-1
j
1
H
V
(b>
(I)
Step-by-step solution
step 1 of 4
For first sjrstem phase margin is very low while for second system
Phase margin is high.
Step 2 of 4
b.
First system has low damping ratio while second has high damping ratio.
Step 3 of 4
For first sjrstem
Gain > Ko
For second system Gain < Ko
Step 4 of 4 ^
First system is Type - 2 since full circle is Enclosed
Second system is Type - 1 since semicircle is enclosed
Problem 6.39PP
The steering dynamics of a ship are represented by the transfer function
V js ) ^
^
S f( s )
n - ( j/0 .1 4 2 ) + l]
f ( j / 0 .3 2 5 + l)(5 /0 .0 3 6 2 + l ) ’
where V is the ship’s iaterai veiocity in meters per second, and 5r is the rudder angle in radians,
(a) Use the Matlab command bode to plot the log magnitude and phase of G( jo)) for K = 0.2.
(b) On your plot, indicate the crossover frequency. PM, and GM.
(c)
Is the ship-steering system stable with K = 0.2?
(c)
Is the ship-steering system stable with K = 0.2?
(d)
What value of K would yield a PM of 30*. and what would the crossover frequency be?
Step-by-step solution
step 1 of 4
K
G (s)= -
- f —^ l + l l
. U .1 4 2 J
UO.325
A
J
o.0362
)
Step 2 of 4
b.
|g M = - 5 .a jg 1
\P M = - 2 3 . r \
at m = 0 .0 5 7 6 ra i/ sec
at
» = 0.087 n jrf/sw
Step 3 of 4
c.
No , ship steeling system is not stable
Step 4 of 4
A
A t PAf = 30* ® = 0.0318raA /s»c
|Aad K = 0.042|
Problem 6.40PP
For the open-loop system
K (s + \)
K C (S ):
determine the value for K at the stability boundary and the values of K at the points where PM =
30“ .
Step-by-step solution
ste p 1 of 4
Step 1 of 4
Write the characteristic equation.
1 + G (j)« (i) = 0
i^ ( j+ lo p '
» * { i+ io ) * + A : ( i+ i) = o
»^(** + 100+20s)+AS+i(r = 0
j ‘ + 2 0 j ’ + 1 0 0 i* + K j + A : = 0
Apply Routh-Hurwitz criterion.
100 K
1
K
20
0
2 00 0 -X
20
2 0 0 0 - a:
20
K
For system to be stable there should not be sign change in the first column.
From ^2 row, the value x becomes,
2000^^0
20
20o o > a :
From j t row, the value
m
becomes,
—
2 m -K -m > o
\€ 0 0 > K
Range of
value at the stability boundary is |0 < /r < l6 0 0 l-
Step 2 of 4
Consider the open-loop system.
s * ( j + 10)^
Convert the function to the bode fonn.
’
s ^ ( ja + \ o f
Phase margin Is,
PM = 1 8 0 » + Z G ( y a > ) L , ^ ^ ,
Substitute 30® for PM in the equation.
30° = 1 8 0 °+ Z G (;a > H
...
. I
__________ = - > » ”
Step 3 of 4 ^
Find the phase frequency at which angle becomes 150® of the transfer function.
Z G 0 « ) = U n - ( l) - 1 8 0 " - 2 , « .- g )
,-(i)-.80--2.»-g)
-I50° =tan-‘
-150'=tan
€)
a> 180'-tan
since 2tan"'(;c)=tan"'^j^^j
i'-© J
Simplify further.
<a
30®=tan"* j - tan"'
i'-© J
( i ) '
i'-O J
30* tan"'
since tan" X-tan"
‘
<11
-a
tan30«»-
A
tan'
— fe)
[-© j
x-20
F IR
tan30®»
-I9 a )^ -I0 0
«i’ -l0 0 < » + 2 0 i»
lan w »
-1 9 « » * -1 0 0
,
n r -8 0 ®
^ 3 0 , ^ 1 9 ^
80®
0.577(80fli-<i>’ )-19< !> *-100 = 0
-0 3 7 7 < »’ - 1 9 ® * + 4 6 .18< »-100 = 0
Solve the equation.
® = -3 5 .3 1 rad/sec» ® = 1.2019±1.860y r a d /s e c
Step 4 of 4 ^
Find the magnitude of the open loop system.
|G(y®)|---------- ,
----------- --
^ / ^ ^ ( 2 0 ® ) * + ( l0 0 - f f l* ) ^
Find the magnitude at frequency
®s—
35^3lad/sec-
------------a/(-3 5 -3 1 )'^ ((2 0 )(-3 5 .3 1 ))V ( i 0 0 -(-3 5 .3 1 )* )
35.324
~ (35.3I)^(498718.44)-1146.7961
35.324
(35.31)(705.387)
= 0.00141822
Therefore to get crossover at that frequency x
jc = —
become,
!—
0.00141822
-70S.108
Therefore, the fc value at which PM s 30® i® 1705.1081
Problem 6.41 PP
The frequency response of a plant in a unity-feedback configuration is sketched in Fig. Assume
that the plant is open-loop stable and minimum-phase.
(a)
What is the velocity constant Kv for the system as drawn?
(b) What is the damping ratio of the complex poles at w = 100?
(c)
Approximately what is the system error in tracking (following) a sinusoidal input of cu = 3
rad/sec?
(d)
What is the PM of the system as drawn? (Estimate to within ±10“ .)
(d)
What is the PM of the system as drawn? (Estimate to within ±10“ .)
Figure Magnitude frequency response
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 6.42PP
For the system
1 0 0 (j/a + l)
'5 ( j+ l) ( j/ f r + l) *
G(i) = -
where b = 10a, find the approximate value of a that will yield the best PM by sketching only
candidate values of the frequency-response magnitude.
Step-by-step solution
step 1 of 1
Step 1 of 1
G(s)=-
^ e r e b=10a.
is alead compensatorwith a=0.1
Gain o f - 2 0 l o g - l O d B occurss at cd^=17.8
Va
cV
= - 5 j==17.8
7>l2
T=0.1776
=-^=5.63
T_____
Problem 6.43PP
For the open-loop system
determine the value for K that will yield PM > 30° and the maximum possible closed-loop
bandwidth. Use Matlab to find the bandwidth.
Step-by-step solution
step 1 of 10
Step 1 of 10
The open loop system is.
j= (s + 2 0 )
The phase margin of the system is.
PM = 180® +^^
The phase of the system is.
* - 1 8 0 ® + t a n " '® - 2 t a n " '—
20
Step 2 of 10
Hence, the phase margin of the system is.
PM = l8 0 * - 1 8 0 » + t ii n - '® - 2 ta n - '
■
-
20
I
= tan *0 - 2 tan —
20
Since the phase margin is calculated at gain cross over frequency
PM * Ian"'
-2 ta n " ‘ —
20
Step 3 of 10
Since the phase margin. PM ^30® >
30“ S t a n " ' - 2 ta n " '- £ •
20
”
tan"'«i - l a n " ' ^ - t a n " ' ^ a 3 0 “
20
a
tan"'
-
20
20
- t a n " '- 2 - a 30“
20
” 20 J
t
a
n " ' - t a n " ' ^ 2 30“
1^20+n
.J,J
20
Step 4 of 10
Further simplify.
1 9 ® ,-^
20
230“
20;
379a.
2 tan 30“
2 0 + ( 2 0 + ® i) » ,
379® .
1
2 0 + ( 2 0 + « ^ . ) ® ^ ] ^ '^
656.447® ^ 2 20 + 20® ^ + aij,
® ? - 6 3 6 .4 4 7 ® .+ 2 0 S 0
step 5 of 10
The roots are,
0 ^ — 25.24
<9^=0.0314
= 25.21
The suitable value for 0 ^ is 25.21.
Hence.
0 ^ ^ 2 5 .2 1 rad/s
Step 6 of 10
The gain of the system is,
a: ( > / ®
|A:c(a)|=
'+ i )
i*(V ® “ + 4 0 0 )
Since the gain of the system at 0 ^ is unity or 0 dB,
L +400
T
(fflj,+ 4 0 0 ) = a: (
^
® ; .(f <
e+
+ 4< 0o 0 )
Step 7 of 10
Substitute 25.21 rad/s for 0 ^ .
(25.21)^ ((2 5 .2 1 )* + 400)
7 (2 5 .2 1 )’ + !
658133.94
25.23
> 2 6 0 8 5 .3 7
Thus, the value o f /< is 126085.371-
Step 8 of 10 ^
Find the maximum possible bandwidth.
K G ( s ) = - ^
' ' j ‘ (s + 20)
AT(a+ l)
“ j* + 4 0 i’ +400s'
®l = 0 .0 3 1 4
> 2 5.21
The bandwidth is,
= 2 5 .2 1 -0 .0 3 1 4
= 20.178 rad/s
Thus, the maximum possible bandwidth is 120.178 rad/sl •
Step 9 of 10
The MATLAB code to
k=26085.37;
num=[k k]:
den=[1 40 400 0 0];
sys=tf(num,den);
bodeplot(sys)
margin(sys)
grid
Step 10 of 10 ^
Thus, the plot of MATLAB is sown in Figure 1.
B o d e D ia g ra m
F re q a c n c y (ra d /s)
Problem 6.44PP
For the lead compensator
Tds + 1
Dc(s) = -
’ aToS+ r
where a < 1.
(a) Show that the phase of the lead compensator is given by
<p= tan-1 {TDw) - ^an-^{aTDu)).
(b)
Show that the frequency where the phase is maximum is given by
I
Td ->A*
and that the maximum phase corresponds to
o«nu = = —7=
and that the maximum phase corresponds to
1 —a
sin^nm =
+o
(c) Rewrite your expression for ojmax to show that the maximum-phase frequency occurs at the
geometric mean of the two corner frequencies on a logarithmic scale;
log«m „ = 5 ( l o g ^ + l o g j ^ ) .
(d) To derive the same results in terms of the pole-zero locations, rewrite
Dc(s) as
Dc(s) =
s+ p
and then show that the phase is given by
such that ....... = v'UTRHence the frequency at which the phase is maximum is the square root of the product of the pole
and zero locations.
Step-by-step solution
step 1 of 9
(a)
Step 2 of 9
Consider the transfer function of lead compensator.
^
a V + l
Substitute J(p for s.
TJa>+l
D , ( jt a } =
a T g j6 f+ l
l+ a T i,6 ) j
From this equation, the value of phase is,
4 = tan*' {Tj^0) - tan*' (or7i,<»)
Thus, the value of phase is.
= tan*' (Tjyoi) - tan*' (grp<p)|
(1)
Step 3 of 9
(b)
Consider following equations.
(2)
=
B = tan"' { a T ^ )
(3)
Substitute equations (2) and (3) in (1).
4 = A - B ......(4)
Consider the trigonometric relationship equation.
^ l+tan(i4)tan(£)
'
Substitute equations (2), (3) and (4) for A, B and A—B.
t ^ U ) ___
ij.f r
Consider the equation of sin ^^-
. 2,
tan*d^
S in V = : ------- T T ......
1+ tan*#
Substitute equation (5) in (6).
r r „ « . ( i- « ) T
[ i+ a r > ' J
. J,
™
^
, r r „ « > ( i- « ) T
1+
Ti+«r„vJ
[r„ a » (l-g )]*
{ l+ o T jf f l’ ) + ( r o < » ( l- a ) )
' l + o ’7^<»* + 2 « r > = +72<»’ ( l - 2 a + o ' )
r „ V ( i- g ) '
1
+
(l + a* )
Take square root on both sides.
—
B
E
f L
^ 1 + o ^ 7 j® * T 7 2 ^ ( l+ a ^
T M l- o )
(7)
^ l + a ’ 72®‘ + r ^ ® ’ ( l + g ’ ^
Step 4 of 9
Let
g f(s in jt)^
^7 s 0
frequency value at maximum phase value.
da
Differentiate equation (7) with respect to
7 i® ( l - a )
^ l + o ’ 7 2 ® * + 7 -„ W (l+ a ’ )
( 4 a ' r „ V + 2 r X l + a ’ ))
,(r« 0 -« ))
</(sinj>)
2 ^ l+ g = r > ‘ + r > ^ ( l + a ‘ )
da
(^U .a ^ 7 S ® ‘ + r „ V ( l + « ' ) ) ’
7 i® ( l - g )
. J l+ a ’ 2 ; ® * + r > ’ ( l + o ' )
( 4 o ^ r „ V + 2 r > ( i+ o * ) )
2 ^1+ o*72 ® ‘ + r > ' ( l + o ^ )
=0
( ^ l + o ^ J S ® * + 7 i V ( l+ a ^ ) ) ’
' ( 4 a ^ r „ V + 2 r > ( l+ a * ) ) '
=
0
2 ^ \ + a ’ T iei)*+ T ia/‘ { l + a ‘ )
■ 2 ( i + a ' r „ V + r > ’ ( i + a ') ) '
(4 o ’ 7 ji» ’ + 2 r j ® ( l + a ’ ))
[2(l+a’r>*+7->’(l+o'))(r„(l-o))]-
=0
(4a=72®‘ + 2 r ji» ’ ( l + a ' ) )
=
0
r , ( l- a ) [ [ 2 ( l+ a * 2 S ® * + r „ V ( l+ a ^ ) ) ] - [ 4 a 'r J ® V 2 r „ W ( l+ a ^ ) ] ] = 0
[ 2 ( 1+ a ’ r > * + r ja i" ( 1+ o ' ) ) ] - [ 4 o ’ 7 j ® '+ 2 rJ ® ’ ( l + o ' ) ] = 0
[2]-[2o'J2®'] = 0
2
‘2aX
___l _
* a«2r<
Tn
1
n>_ = - j —— .
daTo
( 8)
Thus the frequency value at maximum phase is verified and shown below.
1
5 ^
Step 5 of 9
Substitute equation (8) in (7).
T,
s in tL ,
____________
.2 +
l+ a *
I
2 0 + 1+ 0 '
1
^
= -7
1 2o + l+ o
0 -° )
V20+1+0'
0 -°)
Vo+«)’
1 -0
1 -fa
Thus, the maximum phase value is verified and shown below.
\-a
s in ^ ^
l+ a
step 6 of 9
(c)
The maximum frequency is occurs midway between the two break frequencies on a logarithmic
scale.
k , g ® .= lo g ^
.
*
.
1
**® s -7 = r+ *o « -r= r
v*o
>1^*0
Io g « . - . = i.o g ( ^ ) + i|o g ( ^ )
Thus, the required equation is derived and shown below.
log
10gtf> a 9 M
2
fe w
a
step 7 of 9
(d)
Consider the compensator transfer function.
A (*)= « —
' '
s+ p
Substitute j a for s.
ja > + p
The phase angle equation is.
Thus the phase equation is.
(9)
Step 8 of 9
Consider following equations.
> 4 a t a n - 'f ® l......(10)
■“ "(fi)
B
( 11)
■ " " fe )
Now substitute equations (10) and (11) in (9).
4 = A - B ..... (12)
Consider the trigonometric relationship equation.
'
'
l + ta n ( i4 ) ta n ( £ )
Now substitute equations (10), (11) and (12) for A, B and A-B.
ta n ^
4
1+
a
a
h
~h ]
(13)
(1I 4)
step 9 of 9
Let ^ 5 2 ^ = 0.
Differentiate equation (13] with respect to a .
d ta n ^
[ '* (
w r
] ] ( A |p |] [ S
da
2
I
r
'
1+
(ii)](H ‘ R]“(iSi)(R‘ R]°®
( R 'R j't e lf e 'R )
( R 'R )t e ) ° ( R 'R )
® '- R R
F llf l
Thus, the value of frequency at maximum phase is.
r
)
Problem 6.45PP
For the third-order servo system
(K S ) =
50,000
« (i+ IO )(f + 50)*
use Bode plot sketches to design a lead compensator so that PM > 50° and cuBW > 20 rad/sec.
Then verily and refine your design by using Matlab.
Step-by-step solution
step 1 of 1
Step 1 of 1
50,000
s(s+10)(s+50)
PM—10.5" at (1r 28.6 rad/sec
let e=5"
Adds onal phase 1ead = 50 - (-1 0 .5 ')+ y = 70.5'
l+sin^^
l+sin70.5"
Gain of ( - 2 0 1 o * ^ ) = - 2 0 .,6 dB at cc^=58rad/sec
cc^ = - ^ = 5 8 rad/s
y/cLt
|t=0.084|
D(s) =
■■
0.084S+1
0.00395S+1
50.000(0.0845+1)
^'( j + io )(5+50)(0.003955+1)
But still this system cannot have phase margin greater than 50"
Problem 6.46PP
For the system shown in Fig., suppose that c ( j) = ■ s(i + 1 ) ( 5 /5 + I) ‘
Use Bode plot sketches to design a lead compensation Dc(s) with unity DC gain so that PM >
40°. Then verify and refine your design by using Matlab.
What is the approximate bandwidth of the system?
Figure Control system
S te n -h v -< ;te n ^ n liitin n
Step-by-step solution
step 1 of 7
The open loop transfer function is.
C M -
The transfer function of the iead compensator is.
£ )(.)«
Ts+l
,or < 1 and with unity DC gain.
aT s*\
Write MATLAB code to plot the Bode plot using margin to get the phase margin.
» numG=5;
» denG=conv([1 1 0],[0.2 1]);
» sysG=tf(numG,denG);
» margin(sysG)
Step 2 of 7
Get the MATLAB output for the bode plot showing the gain and phase margins.
Bode D iofron
Step 3 of 7
The phase margin of the open loop system, PM s 3.94*.
As the required phase margin is P A /^ 4 0 * . allowing a margin of 5 * , the phase margin to be
compensated by the lead compensator is,
40® -I-5*+ 3.94* = 49*
Refer to Figure 6.54 in the text book for the maximum phase for lead compensation.
For maximum phase lead equal to
4 9
*, the value of — is,
± =8
a - 0 .1 2 5
The new gain cross-over frequency is greater than 2.04 rad/s. Select the maximum gain cross­
over frequency as 10 rad/s.
1
r= -
=10
1
" lo V a
Substitute0.125for a .
lO V O lM
= 0.283
tfT = (0 .I2 5 )(0 .2 8 3 )
= 0.035
Step 4 of 7
Determine the transfer function of lead compensation.
0.2835+1
CM
0.035J + 1
8 (^ + 3.S3)
i + 28.6
Thus, the lead compensation with unity DC gain, D (s) is
8(5 + 3 .53)
5 + 28.6
Determine the transfer function of the system with lead compensation.
4 0 ( 1 + 3.53)
j ( j + l ) f | + l ] ( i + 28.6)
J
200(5 + 3 .53)
5 ( 5 + 1 ) (5 + 5 ) ( 5 + 2 8 .6 )
Step 5 of 7
Write MATLAB code to plot the Bode plot of the feedback system using m argin to verify the
phase margin with the designed values.
» numG=200*[1 3.53];
»den1=conv{[1 1 0],[1 5]);
» den2=[1 28.6];
» denG=conv(den1,den2);
» sysG=tf(numG,denG);
» sys=feedback(sysG,1);
» margin(sys)
Step 6 of 7
Get the MATLAB output for the Bode plot.
Step 7 of 7
Observe from the MATLAB output that the phase margin of the system is 4 |,7 *
Therefore, the design requirements are met with the designed lead compensation.
Band width is defined as the maximum frequency at which the output of the system is attenuated
to a factor of 0.707 times the input.
Observe from the MATLAB output that the magnitude is 0.7 dB at 0.669 rad/s.
Thus, the approximate bandwidth of the system is 0.67 rad/$ •
Problem 6.47PP
Derive the transfer function from Td to 0 for the system in Fig. Then apply the Final Value
Theorem (assuming Td = constant) to detennine whether 0C“ ) is nonzero for the following two
cases;
(a)
When Dc(s) has no integral term:
constant;
(b) When Dc(s) has an integral tenri:
In this case.
constant.
Figure Block diagram of spacecraft control using PID design
Figure Block diagram of spacecraft control using PID design
Step-by-step solution
step 1 of 3
0.9
1d(s)
2 ^ 0 .9 '!
1+1
s+ 2 V
09(s+2)
” !"(ft-2 )+ 1 .8 D(s)
Step 2 of 3
0
(oo)=lmts0 (s)
0.9=(=+2)T,(s)
*-»® (s+2 ) + 1 .8 D(s)
TT
As T4 =constant=K, T4 (s)=—
8
0
, 0.9K(s+2)
(oo) = Imt———^7 - ^
^ ^
1.8D(0)
D(0)
Step 3 of 3
b.
li^tD^(s)s=coiistant=4t^
0 (00)=lmt
.=0
[Hence 9 (00) isZeroj
Problem 6.48PP
The inverted pendulum has a transfer function given by Eq., which is similar to
(a) Use Bode plot sketches to design a lead compensator to achieve a PM of 30®. Then verify
and refine your design by using Matlab.
(b)
Sketch a root locus and correlate it with the Bode plot of the system.
(c)
Could you obtain the frequency response of this system experimentally?
Eq.,
yuu uuieiii i m e iiev^uetiuy ledpui
u iis sy aiei 11 eApei iiiiei iieiiiy r
Eq..
0'{j) _
Mpl
U (s) ~ ( ( / + m pp)(m t + nip) - m jP )s ^ - mpgl{mt + nip) *
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 6.49PP
The open-loop transfer function of a unity-feedback system is
C(*) = -
j(«/5-M )(j/50-H )‘
(a) Use Bode plot sketches to design a lag compensator for G(s) so that the closed-loop system
satisfies the following specifications;
(i) The steady-state error to a unit-ramp reference input is less than 0.01.
(ii) PM > 40“ .
(b) Verify and refine your design by using Matlab.
Step-by-step solution
step 1 of 5
(a)
The open loop transfer function of a unity feedback system is.
C M -
Determine the open-loop gain K that gives a phase margin of 40«.
The steady state error to a unit ramp input is.
s lim ji
F ) F )
1
1
.
0.01
/:2 io o
Take X = 100
Step 2 of 5
Draw the Bode plot of the uncompensated system using MATLAB for K —5num=5;
den=conv([1/5 1 0],[1/50 1]);
sys=tf(num,den);
margin(sys)
BodeDiafmn
Gm - 20.S dB (it IS.S n d h ) . Pm - 47.4 deg (at 3.92 n d ^ )
Step 3 of 5
The value of K equal to 5 gives a phase margin of
47.4®at a cross-over frequency of
3.92rad/s
The compensator transfer function is.
D {s) = a — — - « > 1
' '
a r f+ 1
The low frequency gain must be raised by a factor of 20, so the lag compensation should have
a equal to 20.
Choose the comer frequency
* — to be one octave to one decade below the cross-over
frequency, that is, (0.2)(3.92) = 0.784 rad/s
T
0.784
r = i. 3
Step 4 of 5
Determine the other corner frequency.
1
aT
1
■ (2 0 )(1 .3 )
_J^
°2 6
The desired compensator is.
“i-i-K S ri)
Thus, the desired compensator is
l,5 0 0 jU .0 0 2 6 j + U
Step 5 of 5
(b)
Verify the design using MATLAB.
num=100*[1.3 1];
den=conv([1/5 1 0],conv([1/50 1].[26,1]));
sys=tf(num,den);
margin(sys)
Get the MATLAB output for the Bode plot.
BadeMagraM
Increasing the gain does not result in further increase of phase margin.
Hence, the design is verified and cannot be refined.
Problem 6.50PP
The open-loop transfer function of a unity-feedback system is
G(s) = .
a (« /5 + I)(5 /2 0 0 + l)‘
(a)
Use Bode plot sketches to design a lead compensator for G(s) so that the closed-loop system
satisfies the following specifications;
(i) The steady-state error to a unit-ramp reference input is less than 0.01.
(ii) For the dominant closed-loop poles, the damping ratio ^ > 0.4.
(b) Verify and refine your design using Matlab, Including a direct computation of the damping of
the dominant closed-loop poles.
the dominant closed-loop poles.
Step-by-step solution
step 1 of 2
G (0 = 1
Now, K y= — = m
^ 0.01
Now ?0.4 , PM£1004»40"
P M o f G(s) = 6.54'
Let € = 6 '
Additional phase added = 40^ - 6 .5 4 '+ 6 ' = 39.46'
l-s in ^
l+sin<^
l - s i n 39.46*
1+sin 39.46'
Step 2 of 2
Gain
lin of ^-2 0 1 og
j =-6 .6 dB occxirs at CD^=32.4rad/s
a ^ = - ^ = 3 2 .4
Vott
r = 0.066
.-.D(s) =
0 .0 6 6 S + 1
Hence D (s)G (s)*-
100(0.066s+l)
5 )5 3 ^ ^
And its |^=0.4|
Problem 6.51 PP
A DC motor with negligible armature inductance is to be used in a position control system. Its
open-loop transfer function is given by
C(s) =
SO
■ «(s/5 + l ) '
(a) Use Bode plot sketches to design a compensator for the motor so that the closed-loop system
satisfies the following specifications;
(i) The steady-state error to a unit-ramp input is less than 1^00.
(ii) The unit-step response has an overshoot of less than 20%.
(iii) The bandwidth of the compensated system is no less than that of the uncompensated
system.
(b) Verify and/or refine your design using Matlab, including a direct computation of the stepresponse overshoot.
Step-by-step solution
step 1 of 2
G(s)=-
50
(F^
= in n « ^
K ya200, MP=20%=100e
=>|;=0.456|
PM a 1005=^5.6 ’
original PM=9.04“
Step 2 of 2
let £= 6 “
additional phase =45.S"-9.04'’+6"=42.56"
■1-sincp, _ l-sin42.56
l+sin(p^ l+sin42.56
gain of [-2 0 1 o g -^ |—7.14dB atca^=47.6rad/sec
1
now, cd^= -= -= 4 7 .6
Vat
|t=0.0478|
4(0.0478s+l)
D(S):
(0.00973S+1)
, D (s)G (s)^
200(0.0478s^1)
“ [5 ^]('> '> '> S 7 3 s+1)
PM of componsated system = | 47.5” |
Problem 6.52PP
The open-loop transfer function of a unity-feedback system is
C (J ) =
■
j (H-5/5K1+»/20)*
(a)
Sketch the system block diagram, including input reference commands and sensor noise.
(b)
Use Bode plot sketches to design a compensator for G(s) so that the closed-loop system
satisfies the following specifications;
(i) The steady-state error to a unit-ramp input is less than 0.01.
(ii) PM > 45“ .
(iii) The steady-state error for sinusoidal inputs with a)< 0.2 rad/sec is less than 1.^50.
(iv) Noise components Introduced with the sensor signal at frequencies greater than 200 rad/sec
(iii) The steady-state error for sinusoidal inputs with a)< 0.2 rad/sec is less than 1.^50.
(iv)
Noise components Introduced with the sensor signal at frequencies greater than 200 rad/sec
are to be attenuated at the output by at least a factor of 100.
(c)
Verify and/or refine your design using Matlab, including a computation of the closed-loop
frequency response to verily (iv).
Step-by-step solution
step 1 of 3
G (.)= -
Step 2 of 3
K = K k - i- = 1 0 0
’
0.01
Required PM ^45" , Initial PM =-28.?
I Lag compensator should be i^ lie d I
phase angle = -180*+45^+9=-126'
At -126^ phase angle, 0^=2.76rad/sec
And Gain at 2.76 rad/sec is |30.2dB|
201ogp=30.2
|p=32.36|
Step 3 of 3
Placing upper corner frequency 3 octaves below <9 ,
1 = ^
z
8
D (0 =
3s+l
97s+l
H e n c e D ( s) G ( s) =
and its PM is 47°
100(3s+l)
Problem 6.53PP
Consider a Type 1 unity-feedback system with
C(J) =
J (J -I- 1) ‘
Use Bode plot sketches to design a lead compensator so that Kv = 20 sec-1 and PM 40*. Use
Matlab to verily and/or refine your design so that it meets the specifications.
Step-by-step solution
step 1 of 7
Step 1 of 7
The open loop transfer function is.
K
<?(*)
s{s+ \)
Substit
Substitute 20 for K .
20
<?(*)
j( j+ l)
MATLAB code to calculate the phase margin;
num=[20]:
den=[1 1 0];
sys=tf(num,den);
margin(sys)
Step 2 of 7
Bode plot with phase and gain margin is shown in Figure 1.
Bode Diagram
Om«bifdB(atbifrad/s), Pm ^ I2.8deg(at4.42rad/s)
step 3 of 7 ^
From the bode plot, the phase margin of the system is 12.8* •
Let.
e=e>
The maximum phase contributed by the lead is,
^ - 4 0 « - 1 2 .8 ” +6“
-33.2*
Calculate the value of a .
l+ s in A m
l-sin(33.2*)
■ l+sin(33.2»)
1-0.547
1+ 0.547
= 0.292
Step 4 of 7
Calculate the gain.
dB=-20log|
ii]
=-20log!
( v o . 292 ]
«(-20)(0.267)
= -5.34dB
From the bode plot shown in Figure 1, the gain —5.34 dB occurs at 6.04rad/s
That is,
-6>04rad/s
The zero at — is,
To
Substitute d.04 rad/s for
and 0.292 for a .
1
— « (6 .0 4 )V O !^
= 3.2638
Tn=Z
1
3.2638
*0.3063
Step 5 of 7
The lead transfer function is.
' '
aV +1
Substitute 0.3063 for Tp and 0.292 for a .
^
(0.3063)j+l
(0.292)(0.3063)a + l
0.30635+1
0.08945 + 1
3.426(5+3.263)
5 + 11.18
MATLAB code to plot bode plot with phase margin:
num1=[20]:
den1=[1 1 0];
sys1=tf(num1 ,den1);
num2=3.426*[1 3.263];
den2=[1 11.18];
sys2=tf(num2,den2);
sys=sys1*sys2;
margin(sys)
Step 6 of 7
The Bode plot is shown in Figure 2.
Bode Diagram
Frequency ( n d /s )
Figure 2
Step 7 of 7
From the Bode plot shown in Figure 2, the phase margin is 42.6®Therefore, the lead compensator meets the required specifications.
Therefore, the designed lead compensator is D{s)
3.426(5+3.263)
5 + 11.18
Problem 6.54PP
Step-by-step solution
Network error loading content
<ErrorCode.2>
RELOA D PAG E
Problem 6.55PP
In one mode of operation, the autopilot of a Jet transport is used to control altitude. For the
purpose of designing the altitude portion of the autopilot loop, only the long-period airplane
dynamics are important. The linearized relationship between altitude and elevator angle for the
long-period dynamics is
k(s)
20( 1 -1- 0.01)
ft/s e c
= 77:7 = S{s) f( i2 + 0 .0 1 i-l'0 .0 0 2 S ) deg
The autopilot receives from the altimeter an electrical signal proportional to altitude. This signal Is
compared with a command signal (proportional to the altitude selected by the pilot), and the
difference provides an error signal. The error signal is processed through compensation, and the
result is used to command the elevator actuators. A block diagram of this system Is shown in Fig.
You have been given the task of designing the compensation. Begin by considering a
proportional control law Dc (s) =K.
F in u r e C o n tr o l s v s t o m fo r P r o b le m fi FtFt
You have been given the task of designing the compensation. Begin by considering e
proportional control law Dc (s) =K.
Figure Control system for Problem 6.55
Hm
(a)
Use Matlab to draw a Bode plot of the open-loop system for Dc (s) =K =
(b)
What value of K would provide a crossover frequency (i.e., where |G | = 1) of 0.16 rad/sec?
(c)
For this value of K, would the system be stable If the loop were closed?
(d)
What is the PM for this value of K7
(e)
Sketch the Nyquist plot of the system, and locate carefully any points where the phase angle
is 180° or the magnitude is unity.
(f)
Use Matlab to plot the root locus with respect to K, and locate the roots for your value of K
from part (b).
(g)
What steady-state error would result if the command was a step change In altitude of 1000 ft
For parts (h) and (i), assume a compensator of the fonn
Deis)-
(h)
T p s+ l
a ro S -h l
Choose the parameters K, TD, and a so that the crossover frequency is 0.16 rad/sec and the
PM is greater than 50°. Verify your design by superimposing a Bode plot of Dc(s)G(s)/K on top of
the Bode plot you obtained for part (a), and measure the PM directly.
(i)
Use Matlab to plot the root locus with respect to K for the system, including the compensator
you designed in part (h). Locate the roots for your value of K from part (h).
(j)
Altitude autopilots also have a mode in which the rate of climb is sensed directly and
commanded by the pilot.
(i) Sketch the block diagram for this mode;
(ii) Modiiy the G(s) stated above for the case where the variable to be controlled is the rate of
altitude change;
(iii)
Design Dc (s) so that the system has the same crossover frequency as the altitude hold
mode and the PM is greater than 50°.
S te p -b y -s te p s o lu tio n
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 6.56PP
For a system with open-loop transfer function
10
- Jl(s/1.4)+ ll[|(l/3) + l l ’
design a lag compensator with unity DC gain so that PM > 40°. What is the approximate
bandwidth of this system?
Step-by-step solution
step 1 of 10
Step 1 of 10
Write the open loop transfer function.
10
G (s ).
42
( j + 1 .4 )( 5 + 3 )
j
42
+4.4 j *+4.2 j
Assume lag compensator of unity gain £>^(5) s
Ts+\
a T s* \
step 2 of 10
Draw the bode plot for uncompensated system <7(s)
Bode Diagram
Step 3 of 10
From the bode plot,
phase margin a -2 0 ^ at
a3
G a in i n a i ^ s - 7 . i3 d B
■ 0.44
rad
^
«.«.,rad
at <pa2.0S —
sec
Step 4 of 10
To get the phase margin of 40® the lag compensation needs to lower the cross over frequency.
From the uncompensated bode plot, let the new cross over frequency is
® , „ = 0.8I —
rad
The gain at 0.81—
is,
|G ( M ) | = 10-4
Therefore, the lag compensator needs to lower the gain at
from 10.4 to 1.
Step 5 of 10
Choose the zero breakpoint of the lag a factor of 20 below the cross over, to avoid influencing the
phase at
1
r "
20
7- = ^
0.81
>25
Step 6 of 10
Choose the value of a .
1
T
At < » » — , D ^ { ji o ) s —
1
ct
1
= 10 .4
10.4 for a and 25 for j* in the compensator £>^(5 ) a
a T s* \
25s+ \
2605 <fl
-----0.04
7+ 1
3.846x10“’
*-+l
^ -Q .04r+ 1
0.0038
0.04
Therefore, the lag compensator is
r+ 1
0.0038
Step 7 of 10
Determine the loop transfer function.
r(5 )-A (5 )G (5 )
_ ^ 2 5 5 + n _____ 42
U 6 0 5 + l j5 ’ +4.+4.45*+4.2 j
10505 + 42
2605^+11 4 5 5 ’ + 1096.45*+4.25
Step 8 of 10
Draw the bode plot for compensated system G(s)
Bode D ig ra m
G m « 12.9 dB (at 2.01 rad/s), Pm » 42.5 deg (at 0.806 rad/s)
F ig u re 2
step 9 of 10
The stability margins of compensated system are.
. rad
phase margin s 4 2 ^ at
^ 0 .8 —
®
Gainmaigins 12.9dB ,
sec
,«.«rad
at A»a2.0 —
“ 4.4
sec
Step 10 of 10
Determine the approximate bandwidth.
. rad
phase margin s 4 2 ^ at ^ sO.8 —
®
sec
a „= 2 a .
- 2 { 0 .S )!^
sec
-. 1 .6 ! ^
sec
Therefore, the approximate bandwidth of the system is
, rad
Problem 6.57PP
For the ship-steering system in Problem,
(a)
Design a compensator that meets the following specifications;
(i) Velocity constant Kv = 2.
(ii) PM > 50“ .
(iii) Unconditional stability (PM 0 for all oj < ojc, the crossover frequency).
(b)
For your final design, draw a root locus with respect to K. and indicate the location of the
closed-loop poles.
Problem The steering dynamics of a ship are represented by the transfer function
Problem The steering dynamics of a ship are represented by the transfer function
m
^
K {-is/0.U 2i+ \]
^
i (j /0J25-f
S ,(s )
1)(j /0.0362-H)’
where V is the ship’s lateral velocity in meters per second, and 5r is the rudder angle in radians,
(a)
Use the Matlab command bode to plot the log magnitude and phase of G( jo)) for K = 0.2.
(b)
On your plot, indicate the crossover frequency, PM, and GM.
(c)
Is the ship-steering system stable with K = 0.2?
(d)
What value of K would yield a PM of 30*, and what would the crossover frequency be?
S te p -b y -s te p s o lu tio n
step
1
of 2
[l,0.325j J[l,0.0362j J
K ^=K=2, initialPM =-llf
let €= S'
and
Lag compensation since PM of
uncompensated system is very low,
Phase angle at crossover frequency = -1 8 0 ' + 8 *+ 50= -122'
.-. m,=0.0139 at [G(jcj)=-122®
and gain =|42.8dB ] at oa^=0.0139
201ogp=42.8
■ .H I]
Placing upper comer frequency 3 octave below atg
1
at
i = -i= > T = 583.9
^
'
(80584S+1)
( 5 8 3 .9 S + 1 )
.-.D(s)G(s)=
s\(
*
V lll
'
V l l
L U .3 2 5 J
J [ < 0 . 0 3 6 2 ; ‘J
|PM=54.y I for compensated S3rstem
Step 2 of 2
Roots are at-0.0143±j0.0154, -0.33
Problem 6.58PP
Consider a unity-feedback system with
G (s) =
1
s (i/20 + 1) (i2/1002 + O.Sj /100 + 1) ■
(a) A lead compensator is introduced with a = 1/5 and a zero at 1/T = 20.
How must the gain be changed to obtain crossover at cue = 31.6 rad/sec, and what is the
resulting value of Kv7
(b) With the lead compensator in place, what is the required value of K for a lag compensator
that will readjust the gain to a /Cv value of 100?
fcl Place the, Dole of tfie laa comoensator at 3.16 rad/sec. and determine the zero location that
(c) Place the pole of the lag compensator at 3.16 rad/sec, and determine the zero location that
will maintain the crossover frequency at ojc = 31.6 rad/sec. Plot the compensated frequency
response on the same graph.
(d) Determine the PM of the compensated design.
Step-by-step solution
step 1 of 5
G(s)=f -5—
UOO
100
+l l
)
Step 2 of 5
K(0.05s+1)
(O.Ols+l)
At G^—31.6, Gain —29.6dB
-201og K=-29.6 => |K=30.2|
K y = K = 3 0 .2
Step 3 of 5
b.
K=Ky
KforLag =
100 _
30.2'
Step 4 of 5
c.
pole=3.16 rad/sec
To maintam <d,= 31.6, n ew zero = ^ ^ = -^ ^ = |0 .9 5 4 |
‘
K
3.31 '------- '
Step 5 of 5
d.
PM of compensated design = 61.2^
Problem 6 .5 9 P P
Golden Nugget Airlines had great success with their free bar near the tail of the airplane. (See
Problem) However, when they purchased a much larger airplane to handle the passenger
demand, they discovered that there was some flexibility in the fuseiage that caused a iot of
unpleasant yawing motion at the rear of the airplane when in turbulence, which caused the
revelers to spill their drinks. The approximate transfer function for the rigid body roll/yawl motion,
called the "Dutch roll” mode (Section 10.3.1) is
r(s)
8 .7 5 (4 j ^ + 0 .4 s + 1)
M s ) ^ ( s /0 .0 1 + l ) ( s 2 + 0 . 2 4 S + 1 ) ’
where r is the airplane’s yaw rate and dr is the rudder angle. In performing a finite element
analysis (FEA) of the fuselage stmcture and adding those dynamics to the Dutch roll motion, they
found that the transfer function needed additional terms which reflected the fuselage lateral
bending that occurred due to excitation from the rudder and turbulence. The revised transfer
function is
found that the transfer function needed additional terms which reflected the fuselage lateral
bending that occurred due to excitation from the rudder and turbulence. The revised transfer
function is
r(s)
8.75(4 ^2 + 0 .4 s+ 1)
M s)
(s/0.01 + I)( s2 + 0 .2 4 s + 1)
+ 2{s/o)i, + 1) ’
where rub is the frequency of the bending mode (= 10 rad/sec) and ^ is the bending mode
damping ratio ( =0.02). Most swept-wing airplanes have a “yaw damper," which essentially feeds
back yaw rate measured by a rate gyro to the rudder with a simple proportional control law. For
the new Golden Nugget airplane, the proportional feedback gain /( = 1, where
Eq.1
5r(s) = -Kr(s).
(a)
Make a Bode plot of the open-loop system, determine the PM and GM for the nominal design
and plot the step response and Bode magnitude of the closed-loop system. What is the
frequency of the lightly damped mode that is causing the difficulty?
(b)
Investigate remedies to quiet down the oscillations, but maintain the same low-frequency gain
in order not to affect the quality of the Dutch roll damping provided by the yawrate feedback.
Specifically, invesfigate each of the following, one at a time;
(i) Increasing the damping of the bending mode from ^ = 0.02 to ^ = 0.04 (would require adding
energy-absorbing material in the fuselage structure).
(ii) Increasing the frequency of the bending mode from rub = 10 rad/sec to rub = 20 rad/sec
(would require stronger and heavier structural elements).
(iii) Adding a low-pass filter in the feedback—that is, replacing A in Eq. (1) with KDc (s), where
Dc(s) =
(iv)
s /r p -p
1
Adding a notch filter as described in Section 5.4.3. Pick the frequency of the notch zero to be
at tub, with a damping of ^ = 0.04, and pick the denominator poles to be ( s /100 + 1) 2, keeping
the DC gain of the filter = 1.
(c)
Investigate the sensitivity of the preceding two compensated designs (iii and iv) by
determining the effect of a reduction in the bending mode frequency of-10%. Specifically,
reexamine the two designs by tabulating the GM,PM, closed-loop bending mode damping ratio,
and resonant-peak amplitude, and qualitatively describe the differences in fhe step response.
(d)
What do you recommend to Golden Nugget to help their customers quit spilling their drinks?
(Telling them to get back in their seats is not an acceptable answer for this problem! Make the
recommendation in terms of improvemenfs to the yaw damper.)
Problem
For the open-loop system
KG(s) =
AT(»-f 1)
j2(j
-F10)2’
determine the value for K af fhe stability boundary and the values of K af fhe points where PM =
30” .
Step-by-step solution
S le p t o f21
The transfer function is given by
r(s)
_
8 . 7 5 ( 4 s ^ - f 0 . 4 s -f 1)
( 3 ^ + l) ( 4 ’ + 0.24a + l ) ’
■Where,
r is the airplane’s yaw rate and
is the rudder angle.
Step 2 of 21
The MATLAB program to draw the bode plot for the given open loop function ii
» num =8.75*[4 0 .4 1 ]:
» f l = [ l / D . D l 1] ;
» f 2 = [ l 0 .2 4 1] ;
» d e n = c o n v ( f l, f 2 ) ;
» s y s = tf(n u m , d en)
» bode (sy s)
» g rid
Step 3 of 21 -e.
The Bode plot obtained onexecutionofthe program is
B e d e D iag ram
Step 4 of 21
The gain and phase margins from the Bode plot are
|gj|/= oo|
\PM = ?7.5° at ai = 0.085 ra d ^
Step 5 of 21
The revised system is
r(s )
8.75(4 s ^ + 0 .4 s -F l)
Ifs’ -F 0.24s+
Where,
rad/sec
= 10
^■=0.02
The MATLAB code to plot the Bode plot for the revised open loop system
IS
»
»
»
»
»
»
»
»
num =0.75*[4 0 .4 1 ];
f l = [ l / 0 . 0 1 1];
f 2 = [ l 0.2 4 1 ];
f3 = [l/1 0 0 0.0 4 /1 0 1 ];
d en= conv(fl, conv(f2, f 3 ) );
s y s = tf(n u m , den)
m a r g in (sys)
g rid
Step 6 of 21 -o.
The Bode plot obtained onexecutionofthe program is
H ade D iagram
The gain and phase margins from the Bode plot are
lOilf = U d B a t 4 >= 1 0 rad/s~l
\PM = 97.5° at a>= 0.085 r a ^
The MATLAB codetodrawtheBode magnitude plot of the do sed loop
system IS
»
»
»
s y s l= fe e d b a c k (s y s ,1 )
m a rg in ( s y s l )
g rid
The Bode plot obtained onexecutionofthe program is
Bmle liia g n u n
The low GM is c aused by the resonance b eing close to is stabihty.
From the Bode plot of the do sed-loop system, the frequency of the lightly
damped mode is at =-10 rad/s .
And thisisborneoutbythestep response that shows a hghtly damped o scillation
at 1.6 Hz or 10 rad/s .
(b)
The MATLAB codetodrawtheBodeplotofthe system with the
bending mode damping increased from ^ = 0.02 to
0.04 is
»
»
»
»
f l = [ l / 0 . 0 1 1];
f 2 = [ l 0 . 2 4 1]1
f3 = [ l/1 0 0 0 .0 0 /1 0 1 ];
d en= conv(fl, conv(f2, f 3 ) ) ;
Step 7 of 21
»
»
»
s y s = tfr (num, dan)
m a rg in (sy s)
g rid
Step 8 of 21 -a.
The Bode plot obtained onexecutionofthe program is
R ede Diagram
G m - 7.13 iIB (a l ID ra d /s e c ), Pm - 97A deg (a l D.DII5 rad/sec)
Step 9 of 21
The gain and phase margins fromtheBodeplotare
|g J / = 7 1 3 d B a t a i= 1 0 r ^ ^
lfM = 97.5°at ai=0.085rad^
Step 10 of 21
We see that the (jM has increased considerably because theresonantpeslc
is well below magnitude 1 ; thus the system will be much b etter behaved
Step 11 of 21
The MATLAB codetodrawtheBodeplotofthe system with the
frequency of the bending modeincreasedfrom <z^=1 0 rad/sec
to = 2 0 rad/sec is
» num =8.75*[4 0. 4 1 ] ;
» f l = [ l / 0 . 0 1 1 );
» f 2 = [ l 0 .2 4 1 ];
» f 3 = [ l / 4 0 0 0 .0 4 /2 0 1 ) ;
» den=cx>nv ( f l , co n v ( f 2 , f 3 ) ) ;
» s y s = t f (num, den)
» m a rg in (sy s)
» g rid
Step 12 of 21 A
The Bode plot obtained onexecutionofthe program is
Kudc IN agram
Step 13 of 21
The gain and phase margins from the Bode plot are
|gM =7.17dB ata»= 2 0 rad/s I
\PM = 97.5° at a>= 0.085 l a ^
And again the GM is much improved and the resonant peak is
signifrcantly reduced from magnitude 1 .
Step 14 of 21
@ii)
By picking up
= 1the MATLAB code to draw the Bode plot is
» num =0.7S*[4 0. 4 1 ] ;
» f l = [ l / 0 . 0 1 1 ];
» f 2 = [ l 0 .2 4 1 ];
» f 3 = [ l / 1 0 0 0 .0 4 /1 0 1 ] ;
» f 4 = [ l 1 ];
» d e n = c o n v ( f l, c o n v ( f 2 , c o n v ( f 3 , f 4 ) ) ) ;
» s y s = t f (num, den)
» m a rg in (sy s)
» g rid
Step 15 of 21 A
The Bode plot obtained on execution of the program is
B odcD iagiaB i
Gm - 34.8 d ll (a l 8.62 la d /H v ) , P m - 92.7 deg (a l U.U847 nid/icc)
Step 16 of 21
The gain and phase margins from the Bode plot are
|gM = 34.8 dB at a = 8.62 rad/s I
\PM = 92.7° at a>= 0.0847 r a ^
which are healthy margins and the resonant peak is again, well below
magnitudel.
Step 17 of 21
The MATLAB code to draw the Bode plot with notch Biter with the
transfer function,
1
Uoo J
»
»
»
»
»
»
»
»
num =0.7S*[4 0 . 4 1 ] ;
f l = [ l / 0 . 0 1 1] ;
f 2 = [ l 0 . 2 4 1] ;
f 4 = [ 1 / 1 0 0 1] ;
d en= conv(fl, c o n v (f2 ,c o n v (f4 , f 4 ) ) ) ;
s y s = tf ( n u m , den)
m argin(sys)
g rid
Step 18 of 21 A
The Bodeplotofthe system with the given notch Biter is shown below.
B ode D ia g ra in
G m - S U dB (a l I M r a d /s e c ) . I’m - 97.4 deg (a l D.083 lad A e c)
Step 19 of 21
The gain and phase margins from the Bode plot are
lGA/ = 55.2dB atai = 100 r ^
\PM = 97.4° at ai = 0 085 r a ^
which are healthiest margins of all the designs since the notch Biter has
essentially canceled the bending mode resonant peak.
Step 20 of 21
W
Generally, the notch Biter is very sensitive to where to place the notch zeroes in
order to reduce the hghtly damped resonant peak So if you want to use to notch
Biter, you must have a good estimation ofthelocaBonofthe bending mode
poles and the poles must remain at that location for all aircraft conditions,.
On the other hand, thelowpassBlteris relatively robust to where to place its
break point Evaluation of the margins with the bending mode frequency lowered
by lO^will show a drastic reduction in the margins for the notch Biter and very
httle reduction for the low Biter.
Step 21 of 21
The table comparing the GM, PM, closed-loop bending mode dancing ratio, and
resonant peak anq/litude is
GM
PM
Qosed loop bending
Mode Damping ratio
Resonant peak
Low pass Biter
34.8
(ar = 8.62 rad/s)
92.7°
Notch Biter
55.2
(ai = lOOrad/s)
97.4°
(ai = 0.0847 rad/s)
= 0 .0 2
(ai = 0.085 rad/s)
= 0.04
0.08
0.068
The magnitude plot of the closed-loop system for Low-pass Biter is
B ede liia g n u n
The magnitude plot of the closed-loop system for Notch Biter is
B ode D ia g ra m
(d)
■While increasing the natural damping of the sj/stem would be the best solution, it
might be difficult and e^ensive to carry out. Likewise, increasing the frequency
is e^ensive and makes the structure heavier, not a good idea in an aircraft. Of the
remaining two options, it is a better design to use a low pass Biter because of its
reduced sensitivity to mismatches in the bending mode frequency.
Therefore the best recommendation would betousethelow pass Biter.
Problem 6.60PP
Consider a system with the open-loop transfer function (loop gain)
1
G (i) = -
i ( i + l) ( j/ 1 0 + l ) '
(a) Create the Bode plot for the system, and find GM and PM.
(b)
Compute the sensitivity function and plot its magnitude frequency response.
(c)
Compute the vector margin (VM).
Step-by-step solution
step 1 of 5
G(s) =
Step 2 of 5
a.
GM=20.8dB
PM=45*
Step 3 of 5
b.
1+
-
S(j<D) =
step 4 of 5
Step 5 of 5
c.
VM=Shortest distance firom (-1,0)
The smallest possible circle of center (-1.0) touches l^quist plot at -0.68-4^0.769.
V M = ^ { 0 .1 6 9 f
+(1.0.689)’ =0.829
Problem 6.61 PP
Prove that the sensitivity function S(s) has magnitude greater than 1 inside a circle with a radius
of 1 centered at the -1 point. What does this imply about the shape of the Nyquist plot if closedloop control is to outperform open-loop control at all frequencies?
Step-by-step solution
step 1 of 2
Step 2 of 2
S(s) =
l+D(s)G(s)
1
|l+D(jcD)G(j(D)|
Now distance of any point inside the circle &om its
centre is alwsQrs less then its radius
If any point of D (s)G (s) lies inside the circle then its
distance from centre = |l+D(jco)G(jo)|
■ |l+D(jo))G(j(i))|<l
1
|S(jo)|=
>1 Hence proved
|l+D(jm)G(j<D)|
Problem 6.62PP
Consider the system in Fig. 1 with the plant transfer function
^ i(f/1 0 +
l)‘
(a) We wish to design a compensator Dc(s) that satisfies the following design specifications:
{\)K v = 100,
(ii)
PAf>45“,
(iii) Sinusoidal inputs of up to 1 rad/sec to be reproduced with < 2% error,
(iv) Sinusoidal inputs with a frequency of greater than 100 rad/sec to be attenuated at the output
to < 5% of their input value.
(b) Create the Bode plot of G(s). choosing the ooen-looD aain so that Kv = 100.
(b) Create the Bode plot of G(s), choosing the open-loop gain so that Kv = 100.
(c)
Show that a sufficient condition for meeting the specification on sinusoidal inputs is that the
magnitude plot lies outside the shaded regions in Fig. 2. Recall that
Y
KG
R
l + KG
^
and
E
1
R
l+ K G ‘
—= ■
(d)
Explain why introducing a lead network alone cannot meet the design specifications.
(e)
Explain why a lag network alone cannot meet the design specifications.
(f)
Develop a full design using a lead-lag compensator that meets all the design specifications
without altering the previously chosen low frequency open-loop gain.
Figure 1 Control system
Figure 2 Control system constraints
S te p -b y -s te p s o lu tio n
step 1 of 7
G(s)=-
Step 2 of 7
a.
K»=100
KG(s) =
Step 3 of 7
Step 4 of 7
c.
forcD<Irad/sec and error ^2% , |H(jcD)|=50
For (3>>100rad/sec and attenuation ^5% , |
Step 5 of 7 ^
d.
Lead network will not affect the lower frequency
Step 6 of 7 ^
e.
Lag network will not affect the higher frequency
Step 7 of 7
f
F irrt sq>ply Ls^ compensation to Gain P M ^ 5 ^
Lag networl^
0 .4 6 5 S + 1
1 .3 5 8 S + 1
And lead network =
0 .0 7 8 S + 1
0 .0 2 1 S + 1
10°
( 0 .4 6 5 n - l) ( 0 .0 7 8 s + l)
(1.358s+l)(0.021s+l)
0. 05
Problem 6.63PP
For Example, redo the design by selecting 1/TD = 0.05 and then determining the highest possible
value 1/T/ that will meet the PM requirement. Then examine the improvement, if any, in the
response to a step disturbance torque.
Example
EXAMPLE 6 .2 0
PID Compensation DesignforSpacecraji Attitude Control
A nmplified design for ipnoeciif t Mtitiide control was presented ni
Section 6 J ; however, here we hare a more realistic sHaation that jnchides a
senior lag and a dittuibing torque. Hgure 6£1 defines the qrtfeni.
1. Design a PID controller to have zero steadyesTor to a Gonstantdistnrfaance torque, a PM o f 6S”, and as
a bandwidth as is
reasonaUy possible.
2. Plot the step response versus a coounand input and the slq» reqMOse
to a constant disnubance torque.
3. Pint the elnMwUlnnn f t r r r n n f leainM e. S — . mwI the ■em iti wtv
reasonaUy possible.
2. Plot the step response rewus a command input and the step reqwnae
to a constant disturbance torque.
3. Plot the cloBed-loop frequency response ,
n d th e sensitivity
fu n c tio n ,^ .
4. Determine waw andmona.
5. For a torque distuihtmce from sUar pressure that acts as a sinusoid at
the orbital fate (o» = 0.001 rad/secorn> lOO-minote period), oonunent
on the usefirlness o f this controller to
SolutioiL R iit, let us talm case o f the steady-stme error. For die spacecraft
to be at a steady final value, the total input torque, 7^ 4 -Tf, most equal zoo.
Hieadbre, if 7a 9^ 0, then Te — —7a. H ie only way diis can be true with
no error (a s 0) is for Dr(s) to contain an integiml term. Hence, including
in t^ ra l control in the compensation will meet the steady*state requtremenL
This could also be verified mathematically by use of the Final Vdne Theorem
(see Problem 6.47).
The frequency response o f the spacecraft and sensor; GH, where
0.9
C(a) =
\H(s)
(6.40)
Rgura 6.67
Block diagram o f
spacecraft co n tro l using
FID design,
Bcample 6.20
is shown ioH g. 6.68.H)eBlopescrf—2 (tfa aiis,-4 0 d b p erd eca d e)a o d —3
(—60 db per decade) show that the system would be unstable for any value
o f i t if no derivwive feecl»dc were used. This is clear
o f Bode’s
gain-^>hase relationship, which shows that the phase would be —iS irf w th e
—2 s k ^ and —270" for die —3 slope, wUch would correqiood to a PM of
(P or —90", respectively. Therefore, derivative cootrol is required to bring
the slope to —1 at the crossover frequency that was shown in Secdon 6.5 to
be a requiremeiU for s t^ lity . The problem now is to iMck values for the three
parameters in Eq. (6.48)— K, To, and 7/—that will satisfy the speci6catioiis.
The eariest apiMooch is to worii first on the phase so that PM = 65" is
achieved at a leasmiably high frequency. This can be accomplished primarily
by adpisting To. n otii^ that 7 / has a minor eSect if sufficiently larger than
Rgura $.68
11 1 \
111 \
T ntnT V
Compensation fo r PID
design in Example 6 .20
too
20 10
T
.t>1“
s
1
at
ac101 a01
to
to
a1
1 //
m=
t/
/
,f
-210"
-
4
aoDi
0
-2D
5 sV
\
2 5^
- (»tacS)
I
-IM*
20
(g-1)1
L. V
5 ''j
s.
L
5
1111111___
A. 0(
—D 3(1).^-at
\
tocf)
<
ooi
ai
1 2 5 10
t»(ndtae)
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
loo
Problem 6.64PP
Assume that the system
has a 0.2-sec time delay {Td = 0.2 sec). While maintaining a phase margin > 40°. find the
maximum possible bandwidth by using the following;
(a)
One lead-compensator section
Dc(s) = Kr f- - j l f
s + b*
where b/a = 100.
(b)
Two lead-compensator sections
(b)
Two lead-compensator sections
whereb/a =10.
(c)
Comment on the statement in the text about the limitations on the bandwidth imposed by a
delay.
Step-by-step solution
step 1 of 4
•UM
G(s ) = - —
^ ^ s+ 1 0
Step 2 of 4
Magnitude—201o g - = —20dB at cd<10
Vet
L et CD,= ^ = 7 . 0 7
-J2
D ( s ) = 1 0 o f l^ ^ l
'■ ^
{s ¥ 7 m }
|PM=55°|
Step 3 of 4
0 ^ .1
Let K=1000
and take cc^=2 rad/sec
D ( s ) = 1 0 0 o i^ ! ^ L
^'
(s+2 0 )'
|PM=6 8 “ |
Step 4 of 4
B 'W in bodi cas es is reduced.
Problem 6.65PP
Determine the range of K for which the following systems are stable;
(a) G(s) =
(b)
G(s) -
Step-by-step solution
step 1 of 2
Step 1 of 2
Ke-*
a.
Intersection with real axis at 4oc^—
2
G M =-
_____ 8
K<-
Step 2 of 2
GW =
Kef(s+2)
\ ^ I cosco-isinffl I
G n © )= K — ^ — r
GQ^ = 1.05rad/sec
Ato>=cc^, |G(jo)|=201ogK-7.5dB
20logK<7.5dB
|K<2.37|
Problem 6.66PP
Consider the heat exchanger of Example 2.16 with the open-loop transfer function
-5 i
G (i) = -
(10s + 1 )(6 0 5 + 1 )‘
(a) Design a lead compensator that yields PM > A5“ and the maximum possible closed-loop
(b) Design a PI compensator that yields PM > 45° and the maximum possible closed-loop
bandwidth.
Example 2.16
Equations fo r M odeling a H eat Exchanger
EXAMPLE 2 .1 6
A heat excfanger b shown in R g. 2 J 7 . Steam enters the chamber through
Equations fo r M odeling a H eat Exchanger
EXAMPLE 2 .1 6
A beat exchanger b shown in Hg. 2J37. Steam eaters the chamber tbiDugh
the controUable valve at the top, and cooler steam leaves at the bottom. There
b a coostant Bow of water th ro n g the pipe that winds U u o u ^ the middle
of the chamber so that k picks up beat from the «team- Find the diffemnial
equatioas that describe the dynamics o f the measured water outflow temper­
ature as a functioo o f the area Aj o f the steam-inlet cootrol valve when open.
The sensor that measures the water outflow temperature, being downstream
from the exit temperature in the pipe, lags the temperature by ts sec.
Solntion. The temperature o f the water in the pipe will v a y contuniously
along the p ^ as the heat flows from the steam to the water. The temperature
the steam vrill also reduce in the chamber as it passes over the mare of
pipes. An accurate tbennal model o f tb b process b therefore quite involved
because the actual heal transfer from the steam to the water will be propor­
tional to the local temperatures o f each fluid. For many cmkrol appUcatkms
it b not necessary to have great accuracy because the feedback will correct
for a considerable amount o f error in the modeL Therefore, it makes sense
to combine the spatially varying temperatures into single temperatures Tg
Rgura 2J7
Heat exchanger
X.
and Tw for the outflow steam and water temperatures, respectively. We flwn
MSMine that the heat transfer from steam to water b proportional to the dif­
ference in these temperatures, as given by Eq. (2.81). There b also a flow of
beat into the chamber from the inlet steam that depends on the steam Bow
rate and its temperature according to Eq. (2.84X
qu = wjCw(7W —Tt),
Wg = KgAg, m an flow rtee o f the w—m,
A( = area o f the steam inlet valve.
Kg = Bow coefhcieat o f the inlet valve,
cw = specific beat
the steam,
Tgg = temperature of the inflow steam,
Tg = temperature of the outflow steam.
The net beat flow into the chamber b the difference between the beat from
the hot incoming steam and the heat flawing out to the water. This net
flow determines the rate of temperature change of the steam according to
Eq.(2.82X
Cgtg=AgKgCgg(rgi-Tg)-^(Tg-T,gy,
(2.85)
Cg = iHgCggis the thermal capacity o f the steam in the chamber
R = the thermal resistance o f the heat flow averaged over the
eikiie exchangee
U tew ise, the difierential equatioo describing the water temperature is
1
(2.86)
Wv = mass flow rate o f the water,
Cew= specific heal of the water,
7 ^ = temperteure erf' the inenming wale^
Tw = temperature o f the outflowing water:
l b complete the dynamics, the time delay between the measurement and the
exit flow is described by the relation
vdiere 7 ^ b the measured downstream temperature of the water and b
the time delqr. There may also be a rlehy in the measurement of the steam
temperature Tg, vdiich would be modeled in the same manner.
Equation (2.85) b nonlinear because the quantity Tg b multiplied 1^ the
control input At. The equation can be
about Tgo (a specific value
<rf Tg) so that Tgi —Tf b assumed constant for purposes of ^tproximating
the
tenn, which we will define as ATg. In o rd n to eliminate the
Twi term in Eq. (2.86), it b conveniem to measure all temperatures in terms
of deviation in degrees from Twi. The resulting equatioita are then
c,t, =
CwTw = -
T, + i r , ,
A ltiiou^ the time delay b not a nonlineariQr, we will see in Chapter 3
tlu t operationally, Ta =
Therefore, the transfer function o f the heat
excfamiger has the form
T ,(i)
Ke-V
A,(»)
(T is-H )(n * -H )
' ' = ------------------------.
(2.87)
'
*
S te p -b y -s te p s o lu tio n
step 1 of 3
G (,) =
(10s+l)(60s+l)
Step 2 of 3
Now KnotmettionedsotakeKw 6
NowPM=40*'
to get PM = 45®, L cte= 6 ®
additional phase = l r
_ l-sin(p^
0 ^ -------1 +sintp^
1
cou^.OSrad/sec, ccu=— = ^ . 0 8
T v /i
|T=15.16|
D (,)
_6(15.6s+l)
PM= 4 3 .?
( 1 0 .3 S + 1 )
Step 3 of 3
Let us bring co^ at co=0.04rad/sec
Placing comer frequency at 0.004rad/sec,
D (,)G (0 =
PM=44“
0.02(272s+l)e*
i(10s+l)(60s+l)
Problem 6 .6 7 P P
A feedback control system is shown in Fig. 1. The closed-loop system is specified to have an
overshoot of less than 30% to a step input.
Figure 1 Control system
R
(a)
Determine the corresponding PM specification in the frequency domain and the
corresponding closed-loop resonant-peak value Mr. (See Fig. 2.)
(b)
From Bode plots of the system, determine the maximum value of K that satisfies the PM
specification.
(b)
From Bode plots of the system, determine the maximum value of K that satisfies the PM
specification.
(c)
Plot the data from the Bode plots [adjusted by the K obtained in part (b)] on a copy of the
Nichols chart in Fig. 3, and determine the resonant peak magnitude Mr. Compare that with the
approximate value obtained in part (a).
(d)
Use the Nichols chart to determine the resonant-peak frequency turand the closed-loop
bandwidth.
Figure 2 Transient-response overshoot {Mp) and frequency-response resonant peak {Mr) versus
a»S
PM for f(s) _
Fhae mfgifl
Figure 3 Nichols chart
Step-by-step solution
step 1 of 5
M .=30% =100e’^
%
. |;=0.358|
Step 2 of 5
a.
/ ’Jf£ 1 0 0 { = 35.8"
Step 3 of 5
b.
| g = 7.8|
Step 4 of 5
c.
\M^ = 1.5 by Mchols chart also
Step 5 of 5
d
BandwidA = |4.36 rad/sec|
Problem 6.68PP
The Nichols plots of an uncompensated and a compensated system are shown in Fig.
(a) What are the resonance peaks of each system?
(b)
What are the PM and GM of each system?
(c)
What are the bandwidths of each system?
(d)
What type of compensation is used?
Figure Nichols plots
Figure Nichols plots
Step-by-step solution
step 1 of 4
a.
Mg
(Uncompensated)—l.S
Mg (Compensated)=1.0S
Step 2 of 4
b.
GM (Uncompensated) = 5
GM(Compensated) =10
PM (Uncomp ensated)=40*
PM (Compensated) ^63"
Step 3 of 4
c.
£andwidtli(Uncompensated)=70radf/f
Bandwidth (Compensated) = 3Qrad f s
Step 4 of 4
d.
Lagcompensationisused.
Problem 6.69PP
Consider the system shown in Fig.
(a)
Construct an inverse Nyquist plot of [Y(jo))/E(yojj]-1. (See Appendix W6.9.2.)
(b)
Show how the value of K for neutral stability can be read directly from the inverse Nyquist
plot.
(c)
For K = A ,2 , and 1. determine the gain and phase margins.
(d)
Construct a root-locus plot for the system, and identify corresponding points in the two plots.
To what damping ratios ^do the GM and PM of part (c) correspond?
(d)
Construct a root-locus plot for the system, and identify corresponding points in the two plots.
To what damping ratios ^do the GM and PM of part (c) correspond?
Figure Control system
Step-by-step solution
step 1 of 4
a.
KG(s)=-
3K
(s+l)(s+3)
Step 2 of 4
b.
K is intersection with real axis
Step 3 of 4
c.
K=4;GM=0dBPM=0"
K*2;GM=6.02dB ,PM=18.3"
K=1;GM=12dB J>M=38.1"
Step 4 of 4
d
for K=4. ^
forK= 2 , ^ . 1 8 3
for K=l, 5=0.381
Problem 6 .7 0 P P
An unstable plant has the transfer function
Y(s) _
F(s) ~
^+1
A simple control loop is to be closed around it, in the same manner as in the biock diagram in
Fig.
(a) Construct an inverse Nyquist plot of Y/F. (See Appendix W6.9.2.)
(b) Choose a value of K to provide a PM of 45®. What is the corresponding GM?
(c) What can you infer from your plot about the stability of the system when K < 0 7
(c) What can you infer from your plot about the stability of the system when K < 0 7
(d)
Construct a root-locus plot for the system, and identify corresponding points in the two plots.
In this case, to what value of ( does PM = 45° correspond?
Figure Control system
Step-by-step solution
step 1 of 4
Step 2 of 4
b.
K = 3.S9
Jbr
PM =A ^ .
O M = -5.1SdB
Step 3 of 4
c.
At K < 0 system is unstable
Step 4 of 4
Root locus plot of Y/F
At PM = 4 ?
. ^ = 0.45
Problem 6.71 PP
Consider the system shown in Fig. 1{a).
(a) Construct a Bode plot for the system.
(b) Use your Bode plot to sketch an inverse Nyquist plot. (See Appendix W6.9.2.)
(c)
Consider closing a control loop around G(s), as shown in Fig. 1(b). Using the inverse Nyquist
plot as a guide, read from your Bode plot the values of GM and PM when K = 0.7,1.0.1.4. and 2.
What value of K yields PM = 30“?
(d)
Construct a root-locus plot, and label the same values of K on the locus. To what value of ^
Ho p s p a n h n a ir nf PM /f^M v a lu p a m rrpR nnnri? C n m n a rp (^vpraiiR P M with th p rniinh
(d)
Construct a root-locus plot, and label the same values of K on the locus. To what value of ^
does each pair of PM/GM values correspond? Compare ^ versus PM with the rough
approximation in Fig. 2.
Figure 1 Control system
Figure 2 Damping ratio versus PM
Phue niMgtn
Step-by-step solution
step 1 of 4
Step 2 of 4
Step 3 of 4
K=0.7
K=1.0
K=1.4
K=2.0
;
;
;
;
GM=15.1dB,
GM=12dB ,
GM=9.12dB,
GM=6.02dB,
PM=54.7"
PM=44.1®
PM=33.1*
PM=21.4"
AtK=1.54, PM=30“
Step 4 of 4
d.
The s^roximation of
PM
---- is valid in all die cases
100
Problem 7.01 PP
Write the dynamic equations describing the circuit in Fig. Write the equations as a second-order
differential equation in y(t). Assuming a zero input, solve the differential equation foryffj using
Laplace transform methods for the parameter values and initial conditions shown in the
figure.Verify your answer using the initial command in Matiab.
Figure Circuit
£.= 1H
Jt= 2 n
y(0
?(U= 1V, y(g=o
>(U=lV,3Kg=0
Step-by-step solution
Step 1 of 10 -
The given circuit is
L=1H
----------—
R = 2 ii
-------------- 0
WV—
u(t)
m
C =1F “
o
0
y ( 'o ) = iv
/(/„ ) =
o
Step 2 of 10
From the figure.
The current through the Cc^acitor is
iW =
^'
dt
DifTerentiating equation (1) we get
( 1)
( 2)
dt
The voltage across the inductor is
-di
V= L —
dt
Writing mesh equation to the first loop,
- u ( t ) + L — + S i( t) +,y(i) = 0
dt
Therefore, we get
dt
L
^ (0
L
dt
L
L\
dt
L
L dt
1 .A
d t)
Now substitute equation (2) in the above equation,
dt
L
L dt
(3)
U ^ '
Step 3 of 10
Given that
£ = 1H
C = 1F
Substitute the given values of £ , i? and C in equation (3),
1 ^
dt
Rearranging the equation,
y { l) + 2 y [t) + y \ t ) = u (t)
Taking L^lace transform on both sides,
tY l ^ s ) - ^ [ h ) - y '[ h ) + 2 s r ( s ) - 2 7 ( i ,) + r ( s ) = U{s)
Substituting the initial conditions shown in the figure,
sV (s) - s - 0 + 2 s r (s) - 2 + r (s) = 0
(s“ + 2 s + l) r ( s ) = s + 2
r(s)=
7 + 2 s +1
'
r(s )= _ £ ± ij
Step 4 of 10
Using partial fi^ction e^ansion method,
s+ 2 _ A ^ B
(s + l)’
j4( s + 1 ) + J
( - + ') '
Equating the coefficients we get
A =\
^+5=2
5 =1
So, we have
g+ 2
1
^
1
(s + l)“ s + 1 (s + 1 ) '
Taking inverse l^ la c e transform
So, we get
Step 5 of 10
ilie transfer function of the given sjrstem is
r(s)=
Step 6 of 10
To verify the solution using MATLAB, re-write the differential equation in state space
form.
=f1,-1
“
- 2 ) \j,),
£
s ai^ + bu
^ = [’ %
= tX
v&ere
X = [y
y f
Step 7 of 10 ^
The MALAB program to verify the answer is
»
»
»
»
»
»
»
»
»
»
»
»
a = [ 0 ,l;- l,- 2 ] ;
b = [ 0 ; l} ;
c = [ l,0 ] ;
d = [0 ];
s y s = s s (ar h,
d) ;
x o = [1 ;0 ];
[ y / t , x ] * i n i t i a l ( s y s ,x o ) ;
p lo t ( t,y ) ;
x l a b e l ( ^Tinie ( s e c ) * ) ;
y l a b e l ( *y ( t ) * ) ;
t i t l e (* I n i t i a l c o n d i tio n r e s p o n s e * ) ;
g rid ;
Step 8 of 10
The MATLAB ou^ut is
Initial ceadilion nKuiHise
Step 9 of 10
The MATLAB code to plot the response obtained using Ls^lace transform is
»
»
»
»
syas t
t= 0 ; 0 . 0 0 0 1 ; 8 ;
y = e x p (-t)+ t. * e x p ( - t) ;
p lo t (t,y ) ;
The MATLAB ou^ut is
Response nsing l.aplace Transfarm
Step 10 of 10
The response obtained using i n i t i a l command and that obtained using Lt^lace
Transform are identicaL Hence the solution is verified.
Problem 7.02PP
A schematic for the satellite and scientific probe for the Gravity Probe-B (GP-B) experiment that
was launched on April 30,2004 is sketched in Fig. Assume that the mass of the spacecraft plus
helium tank, m1, is 2000 kg and the mass of the probe, m2, is 1000 kg. A rotor will float inside
the probe and will be forced to follow the probe with a capacitive forcing mechanism. The spring
constant of the coupling k is 3.2 x 106. The viscous damping b is 4.6 x 1 03.
(a) Write the dynamic equations of motion for the system consisting of masses ml and m2 using
the Inertial position variables, y1 and y2.
(b) The actual disturbance u is a micrometeorite, and the resulting motion is very small.
Therefore, rewrite your equations with the scaled variables
z1 = 1063^1, z2 = 106y2, and v= lOOOw.
aIicicivjic, lev*I lie yuui ei^uauui lo win i u k si/aieu V
z1 = 1063^1, z2 = 106y2, and v= lOOOw.
(c) Put the equations in state-variable form using the state X = (z| i | Z2
z2, and the input an impulse, u= 10-36ffj N sec on mass ml.
(d)
Using the numerical values, enter the equations of motion into Matlab in the form
x = Ax-l-Bv
y = OL+Dv.
and define the Matlab system: sysGPB = ss(A,B,C,D). Plot the response of y caused by the
impulse with the Matlab command impulse(sysGPB). This is the signal the rotor must follow.
(e) Use the Matlab commands p=eig(F) to find the poles (or roots) of the system and z
=tzero(A,B,C, D) to find the zeros of the system.
Figure Schematic diagram of the GP-B satellite and probe
i
/•“
Step-by-step solution
ste p 1 of 14
(a)
Step 2 of 14
The coordinates of the two m asses are
jtj
and y is the displacements of the masses from
their equilibrium conditions.
Step 3 of 14
Refer Figure 7.84 in textbook and draw the free body diagram to identify the direction of the
forces on the object.
Step 4 of 14
Consider
and fixed and increase
from 0.
The it spring will be compressed producing its spring force.
Step 5 of 14
The free body diagram 1s shown in Figure 1.
Step 6 of 14
b ih r h )
Figure 1
Step 7 of 14
Consider the Newton’s law of equations of motion for any mechanical system.
F = m a ......(1)
Where,
F is the vector sum of all forces applied to each body in a system
a is the vector acceleration of each body with respect to an inertial reference
frame.
m is the mass of the body.
Refer Figure 1 and write the equation of motion.
“ ij’i
-
.
Write the equation of motion for the mass
" W j + *(> ’2 - J'i )+M.V 2 - > ’i) = 0
* y ’2 = - * ( > ’2 - > l) - * ( ^ 2 - .) 'l)
(3)
Thus, the differential equation for the mechanical system is
h j > i = - * U - y 2 ) - * ( > ’| - > ’2 ) ^ 3nd
) - * ( > ; -y ,)]
ste p 8 of 14
(b)
Consider the value of
z,=10*;r.
y.-lO-z,... ... (4)
y.-lO-i,... ... (5)
#,-10-2,... ... (6)
Consider the value of
Z, = 10*.v,
yj.lO-z,
>,=10-4*,
j!,. 10-42,
....(7)
....(8)
....(9)
Consider the value of
v=U000ir
1
1.000
( 10 )
Consider the mass value of spacecraft plus helium tank.
» H = 2 ,0 0 0 /ig ...... (11)
Cohsider the mass value of probe.
I», = l,0 0 0 te ...... (12)
Cohsider the spring constant value.
* = 3 .2xI0‘ ...... (13)
Consider the viscos damping value.
* - 4 .6 x 1 0 ’
(14)
Step 9 of 14
Substitute equations (4) to (14) in equations (2) and (3).
-(3.2xl0‘)(l0-‘z,-10-‘z,)
(2.000)
10-‘ z,
(2.000)
10-‘i:,
-(4.6xl0’)(l0-i.-10-‘i.) 4 ^ v
-(3.2xl0’)(l0-‘)(z,-z.)
-(4.6xl0’)(l0-)(z,-i,) +^ v
10*
^-3.2(z, -z,)-(4.6xl0-’)(z, -* 2)+ ij^ v j
" (2,000)
z, =-l,600(z,-z,)-2.3(z,-z,)+0.5v...... (15)
(1.000)10-*z, =
-(3.2xl0‘)(l0'*zj -10‘‘z,)'
-(4.6xl0’)(l0-‘i,-10-‘z,)
- ( 3 . 2 x 10‘ ) (1 0 -4 )(z, - z,) ‘
(1,000)10-‘ 2,
-(4.6xl0’)(l0-‘)(i.-i.)
10*
h
- * | ) - ( “ «X 10'’)(Z2 - i | ) ]
zj =-3,200(zj-z,)-4.6(z,-z,)
(16)
Hence, the equation with the scaled variables are
s —l,6 0 0 ( z ,- Z j) - 2 .3 ( i, —
and |z, = - 3 , 2 0 0 ( z a - z , ) - 4 . 6 ( i a ^
Step 10 of 14
(c)
Consider the value of z ,.
...... (17)
Consider the value of x^...... (18)
Consider the value of x^.
..... (19)
=
Consider the value of
........( 2 0 )
=
Consider the value of Zj.
........... ( 21 )
Consider the value of x^* 4 = ^ ...... (22)
Consider the value of x^* 4 = i j ...... (23)
Consider the value of
^4=^2 ..... (24)
Substitute equations (17) to (24) in equations (15) and (16).
i , = - 1 .6 x l0 ’ ( z ,- .) ^ ) - ( 2 .3 ) ( z ,- z ,) + 0 .5 i i
X, = - ( 3 , 2 0 0 ) ( z , - z ,) - 4 .6 ( z ,- z ,)
(25)
(26)
Write the state matrix form from equatiohs (18), (22), (25) and equatioh (26).
0
1
0
0
-1,600 -2.3 1,600 2.3
0
0
0
1
3,200 4.6 -3,200 -4.6JL*4J
0
1
0
0
-1,600 -2.3 1,600 2.3
0
0
0
1
o'
x+
0.5
(27)
0
0
Consider the output equation.
y=
...... (28)
Modify equation (28) by equation (21).
Consider the output equation.
;- = [0 0
1 0]
,y = [0 0
1 0 ]* + 0
+0
(29)
0
Hence, the state variable form is irs
|y = [0
0
1
-1,600 -2.3
0
0
1,600
2.3
0
0
0
1
3,200
4.6
-3,200
-4.6
0
x+
0.5
0
U
0
1 0 ]x + 0 |.
Step 11 of 14
(b)
Write the MATLAB program from equations (27) and (29) for plot the response of y caused by the
impulse.
A= [0 1 0 0;-1600 -2.3 1600 2.3;0 0 0 1;3200 4.6 -3200 -4.6];
B=[0:0.5:0;0]:
c =[0 01 0]:
D=[0]:
sysGPB=ss(A,B,C,D);
t=0:0.001:1;
y=impulse(sysGPB,t):
plot(t,y)
The output of the MATLAB program is shown in figure.
Step 12 of 14
Hence, the response of the output y caused by the Impulse is
by using numerical value
of A, B, C and D.
Step 13 of 14
(e)
Write the MATLAB program from equations (27) and (29) to identify the location of poles and
location of zero of the system.
clc;
A= [0 1 0 0;-1600 -2.3 1600 2.3;0 0 0 1;3200 4.6 -3200 -4.6];
B=[0:0.5:0;0]:
c =[0 01 0]:
D=[0];
p=eig(A)
z=tzero(A,B,C,D)
Step 14 of 14
The output of the MATLAB program is given below.
p = -3.4500 +69.1961 i
-3.4500 -69.19611
-0.0000 + O.OOOOi
-0.0000 - O.OOOOi
z = -695.6522
Hence, the poles and zero locations are obtained by using MATLAB.
Problem 7.03PP
Give the state description matrices in control-canonical fonri for the following transfer functions:
(a) G(J) =
"■> « ' ) = wmTTT-
"F+3+2"
(d) C(j) =
(d) C(i) =
Step-by-step solution
ste p 1 of 8
(a)
Consider the numerator part of the gain.
( 1)
=
Consider the denominator part of the gain.
a {s)= s "+ 0,$'^ +ajS*~^ +...+a^
(2)
Consider the general canonical form of the equation.
cM =
(3)
a ( j)
Substitute equations (1) and (2) in equation (3).
S +fljJ
+O2S
+*** + fl.
ste p 2 of 8
Write the state description matrices in control canonical form.
-<h .......... -a .
A ,=
1
0
..........
0
1
0
0
1
•••
0
...........
1
0
0
B ,=
0
(5)
0
(6)
b, ........... *.]
C .= [A
(7)
£)^=Seperate valuefromnumeratoraiiddeiiominatorelse 0
(8)
Consider the gainof G (j)-
1
2s+\
1
2{s+0.5)
G (*) =
0.5
G(s) =
(9 )
j+ 0 .5
Write the state descriptioh matrices in controi canonicai foim from equations (4) to (9).
A . =[0.5]
B, = [l]
C,
=[0.5]
D.
=0
A , =[0.5]
Hence, the value of state description matrices In control canonical fonn is
B, = [l]
C . = [0.5]
D.
=0
Step 3 of 8
(b)
Consider the gainof G ( 4)
G(«) =
_ £ a
10 )
I
=2 0 i^
4 + 10
201+60
G (4 )
4 + 10
204 + 200-140
4 + 10
20(4+10)-140
4 + 10
G (4) = 2 0 - ^
. (
10 )
4 + 10
ste p 4 of 8
Write the state description matrices in control canonical form from equations (4), {5). (6). (7). (8)
and (10).
A , = [-1 0 ]
B, = [l]
C,
= [-1 4 0 ]
D =20
A , = [-1 0 ]
Hence, the value of state description matrices in control canonical fonn is B, = [l]
C , = [-140]
D,
=20
Step 5 of 8
(c)
Consider the gainof G ( f
84 + 1
G (4 )
4*+34 + 2
( 11)
Write the state description matrices in control canonical form from equations (4), (5), (6), (7), (8)
and (11).
-3 -21
1
oj
pj
8 1]
D - 0]
A ,=
Hence, the value of state description matrices in control canonical fonn is B . -
;i
c , = 8 1]
0]
Step 6 of 8
(d)
Consider the gainof G (4)
. (
12 )
Write the state descriptioh matrices in controi canonicai form from equations (4), (5). (6). (7). (8)
and (12).
’ -2
-2
O'
1
0
0
.0
1 0
r
0
0
0
c .=
1 7]
»]
A ,=
Hence, the value of state description matrices in control canonical fonn is
-2
-2
1
0
O'
0
0
1
oJ
B .= 0
0
Q = 0 1 7]
0]
Step 7 of 8
(e)
Consider the gainof G(4)
(4 + 10)(4^+4 + 25)
^
^ '4 ^ ( 4 + 2 ) ( 4 '+ 4 + 36)
, ^
4*+ 114^ + 354 + 250
(13)
' * ' ° 4 ’ +34*+384"+724^
ste p 8 of 8
Write the state description matrices in control canonical form from equations (4), (5), (6), (7), (8)
and (13).
-3
-38
A .=
0
0
0
0
0
0
0
0
1 0
C.=
I
11 35 250]
o ,=
Hence, the value of state description matrices in control canonical fonn is
Problem 7.04PP
Use the Matlab function tf2ss to obtain the state matrices cailed for in Problem.
Give the state description matrices in control-canonical fonn for the following transfer functions:
Problem
~ 2l+l*
"■> « ' ) = wmTTT-
"F+3+2"
IC|
=
(d) C(i) =
Step-by-step solution
(a)
Consider the value of gain <7(^)
' ’ 2^+1
2 ( i+ 0 .5 )
C ( ,) =
0.5
( 1)
j+ 0 .5
Write the MATLAB program from equation (1) to find the state description matrices.
num=[0.5];
den = [1 0.5];
[A,B,C.D]=tf2ss(num,den)
Step 2 of 11 ^
The output of the MATLAB program is given below.
A=
-0.5000
B=
1
C=
0.5000
D=
0
Hence, the value of state description matrices is obtained by MATLAB.
(b)
Consider the value of gain G (j)-
G (« ) =
.m
I 10 J
f +10
C (5 ) =
20f + 60
(2)
5 + 10
Write the MATLAB program from equation (2) to find the state description matrices.
num=[20 60];
den = [1 10];
[A,B,C,D]=tf2ss(num,den)
Step 4 of 11
The output of the MATLAB program is given below.
A=
-10
B=
1
C=
-140
D=
20
Hence, the value of state description matrices is obtained by MATLAB.
Step 5 of 11
(c)
Consider the value of gain G(5)85 + 1
»’ + 3 i + 2
C (5 )
(3)
Write the MATLAB program from equation (3) to find the state description matrices.
num=[8 1];
den = [13 2];
[A,B,C,D]=tf2ss(num,den)
Step 6 of 11
The output of the MATLAB program is given below.
A=
-3 -2
10
B=
1
0
C=
81
D=
0
Hence, the value of state description matrices is obtained by MATLAB.
(d)
Consider the value of gain G(5)5+7
G (4
5 ( 5 ^+ 25 + 2)
G (« ) =
5+ 7
■(4)
5* + 25*+25
Write the MATLAB program from equation (4) to find the state description matrices.
num=[1 7];
den = [1 2 2 0];
[A,B,C,D]=tf2ss(num,den)
Step 8 of 11 ^
The output of the MATLAB program is given below.
A=
-2 -2 0
1 00
0 10
B=
1
0
0
C=
0 17
D=
0
Hence, the value of state description matrices is obtained by MATLAB.
(e)
Consider the value of gain G(5)( s + 1 0 ) ( i’ + j+ 2 S )
G(;
j ^ ( s + 2 ) ( j '+ i + 3 6 )
, ,
(s + 1 0 ) ( j’ + s + 2 5 )
Step 10 Of 11
Write the MATLAB program from equation (5) to find the state description matrices.
num=conv{[1 10],[1 1 25]);
den = conv([1 2 0 0],[1 1 36]);
[A,B,C,D]=tf2ss(num,den)
Step 11 of 11
The output of the MATLAB program is given below.
A=
-3 -38 -72 0 0
10000
01 0 0 0
00 100
0001 0
B=
1
0
0
0
0
C=
0 1 11 35 250
D=
0
Rewrite the output with appropriate notations.
-3
-3 8
-7 2
0 O'
0
0 0
0
0 0
K =
1
0 0
I
0
0
0
0
p
C.
= 0
D.
= 0]
1 11 35
250]
Hence, the value of state description matrices is obtained by MATLAB.
Problem 7.05PP
Give the state description matrices in modal carionical form for the transfer functions of Problem.
Make sure that all entries in the state matrices are real valued by keeping any pairs of complex
conjugate poles together, and realize them as a separate subblock in control canonical form.
Give the state description matrices in control-canonical form for the following transfer functions;
Problem
"■> « ' ) = wmTTT-
"F+3+2"
"F+3+2"
(d) C(i) =
Step-by-step solution
ste p 1 of 10
(a)
Consider the numerator part of the gain.......(1)
Consider the denominator part of the gain.
2
o ( i) = i" + o ,i" ''+ < i » ''’ + ...+ a ,
(2)
Write the general canonical form of the equation.
.m
a(s)
(3)
Substitute equations (1) and (2) in equation (3).
C (s) =
s" +a,s”"'
(4)
+ —+a.
Write the state description matrices in control canonical form.
-<h ........... - 0,
-« j
1
0
...........
1
0
..
0
0
0
1
...
0
............
1
0
(5)
1
0
B ,= 0
...... (6)
0
C .= [A
6,
........... i , ]
(7)
£)^=Seperate value&omnumetatoraiiddeiiominatorelse 0
(8)
Consider the value of G (j).
GW =—
'
’
2s+\
1
2 ( j+ 0 .5 )
G(s) =
0.5
(9)
j+ 0 .5
Write the state description matrices in control canonical form from equations (1) to (9).
A , = [-0 .5 ]
B,
=[l]
C,
=[0.5]
D=0
A . = [-0.5]
Hence, the value of state description matrices in control canonical form is B .
C.
= [l]
=[0.5]
0=0
Step 2 of 10
(b)
Consider the value of G (j)
G (*) =
f e - ')
.jx l
■ (H ?)
s + lO
201+60
G (*)
i + 10
20 1 + 2 0 0 -1 40
j+ IO
2 0 (j+ 1 0 )-1 4 0
i + lO
.(10)
G (5) = 2 0 - ^
s+ 1 0
Write the state description matrices in control canonical form from equations (1) to (8) and
equation (10)
A , = [-1 0 ]
B, = [l)
C,
= [-140]
D,
=20
Hence, the value of state description matrices in control canonical form is
A ,= [ - 1 0 ] ,B , = [I]
C, = [-140]and£),=20
Step 3 of 10
(c)
Consider the value of G W
G(5)
8>+l
»’ + 3 i + 2
GW =
8s + I
( i + l ) ( j + 2)
.(11)
Apply partial fraction in equation (11).
*»+l
B
A
( i + l ) ( * + 2 ) ° ( i + l ) '^ ( j + 2 )
* '
Find A from equation (12).
,
8s + l
(s + l)(» + 2 )
'" i
8 s+ I
-8 + 1
-1 + 2
A = -7 ...... (13)
Find B from equation (12).
„
8s+l
(s + l) ( j + 2 ) <
" 1
85 + 1
■ ( s + 1)
-1 6 + 1
-2 + 1
B = 1 5 ...... (14)
Substitute equations (13) and (14) in equation (12).
-7
85 + 1
(5 + 1)(5 + 2)
15
7+7
(5 + 1)
.(15)
(5 + 2 )
Draw the block diagram from equation (15).
Figure 1
ste p 4 of 10
Write the general form of state space equation.
x = A x+B tr ...... (16)
Write the state equation in control canonical form from figure 1.
...... '
Write the general form of output equation.
y = C x+ D ...... (18)
Write the output equation in control canonical form from figure 1.
k= [-7
1 5 ][^ ]+ 0
y = [ - l 1 5]*+ 0
(19)
Write the state
tate description
de:
matrices in control canonical form from equations (16), (17), (18) and
(19).
i ]
:]
-7
D.
15]
0]
-1
A, =
Hence, the value of state description matrices in control canonical form is B . -
[o
01
-2 j
;]
c , = -7
15]
0]
Step 5 of 10
(d)
Consider the value of G (5)
5+7
G (5 ).
(20)
5 ( 5 ’ + 25 + 2 )
Apply partial fraction in equation (20).
5+7
A
5(5* + 25 + 2)
s
Bs + C
^ 5 ’ +25 + 2
Determine A value from equation (21).
A=
5+ 7
5 ( 5’ + 2 5 + 2 )
5+7
' ( 5' + 25 + 2)
A = j ...... (22)
Determine the value 6 and C from equation (21).
j+ 7
a (5’
+25 + 2)+(B 5+C )5
5 ( 5' + 2 5 + 2 )
5 (5 ' + 2 s + 2 )
s + 7 —s^A+ 2sA+2A+Bs^+ C s ...... (^^)
Write the coefficient of
from equation (23).
A + B = 0 ...... (24)
Substitute equation (22) in equation (24).
j + B =0
(25)
- I
Write the coefficient of s from equation (23).
l = 2A + C ...... (26)
Substitute equation (22) in equation (26).
C = l-7
C = - 6 ...... (27)
Substitute equations (24), (25) and (27) in equation (21).
7
7
-5+6
5+7
_____________
, 2 _ _ 2 ............ ...... (28)
5(5’ + 25 + 2)
5^+25 + 2
5
Draw the block diagram from equation (28).
Figure 2
ste p 6 of 10 .+■
Write the state equation in control canonical form from figure 2.
0
*1
0
0
0
0
1
0
x= 0 -2
0
1
4)
42 + 1
0
.’h.
-2
= 0 -2
0
r
0
- 2 x+ 1 If ....
1
0
(29)
0
Write the output equation in control canonical form from figure 2.
,.[ Z
-z
. ]
+0
,.[Z -Z .] x + 0
(30)
Step 7 of 10 ^
Write the state description matrices in control canonical form from equations (16), (17), (29) and
(30).
0
0
0
0 -2
-2
0
1
0
1
B,
=
C.
1 - I =J
2 2 J
A =
Hence, the value of state description matrices in control canonical form is:
Step 8 of 10
(e)
Consider the value of G (5)
(4 + I0 )(5 ^+ 5 + 25)
(31)
' ^~5^(5 + 2)(5’ +5 + 36)
Apply partial fraction in equation (31).
(5 + 10)(5’ +5 + 25)
As+ B ^ C ^ Ds+E
5'
^ { s + 2 )^ ^ + s + 3 6
5'(5 + 2)(5’ +5+36)
(32)
Determine A value from equation (32).
C
(5 + 10)(5^+ 5+ 25) ,
5* ( 5 + 2) ( 5 * + 5+36]
'" i
(5+ 10)(5^+5 + 25)1
5 ^( 5 “ + 5 + 3 6 )
L
(-2 + 1 0 )(4 -2 + 2 5 )
4 (4 -2 + 3 6 )
, (8)(27)
4(38)
C = 1.421 ...... (33)
Modity equation (32).
(4|5 + B )(5 + 2)[5* + 5 + 36)
+c(5= )(5= +5 + 36) + (fl5 + £)(5* )(5 + 2)
(5 + 10)(5’ +5 + 25)
*(5 + 2)(5*+ 5 + 36)
5®(5 + 2)(5*+ 5 + 36)
'(A5 + B )(5 + 2 )(5 '+ 5 + 36)
( s + 10)(5*+5+25) =
+ 0 ( 5 ^)(5^ + 5 + 36)+(Z)5 + £ )(5 ^)(5 + 2)
A s' + As’ +36As^ + 2As^+2As’ +T2As
5’ +ll5'+355 + 250-. +a5’ + «s"+36B5+2fl5’ + 2a5+72B+C5’
+ 0 ’ + 36C 5'+ D 5 * + 2 C s ’ + £ 5 ’ + 2 £ s '
'A5*+ 3 .45’ +38^5*+72X5
5’ +ll5’ +355 + 250 == +B5’ +3B5’ +38B5+72B+C5*
(34)
+Cs’ +36Cs* + Q s' +2Ds' + Es'+2Es^
Write the coefficient of constant from equation (34).
250 = 72B
B = 3.472 ...... (35)
Write the coefficient of s from equation (34).
35 = 72X +38B
(36)
Substitute equation (35) in equation (36).
35 = 72X+38(3.472)
X = -1.346 ...... (37)
Write the coefficient of
from equation (34).
ll = 38X +3B +36C +2£
(38)
Substitute equation (33). (35) and (37) in equation (38).
l l = 38(-1.346)+3(3.472)+36(1.421)+2£
£ = 0.288 ...... (39)
Write the coefficient of 5* from equation (34).
l = 3 X + B + C + 2 f l+ £
(40)
Substitute equation (33). (35). (37) and (39) in equation (40).
1= 3 ( - l .346)+3.472+1.421+2£>+0.288
D = -0 .0 7
(41)
Substitute equations (33), (35), (37). (39) and (41) in equation (32).
(5 + 10)(5^+5 + 25)
-1,3465 + 3.472
1.421
s‘
5 '( 5 + 2 ) ( 5 ’ + 5 + 3 6 )
-0.075 + 0.288
( 4 + 2)
5*+ 5+ 36
ste p 9 of 10
Draw the block diagram from equation (42).
Figure 3
ste p 10 of 10 A
Write the state equation in control canonical form from figure 3.
"4l
0 0
0
0
0
4j
1
42
1 0
0
0
0
4i
0
= 0 0 -2
0 0
4.
0
0 0
.45.
0
-36
4.
0
.45.
0
0 0
0
1 0
4i + 1
1
0
-1
1
0
0
0
r
0
0
0
JC= 0 0 - 2
0
0
0
0 0
0
-1
-36
1
0 0
0
1
0
0
x+ 1 tt
(43)
Write the output equation in control canonical form from figure 3.
y = [-1.346 3.472 1.421 -0.07
0.288]
y = [-1.346 3.472 1.421 -0.07
0.288]x+0
+0
(44)
Write the state description matrices in control canonical form from equations (16), (17), (43) and
(44).
0 0
0
0
0 ■
1 0
0
0
0
A .= 0 0 - 2
0 0 0
0
0
-1
-36
1
0
0 0
0
0
B ,= 1
1
1
p
c . = -1.346 3.472
1.421
2>c = 0]
Hence, the value of state description matrices in control canonical form is
Problem 7.06PP
A certain system with state x is described by the state matrices
-[= -]■
-=[1]-
C= [
D =0.
1 0 ],
Find the transformation T so that if x = Tz, the state matrices describing the dynamics of z are in
controi canonical form. Compute the new matrices A, B, C, and D.
Step-by-step solution
ste p 1 of 2
Step 1 of 2
(a)
Consider the state description matrix.
!l.
(1)
■[:;;]
B
C = [l 0]
(3)
D = 0 ......(4)
Write the general formula for ^matrix.
A = T 'A T ..... (5)
Where,
X is transformation matrix.
Write the general formula for entire transformation matrix,
t,A
t
T"' =
.
......(6)
Consider the general formula for inverse of the transformation matrix,
f . = [0 0 ... l]C -‘
.
(7)
Consider the general formula for controllability matrix. C ■
C =[b
AB
... A - ’ b ] ..... (8)
Calculate AB^rom equations (1) and (2).
;
k
oil 3
- [ 'J
AB = | . I ......(9)
Substitute equations (2), (9) in equation (8).
^^=[3 J
Calculate t^from equations (7) and (10).
-[I
t]
Write the general formula for t,t ,= t j A ......(12)
Calculate t|from equations (1), and (11) in equation (12).
'■[I
tK
;i
(13)
- [ ? I]
Calculate entire transformation matrix, j - t from equation (6). (11) and (13).
3"
5 5
3 -1
L5 5.
(14)
Calculate transformation matrix, x ftoid equation (14).
T =L
J ......(15)
Substitute equations (1), (14) and (15) in equation (5).
d
1
K ® :i
(16)
Consider the general formula for
q
.
B s T ' B ...... (1^)
Substitute equations (2) and (14) in equation (17).
ll
Bs
I
5 5
3 ^
5 5J
(18)
Write the general formula for
q
.
C = C T ..... (19)
Step 2 of 2
Substitute eauations (3) and (15) in eauation (19).
"i[! 3
C = [l 3] ......(20)
Write the general formula for ^ .
D ^ D ...... (21)
Substitute equations (4) in equation (21).
D ^O
A=
-2 -2 ]
.1
Hence, the value of new matrices are B =
i]
c = > 3]
0
oJ
Problem 7.07PP
Show that the transfer function is not changed by a linear transformation of state.
Step-by-step solution
ste p 1 of 2
Assume the original system,
X = PX-\-Ou
y = H X -\-Ju
a(s) = H(si-py'a+j
Assume a change of state X to Z using the non singular transformation T,
y = n ^ + ju
0 (s ) = H ( S l - F y 'o + J
Assume a change of state X to Z using the non singular transformation T,
X=TZ
The new system matrices are,
A= T^F T
B =
C = HT
D = J
Step 2 of 2
The transfer function is,
0 ,( S ) = C { S l-A y 'B + D
= H T [ S I - T - 'F T y ^ ’r ^ a + J
a,{s)= h t Is t t ^-t ' ftY't -'o+j
= H { S I - F y ''a + J
I
= a{s)
=o{s) I .
Problem 7.08PP
Use block-diagram reduction or Mason’s rule to find the transfer function for the system in
observer canonical form depicted by Fig.
Figure Observer canonical form of a third-order system
Step-by-step solution
Step-by-step solution
ste p 1 of 3
Step 2 of 3
Mason’s gain formula:
TrMsfer Function = . ^ 4 = V
U {s)
tl
A
Where K = Number of Forward Paths
C/(s) "
A
where J/;^»Gainof Jt* forward path
A = l-(Sum of IndiTidual Loop Gains)+(Sum of Products of two non Touching Loop Gains)
- (Sum of Products of three Non Touching Loop Gains) + -----Aj^ = Nodiing but A , not
touching the ^ forward path.
Step 3 of 3
From the figure.
s
where
, £3 ,
are the loop gains.
i j + AjS +
+a^s +OiS^
s
Problem 7.09PP
Suppose we are given a system with state matrices A. B, C (D = 0 in this case). Find the
transformation T so that, under Eqs. (1) and (2), the new state description matrices will be in
observer canonical form.
Eqs. (1)
A - T “ 'AT,
Eqs. (2)
C - c r . fi-D .
Step-by-step solution
Step-by-step solution
ste p 1 of 4
(a)
Write the observer canonical form of the state space equations.
...... (1)
Write the output equation.
y - C . * , ...... (2)
Write the state description matrices in observer canonical form.
-Oi
A .=
1
0
0
1
0
0
0
0
0
0
0
0
0
0
c .= 1 0
(3)
1
0
0
(4)
0]
..... (5)
D .=
Refer equation 7.21 in the textbook and write the equation for ^ matrix.
A = T - 'A T ...... (6)
Where,
X is transformation matrix.
Step 2 of 4
Write the general equation for transformation matrix.
T = [t,
t,]
(7)
Modify equation (6).
;T-i
^ = AT
TA = A T ..... <®)
Consider the new state description matrix is in observer canonical fonn.
Substitute equations (3) and (7) in equation (8).
[t,t, t,]
-o,
I O'
-a ,
0 1 =A[t, t, t,]
0 0
-a,
[“ *1^
*1
^*2
^ * 2] ......
Write the matrix equation from second and third columns from equation (9).
t j - A t , ..... (10)
...... (11)
Substitute equation (10) in equation (11).
t,=AAt,
...... (12)
Substitute equations (4) and (7) in equation 7.22 in the textbook.
[1 0 0]=C[t,
t, t,]
[1 0 0] = [t,c tjC tjC ] ...... (13)
ste p 3 of 4
Write the matrix equation from second and third columns from equation (13).
t j C ^ O ...... (14)
t,C = 0 ...... (15)
t,C = I ...... (16)
Substitute equation (12) in equation (16).
tjCA’ . l ...... (17)
Substitute equation (10) in equation (15).
t j C A ^ O ...... (18)
Substitutes equations (11), and (12) in RHS of the equation (13).
[1
0] = [t,C A *
0
[1 0 0] = t, CA’
t,CA
CA
t,C ]
C]
c
CA
CA’
t,o
t, = [0 0 i f o - '
.(19)
ste p 4 of 4
c
CA
CA’
Hence, from the results the transformatioh matrix values are |t , = A t , |. |t , = A t , |a
t ,= [ 0
0 1]’ 0~'
Problem 7.1 OPP
Use the transformation matrix in Eq. to explicitly multiply out the equations at the end of Example.
Eq.
T = [ - t
? ] ’
']•
Example
Tran^ormation o f Themud Systemfrom Control to Modal Form
EXAM PLE 7.9
Rod thee
modal fonn of Eq. (7.14).
s ia Eq. (7.1 ^ into the
StdvUon. Aoconfing to Eqi. (7 J 4 ) and (7.35), we need fint to find die
eigenveclois and eigenvalues o f the Af maliu. Wb take the eigenvectofi to be
[S]
aaUL'Wei
es oi me Af maanL
>ve nice me eigenvecton lo He
i5I
[s]
aJ6 a)
-7 f|i -1 2 ^1 —pliu
036b)
(7J6c)
'll
Suhstiniting Eq. (7.36c) into Eq. (7.36b) results in
- 7 p ll i- l 2 U i = p * * 2 l.
^ t i i + 7 ^ 1 + 12f2i = 0,
aJ7 a)
(7.37b)
(737c)
p ^ + lp + l l- O ,
- 3 ,- 4 .
(7.37d)
Wb have found (again!) that the eigenvalues (poles) a e - 3 and- 4 ;
fartfaennore. Eq. (7.36c) tdls os that the two eigeuvecton are
r-^ i
L ft.
]
where til and tn a e arbitrary nonzero scale focton. We arant to select the
two scale focton such tha both elements o f Bg| in Eq. (7.14a) are mhy. Hie
equation for Bn in terms o f Br is TB* » Br. and its solution b I2i = ^1
and l22 — !• Therefore, the traMforinatioo matrix and its iovene^ are
"■[-t 1]- ""-[I *]■
a.38)
Elementary matrix muhqiUcation abows that, using T as defined fay
Eq. (738), the matrices o f Eqs. (7.12) and (7.14) are related as follows:
Am^T~'A€T,
B in = T “ *Bc.
C , = CfT.
D»=Dc.
*1b fied the iavMse e f a 2 X 2 ■
M l '22.'* d w ^ die s ^ o f * e **i r H d d ie ‘7 P I
[slisE4.(7JB)l.
These computations can be carried out by using the following Matiab
T -[4 -3 ; -11];
Am-inv(T)*Ac*T;
Bm«1nv(T)*Bc
Cm-CcT;
Dm-Dc;
Step-by-step solution
Step 1 of 5
According to equation (7.30)
Transformation Matrix
-= [; a
And die equations are
A . = r% T
B, = T^B
C . = C.T
D , = D,
Step 2 of 5
and
A = 0'
1 3
•7 -12
1 4
1
0
][-. -.1
■28+12 21-121
4+10 - 3 - l o J
1 4
1 3‘ •16 9 ]
4 - 3J
1 4
-16+12 9. - 9-1
-16+16 9 - I 2 J
1 3'
o'
-4
4 =
0
Ans.
-3
Step 3 of 5
aT%
1 3
1 4
'l+ O '
a
1+0
B .=
A ns.
Step 4 of 5
c .- c ^ r
■" 4 - . V]
= [4
= [2
IC, = [ 2
- 3 +2]
- 2
- 1]
-1]| A?is.
Step 5 of 5
= 0| Ans
a» )
Problem 7.11 PP
Find the state transformation that takes the observer canonical form of Eq. to the modal
canonical form.
*• = [ - 1 2
J]-
»- = [ 2 ] -
Co = [
0 ],
Do = 0.
I
Step-by-step solution
ste p 1 of 4
Step 1 of 4
As
4 , = r ‘ 4 ,T
CD
(2)
(3)
(4)
■Where
and
are model matrices
and
are in control Canonical
form and
Step 2 of 4
-O j
-
1
0
0
0 ■■■
4 .=
1
0
(5)
Step 3 of 4 ^
If 4) >
Qi >
^ are in observer Canonical form,
■Where
4
1 0
... o'
... 0
0
... 1
... 0
-1^
1 0 0
-a,
0
=
1 0
Co = [1 0 0 . .
0]
A]
A=
(6)
’
kJ
A=o
Step 4 of 4
Comparing Equations (5) & (6)
We can say that,
C c= s;
£)£.=A .
Substituting the above equations into equations
B , = T - 'c ;
c.
= b;
a .= A
(7)
Problem 7.12PP
(a) Find the transformation T that will keep the description of the tape-drive system of Example
7.10 in modal canonical form but will convert each element of the input matrix Bm to unity.
(b) Use Matlab to verify that your transformation does the job.
Example 7.10
Using Matlab to Fhui Poles and Zeros ofTape-Drrve System
EXAMPLE 7.10
Find the dgeDvahies of die qrttem mMiix described below for the tape-drive
conind (see Fig. 3.S0). Abo, compute the tramformatioa o f the equations
of the t^ie drive in their giwea fm n to modal canonical fonn. The system
sare
■0
2
0
-ai -(U5 ai
A»
'O'
®1
0
. B= 0
0
0
2
.A
’ 0
2
0
0
0 *
'o'
-ai -035 0.1 ai a?5
0
0
0 0
0
2
. B« 0
0.4
a4 -a4 - 1.4 0
0
1
0
-04)3 0
0
-1
0
ai
0.75
0
a.4(»
C2 * [OD
OlO 1.0
Oil Oil]
Servomotor
C3 s [OJ
0.0 OJ
Oil Oil]
Posiboii M leaiVwfite head as oatpal,
C r - [ -0 .2
ontpm,
- O J 0.2 0.2 0.0] Ibnsion output,
D=Qja.
The sM e vector b defined as
xi (tape positiao at capstan)
« i (speed o f the drive wheel)
JK3 (posttioii of the tape at the head)
» 2 (ouQNit speed)
The matrix C3 corresponds to malrii^ X3 (die position o f the tape over the
lead/write head) the output, and the matrix Ct corresponds to making tension
the output
Sclution. To compute the eigenvalues using MMhb, we write
P-eig(A),
which results in
-0.6371+0.6669i
-0.6371 -0.6669i
0.0000
-0.5075
-0.9683
Ps
Notice that the system has all poles in the left half-plane (LHP) except for
one pole at the origin. This means that a stq> iiqiut will result in a ramp
output, so we conclude that the system has Type 1 behavior.
To transform to modal fmin, we use the Matlab ftmctitm canon:
sysG-ss(A,B,G,D);
[sysGm,Tl]-canon(sysG/modal');
[Am,Bm,Cm,Dm]<sdata(sysGm)
The result o f this ctMnputatitMi is
-0.6371
-0.6669
0.0000
0.0000
0.0000
Am = A* =
0.6669
-0.6371
0.0000
0.0000
0.0000
0.0000 0.0000
0.0000
0.0000 0.0000
0.0000
0.0000 0.0000
0.0000
0.0000 -0.5075
0.0000
0.0000 0.0000 -0.9683
Notice that the cmnplex p d es ^^>ear in the 2 x 2 block in the uf^ier-left
comer o f Am, and the real poles fall on the main diagonal o f this matrix. The
rest o f the conqNitatkxis from canon are
0.4785
-0.6274
-1.0150
-3.5980
4.9133
Bm= B«:
Cm= C« = [ 1.2569 -1.0817 -2.8284 1.8233 0.4903 ],
Dm= />* = 0,
-0.3264
-0.7291
-1.3533
-2.3627
0.2077
-0.3439
0.1847
-0.1844
0.3353
-0.0017
TI = T-* =
0.7741
0.3439
-0.1847
0.0969
-0.1692 -0.3383
-0.3353 -1.0161
0.0017
0.0561
0.4785
-0.6247
-1.0150
-3.5980
4.9133
It happens that canon was written to compute the inverse o f the transforma­
tion we are w<xidng whh (as you can see from TI in the ixevious equaticm),
so we need to invert our Matlab results. The inverse is computed from
T = ln v a i)
and results in
03805
-a4112
11334
03317
0.0130
Eigenvalues
0.8697 -18284
1.3406 a4714
-0.1502 0.0000 -0.3402 -0.2282
-3.0330 -18284
2.3060 03093
1.6776 0.0000 -0.58SI -0.2466
-a0114 -0.0000 0.0207 01160
The eigenvecton computed with [V,P)i^io(F) are
-0.11684-0.19251
-0.0270-0.1003/
0.8797
-0.28024-0.29331
aoo40 4-o.ooi0i
-a il6 8 -a i9 2 S i -0.7071
-a02704-O.I003i -0 .0000
0.8797 -0.7071
-01802 - 0.2933i -aOOOO
a o o 4 0 -o x» io i aoooo
0.4871 03887
-0.1236 -0.2850
0.8379 0.6360
-01126 -03079
0.0075
01697
Notoce tbit the first two cohunm o f the real tmsfonnation T are com­
posed of the real mid the imaginay ports o f the first mgeiivector in the first
cohium of V. It n dib step that causes the complex roots to appear in the 2 X2
Mock in die upper left the A* matrix. The vectors in V are norraalired
to unit ler^th, whidi results in Donnormalized values in Bm and € « . If we
found it desirable to do sex we could readily find further transformatioiis to
make each efement o f Bm equal 1 or to mterchange die order in adiich the
poles appear:
Step-by-step solution
Step 1 of 4
(a)
For transformation
B,
=
1^ 'a
matrix Ti, S.t,
= [1 1 1 1 i f
Ti is full rank matrix.
So,
a = T,B,
And,
%. = T N = [«ili
....
0 = T ^ B , = T N B , = T,
H = T 'a
And,
» = [»i
«i
....
Hence,
= T N = Tdtagin^n^ .....
Step 2 of 4
(b)
Using Matlab,
1.4708 -0.17221 -1.4708 -1.5432 -3.323
0.7377
2.8933 -0.7377 -2.9037 -0.1282
-0.7376 -5.4133 0.6767 -1.3533 -4.059
0.9227 -6.5022 -0.9227 -2.7962 -9.9016
-0.0014 0.1663
0.0014 0.0449 3.9341
T
g
N = diag{n)
Step 3 of 4
: T*N
■-0.3883 0.0247 2.8708 -4.8234 2.3162'
-0.0897 -0.0129 0.0000 1.2239 -1.1214
2.9239 0.0000 2.8708 -8.2970 2.5023
-0.9314 0.0376 0.0000 2.1053 -1.2115
0.0133 0.0001 -0.0000 -0.0745 1.0612 _
B. =
1 = '^
1 1 1 if
ste p 4 of 4
■70.6371 0.0257 -0.00 0.000
-0.00 ■
-17.2941 -0.6371 0.00 -0.000 -0.00
-0.00
0.000 -0.00 0.000
00.00
-0.00
0.000
0.00 -0.5075 -0.00
-0.00
0.000
0.00
0.000 -0.9683
Problem 7.13PP
(a) Find the state transformation that will keep the description of the tapedrive system of Example
7.10 in modal canonical form but will cause the poles to be displayed in Am in order of increasing
magnitude.
(b) Use Matlab to verify your result in part (a), and give the complete new set of state matrices e
. A ,. B, .C, and .D.
Example 7.10
EXAMPLE 7.10
Using M atlab to Find Poles and Zeros ofTape-Drive System
Find the dgeDvahies of die qrttem mMiix described below for the tape-drive
conind (see Fig. 3.S0). Abo, compute the tramformatioa o f the equations
of the t^ie drive in their giwea fm n to modal canonical fionn. The system
contnd (see Hg. 3.S0). Abo, compute the tiansfonnatioa o f die equations
of the t^ie drive in their given fnm to modal canmical fbnn. The system
' 0
-a i
0
A»
0.4
0
2
-0 3 5
0
a4
-0 4 B
0
0.1
0
-a 4
0
0
ai
2
- 1 .4
0
0 *
a75
0
0
-1
'0'
0
0
0
.1,
a.40)
C2 * [OD OlO 1.0 Oil Oil]
Servomotor
C3 s [(L5 0.0 OJ
Posidoii MleaiVwfite head as ootpal,
C r - [ - 0 .2
Oil Oil]
ontpm,
- O J 0.2 0.2 0.0] Ibnsion output,
D = 0 .0 .
The sMe vector b defined as
xi (tape positiao at capstan)
« i (speed o f the drive wheel)
JK3 (posttioii of the tape at the head)
(ouqwt speed)
The matrix C3 corresponds to malrii^ X3 (die position o f the tape over the
lead/write head) the output, and the matrix CTConespoods to making tension
the output.
Sclution. To compute the eigenvalues using Mathb, we write
P««ig(A),
which results in
- 0 .6 3 7 1 + 0 .6 6 6 9 i
- 0 .6 3 7 1 - 0 .6 6 6 9 i
0.0000
Ps
-0 .5 0 7 5
-0 .9 6 8 3
Notice that the system has all poles in the left half-plane (LHP) e x c ^ for
one pole at the origin. This means that a stq> iiqNit will result in a ramp
output, so we ctmclude that the system has Type 1 behavior.
To transform to modal fm n , we use the Matlab fimctitm canon:
sysG-ss(A,B,G,D);
[sysGm,Tl]-canon(sysG/modal');
[Am,Bm,Cm,Dm]<sdata(sysGm)
The result o f this ctMnputatitm is
' -0.6371
-0 .6 6 6 9
0.0000
0.0000
0.0000
Am = A « =
0.6669
-0 .6 3 7 1
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
-0 .5 0 7 5
0.0000
0.0000 '
0.0000
0.0000
0.0000
-0 .9 6 8 3
Notice that the cmnplex p d es ^^>ear in the 2 x 2 block in the uf^ier-!
comer <^A«, and the real poles ftdl on the main diagcmal o f this matrix. The
rest o f the conqwtatkMis from canon are
0.4785
- 0 .6 2 7 4
- 1 .0 1 5 0
- 3 .5 9 8 0
4.9133
Bm= B«:
= [ 1.2569
- 1.0817
- 2.8284
1.8233
0.4903 ],
= 0i
TI =
-0 .3 2 6 4
-0 .7 2 9 1
-1 .3 5 3 3
-2 .3 6 2 7
0.2077
■ - 0 .3 4 3 9
0.1847
-0 .1 8 4 4
0.3353
- 0 .0 0 1 7
_
0.7741
0.0969
- 0 .3 3 8 3
-1 .0 1 6 1
0.0561
0.3439
-0 .1 8 4 7
-0 .1 6 9 2
-0 .3 3 5 3
0.0017
0.4785
-0 .6 2 4 7
-1 .0 1 5 0
-3 .5 9 8 0
4.9133
It haf^ieiis that canon was writtra to ctmqMite the inverse o f the transfcxmatimi we are w<xidng with (as you can see from TI in the inevious equaticm),
so we need to invert our Matlab results. The inverse is computed from
T = lnv(TI)
and results in
03805
08697
- 0 4 1 1 2 -0.1502
11334 --3.0330
03317
1.6776
0.0130 - 0 0 1 1 4
T= T =
Eigenvalues
-1 8 2 8 4
0.0000
-1 8 2 8 4
0.0000
- 0.0000
1.3406
-0.3402
2.3060
-0.58SI
0.0207
a4714 ■
-0.2282
0J093
-0.2466
01160
The eigenvecton compuled with [V,P)i4 ig(F) me
' -0.1168 + 0.1925/
-0 .0 2 7 0 -0 .1 0 0 3 /
V=
0.8797
-0.2802 + 0.2933/
00040 + 0.00101
V =
-0 1 1 6 8 -0 1 9 2 5 /
-0 0 2 7 0 + 0 1 0 0 3 /
08797
-0.2802 - 0 2 9 3 3 /
00040 -OOOlOi
-0.7071
- 0.0000
-0.7071
0.4871
-0.1236
aS379
-O O O O O
-0.2126
ooooo
0.0075
0J887
-0.2850
0.6360
-0 J 0 7 9
01697
Notice dmt the first two cohimm o f the real tmsfonnadon T are com­
posed of the real mid the imaginary ports o f the first mgenvector in the first
cohirrm of V. It b dib step that causes the complex roots to appear in the 2 X2
Mock in die upper left the A* matrix. The vectcrs in V are normalired
to unit ler^th, whidi results in normoimalized values in Bm and € « . If we
found it desirable to do sex we could readily find further transformations to
make each efemeM o f Bm equal 1 or to interchange die order in adiich the
poles appear:
Step-by-step solution
Step 1 of 4
(a)
Refer Example 7.10 in the textbook and find the state transformation of the matrix.
To display the poles of Am in the order of increasing magnitude, rearrange the eigen vector T in
the order of increasing magnitude.
Consider the eigen vector of T.
0.3805
0.8697
- 2 .8 2 8 4
1.3406
0 .4 7 1 4 '
-0 .4 1 1 2
- 0 .1 5 0 2
0.0000
- 0 .3 4 0 2
-0 .2 2 8 2
2.1334
- 3 .0 3 3 0
- 2 .8 2 8 4
2.3060
0.5093
0.3317
1.6776
0.0000
-0.5851
-0 .2 4 6 6
0.0130
- 0 .0 1 1 4
- 0 .0 0 0 0
0.0207
0.2160
Let the eigen values of the matrix T is given below.
-0 .6 371+ 0.6669/^
- 0 .6 3 7 1 -0 .6 6 6 9 /
P * -0.0000+0.0000/
-0 .5 0 7 5 + 0.0000/
-0 .9 6 8 3 + 0.0000/
The row values of P gives the corresponding column values of the matrix T.
Arrange the eigen values in increasing order.
>5 ( - 0 .9 6 8 3 + O.OOOOi) > P, (-0 .6 3 7 1 + 0 .6 6 6 9 /)
> Pj (-0 .6 3 7 1 - 0 .6 6 6 9 /) > /J (-0 .5 0 7 5 + 0.000 0 /) >
P ,(-0.0000 + 0.0000/)
Hence, the matrix must be arranged as shown below.
r2= ['5
'2
U
Where,
Tj is the rearranged eigen vector T in the order of increasing magnitude.
/i,/j,/j,(j,a n d /ja re the respective column of matrix T.
Therefore the rearranged eigen vector T in the order of increasing magnitude is given below.
■0.4714
0.3805
0.8697
1.3406
-0 .2 2 8 2
-0 .4 1 1 2
-0 .1 5 0 2
-0 .3 4 0 2
-2 .8 2 8 4 '
0.0 0 0 0
0.5093
2.1334
-3 .0 3 3 0
2.3060
-2 .8 2 8 4
-0 .2 4 6 6
0.3317
1.6776
-0.5851
0.0 0 0 0
0.2160
0.0130
-0 .0 1 1 4
0.0207
- 0.0 0 0 0
Step 2 of 4
(b)
Write the matlab program to verify the result and to obtain the new set of state matrices,
cic
A=[0 2 0 0 0;-0.1 -0.35 0.1 0.1 0.75;0 0 0 2 0;0.4 0.4 -0.4 -1.4 0;0 -0.03 0 0 -1];
B=[0:0;0:0;1]:
c2=[0 0 1 0 0]:
c3=[0.5 0 0.5 0 0];
cT=[-0.2 -0.2 0.2 0.2 O.Oj;
D=0;
[P]=eig(A);
sysG=ss{A,B,c3,D):
[sysGm,TI]=canon(sysG,'modal'):
[Am,Bm,Cm,Dm]=ssdata{sysGm):
TI=[-0.3439 -0.3264 0.3439 0.7741 0.4785; 0.1847 -0.7291 -0.1847 0.0969 -0.6247; -0.1844
-1.3533 -0.1692 -0.3383 -1.0150;0.3353 -2.3627 -0.3353 -1.0161 -3.5980; -0.0017 0.2077 0.0017
0.0561 4.9133];
T=inv(TI):
% [V,P]=eig{P)
[f,indices]=sort(abs(P));
T2=T(:,indices):
n=T2\B;
T3=T2*diag{n);
Am2=T3\A*T3
Bm2=T3\B
Cm2=c3*T3
dm2=0
Step 3 of 4
The output obtained on executing the code is given below;
The Eigen values of matrix are displayed.
P=
-0.6371 + 0.66691
-0.6371 - 0.66691
0.0000 + O.OOOOi
-0.5075 + O.OOOOi
-0.9683 + O.OOOOi
Step 4 of 4
The new set of state matrices are shown below.
Am2 =
0.0000 0.0002 0.0000 -0.0001 -0.0002
-0.0000 -0.5075 -0.0000 -0.0000 -0.0000
-0.0000 0.0002 -0.6371 -0.8707 -0.0040
0.0000 0.0001 0.5108-0.6371 0.0016
-0.0000 0.0000 -0.0000 -0.0000 -0.9683
Bm2 =
1.0000
1.0000
1.0000
1.0000
1.0000
Cm2 =
2.8705 -6.5593 0.6014 0.6755 2.4120
dm2 =
0
Hence the results are verified new set of state matrices are obtained using matlab.
Problem 7.14PP
Find the characteristic equation for the modal-form matrix Am of Eq. (1a) using Eq. (2).
Eq. (la)
Eq. (2).
det(/r(l —A) = 0.
Step-by-step solution
ste p 1 of 1
ste p 1 of 1
■
i r j
Characteristic Equation is:
det ^ 7 - .4 ] = 0
s
0
h :-J
0
rs+ 4
[ 0
0 ]
S + 3J
Characteristic Equation is:
(s+ 4) (s+3) = 0
+ 7 5 + 12 ^ 0]
Problem 7.15PP
Given the system
* = [_ 2
with zero initial conditions, find the steady-state value of x for a step input u.
Step-by-step solution
ste p 1 of 1
Write the general form of state space equation.
Write the general form of state space equation.
± = A x + B u ...... (1)
Consider the value of t a t steady state condition with zero initial condition.
i = 0 ..... (2)
Consider the value of wat unit step input condition.
a = I ...... (3)
Consider the value of x at steady state condition.
x=
...... (4)
Consider the value of matrix
a
•
b
•
"=L-2 -.J
Consider the value of matrix
...... (6)
Substitute equations (2), (3) and (4) in equation (1).
0 = A x ,+ B
-B = A x,
-B
A
-B A -‘ - X . ...... (7)
Substitute equations (5) and (6) in equation (7).
- m
- j'
Hence, the steady state value of x is
Problem 7.16PP
Consider the system shown in Fig.
(a) Find the transfer function from U to Y.
(b) Write state equations for the system using the state-variables indicated.
Figure A block diagram
Step-by-step solution
ste p 1 of 6
Step 2 of 6
=z
U [s)
where
Afg = Gain of 1 ^ Forward path
A s 1- (Sum of Individual Loop Gains) + (Sum of Products of two Non Touching Loop
Gains) - (Sum of Products of Three Non Touching Loop Gains) +.........and A;^=
Nodiingbut 'A' not touching the
forward path.
1
i/ « _L
2s (s +4) ’
^
s+ 4
-1
A*
2s*(s+4) ’
^
s(s+ 4 )
1
1
2s^(s+4)
s(s+ 4 )
A, = 1, A, = 1
Step 3 of 6
F (s)
2s(s+4)
s-l-4
1
2 s^ (s+ 4 ) "*"3 (5 + 4 )
■■■ U { ^
y (s) _
U [s)
2s+l
2s(s+4)
2s^(s+4)
2s^(s+4)+l+2s^
s* (2s+l)
s “(2s+l)
2s'+ 8 s’ + 2s=+l
. m
ste p 4 of 6
b) From the Figure,
s+ 4
sJf* (s) + 4 ^ 4 (s) = U - JSfj (s)
= -
4*4 - X 3 + u
(1)
I3 =
Et3(s) = I j( s ) + I ,( s )
Xj = X2 +X4
(2)
Step 5 of 6
sX3(s) = jr ,( s )
(3)
^ = *1
‘
2s
= O.Sjc.
^ =
(4)
+
(5)
Step 6 of 6
A
A
'0
0
1
0
0
0
A
0
1
0
1
.A.
0
0
- 1
y = [l 1 0 1]
0
0.5
xa
- 4
X2
Problem 7.17PP
Using the indicated state-variables, write the state equations for each of the systems shown in
Fig. Find the transfer function for each system using both block-diagram manipulation and matrix
algebra [as in Eq.].
Figure Block diagrams
I I ,l.
II
4 I^
I 1 I'l
(b)
Eq.
G(*) = ^
= C ( i I - A r ' B + D.
Step-by-step solution
ste p 1 of 14
Step 2 of 14
BLOCK DIAGRAM REDUCTTON:
The above block diagram can be redrawn as:
Step 3 of 14
Step 4 of 14
2(s+2.5)
(J+3X-V+5)
s+ 4
Step 5 of 14
Step 6 of 14
r (s)
■y (s )
ff^-l-lCte + 20
(s+ 3 )(s+ 4 )(s+ 5 )
Matrix Algebra:
From the given block diagram:
U
s+A
EXi[s)+ AX^{s)^U [s)
X, = - 4 x , +u
(1)
Step 7 of 14
^
s+3 ’
aJTj ( s ) + 3JSTj (s) = sJTj ( s ) + 2 JTj (s)
* - 3 ;i^+ X 3+2 x3
= -3x2
A-u + 2j^
Xj = a^+X j-S xj
(3)
(4)
SX3+X,
-5 1 1
^1'
Xj = 0 -3 -2
0 0 -4
.^3.
0
X2 -f 1 u
1
(5)
y = [1 0 1]
Step 8 of 14
The above matrix equations are in the form of
X = iTx+Gb
y= H x+ Ju
where 7 = 0
-5
r
0
0
-3 - 2
0 -A
where F =
Y[s)
U(s)
‘0*
0 =
J=0
« = [1 0 1],
1
s“+10s+20
(s+ 3 )(s+ 4 )(s+ 5 )
Step 9 of 14
1
1
i +10
^+ 2
,p r
4 1 2
ste p 10 of 14
Step 11 of 14
Step 12 of 14
r(s ) _
s(s+ l)(s’ +3s+4)
U{s) ” s(s+3)(s+10)(s^ + 2s+4)
(s+ ll)(s’ + 3s+4)
(s+3)(s+ 10)[s^+2s+4j
^
(s)
'■ '
s+10
SX4+IOX4 =u
x^ = - 10 x4 +tt
0)
jr5 = (A ,+ U - j^ ) —
S +2
X + 2 x3 - x^ + U - X 2
6 3
X3 =
2X3+ X4 -
(2)
4
a^= -X 4
X j=-3xj +X2
y = x i-x ^
(4)
(5)
S tep 1 3 o f1 4
■-3
0
0 0
^1
xa
3-= [1 0 - 1
1 0
0 4
1 - 2
0 0
O'
0
1
-10
^1
Xa
0]
and
£W
H (s i-F y ^ a + j
where
F *
-3
1
0
0 ■
o'
0
0
4
0
0
0
-1
-2
1
1
0
0
0
-10
1
0 =
Step 14 of 14
«=
r(s)
[1
0 - 1
_
0
]
j
=0
(s+ n )(s^ + 3 s+ 4 )
U{s) ~ ( s + 3 )(s + 1 0 )(s ’ + 2 s+ 4 )
>'2
1
5+3
Problem 7.18PP
For each of the listed transfer functions, write the state equations in both control and observer
canonical form. In each case, draw a block diagram and give the appropriate expressions for A,
B. and C.
(a) C(s) =
(control of an inverted pendulum by a force on the cart).
(b) (b)
=
Step-by-step solution
diep^by-^iep* S olliiion
Step 1 of 10
(a)
Consider the numerator part of the gain.
b {s)= b ^^+ b2S^^+ ...+ b , ...... (1)
Consider the denominator part of the gain.
a ( j ) = y + < i ^ " " '+ a j j ^ ^ + ...+ o ,...... (2)
Write the general canonical form of the equation.
■(3)
Substitute equations (1) and (2) in equation (3).
G (*) =
s"
+ 02S"~*
(4)
+
Write the state description matrices of controller canonical form.
-<h .......... -a .
1
0
..........
0
1
0
0
1
•••
0
...........
1
0
0
0
(5)
0
(6)
B,=
C .=[A
.......... *.]
b,
(7)
£), =Sqwrate valuefromnumeratoraiiddeiiominatorelse 0
(8)
ste p 2 of 10
Consider the value of gain.
C (* ) = 4
^
(9)
s -s
Write the general form of state space equation.
jc = A x + B u ...... (10)
Write the state equation in control canonical form from equations (4) to (10).
0
’■*1
1 0 0
_
0
•*3
I
*1
0 0 0
0
+
1 0 0
0 0
I
1 0
1 0
1 0
0
0
’h
0
*4.
0
0
0
0
0
0
0
(11)
0
1 0
Write the general form of output equation.
y = C x+ D ...... (12)
- 2 ] x + 0 ...... (13)
;- = [0 1 0
d
0 0
0 0 0
Hence, the state equations of controller canonical fonn are X s
0
0 0
0 0
|;- = [0 1 0
1
0
x+
U
0
0
0
- 2 ] I |.
Step 3 of 10 ^
Draw the block diagram for controller canonical form from equations (11) and (13).
Figure 1: Block digram for controller canonical form
Hence, the block diagram for controller canonical form is shown in figure 1.
Step 4 of 10 ^
Write the observer canonical form of the state space equations.
=
...... (14)
Write the output equation.
...... (15)
Write the state description matrices in observer canonical form,
-o ,
1
0
••• 0
-a,
0
1
0
=
A .=
0
0 0 1 0
■ 0
-a ,
0
(16)
0
1
0 .......... 0
.(17)
B .=
C ,= [ l
0
0
••• 0 ] ...... (18)
J J .= 0 ...... (19)
Write the state equatiohs ih observer canonical form from equations (4), (9). (14). (15). (16). (17).
(18) and (19).
0
1 0 O'
_10
o'
•'2 + 1
0
0 0 0 1 *1
0 0 0 0, .■*4.
-2^
.*4.
0
1 0
1 0
1 0
0
o'
1 0
1
0
0 0
1
0
0
0 0
0
-2
;> = [1 0
0
0 0]
^- = [1 0
0
0 O j*.
(20)
.■*4.
(21)
0
Hence, the state equations of observer canonical form are
|y = [l
0
0
o'
0
0
0 0
0
1
0 0
0
0
1
x ,+
0
-2
0 0 O " ^
Step 5 of 10 ^
Draw the block diagram for observer canonical form from equations (20) and (21).
Figure 2: Block diagram for observer canonical form
Hence, the block diagram for observer canonical form is shown in figure 2.
Step 6 of 10
(b)
Consider the gain value.
3s+4
s’ *2s+ 2
(22)
Write the state equation in control canonical form from equations (4) to (9) and equation (22).
?]'•[:>...
Write the general form of output equation.
y = C x* D
.y=[3 4 ] * - f 0 ...... (24)
Step 7 of 10
Hence, the state equations of controller canonical fonn are: X S
-2
-2
1
0
X+
1
0
u
b =[3 4]x|.
Draw the block diagram for controller canonical form from equations (23) and (24).
Figure 3; Block diagram for controller canonical form
Hence, the block diagram for controller canonical form is shown in figure 3.
Step 8 of 10
Write the state equations in observer canonical form from equations (4), (14), (15), (16), (17),
(18), (19) and (22).
(25)
"•=[-2
J-=[1
0]
.r=[i o]j(,
(26 )
Hence, the state equations of observer canonical form are:
jj-li
-2
1
-2
0
OKI
Step 9 of 10 ^
Draw the block diagram for observer canonical form from equations (25) and (26).
Figure 4; Block diagram for observer canonical form
ste p 10 of 10 ^
Hence, the block diagram for observer canonical form is shown in figure 4.
Problem 7.19PP
Consider the transfer function
yw
»+6
+ +
(a) By rewriting Eq. in the form
m s}
s + 3 \s + 2)
find a series realization of G(s) as a cascade of two first-order systems.
(b) Using a partial-fraction expansion of GfsJ, find a parallel realization of G(s).
(o) Osrrlg~a p^n'iar-iractiuh expartsiijifijnj(:>;, iinu a paraiiei reaiizaiion ui u{sj.
(c) Realize G(s) in control canonical form.
Step-by-step solution
ste p 1 of 6
(a)
Series realization of <?(#):
Consider the value of gain G(.r)(1)
Write the series realization of G{«)from equation (1).
(2)
Draw the diagram of series realization from equation (2).
yi = u2.
j+6
M = M io -
s+ 2
1
5+ 3
y= y2
Figurel
Step 2 of 6
Write the state description matrix from figure 1.
i, = - 2 j i + i/,
i,
(3)
= - 3 x , + U j ...... (4)
y , = - x , + u , ...... (5)
...... (6)
« = « , ...... (7)
......(8)
y = y ,
» , = y , ...... (9)
Substitute equation (9) in equation (4).
x ^ — 3 x ^ + y t ......(10)
Substitute equation (5) in equation (10).
...... (11)
*2 — X ,- 3 X 2 + U ,
Substitute equation (7) in equation (11).
* - ;t,- 3 jC 2 + « ...... (12)
Substitute equation (7) in equation (3).
i l = - 2 j c , + « ...... (13)
Write the state description matrix from equations (12) and (13).
[a-Kaa*[ii« ] ■ Substitute equation (6) in equation (8).
y=
...... (15)
Write the output matrix from equation (15).
,i[^ ]
y=[0 l];c ...... (16)
Hence, the state description matrix in series realization of C{4) at cascade connection are
-2
0
-1
-3
and |y = [0
l]jc].
Step 3 of 6
(b)
Parallel realization of
Apply partial fraction in equation (2).
s+6
(j+ 3)(j+ 2 )
A
B
(5 + 3) (j+ 2 )
.(17)
Find A from equation (17).
„
5+6
(s + 3 ) ( ^ + 2 )
-3 + 6
“ (-3 + 2 )
<"’L
A - - 9 ......
Find B from equation (17).
s+6
.(s+2)
(s+3){s+2y
B = 4 ...... (19)
Substitute equations (18) and (19) in equation (17).
s+6
-3
4
r+ 7
( 5 + 3) ( j + 2 )
( j+ 3 )(j+ 2 )
(20)
Draw the block diagram from equation (20).
Figure 2
Step 4 of 6
Write the state equation in control canonical form from figure 2.
K ] - [ : :E 1*[:1"[• +H 1"
=I -
- |J f + l . |w ...... (21)
Write the output equation in control canonical form from figure 2.
3- = [-3
4 ]*
(22)
Hence, the state description matrix in parallel realization of C{4) are
-3
0
0
-2
and |y = [-3
4 ]x |
Step 5 of 6
(c)
Consider the numerator part of the gain.
b{s) = b^^+b2S^^+ ... +b, ...... (23)
Consider the denominator part of the gain.
...... (24)
Write the general canonical form of the equation.
■(25)
Substitute equations (23) and (24) in equation (25).
+b2s"~^+’” +b^
C (s) =
(26)
Write the state description matrices in control canonical form.
-<h ........... -a,
0 ....... 0
1 0
0
0 1 •• 0
........ 1 0
1
c
(27)
I
0
0
...... (28)
0
C .= [A
^
».]
(29)
D^sSeperatevaluefromnumeratoranddeiiominatorelse 0 ...... (30)
Consider the value of G (4).
C (5)
5+ 6
(31)
5*+55 + 6
Write the state description matrices in control canonical form from equations (23) to (31).
(32)
C .=[6 1] ...... (34)
O ,= [0 ]
(35)
Write the general form of state space equation of canonical form.
x = A x + B ti ...... (36)
Write the state equation in control canonical form from equations (32), (33) and (36).
KHrim i...
Write the general form of output equation of canonical form.
y =Cx+D ...... (38)
Write the general form of output equation from equations (34), (35) and (38).
y^[6
IJI *1+0
(39)
Draw the block diagram of canonical form from equations (37) and (39).
F ig u re s
Step 6 of 6
Hence, the state description matrix in parallel realization of C{4) =
-5
-6 '
1
0
+
.* 1.
I
0
u and 3’ = [« ']
+0
„
Problem 7.20PP
Show that the impulse response of the system {A. B, C. D) is given by
h(t) = CeAB + D5(t),
where eAf is the matrix exponential defined by
-Al
/.
AV
\
” A‘l*
Step-by-step solution
ste p 1 of 1
Step 1 of 1 ^
Consider the state variable equations for the inhomogeneous case with a forcing function input
» (/)
i(0 = A (/)x (0 + B (r)« (/)
(1)
The solution of {1) is
x(») = ®(/,»„)i„+|®(/,r)B(r)ii(r)</r
(2)
Differentiate equation (2) with respect to dt
4 * (0 =
+ A(/) J®((,r)B(r)i<(r)</r+B(/)ii(/)
For the time-invariant case.
Where,
e " = (l+ A /+ ^ + ...]
The state solution is
x(0 = e*<'-'>x,+|e*<'-''Bii(r)</r
h
y (l)= C x (t)+ D u {t)
y { ,) = C e*<'''’x,+|e*<'‘'>Bii(r)</r + Du(t)
j-(/) -Ce'^"*>x,+Cje*<'-'*Bii(r)</r+D»(/)
h
Assume x ^ sQ
h
y { l) = ^ b ( l - T 'p u ( i ) d i
h
Where,
A(/) = Ce*'B+/M(/)
Thus, the impulse response of the system is |*(<) = Ce"B-H?^(t)|
Problem 7.21 PP
Consider the plant described by
* = [?
- ! 4 ] * + [ 2 ] “-
y = [ 1 3 ]x.
(a) Draw a block diagram for the plant with one integrator for each state-variable.
(b) Find the transfer function using matrix algebra.
(c)
Find the closed-loop characteristic equation if the feedback Is
rn iu [ii^'‘ur6^bu-iuup urieiiauieiisuu equciuuii ii m e leeuu<du^ is
{\)u=-[K1 K2]x;
(ii) u = -Ky.
Step-by-step solution
ste p 1 of 9
(a)
The plant is described by,
:;i] ;>
K
And,
3]»
y= [i
■[. .iH
The state equations are,
...... (1)
= 7 jc ,- 4 j ^ + 2 tt...... (2)
The output equation Is,
y=x^+3xj
(3)
ste p 2 of 9
From equations {1), (2), and (3), draw the block diagram for the plant.
The block diagram for the plant is shown in Figure 1.
F^ve 1
Therefore, the block diagram of the plant is shown in Figure 1.
Step 3 of 9
(b)
Write the expression for the transfer function.
Consider the state description,
x = Ax+Bu
The output description is,
>>= C x + D tf
Hence From equations (1), (2), and (3),
The matrices are.
A=
B=
C=
Step 4 of 9
Calculate the matrix ( 5 l - A )
:h ; : i
fj-o
0 - 1]
“ [o -7
i+ 4 j
=[*
-7
s -+'1
4]
Step 5 of 9
, y i _ ««ljointof(rf-A )
~
^
|iI-A |
1
p+4
1]
-7 i+ 4 |
________ 1
1
p+4
p+4
1]
“ j ’ + 4 s -7 [ 7
jJ
1]
Now, calculate the transfer function of the plant.
G(j ) = C( sI-A )"'B + D
«+4
I
7
f
■[> % i ’ + 4 j - 7
+0
i+ 4 + 2 '
7 + 2i
*+6l
“ T r ir y C
^1 2j + 7
j+ 6 + 3 (2 i+ 7 )
j *+
4j - 7
_ j+ 6 + 6 j+ 2 1
j* + 4 s-7
I s +21
i'+ 4 * -7
Therefore, the transfer function G { .) of the plant Is
75+27
5 ^ + 4 5 -7
Step 6 of 9
(C)
(i)
The state feedback is.
u=
- [ a:,
a:,] i
Calculate the matrix ^5! —A + B K ) •
5 0]
.0
_
■ 5+JC,
- 7 + 2JST,
-l+ ^ T j 1
5 + 4 + 2 /:J
Step 7 of 9
To calculate the characteristic equation equate determinant of matrix ( > U - A + B K ) to zero.
Calculate the characteristic equation.
|/ U - A + B K | = 0
I i+AT,
-1 + AT, I
I-7+2A ,
i+ 4 + 2 A ,|“
(j+ ir ,) ( j+ 4 + 2 /r ,) - ( - i+ A : .) ( - 7 + 2 ^ r ,) = o
s^+4s+2K^ + K ^+4K , +2K,Kj - 7 + 2AT, + 7JC, -2AT,A:, = 0
»’ + (/:, + 2iT, + 4 ) i +(6A:, +7AT, - 7 ) = 0
Therefore, the characteristic equation is | i‘ +(A:, + 2 A ,+ 4 )n -(6 A | + 7A r,-7) = o|.
Step 8 of 9
(ii)
The state feedback is.
u = - A>
=-js:[i 3]
.-[A
3iT]
Calculate the matrix ^5! —A + B K ) •
5 l- A + B K » 5
[;:h;-h;]'*
j
0]
p
s j ” [7
ro
1
]
r
-4J'^[2A:
s+ K
-1 + 3 A ]
-7 + 2 A
s+ 4 + 6 ^ :J
3 /:]
6A:J
Step 9 of 9
To calculate the characteristic equation equate determinant of matrix (A I-A + B K ) to zero.
Calculate the characteristic equation.
d e t[A I-A + B K ] = 0
|A I-A -l-B K |> 0
I s+K
-1+3AT
I-7+2AC *+ 4 + 6A :|"
(j+ X ) ( i+ 4 + 6 X ) - ( - l+ 3 3 : ) ( - 7 + 2 A r ) = 0
s^+4s + 6Ks+Ks+4K + 6K‘ - l + 2K + 2 \K -6 K ^ = 0
s ^ + {lK + 4 )s + {2 1 K -l) = 0
Therefore, the characteristic equation is |^’ + (7 A + 4 ) j + ( 2 7 A : - 7 ) = ^
Problem 7.22PP
For the system
design a state feedback controller that satisfies the following specifications:
(a) Closed-loop poles have a damping coefficient ^ = 0.707.
(b) Step-response peak time is under 3.14 sec.
Verify your design with Matlab.
Steo-bv-steo solution
Step-by-step solution
Step 1 of 5
The system is.
‘ ■[i
Here,
F=
[ i: ]
And,
G =|
Write the expression for peak time.
Substitute 3.14 s for
in the equation to calculate the damping frequency.
3.14s« —
*
3.14 s
«1 rad/s
Step 2 of 5
Write the expression for damping frequency.
^4 ■
and 0.707 for ^ in the equation to calculate the natural frequency.
Substitute 1 rad/s for
-0.707'
___ 1 _
“ 0.707
■ 1.414 lad/s
Write the general second order characteristic equation,
s* +
®0
Substitute0.707for ^ .a n d 1.414rad/s for
»'+2(0.707)(1.414)i+( 1.414)’ = 0
s '+ 2 t + 2 = 0
(1)
Step 3 of 5
The linear state feedback expression is,
tt = -K x
Here.
K is the state feedback gain
K = [JC,
AT,]
Calculate the value of the matrix, F - G K
0
F -G K =
1
-6 -S
_r ®
'
■ [-6
-5
JC
,K,i
0
_r 0
0]
- 5 - atJ
Determine the matrix. ±
XS F - G K ]
0
-6
K, - 5 - j r J *
The closed-loop characteristic equation is,
s' +
( 5 + a:,) s + (6 + a: ,) = o ......(2)
Compare equations (1) and (2).
S +K^ = 2
K jm -3
And.
6+K, = 2
K ,m -4
The state feedback gain is,
K =H
-3]
The state feedback controller, u Is F H
-3 1 4
Write the MATLAB code to plot the step response.
Step 4 of 5
» num=1;
» den=[1 2 2];
» sys=tf(num,den):
» t=0:0.0001:6;
» y=step{sys,t);
» plot(t.y);
» grid;
» xiabel t
» ylabel y(t)
Step 5 of 5
The following is the unit step response.
t
From the unit step response the peak time Is,
/, = 3 . I4 s
The peak time obtained is at 3.14 s. Hence, it is verified using MATLAB.
Problem 7.23PP
(a) Design a state feedback controller for the following system so that the closed-loop step
response has an overshoot of less than 25% and a 1% settling time under 0.115 sec:
*=[o
-10 ]* + [? ]■ ■ •
(b) Use the step command in Matlab to verify that your design meets the specifications. If it does
not, modify your feedback gains accordingly.
Step-by-step solution
Step-by-step solution
ste p 1 of 3
(a)
Given tliat
X =
i:.
For tile overshoot specification, we have
= e'^
Thus, we get ^ s 0.4.
For tile lYo setting time specifications, we use
4.6
'Hierefbre, we get =
1
= 0.01.
ft
46
Thus, we get ^ 's O .4 , and<JL = —
ft
Step 2 of 3
(b)
Write the MATLAB program to verify the d e s ^ .
P =
-1 0 ];
® * [ 0 ;1 ] ;
H = [1 ,0 ] ;
J = 0;
Z e ta = 0 . 4 ; % Tweak v a l u e s s l i g h t l y s o t h a t s p e c s a r e m e t.
t s = 0 .1 1 5 ;
wn = 4 . 6 / ( t s ^ z e t a ) ;
p = r o o t s ( [ l , 2 * ze ta* w n , w n * ^ ]);
k = p l a c e ( F ,G ,p ) ;
s y s C l= s s (F-G *k, G, H, J)
s te p ( s y s C L ) ;
Write the ouQiut obtained on execution of tiie program.
xl
x2
xl
0
-9 8 0 3
x2
1
-80
ul
xl
0
yl
xl
1
x2
0
ul
yl
0
C o n tin u o u s -tim e m odel.
Step 3 of 3
Sketch the step response plot obtained.
Hence, the d e s ^ is verified usit^ MATLAB.
7.24PP
Problem 7.24PP
Consider the system
t
(a) Design a state feedback controller for the system so that the closed-loop step response has
an overshoot of less than 5% and a 1% settling time under 4.6 sec.
(b) Use the step command in Matlab to verify that your design meets the specifications. If it does
not, modify your feedback gains accordingly.
Steo-bv-steo solution
Step-by-step solution
Step 1 of 8
Consider the system shown in equation (1).
■-1 - 2
x= 0
-1
1
0
'2
-2 '
1 jr+ 0
-1
-(1)
1
(a)
It is clear that the overshoot is less than 5%.
Consider the maximum overshoot 5% and calculate the value of damping coefficient ^ .
% O S= e^
Substitute the value of ovemhoot 5% in equation (2).
ln5
(1.609)’ ( l - i - ’ ) = (i-;r)’
2 . 6 - 2 . 6 f ' - 9 .8 6 9 f '
12.46^” - 2 .6
f = 0.456
Step 2 of 8
It is clear that the response of the closed loop system has 1% settling time under 4.6 seconds.
Write the formula of settling time equal to 1%.
In (o .O lV l-f’ )
( “>.
'
(3)
4.6
Substitute the value of 7^equal to 4.6 seconds and the value of ^ equal to 0.456 in equation (3)
and calculate the value of
.
4.6
4.6
0.456 a>.
4.6 s
4V. = 2.19
Write the second order characteristic equation for the system using the values of ^ and
■¥
»0
+ 2(0.456)(2.19)»+(2.19)* = 0
Thus, the characteristic equation of the system is
+ 2 f +4.79 s 0
Use ITAE criterion fro step input and calculate the third order characteristic equation.
si*
+2
.
+
ffl,’ ■
0
Ths, the third order characteristic equation is,
s’ +3.83*’ + 10.31*+ 10.50 = 0
Step 3 of 8
It is clear from the equation (1) that,
-2
-1
-2
F= 0
-1
1
1
0
-1
Determine the characteristic equation of the system with state feedback controller, K.
det * I - ( F - G K ) ] = 0
5
0
0
det' 0
5
0
0
0
5
det
det
■_1 _2
-
5
0
0
5
o'
0 -
0
0
5
-1
I
0 -1
-2 ‘
-1
I
I
0
-I
0
-1-2AT,
0
0
0
5
0
l-K ,
0
[2*:,
2K,
c
0
K1
1^2
-
=0
2K2T\
0
=0
- 2 - 2 K 2 -2 - 2 K j
-1
I
2+2K,
2+2^2
5+1
[ - l + K,
^ 3]
1^2
1
-2
5
0
-
0
5 0
5-M + 2A:,
'2
1
-I
0
det
-2
0
-1
K2
=0
s + l + K^
Thus, the determinant of the matrix is
r + 5 ’ ( 3 + 2 i^ + ^ ) + 5 ( 5 + 2 ^ + i^ + 4 ^ ) + ( 5 - 2 ^ + 3 ic 2 + 3 ^ ) = 0
Step 4 of 8
Compare the determinant of the matrix with the characteristic equation of the system.
**+3.83*’ + 10.3 1*+10.50 = 0
It is clear that.
3+ 2 * i+ A i= 3 .8 3
5 + 2 * ,+ * ,+ 4 l^ = 10.31
5 -2 * ,+ 3 * 2 +3Ai = 10.50
The values of *,,*J,*^ are -0.269,0.375,1.368
Thus, the state feedback controller for the system is [[-0.269
0.375
1.368]| -
Step 5 of 8
(b)
Calculate G (*) ,
C (* )
-n
r* I-F + G * :
-G l
H
. j
d e t[ * I -F + G K ]
Assume that y - x ^ . thus j / = [! 0
Thus 0 ( 5 ) is obtained as
0]x
5+0.462
2.75
6.73
2
0
5+ 1
-1
0
-1.269
1
0.375 5+2.368
0
0
1
0
Simplify the matrix and obtain G (5) •
G (*) = 2*’ +0.006»-1.994
Calculate the transfer function of the system.
r '( * ) =
G (4
1+ G (*)J9(*)
2*’ +0.006*-1.994
*’ +3.83*’ + 10.31* + 10.50
Step 6 of 8
Obtain the step response for the system with transfer function r* (5 ) •
Write the following code in MATLAB to obtain the step response.
» num=[2 0.006 -1.994];
»d en = [1 3.83 10.31 10.50];
» sys1=tf{num,den);
» t=0:0.0001:6;
» y=step{sys1,t);
» plot(t,y);
» grid;
» xiabel t
» ylabel y(t)
Step 7 of 8 •
The plot of the step response is as shown in Figure 1.
Figure 1
Step 8 of 8
It is clear from the Figure 1 that the design meets the required specifications of overshoot less
than 5% and the settling time of 1% under 4.6 seconds.
Problem 7.25PP
Consider the system in Fig.
Figure
(a) Write a set of equations that describes this system in the standard canonical control form as
j = Ax + Bt/ and y = Cx.
(b) Design a control law of the form
« = -[* •
which will place the closed-loop poles at s = -2 ± 2j.
which will place the closed-loop poles at s = -2 ± 2j.
Step-by-step solution
ste p 1 of 4
(a)
Consider the numerator part of the gain.
=
...... (1)
Consider the denominator part of the gain.
=
+.»+<!,
(2)
Write the general canonical form of the equation.
. m
(3)
a(s)
Substitute equations (1) and (2) in equation (3).
C (s) =
(4)
Write the state description matrices of controller canonical form.
-<h .......... -a ,
1
c
0
..........
1
0
0
0
1
•••
0
...........
1
0
(5)
0
I
0
0
........(6)
0
C ,= [A
6* ........... M
(7)
D^sSeperatevaluefromnumeratoranddenominatorelse 0 ... (8)
Consider the value of gain, G(5)-
(9)
^ (4 = 7 ^
Write the general form of state space equation.
jv = Ax+B » ...... (10)
Write the state equation in control canonical form from equations (4) to (10).
R]-[:
*■[! ^ ' * [ j ...
Write the general form of output equation.
y = C x* D
^- = [1 0 ] *
(12)
Step 2 of 4 ^
Hence, the state equations of controller canonical fonn are:
0 -4
1
b = [i
0
o]4
ste p 3 of 4
(b)
Write the matrix A from equations (10) and (11).
^=[>
Write the matrix B from equations (10) and (11).
»=[o]
Consider the law of the form.
« = -[K ,
JC,]
(15)
Find (.si—A + B K ) from Equations (13) and (14)
s
0
0
s
s 0
0 s
0
1
-4 ]
fjC,
AT,]
o j [o
oj
(16)
Find d e t(s I-A + B K ) = 0 from equation (16).
d e t(5 l-A + B K )= (5 + A :,)s -(-I){ 4 + /:j)
d e t ( s I - A + B K ) = j * + ^ ^ + 4 + ^ j ...... (17)
Step 4 of 4 ^
Consider the closed loop systems satisfy the
det(sI-A*FBK) =0 condition.
Find desired characteristics polynomial of the system
The two poles are placed at -2+ 2J and —2—2J( s + 2 - 2 y ) ( s + 2 + 2 y ) = 3 ’ + 4 s + 8 ...... (18)
Compare the coefficients of s from equations (17) and (18)
3 * + /f ,j + 4 - 4 A : , = s ' + 4 j + 8
if, = 4 ...... (19)
Compare the coefficients of constant from equations (17) and (18)
4 + JC j= 8
^ j - 4 ...... (20)
The design of state-feedback gain matrix, from Equations (19). (20) are [AT, s 4 |a n d |AT2 = 4 |.
Problem 7.26PP
Output Controllability. In many situations a control engineer may be interested in controlling the
output y rather than the state x. A system is said to be output controllable if at any time you are
able to transfer the output from zero to any desired output y* in a finite time using an appropriate
control signal u*. Derive necessary and sufficient conditions for a continuous system (A, B. C) to
be output controllable. Are output and state controllability related? If so, how?
Step-by-step solution
ste p 1 of 3
Consider the linear controllina svstem. Assume the initial state to reauired final state has finite
Step 1 of 3
Consider the linear controlling system. Assume the initial state to required final state has finite
time and with finite input. For this case, the state is measurably small while using impulsive
inputs.
Write the mathematical equation for the control signal {u) with respect to time.
Where,
is the derivatives of the delta function up to
SitSv'SM
order.
scalars up to n*order.
Write the control signal (u) equation with respect to scalars as considered.
“ '= [* 1
Where,
II* is the control signal at infinite time.
Assign x is the input at initial state and y Is the output at the final state.
Write the input equation at the initial state for the control signal n*.
x ( 0 + ) - j ( 0 - ) = C jf* ...... (2)
Where,
C is the controllability matrix.
x (0 + ) a n d x (0 -)a re the input state variables at various interval.
In the equation (2), the control signal n*will move the system to arbitrary state when the
controllability matrix C is non-singular. The controllability matrix C is necessary to achieve the
controllability of the state. Therefore, it is required to consider the inverse of the controllability
matrix C in the equation.
C [ x ( 0 + ) - i ( 0 - ) ] = C C ji ‘ ...... (3)
Where,
C is the inverse of controllability matrix.
Hence, equation (3) gives the design to achieve the controllability at the Input state.
Step 2 of 3
Now, assume to achieve the controllability of the state at the output.
Refer equation (3) and write the controllability equation at the output state.
[ y ( 0 + ) - > ’( 0 - ) ] = C C ji‘
(4)
Where,
y(0*f) andy(0-)a> 'e the output state variables at various interval.
Assume there Is no loss In the output and rewrite equation (4). If there is no loss then y ( 0 - ) = 0
; - ( 0 + ) = C C j <‘ ...... (5)
Equation (5) gives the controllability equation at the output state.
Consider the continuous system (A ,B ,C ) , and write the controllable equation for this
continuous system at the output state.
y ( 0 + ) = [ C B C A B ....C A " ^ B ] « ' ....... ( 6 )
Here, the system is controllable in the output only if the condition ^CB C A B .... CA**'B Jis in
full rank.
Therefore, the equation for the controllability at the output state is
y ( 0 + ) = [ C B C A B .... C A ’ ' B ] ii‘ |
Step 3 of 3
Equation (6) is always true for SISO (Single Input Single Output) system if the transfer function
greater than zero.
Therefore, it can be concluded that state controllability express the o u ^ u t controllability but
output controllability does not express die value of state controllability.
Problem 7.27PP
Consider the system
0
-I
s
0
4
-4
7
0
0
0
3
0 ■
■0 ■
0
0
x+
15
-3
0 _
(a)
Find the eigenvalues of this system. {Hint Note the block-triangular structure of A . )
(b)
Find the controllable and uncontrollable modes of this system.
(c)
vTB
For each of the uncontrollable modes, find a vector v such that
= 0. v T A
= /\v 7 .
(c) For each of the uncontrollable modes, find a vector v such that
vTB
= 0, v T A
= /\v 7 .
(d) Show that there are an infinite number of feedback gains K that will relocate the modes of th
system to -5, -3, -2, and -2.
(e) Find the unique matrix K that achieves these pole locations and prevents Initial conditions on
the uncontrollable part of the system from ever affecting the controllable part.
Step-by-step solution
There is no solution to this problem yet.
G et help from a C hegg subject expert.
ASK AN EXPERT
Problem 7.28PP
Two pendulums, coupled by a spring, are to be controlled by two equal and opposite forces u.
which are applied to the pendulum bobs as shown In Fig. The equations of motion are
H^9i
= -k^(6i - 9 i ) - mgl9\ -
III,
= —ka^{92 —0i) —mgl92 + Ih<
Figure Coupled pendulums
(a) Show that the system is uncontrollable. Can you associate a physical meaning with the
controllable and uncontrollable modes?
(b)
Is there any way that the system can be made controllable?
Step-by-step solution
ste p 1 of 3
(A)
Using the state vector.
X
> 01 0, e j
1
0
0
0
0
Ka^
0
0
-
-1
0
ml^
0
■ 0■
0
1
(K<^ a\
x+
ml
0
1
0
ml_
Step 2 of 3
The controllability matriK is determined as.
FO
C =
0
-1
—
-1
„
F ^a
1 ( K a '^ a \ K a ‘
0
ml
ml
o'! r a ’
1 (K a ‘
1
—
„
0
ml
1
ml
a cc p
HFF
1
fn t
m l \ ml^
a 'l
1J
.
Step 3 of 3
(B)
YES.
0
ml
0
1)
0
nFF
Problem 7.29PP
The state-space model for a certain application has been given to us with the following state
description matrices;
'0.174
0
0 0
0.1S7 0.645 0 0
0
1 0
0
0
0
1 0
0
0
0 1
O'
■-0.207'
0
-0.005
0
0 4 B=
0
0
0
0
C = [ 1 0 0 0 0 ].
(a) Draw a block diagram of the realization with an integrator for each statevariable.
(b)A student has computed det C = 2.3x 10-7 and claims that the system is uncontrollable. Is
the student right or wrong? Why?
(b)A student has computed det C = 2.3x io -7 and claims that the system is uncontrollable. Is
the student right or wrong? Why?
(c) Is the realization observable?
Step-by-step solution
ste p 1 of 7
(a)
Write the general form of state space equation.
x =
...... (1)
y = C x + D ...... (2)
Consider the matrix A.
0.174
0
0 0
0.157 0.645 0 0
0
1 0
0
0 10
0
0
0
■(3)
0 1
Consider the matrix B.
’-0.207*
-0.005
0
Bs
(4)
0
0
Consider the matrix C.
C = [l
0 0 0 0]
(5)
Consider the matrix D.
O = [0]
(6)
Write the state equation in control canonical form from equations (1), (3) and (4).
*0.174 0 0 0 o'
0.157 0.645 0 0 0
= 0
1 0 0 0
0
0 1 0 0
0
0 0 1 0^
-*»
■«4
■-0.207'
-0.005
+ 0
0
0
■*1
0.174 0 0 0 0*
-0.207
0.157 0.645 0 0 0
-0.005
0
1 0 0 0 JC4
0
0
0
0 0
0
0
0 0 1 0
0
■(7)
U ■■■■
Write the output equation in control canonical form from equations (2), (5) and (6).
+0
;- = [l 0 0 0 0]
^ =
[1 0 0 0 0 ] * + 0 .........( 8)
ste p 2 of 7 ^
Draw the block diagram for controller canonical form from equations (7) and (8).
Figure 1
ste p 3 of 7
Hence, the block diagram of realization with an integrator for each state variable is IdrawnI -
Step 4 of 7
(b)
The general fonnula for controllability matrix, C is.
C =[b
AB
... A - ’ b ] ..... (9)
Calculate AB^rom equations (3) and (4).
0.174
0
0 0
0 ■-0.207*
-0.005
0.157 0.645 0 0 0
0
ABs
1 0
0
0 0
0
1 0
-0.036018
-0.035724
ABs
-0.005
( 10 )
0
0
Step 5 of 7
Calculate
equations (3) and (10).
A 'B = A B x A ......(11)
Substitute equations (3) and (10) in equation (11).
0 0 0
-0.036018
0.157 0.645 0 0 0
-0.035724
0.174
0
0
A ’B -
0 0 0
1
I
-0.005
0
0
0 0
0 1 0
-6.267x10’
-0.0287
A 'B =
-0.035724
( 12 )
-0.005
0
Calcuiate
a ^BIiiiiii
equations (3) and (12).
A’B = A‘B xA ..... (13)
Substitute equations (3) and (12) in equation (13).
0.174
0
0 0 0
0.157 0.645
A*B =
-6.267x10’’
0 0 0
-0.0287
-0.035724
0
1
0 0 0
0
0
0
0
I
0 0
0
I
-0.005
0
0
-1.09x10’’
-0.0195
A*B =
.(14)
-0.0287
-0.035724
-0.005
Calculate
a ^B ^^^
equations (3) and (14).
A’B = A’B xA ..... (15)
Substitute equations (3) and (14) in equation (15).
0.174
0
0 0 O' -1.09x10’’
0.157 0.645 0 0 0
A*B =
1
0
-0.0195
0 0 0
-0.0287
-0.035724
0
0
1 0 0
0
0
0
1 0
-0.005
-1.8966x10"*
-0,01275
A^B>
.(16)
-0.0195
-0.0287
-0.035724
Substitute equations (4), (10). (12). (14) and (16) in equation (9).
-0.207
-0.036018
-6.267x10’’
-1.09x10’’
-1.8% 6xl0-*
-0.005
-0.035724
-0.0287
-0.0195
-0.01275
0
-0.005
-0.035724
-0.0287
-0.0195
0
0
-0.005
-0.035724
-0.0287
0
0
0
-0.005
-0.035724
Write the MATLAB program to find detennination of Matrix CA=[-.207 -0.036018 -6.267*1
-1.09*10^-3 -1.8966*10''-4;
-0.005 -0.035724 -0.0287 -0.0195 -0.01275;
0 -0.005 -0.035724 -0.0287 -0.0195 ;
0 0 -0.005 -0.035724 -0.0287;
0 0 0 -0.005 -0.035724;];
det(A)
The output of the MATLAB program is given below,
ans =
-2.2828e-07
Hence, value of |det|C| = -2.282xlQ-’ | . This value is not equal to zero. So this is controllable
system. The student computation is wrong.
Step 6 of 7
(C)
Determine the system observabiiity OWrite the observabie condition from description matrices.
C
CA
.(17)
CA’
CA’
CA’
Step 7 of 7
Calcuiate CAftoni equations (3) and (5).
0.174
0
0
0 0
0.157 0.645 0 0 0
C A -[1
0 0 0 0]
0
0
0
0 0 0
1
0
0
1 0 0
0 1 0
CA = [0.174 0 0 0 0 ] ...... (18)
Calculate
equations (5) and (18).
’0.174
0 0 0
0
0.157 0.645 0 0 0
CA’ =[0.174 0 0 0 0]
0
1
0
0
0
0
CA’ = [0.030276 0 0 0 O]
Calculate
0 0 0
1 0 0
(19)
equations (5) and (19).
*0.174
0
0 0
0.157 0.645 0 0
CA’ =[0.030276 0 0 0 0]
CA’ =[5.268x10"’
0 0 0
0
1
0
0
0
0
0 0
o ] .....(20)
Calculate C A ^^°^ equations (5) and (20).
0.174
0
0.157 0.645
CA’ =[5.268x10”
0 0 0 O]
CA’ = [9.166x10’ * 0 0 0
0
0
0
1
0
0
0 0 0
0 0 0
0 0 0
1 0 0
0 1 0
o ].....( 21 )
Substitute equations (5), (18). (19). (20) and (21) in equation (17).
1
0.174
0 =
0.030276
5.268x10’’
O.lOOxlO”
0 0 0 0
0 0 0 0
0 0 0 0
. ( 22)
0 0 0 0
0 0 0 0
Determine the determination ^matrix from equation (22).
det|O|=0
Hence, value of det|O| = 0- So this is (not observable system).
Problem 7.30PP
Staircase Algorithm (Var) Dooren et al., 1978):Any realization (A.B.C) can be transformed by an
orthogonal similarity transformation to (A, B, C), where A is an upper Hessenberg matrix
(having one nonzero diagonal above the main diagonal) given by
' * ai
•
O '
* *
'.
0
a = t ’'at =
* ♦
■0■
, B = T’'B =
Oj,_l
0
. «l .
where g1 # 0, and
C = CT = [ c
,
Cl
-
c.
1, T U T r
Orthogonai transformations correspond to a rotation of the vectors (represented by the matrix
columns) being transformed with no change in length.
fat Prove that if ai =0 and ai+1...... an-^ # 0 for some /. then the controllable and uncontrollable
Orthogonal transformations correspond to a rotation of the vectors (represented by the matrix
columns) being transformed with no change in length.
(a) Prove that if ai =0 and ai+1...an-1 ^ 0 for some /, then the controllable and uncontrollable
modes of the system can be identified after this transformation has been done.
(b) How would you use this technique to identify the observable and unobservable modes of (A,
B,C)?
(c) What advantage does this approach for determining the controllable and uncontrollable
modes have over transforming the system to any other form?
(d) How can we use this approach to determine a basis for the controllable and uncontrollable
subspaces, as in Problem?
This algorithm can also be used to design a numerically stable algorithm for pole placement [see
Minimis and Paige (1982)]. The name of the algorithm comes from the multi-input version in
which the ai are the blocks that make A resemble a staircase. Refer to ctrbf, obsvf commands in
Matlab.
Problem
Consider the system y+ 3y+ 2y= u+ u.
(a) Find the state matrices Ac, Be, and Cc in control canonical form that correspond to the given
differential equation.
(b) Sketch the eigenvectors of Ac in the (x1, x2) plane, and draw vectors that correspond to the
completely observable (xO) and the completely unobservable (xO) state-variables.
(c)
Express xO and xCin terms of the observability matrix O.
(d) Give the state matrices In observer canonical form and repeat parts (b) and (c) in terms of
controllability instead of observability.
Step-by-step solution
There is no solution to this problem yet.
G et help from a C hegg subject expert.
ASK AN EXPERT
Problem 7.31 PP
The normalized equations of motion for an inverted pendulum at angle 0 on a cart are
where x is the cart position, and the controi input u is a force acting on the cart.
(a) With the state defined as X—[ ^
^ ^ ^ jT find the feedback gain K that places the
closed-loop poles at s = -1,-1,-1±iy. For parts (b) through (d), assume that ^ = 0.5.
(b) Use the SRL to select poles with a bandwidth as close as possible to those of part (a), and
find the control law that will place the closed-loop poles at the points you selected.
(c) Compare the responses of the closed-loop systems in parts (a) and (b) to an initial condition
of 0 = 10°.You may wish to use the initial command in Matlab.____________________________
(c) Compare the responses of the closed-loop systems in parts (a) and (b) to an initial condition
of 0 = 10°.You may wish to use the initial command in Matlab.
(d) Compute Nx and Nu for zero steady-state error to a constant command input on the cart
position, and compare the step responses of each of the two closed-loop systems.
Step-by-step solution
ste p 1 of 4
A)
iie t[ S I - F + a K ) = a .,{ S )
Con^aring coefficients yields
_ 10-8P
1 -p
1-P
A ,=
1 -p
^ = [12
16
6
12]
Step 2 of 4
B)
/f = [0010]
% Symmetric root locus
i= [ O ;0 ’G];
c = [0*/?.O];
d= 0
vlocus (a, b. c, d);
* = [13.5 18.36 3.9 13.98]
-
0)
Step 3 of 4 ^
C)
The initial condition response to 6 (0)s10°for both control design in (a) and
( i)
Step 4 of 4
D)
i
Yields
W i
W^=[0 0 1 o f
AT. = 0
Problem 7.32PP
An asymptotically stable Type I system with input rand output y is described by the closed-loop
system matrices (A. B. C, D = 0). Suppose the Input is given by the ramp r = at, for f > 0. Show
that the velocity error coefficient is given by
i- t
, = [ c a -* b ]
Step-by-step solution
ste p 1 of 13
Refer equation 7.45 in the textbook and write the closed loop transfer function of the system.
S te p i of 13
Refer equation 7.45 in the textbook and write the closed loop transfer function of the system.
r(j)=C(iI-A)'B+D
ste p 2 of 13
Substitute
r(s) =
rewrite the equation.
(1)
C ( j I - A ) “'B
ste p 3 of 13
The system is asymptotic stable. So in equation (1), A is Invertible.
Step 4 of 13 ^
Consider the term (rf-A )-' . Since /> 0 apply Taylor’s series and rewrite the equation.
(j I - A ) ( - A '
-
-...) = 1
(j I - A ) '' ■ (- A - '- j A - '- s^A-’ - . . . ) ...... (2)
Step 5 of 13
Apply equation (2) In equation (1).
r( j) =c[(-A-' - jA-* - j'A-* -...)]b
r ( j) = [- C A - 'B - » C A - = B - i'C A - ’B - .. .] ...... (3)
Step 6 of 13
Consider r - a J and convert the equation to s-domain.
*(4= 7
w
Step 7 O f 13
Consider the general emor equation for the transfer function.
£ ( j)= [i-r (j)] « (j)
(5)
Step 8 of 13
Apply equation (3) and (4) In equation (5).
£ ( j ) = ^ l-[-C A -'B -jC A -“B - j'C A ^ B - ... ] ^ ^ jj
£ ( j ) = [ l + CA-'B+ sCA-*B+ i'C A -’B + - ] ^ 7 j ....
(6)
Step 9 of 13
Consider ramp Input is given to the system for infinite time Interval.
Step 10 of 13
Apply final value theorem and write the equation.
Step 11 of 13
Substitute equation (6) in the equation and apply limit to the equation.
e .. foo) = lim j . [ l + CA-'B+ iC A -’B +«’CA-‘B + . . . ] ^ ^
*lim j . [ l + CA-'B+ iC A -’B + i’CA-’B + . . . ] ^ ^
r-*0
j
j
slim ^ j+ ^ C A -'B + a C A -’ B +o!C A-’B + ...j^ jj
«-»0
Step 12 of 13
The velocity error coefficient is defined as the ratio of the input rate to the steady state error.
Mathematically it is expressed as,
K =
w (")
Step 13 of 13
Substitute the value of e ^ (oo) in the equation.
K , -------^
oCA-'B
1
"C A -’B
-[C A -’B]"'
Hence the condition I f , -[C A -’B]"' is proved
Problem 7.33PP
Prove that the Nyquist plot for LQR design avoids a circle of radius one centered at the -1 point,
as shown in Fig. Show that this implies that
< GM <
the “upward” gain margin is GM =
and there is a “downward” GM = Vz, and the phase margin is at least PM= ±60°. Hence the LQR
gain matrix, K, can be multiplied by a large scalar or reduced by half with guaranteed closed-loop
system stability.
Figure Nyquist plot for an optimal regulator
Step-by-step solution
step 1 of 2
We have
a , (S) = det [5 1 - (if - Gi*:)]
= det 151- if [l + (51 - F
- GJi:]|
= dct (51- if ) dtt | l + (51 - if )■' os:])
= I3(5)[l+ i:(51-if)‘"]a
a , (S)a, (-5) = i3 (5 )fl(-5 )[l+ ^ (5 1 -^ )‘‘o] x [l+ i:(-5 1 -if)‘ ‘a]
= l+ C 0 .(5 )0 ,(-5 )
Step 2 of 2
SettingS = jQ ,w e get
= | D ( > ) f [ l + p |0 ( > ) f ]
But
|i+ i< (> B i-if)‘‘m|&i
Let us re - write the loop gain as sum of its real and imaginary parts
L ( ja ) = K ( ja l- p y 'o
= Re (£ [ j a ) ) + j l ^ (£ [jd>))
Finally
{[Re (£iy (b)] + l} ’ + [in [ l (jffi))f a 1
Problem 7.34PP
Consider the system
*= [r'
]•«'=“
and assume that you are using feedback of the form u = -K x+ r, where r is a reference input
signal.
(a) Show that (A, C) is observable.
(b) Show that there exists a K such that (A - BK, C) is unobservable.
(c)
Compute a K of the fonn K = [1, K2] that will make the system unobservable as in part (b);
that is, find H2 so that the closed-loop system is not observable.
(c)
Compute a K of the fonn K = [1, K2] that will make the system unobservable as in part (b);
that is, find H2 so that the closed-loop system is not observable.
(d)
Compare the open-loop transfer function with the transfer function of the closed-loop system
of part (c). What is the unobservability due to?
Step-by-step solution
step 1 of 4
(a)
Write the matrix A.
*-[? ;i
Write the matrix B.
“ -[o]
Write the matrix C.
C = [l
2]
(3)
Consider the law of the form.
Modify equation (4).
» = -[«,
(5)
Determine the system observability Q.
Write the observable condition from description matrices.
°=[^l
Find C A froni equation (3) and (1).
C A -[.
2 ]p
;]
C A -[0
1]
(7)
Substitute equations (3) and (7) in equation (6).
» -[;:]
»>
From equation (8), the O is nonsingular. Hence, (A.C) is lobservabiel-
Step 2 of 4 ^
(b)
Find C ( A - B K ) from Equations (1). (2) and (3).
C (A -H K ).[I
2 ]([-’
j] - [ i] [ A ,
AT,])
;]-P '■])
-t' C " ' ".'■])
c (A -B K )= [-^ r,
1 -^ :,]
(9)
Substitute equations (3) and (9) in equation (6).
-U
.- J
™
Determine the det(0 )fro m equation (10).
d e t(0 ) = I - A : , + 2 J : , .......( 10)
Consider the value of d e t(O ) is zero.
l - i C , + 2 X , = 0 ...... (11)
Hence, the value of l - K , * 2 K , = 0 ,th e n (A - B K , C) is lunobservabtel
Step 3 of 4 ^
(c)
Consider the value of Jf, •
.........( 12 )
Find
from equations (11) and (12).
1-A T ,+ 2 (1 ) = 0
- iC , — 2 - 1
AT,- 3 ...... (13)
Hence, the value of Jf,at unobservable system is [ ^ .
Step 4 of 4 ^
(d)
Write the general form of open loop transfer function of the system.
G „ (j ) = C ( j I - A ) ‘' B
(14)
Substitute equations (1), (2) and (3) in equation (14).
;]-[?
4 ; :]-P 31;]
-
4
■ t'
^ li'+ 2 i- ll0
C „ ( s ) = [l
2]
- . TH
i
*"+ 2 5 -1
i+2
Write the general form of closed loop transfer function of the system.
C rf(i) = C ( f I - A + B K ) ’ ' B
(16)
Substitute equations (1), (2) and (3) in equation (16).
»-<.>-[. < . [ ; ; ] - [ ? ;].[;]i* . 4 [ ; ]
Substitute 1 for Jf, and 3 for
(17)
in equation (17).
<•[;:]-[? ;h ;]i. n
=>■ 4 : :]-[t ;]•[; f t ]
- 4 '^ - f t f t ]
- 4"’ ft]
[t-t-2
-2 + 2 5 + 6 ]
f"+ 3 s + 2
fj + 2
j
2 i+ 4 ir i'
’ +3j + 2
[o
j+ 2
j* + 3 i + 2
j+2
‘ (4 + 1 )(4 + 2 )
(18)
(4 + 1)
Compare equations (15) and (18), the pole is [cancelled w ith o n e 2ero| in unobservable system.
Hence, the closed loop mode is unobservable from the output function.
Problem 7.35PP
Consider a system with the transfer function
W
=
(a) Find (Ao, Bo. Co) for this system in observer canonicai fonri.
(b) Is (Ao, Bo) controllable?
(c) Compute K so that the closed-loop poles are assigned to s = -3 ± 3/.
(d) Is the closed-loop system of part (c) observable?
(d) Is the closed-loop system of part (c) observable?
(e) Design a full-order estimator with estimator error poles at s = -12±12/
(f) Suppose the system is modified to have a zero;
9(5-H )
G,(i) =
j2
-9 ■
Prove that if u = - K x + r, there is a feedback gain K that makes the closed-loop system
unobservable. [Again assume an observer canonical realization for G1 (s).]
Step-by-step solution
step 1 of 9
(a)
Consider the numerator part of the gain.
b { s ) = b ^ ^ + b 2 S ^ ^ + .. .+ b , ...... (1)
Consider the denominator part of the gain.
a { s ) = s " + 0,$ '^
+ ...+ a ^
(2)
Write the general canonical form of the equation.
Substitute equations (1) and (2) in equation (3).
...... m
Write the observer canonical form of the state space equations.
=
...... (5)
Write the output equation.
y = C ^ x ^ + D ...... (6)
Step 2 of 9
Write the state description matrices in observer canonical form,
-o ,
1
0
-a ,
0
1
=
0
A .=
0
0
■ 0
- a,
••• 0
0
0
0
1
0
(7)
0
1
...........
0
(8)
B .=
C ,= [ l
0 0 ••• 0 ] ......(9)
£), =Seperate valuefiomnumeratoraiiddeiiominatorelse 0
(10
Consider the value of gain G {s)-
(11)
Write the state equation in observer canonical fonn from equations (1)to (11).
R ]-[:
•fo n r®i
« h?+ « w ......
[9
Oj
[9 j
Write the output equation in observer canonical form from equations (1) to (11).
y = C x *D
^- = [1 0 ] *
(13)
Step 3 of 9
Write the
value from equations (5) and (12).
Write the
value from equations (5) and (12).
fo l
■(15)
Write the C , value from equations (9) and (13).
C ,= [ l
0]
(16)
Hence, the value of
0
A„ =
9
|C .= [1
9
0 j|
0]|
step 4 of 9
(b)
Write the general formula for controllability matrix. C ■
C = [b,
A .B .
... A - ’B ,]
(17)
Calculate A.B^from equations (14) and (15).
*•“ •= [9
0.
......
Substitute equations (15) and (18) in equation (17).
^^=[9 3
Determine the value of det|C(from equation (19).
det|C|»-81
Hence, value of |det|C| = -^ I| - This value is not equal to zero. So this is controllable system.
Step 5 of 9
(c)
Consider the law of the form.
+r
a = -[Ar,
(20)
-f B ,K ) from Equations (14) and (15)
Find
( .i- A ..B ,K ) = ,[;
(rf-A ,+ B ,K ) =
(rf-A .+ B ,K ) =
s
0
0
p
s
9 0
s
0
0
0
s
9 0
0
I
0
0
9K,
9K,. ]
r
( 21 )
( ,i- A .B K ) = [_ ,;,^ ^
Find det(5l-A -l-B K )sO from equation (21).
d e t(jI-A -fB K )=»*-i-9A :,s-9-F 9A :,......(2 2 )
Consider the closed loop systems satisfy the det(5 l-A * F B K ) = 0 condition.
Find desired characteristics polynomial of the system
The two poles are placed at -3 + 3 y ’ and - 3 —3y-
(i-F 3-3y)(s-F3-F3y)=i’ - f 6 j- H 8 ...... (23)
Compare the coefficients of s from equations (22) and (23)
9 A j= 6
A , = | ...... (24)
Compare the coefficients of constant from equations (22) and (23)
-9 + 9 A .-1 8
A, = 3 ...... (25)
The design of state-feedback gain matrix, from Equations (14). (25) are AT, = 3 and A
K . -- 3I
Step 6 of 9
Write the MATLAB program to verify the value of K:
A= [0 1;9 0]:
B=[0;9]:
C=[1 0]:
D=0;
Po=10;
Ts=1;
z=(-log(Po/100))/(sqrt(pi'‘2-^log(Po/100)'‘2)):
wn=4/(z*Ts);
[numt.dent]=ord2(wn.z):
r=roots(dent):
poles=[-3■^3*i -3-3*1];
K=acker(A,B,poles)
The output of MATLAB program:
K=
3.0000
0.6667
Step 7 of 9
(d)
Determine the system observability QWrite the observable condition from description matrices.
‘’• [ c l ]
"
Calculate C^A^from equations (14) and (16).
c A -(i
• ( ; ; ]
C .A .- [ 0
1]
(27)
Substitute equations (16) and (27) in equation (26).
“]
‘ ’ = [0
Determine the determination ^m atrix from equation (28).
det|0|=l
Hence, value of l«‘ » t|o |= o | . So this is [obsCTvable system|.
(e)
Write the MATLAB program to verify the value of full order estimator L at —12±12y poles.
A= [0 1;9 Oj:
B=[0;9]:
C=[1 Oj:
D=0;
Po=10;
Ts=1;
z=(-log(Po/100))/(sqrt(pi'‘2-^log(Po/100)'‘2)):
wn=10*wn;
[numt.dent]=ord2(wn.z):
r=roots(dent):
poles=[-12■^12*i-12-12*1];
L=acker(A',C'.poles)'
The output of MATLAB program:
L=
24
297
Thus the values of full order estimator at —12±12ypolesare
=24|and \L^ s 2 9 7 |.
Step 8 of 9
(f)
Consider the value of gain. ^ ( 5 ). ’ (4 + 1)
i '- 9
(29)
c(*)=
Write the state description matrices in observer canonical form from equations (4). (5). (7), and
(29).
.(30)
- [ : ;i
Write the state description matrices in observer canonical form from equations (4). (5). (8) and
(29).
(31)
B .=
Write the C , value from equations (9) and (29).
C ,= [l
0]
(32)
Calculate the value of A , —B^Kfrom equations (30) and (31).
• - • “ ■
'- J a
K
Calculate the value of C , ( A . - B^K)from equations (32) and (33).
c .K - .u n - |.
c
. 1^
. ( a . - b . k ) = [ ^ a :,
,
-
3
1-9AT,]
(34)
Modify observable condition in equation (26).
’
[c .( A .-B .K )]
...... (35)
Write the observability matric from equations (34) and (35).
oJ '
[-9AT,
“ 1 ...... (35)
1-91C J
Calculate det|0|from equation (35).
det|0| = I - 9 A : j ...... (36)
If detjO jis zero, the system is unobservable.
Consider the value of det|0| is zero.
i
- 9 a: , « o
■(37)
Determine d e t(5 l-A ^ ■ !‘ B ,K )fro m equations (30)and (31).
det(*I-A,+B,K) =det[.[j ®]-[®
5+9AT,
- l+ 9 A j
L-9-f9AT,
L"
S-I-9J:,
=det['
= ( i ■ ^ 9 ^ :, ) ( * -I-9AT, ) - ( ^ + 9 A , ) ( - l + 9AT, )
= * ’ ■l■9A:,^-^9^:,^-t-SlA.^, -9+818T, ■l■9A:, -818T,8r,
d e t ( j I - A , - i - B , K ) = j ’ -f9/:,»-f9A :,i-9-F81A r,-F9A :,...... (38)
Substitute equation (37) in equation (38).
d e t ( s l - A , + B . K ) = « '+ 9 ^ 5 + 9 8 :,i - 9 + 8 l | + 9 i r ,
= j ' - i -j -i-98:,j - 9 - i-9-f 9A:,
= « ’ + *-F 9 ir,» + 9 X ,
d e t ( jI - A ,- f B . K ) = ( j- i- 9 A r , ) ( * - H ) ...... (39)
Step 9 of 9
From equations (39), the reason for the unobservable system is
| - l v a lu e o f pole is cancelled b y zero o f th eg ain G {s)\-
Problem 7.36PP
Explain how the controllability, observability, and stability properties of a linear system are
related.
Step-by-step solution
step 1 of 2
The concepts o f controllability and observability are closely related to the cancellation of
a pole and a zero in the S3rstem transfer function.
Controllability and observability are the properties which describe structural features of a
dniamic system. These properties p l ^ an in:portant role in modern control system
apoic ano a zero m uie system tiansier [unction.
Controllability and observability are the properties which describe structural features of a
d3mamic system. These properties play an important role in modern control system
design theory.
A system is said to be completely controllable, if every state
can be affected or
controlled to reach a desired state in finite time by some unconstrained control u (t).
The Ii^ u t and output o f a system are always physical quantities, and are normally easily
accessible to measurement we therefore, need a subsystem that performs the estimation
of state variables based on the information received fi'om the input u(<) and output ,y(<).
This subsystem is called an observer whose design based on observability property o f the
controlled system
The platn is said to be conq^letely observable if all the state variables in X (<) can be
observed from the measurements o f the output
d(^). And the ii ^ u tu ( ± ) .
Step 2 of 2
For the Linear system given by
X { t) = A X {t) + £ U ( t)
y ( t) = C X {t)+ D U {t)
0)
Ifthere exists an input u (0 ,^ ), which transfers the initial state .AT(O) to the statX ’ in a
finite time
the state X (0) is said to be controllable. I f all the initial states are
controllable (or) simply controllable. Otherwise, it is said to be uncontrollable.
For the Linear System given by equation (1), If the knowledge of the output' Y’ and the
ii^u t u (t) over a time interval o f time (O,^) suffices to determine the state X ( 0 ) , The
state X (O) is said to be ob servable. I f all initial states are observable, the system is said
to be con^letely observable or singly, observable. Otherwise, the system is aid to be
unobservable.
Problem 7.37PP
Consider the electric circuit shown in Fig.
(a) Write the internal (state) equations for the circuit. The input u(t) is a current, and the output y
is a voltage. Let x1 = iL and x2 = vc.
(b) What condition(s) on R, L, and C will guarantee that the system is controllable?
(c)
What condition(s) on R, L, and C will guarantee that the system is observable?
Figure Electric circuit
Step-by-step solution
step ^ of 7 ^
I
Step 2 of 7
As there are two energy storage elements, at least two state variables are required to
describe the system
Let
^ i= h
X2=V(j
at node,
Cdvc
U = tr + ------‘
dt
u = X i+ C i^
:.X2 = —
^
+ —
C
(1)
C
Step 3 of 7
Applying RVL,
+ V c + ic R = 0
where
ic = “ -*s
Writing the above eq i»tioa In terms of state variables.
+/? [a - i ^ ] = 0
- x ^ R - L x i +X 2 + R u - XiR = 0
.
-2i?
‘
L
and
_^1
'
L '
(?)
L
y = ic ^
= u R - iiR
= u R ~ XiR
(3)
Step 4 of 7
From Equ(l), (2 )& (3 )
Xi
Xj
—
s.
- 22
r
L
-1
—
. c
L
^
0
R'
L
1
UJ
(4)
.c .
Step 5 of 7
-2R
r
L
L
F=
=1
Lc
0
'R '
a=
.c .
J
0]
H = [-R
j =R
-2 R
r
R'
L
L
L
0
1
C.
FO-.
. c
2/?
L
1
r
L
_ R
. LC
Step 6 of 7
i
T = [a
jio ] ]
1
LC
LC
IV be
VC completely controllable,
As for the system to
-R ^
1
I?C ~ C
-RK
.
;e0
LC
1
—
2^
K + A
rC
^
^ 0
-R ^ C -L -¥ 2 R ^ C
^
- i >0
step 7 of 7
H’
=
■-2^
-V
L
1
c
. L
0
2R^'
L
-R
L .
F
2R}
L
L
d e tr= —
L
> 0 ;
R }>
\
Problem 7.38PP
The block diagram of a feedback system is shown in Fig. The system state is
-[?]•
and the dimensions of the matrices are as follows;
A = /?*/).
L = /?x 1.
B = /?X 1.
x = 2/)x 1,
C = 1 X n,
f = 1 X 1,
K = 1 X n,
y = 1 x1.
Figure Block diagram for Problem 7.38
(a) Write state equations for the system.
(b)
Let X = Tz, where
T=
- [ ! ?] ■
Show that the system is not controllable.
(c)
Find the transfer function of the system from r to y.
S tep-by-step solution
step 1 of 5
(a)
Refer figure 7.91 in the textbook and write the state equations of the system.
s A z -B K x ^+ B r
i,- L C x + ( A - L C - B K ) x , + Br
-[.i
Refer figure 7.91 in the textbook and write the output equation of the system.
y = C x,
>- = [C
(2)
A
-B K
LC A -L C - B K
Hence, the state equations of the system are
0]
Step 2 of 5
(b)
Consider the system state x .
x = Tz ...... (3)
'■[:
...
Where.
X is transformation matrix.
- . ] ......
Write the general state space equations.
i = A x + B r ......(6)
^- = C x ...... (7)
Write state description matrix from equations (1). (2). (6) and (7).
* - [i
•■[I]
I”
C = [C 0] ..... (10)
Write the general formula for ^m atrix.
A = T - 'A T ..... C'')
Step 3 of 5
Substitute equations (4), (5) and (8) in equation (11).
-_ j_ ri
o ifA
-BK
- i J [l c
_J_r
a
iri oi
- l c - b k J [i
¥l
0]
- iJ
-IB K
“ i4 a i- l c
J^r
A I'- I 'B K
I"B K
” i * [ a i ’ - l i “c - a i * + l i ' c
J^F a I ’ - I ^ B K
I’BK
0
-
BK 1
"I
( 12)
A-LcJ
0
I
A I“ - L I* c J
A l’ - L l'c J
[A - B K
"[
-iJ
- ib k - a i + l ic + b ik J [i
AI
Write the general formula for g .
B = T‘‘B ......
Substitute equations (5) and (9) in equation (13).
if*
®
= i- L i
-I
(14)
Consider
nsider the value of |3 is equal to one.
(15)
Where.
I is unit matrix.
Write the general formula for q .
C = C T ......C®>
Step 4 of 5
Substitute equations (4) and (10) in equation (13).
•(; i]
....(17)
(17)
C:=[c
= [C 0]] ......
Write
frite the new coordins
coordinate system from equations (12). (15) and (17).
fA -B K
* =[
BK 1
r ...... (18)
A-LcJ*
0
.y=[C 0]z
(19)
The general fonnula for controllability matrix, C 'S-
C = [b
AB ... A - ’b ] ...... (20)
Calculate > ^ fro m equations (8) and (9).
'A
-B K
[lc
ir B l
A -L C -B K J L B
AB = f " » - “ ; ' ‘ l ...... (21,
[ a b - b *k J
Substitute equations (9), (21) in equation (20).
c = fB
...... (22,
[b
a b - b ’k
J
Calculate detjClfrom equation (22).
det|C| = 0 ...... (23)
Therefore the value of
Mc\=o\
Hence, the system is uncontrollable and the system is
decomposed in to controllable and uncontrollable parts in equations (18) and (19).
Step 5 of 5
(c)
Write the formula for the transfer function of the system.
r(s) = C [ l I - ( A - B K ) J 'B
(24)
Substitute equations (12), (15) and (17) in equation (24).
r(j) = C [x I- { A -B K )J 'B
[ ( j- A + L C )
.r c
‘
m
”0
I
(x -A -^ 2 B K )|
^(x’ -(2 A -L C -2 B K ) x+ A*-ALC-2ABK + 2BKLC)LoJ
( j - A + LC)
0
0
( i -A * 2 B K )
= [C
' '
‘
0 )7 :7 -
a
^( x '-(2 A - LC - 2B K )i + A '- ALC - 2ABK + 2BKLC)
B (x-A +LC )l
[--------- ---------- ---------------------------- !■
"(j ’ -( 2 A - L C -2 B K ) j + A '-A L C -2 A B K + 2BKLC)
r f A ________________ BC(x-A+LC)_______________
^ ' (x‘ -(2A -L C -2B K )x+ A ‘ -ALC-2ABK + 2BKLC)
Hence, the transfer function of the system is.
r(x)=
BC(x-A+LC)
(x*-(2A - LC - 2BK) j + A '- ALC - 2ABK + 2BKLC)
Problem 7.39PP
This problem is intended to give you more insight into controllability and observability. Consider
the circuit In Fig., with an Input voltage source u(t) and an output current y(t).
(a)
Using the capacitor voltage and inductor current as state-variables, write state and output
equations for the system.
(b)
Find the conditions relating R^. R2, C. and L that render the system uncontrollable. Find s
similar set of conditions that result In an unobservable system.
(c)
Interpret the conditions found In part (b) physically in terms of the time constants of the
system.
system.
(d)
Find the transfer function of the system. Show that there is a pole-zero cancellation for the
conditions derived in part (b) (that is, when the system is uncontrollable or unobservable).
Figure Electric circuit
Step-by-step solution
step 1 of 10
y«)
Step 2 of 10
From the above circuit,
(1)
y ( ‘ ) = ic + h
where ic = C ^
”
dt
= C J fi
(2)
i n l y i n g KVL to the circuit.
(3)
Writing the above equation in terms o f state variables.
-C X ^ - X i + LX^ +
(■4)
=0
The second loop equation is:
- t t + Xj + io ^ i = 0
= ~Xj + tt
(5)
ia = — + —
s,
s,
Step 3 of 10
Equating Equations (2) & (5),
■ = ----- Xi + -----' u
.. Xi
^
R^C ^
(6)
Step 4 of 10
From Equations (4) & (6),
fl
1 1
„
1
0)
step 5 of 10
From Equations (1) & (5),
;y = ^ +
— +x^
(8)
From equations (6),(7) & (8)
■ -1
R^C
a-
1
o '
R^C
UJ
1 - A
L R^L
1
(9)
^
■[-i ']H
5}
Step 6 of 10
1
1
0
R^C
RG=
R^C
1_ 1
L
-A
R^L
.
1
A.
fe )’
3^- 1
I ^
■ Jt,
- [W
A cJ
^A C
'
-1
B,
B }LC
Step 7 of 10
For the system to be uncontrollable,
R^-\-¥R^R^LC
1
--------B ^ -\+ B iB ^ L C = L
|fi^ ( i+ ^ Z c ) 7 l+ Il
Step 8 of 10
1
c)
^ 1 -1 '
F ' H ’ ' = ~R^C
r
R^L
0
1_
-R^
1'
L a
. a
R^C
R^L
A
- L + B f i{ B i- \)
B^LC
-A
7* =
H ’ ’ F ’’
-B ,
1
^
/ f ]
j;iC (fii-l)-£
__1_
_ jj[C (i^ -l)-£
Bi
B^LC
B ^B ^LC -B jC + B f;+ L
For the sjrstem to be unobservable,
B iB ^ L C -B ^ C -^ B iC + L C = 0
B f H ^ B i- B ^ L - i) = LC
Step 9 of 10
det
T .F :
s l- F
-O
H
J
d e t[fi/-F ]
DENQMINATQR:
___l_
~R^C
^ 1 -1
R,L
S + -L
R^C
0
Iz A
RiL
S+R 2
= s^ + —^ s + J ^ s + - ^
R^C
^
R^C
R^C
R^C
Step 10 of 10
NU1£ERATQR:
1
S + -----
R^C
det
1 -R ,
°
s .B ,
R^L
B fl
- 1
1
A
•
=
{ ie .c s + i'l
"1
k \
1 ] +A
[ lA
R ,]
[
r, c
B,L
« ^ C + l+ s C + J ^ C + C
^ l- ^ i+ £ s + j! ,£
)
B^C
^ si!i‘Z.C +
j
1~
£ ± 5 l± 1 ] +
A J
AC
+ i± ^ l
r, l
J
^LC
+ s j^ £ C + j! i^ Z . C + jj^ £ C + l- iii+ £ s + /^ Z .
B lL C
Problem 7.40PP
The linearized dynamic equations of motion for a satellite are
i = Ax + Ba,
y=
c*.
Where
■0
W
0
0
A=
1 0
0
0
0
0
-2o» 0
0 '
■0 0 ■
2a>
1 0
. B=
1
0 0
0 1
0
« _ r I
[ o
0
0
0 1
0 1
Q J*
- [ S] The inputs u^ and u2 are the radial and tangential thrusts, the state-variables x1 and x3 are the
radial and angular deviations from the reference (circular) orbit, and the outputs y1 and y2 are
the radial and angular measurements, respectively.
The inputs u^ and u2 are the radial and tangential thrusts, the state-variables x^ and x3 are the
radial and angular deviations from the reference (circular) orbit, and the outputs y1 and y2 are
the radial and angular measurements, respectively.
(a)
Show that the system is controllable using both control inputs.
(b)
Show that the system is controllable using only a single input. Which one is it?
(c)
Show that the system is observable using both measurements.
(d)
Show that the system is observable using only one measurement. Which one is it?
Step-by-step solution
step 1 of 9
(a)
Write the state description matrix.
0
As
0
0
0
0
0
0
1
-le a
0
0
0
0
(1)
0
1 0
Bs
0
3<»’
(2)
0
0
0
1
:::]
'-K1
(3)
(5)
The general formula for controllability matrix, C 's.
C=[b
AB
...
A - ’ b ] ...... (6)
Calculate > ^ fro m equations (1) and (2).
AB«
0
1
0
3(»*
0
0
0
0
0
0
0
0
1
0
1
0
ABs
(7)
0
-2 ®
0
Substitute equations (2) and (7) in equation (6).
0
0
1
0
1 0
0
lea
0
0
0
1
0
1
-le a
0
■(8)
From equation (8), the rank of matrix Is [ ^ . So this system Is controllable using both control
inputs.
Step 2 of 9
(b)
Write the BiValueatzero upvalue.
O'
B,
...... (9)
Modify equation (6).
C ;= [b ,
AB ,
...
A - ’ B ,]
(10)
Calculate A B jfrom equations (1) and (9).
0
1 0
0
0
A B ,*
0
0
0
0
-2 ^
2d>
0
1
0
0
1
0
A B ,*
■ (11)
0
-2a)
Calculate A^B,from equations (1) and (11).
A * B ,s
0
1
0
0
~ r~
3<»’
0
0
lO)
0
0
n
0
0
=2&
Q
1
n
—la )
0
0 1
A * B ,s
(12)
-le a
0
Step 3 of 9
Calculate A ’ B,from equations (1) and (12).
r 0
A ’ B ,=
1
0
3<»’
0
0
2fl)
0 ]r 01
-fl)*
0
0
0
1
-2fl)
0
-2 fl)
0
0
0
r -fl)
0
A > B ,=
...... (13)
0
2a>‘
Substitute equations (9), (11), (12) and (13) in equation (10).
0
1
0
1
0
-o ’
0
0
0
-2a>
0
0
-2 «
0
2®’
.(14)
Step 4 of 9
Writethe B^ value at zero u,value.
0
...... (15)
B ,s
Modify equation (6).
C i= [B ,
ABj
...
A -% ]
(16)
Calculate A B , from equations (1) and (15).
1
0
0
0 2a)
0
0
1
-2<»
0
0
A B ,*
0
(17)
A B ,-
Calculate A^B,from equations (1) and (17).
A ’ B ,=
0
1
0
0
0
3fl)*
0
0
2fl)
2fl)
0
0
0
1
1
0
-2fl)
0
0
0
2fl)
0
A ‘ B ,=
(18)
0
-4fl)*
Calculate A ^ jfrom equations (1) and (18).
A > B ,=
0
1
0
0
3®’
0
0
2fl)
2fl)
0
0
0
0
1
0
0
-2fl)
0
0
-4fl)*
0
-2fl)*
A > B ,=
(19)
^ f l) *
0
Substitute equations (15), (17), (18) and (19) in equation (16).
0
0
2d)
0
0
2fl)
0
- la /
...
0
- to *
...
-4fl)^
0
...
0
1
0
(20)
From equations (14) and (20), the rank of matrix in C, matrix is [^a n d rank of matrix in Q
matrix Is.
So this system is controllable only in |B,|value at zero u,value.
Step 5 of 9
(c)
Write the general formula for observability matrix OWrite the observable condition from description matrices.
C
CA
(21)
Step 6 of 9
Calculate CA^rom equations (1) and (3).
CA
CA
fl
0 0
1
0
o '
3d)*
0
0
2d)
1 0
0
'[ o
■0
0
fo
1 0
0]
'[ o
0 0
ij
0
0
0
1
0
-2d)
0
0
(22)
Substitute equations (3) and (22) in equation (21).
1 0 0 O'
0 0 10
Os 0 1 0 0
0 0 0
(23)
1
Therefore, from equation (23), the rank of matrix is [ ^ . So this system is observable using both
measurements.
Step 7 of 9
(d)
Writethe C, value from equation (3).
C ,= [ l
0 0 0 ] ...... (24)
Modify equation (21).
^ C,
C,A
(25)
o , = C,A'
C,A*
Calculate C.Affom equations (1) and (24).
C ,A -[1
■0
1
0
0 '
3®’
0
0
2d)
0 0 0]
C ,A - [ 0
0
0
0
1
0
-2d)
0
0
1 0 0 ] ...... (26)
Calculate C,A*from equations (1) and (26).
0
C,A’ = [0
I
1
0
0 0
0 -2d)
0 0]
C , A '= [ 3 o ’
0
0 o'
0 2d)
0 1
0 0
3d)*
2 < » ]...... (27)
0
Calculate C ,A ’ from equations (1) and (27).
C ,A ’ =[3ffl^
C,A’ = [o
0
■0 1
3®’ 0
2«i]
0 0
0 -2d)
0
-o ’
0
0
0
0
0'
2d)
1
0
0 O ] ......(2 8 )
Substitute equations (24), (26), (27) and (28) in equation (25).
■
0
0
0
I
0
0
0
0
2d)
-d)*
0
0
0
o> = 3d)*
0
...... (29)
Step 8 of 9
Write the C, value from equation (3).
C , = [0 0 I
0 ] ...... (30)
Modify equation (21).
c,
C,A
0 , = C ,A ’
...... (31)
C .A ’
Calculate C ,A from equations (1) and (30).
C ,A - [ 0
0 I 0]
0
1
3o’
0
0
0 0 '
0
0
0 1
0
-2 o
0 0
2o
C ,A = [0 0 0 1 ] ...... (32)
Calculate C ,A ’ from equations (1) and (32).
C ,A ’ = [0
C ,A ’ =[0
0 0
-2 o
1]
0
1
0
3o’
0
0 2o
0
0
0
0
1
0
-2 o
0
0
‘
0 0 ] ...... (33)
Step 9 of 9
Calculate C ,A*from equations (1) and (33).
C ,A ’ =[0
-2 o
C ,A ’ = [-6 o *
0
■0
3d)^
0]
0
0
1
0
0
-2d)
0
0
0
0
0'
2d)
1
0
0 0 -4 o ’ ] ...... (34)
Substitute equations (30), (32), (33) and (34) in equation (31).
o ,=
■0 0 1 0
0 0 0 1
0 -2d) 0 0
-6®* 0 0 -t®’
.(35)
Therefore, from equations (29) and (35), the rank of matrix in
in 0,matt1xis.
matrix is [^a n d rank of matrix
So this system is observable only in j^ jv a lu e .
Problem 7.41 PP
Consider the system in Fig.
Figure Coupled pendulums
K = bt
A,
- jr ( 0 , - 0 2 ) + ^
^ = - « ^ + jT(0| - 02) - fhnl
K=kd
0, =
^
-
<r(0, - 02>+ /7ml
+ IT(0| - 02) - fhnl
(a) Write the state-variable equations for the system, using [ej
$2
vector
and F as the single input.
(b) Show that all the state-variables are observable using measurements of 01 alone.
(c)
Show that the characteristic polynomial for the system is the product of the polynomials for
two oscillators. Do so by first writing a new set of system equations involving the state-variables
.
■yi ■
yi
■01+^ ■
^ O l-0 2 ,
Hint If A and D are invertible matrices, then
[^ •o r= [r‘ ^]■
(d)
Deduce the fact that the spring mode Is controllable with F but the pendulum mode is not.
S te p -b y -s te p s o lu tio n
step 1 of 4
A)
The State vector
J T = [e .
e,
6.
0
0
1 O'
0
0
0
1
0
0
0
0
K
X
-F
Step 2 of 4
B)
We have
r= [l 0
0 0]X
Observability
e=
H
1
HF
0
0
0
HF^
~ - ( ® “ -i-j?:) i :
0
0
HF^
0
-(® ^+ j^ r)
K
0
0
1
0
0
The state is observable with
Step 3 of 4
Q
Let
X =
y if
0 ,+ ^
^+6,
0 1
9 ,-e ,
S i- S , ]
0
0
-<s“
0
0
0
0
0
0
1
0
0
x +
-(o>“ + 2 ir )
The characteristic equation o f the system is
det [ if l - i? ) = ( s ’ + «?)
+ o’ + 2 ^)
Step 4 of 4
D)
There is no coi^ling between the spring mode (6^—@2) and the pendulum mode
(S i+ a .)
Problem 7.42PP
A certain fifth-order system is found to have a characteristic equation with roots at 0, -1, -2. and
-1 ± iy. A decomposition into controllable and uncontrollable parts discloses that the controllable
part has a characteristic equation with roots 0 and -1 ± 1). A decomposition into observable and
nonobservable parts discloses that the observable modes are at 0, -1 , and -2.
(a)
Where are the zeros of b(s) = Cadj(sl - A)B for this system?
(b)
What are the poles of the reduced-order transfer function that includes only controllable and
observable modes?
Step-by-step solution
Step-by-step solution
step 1 of 13
(a)
Step 2 of 13
Consider the equation of the system.
i(>)=Cadj(sI-A)B
Step 3 of 13
The equation is controllable at points (0 » -l± y)and it is observable at
Step 4 of 13
To obtain the zeros of the system, the location of the zeros must be both controllable and
observable which occurs only in stable system.
Step 5 of 13
For a stable system, the location of poles and zeros must be in left half plane. When the poles
and zeros move towards the boundary, It Indicates that the system is going to unstable mode.
Step 6 of 13
Here all the points lie on left half plane so the zeros of the system Is at ( - l . - 2 , - l ± y )
step 7 of 13
Therefore the zeros of the system are at the points |( -l^ -2 ,-l± 7)|.
Step 8 of 13
(b)
Step 9 of 13
Consider the equation is controllable at points (0,-1 ±y')and it is observable at (0 ,-1 ,-2 )
Step 10 of 13
To obtain the poles of the system, the location of the poles must be both controllable and
observable which occurs only in stable system.
Step 11 of 13 X
For a stable system, the location of poles and zeros must be in left half plane. When the poles
and zeros move towards the boundary, it indicates that the system is going to unstable mode.
Step 12 of 13 ^
Here all the points He on left half plane and there is no real pole. So the poles of the system are
at origin.
Step 13 of 13
Therefore the poles of the system are at
[origin|.
Problem 7.43PP
Consider the systems shown in Fig., employing series, parallel, and feedback configurations,
(a) Suppose we have controllable-observable realizations for each subsystem;
X/ =
Ax/ + B/u/,
y/ = Cixi, where / = 1.2.
Give a set of state equations for the combined systems in Fig.
(b) For each case, determine what condition(s) on the roots of the polynomials Ni and Di is
necessary for each system to be controllable and observable. Give a brief reason for your
answer in terms of pole-zero cancellations.
Figure Block diagrams; (a) series; (b) parallel; (c) feedback
.................*1 J~r7r~w5n j,
Figure Block diagrams; (a) series; (b) parallel; (c) feedback
Step-by-step solution
step 1 of 7
(a)
Series connection:
Consider the controllable and observable realizations for any system,
i , = A , * , + B , » , ...... (1)
...... (2)
Where,
i = land2.
Refer figure 7.94 (a) in the textbook and write state equations for series connection from
equations (1) and (2).
X, = A , x ,+ B , h
(3)
x , = A , x ,+ B , C , X | ...... (4)
Write the state equation of series connection in matrix form from equations (3) and (4).
(5)
Write the output equation of series connection.
y = c , x , ...... (6)
Write the output equation of series connection in matrix form from equation (6).
-[o
....
Hence, the state equations of the series connection is i r s
0
' A,
B ,C ,
B ,'
JC+
0
A.
u
y=[0 c,]
step 2 of 7
Parallel connection:
Refer figure 7.94 (b) in the textbook and write state equations for parallel connection from
equations (1) and (2).
X, = A , x ,+ B , h
(8)
X,=A2AC2 + B,U ...... (9)
Write the state equation of parallel connection in matrix fonn from equations (8) and (9).
Ki-i*.' :][;]•&
( 10 )
Write the output equation of parallel connection.
> = C | X , + C jX j...... (11)
Write the output equation of parallel connection in matrix form from equation {11).
■ ( 12 )
'■ [ 4
' ■ f t ] ......
Hence, the state equations of the parallel connection is
y = [C,
A,
0
0
A,
B,
C ,]
Step 3 of 7
Feedback connection:
Refer figure 7.94 (c) in the textbook and write state equations for feedback connection from
equations (1) and (2).
X, ■ A ,x i - A ,C jX j+ B , r ...... (13)
x , = A , x ,+ B , C , X | ...... (14)
Write the state equation of feedback connection in matrix form from equations (13) and (14).
.
(15)
Write the output equation of feedback connection.
y = C ,*, ...... (16)
Step 4 of 7 ^
Write the output equation of feedback connection in matrix form from equation (16).
.|[^-]|
^,.[c,
-=|C, oil
(17)
Hence, the state equations of the parallel connection is irs
y = [C,
0]
■ A,
B,C,
-A ,C ,'
A,
'“ ' r
0
0
step 5 of 7
(b)
Refer 7.94(a) in the textbook and write the transfer function of the series connection.
=
(18)
a (s)
^ f o r
A (*)
'
y(s)
Af,MAT,(5)
U (s)
D ,( ,) D ,( s )
and ^ f o r
A W
CM ) in equations (18).
' '
(19)
Therefore, from equation (19), there are no cancellations in poles and zeros part of the transfer
functions in series combination.
For observability the parts [AT^(j)and£); W should be coprime o th e rw ise Q (j) | is covered
from the output.
For controllability the parts |
(j) a n d P 2( j ) should be ooprime o th e r w i s e ( j ) ~ | is covered
from the output.
Step 6 of 7
Refer 7.94(b) in the textbook and write the transfer function of the parallel connection.
(20)
M l
Substitute f ' . M for <7, ( 5) and
^ (s )
'
A W
y(s)
M ) ^ jv ,( s )
U (s )
D ,(s )
y(s)
A W A W i-JV jW P,W
uW
G,(s) in equations (20).
A W
A W
a
■ ( 21 )
W
Therefore, from equation (21), there are no cancellations in poles and zeros part of the transfer
functions in parallel combination.
For observability and controllability the parts |D ,(5)andZ)2(5) should be coprimel
Step 7 of 7 ^
Refer 7.94(c) in the textbook and write the transfer function of the feedback connection.
R(s)
(22 )
1+ G . W A W
Substitute
for GiW®'''^
A UJ
M
r(s)
G, W 'n
(22).
A
l
QW
^ W 'l . A W A W
AW AW
M l
AW
a w a w
-^a
w a w
AWAW
AW
AWAW
AWAWAW+AWAW
=___ AWAW___
,23)
*w
a w a w
+a
w a w
Therefore, from equation (23), there are no cancellations in poles and zeros part of the transfer
functions in feedback combination. For observability, | AT, ( ^ ) and D j (y ) should be coppme|and
controllability [Af2(j)andZ ^ W should be coprime|
Problem 7.44PP
Consider the system y + 3 y + 2 y = u + u .
(a)
Find the state matrices Ac, Be, and Cc in control canonical form that correspond to the given
differential equation.
(b) Sketch the eigenvectors of A c in the {x1, x2) plane, and draw vectors that correspond to the
completely observable (xO) and the completely unobservable (xO) state-variables.
(c)
Express xO and xCin terms of the observability matrix O.
(d)
Give the state matrices in observer canonical form and repeat parts (b) and (c) in terms of
n n n tm llflh ilih / in<%tpaH n f nh.<?prvahilifv
(d)
Give the state matrices in observer canonical form and repeat parts (b) and (c) in terms of
controllability instead of observability.
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 7.45PP
The dynamic equations of motion for a station-keeping satellite (such as a weather sateiiite) are
x — 2ajy — 3af^x = 0t
y + 2ew: = tt,
where
X = radiai perturbation,
y = longitudinal position perturbation,
u = engine thrust in the y-direction,
as depicted in Fig. If the orbit is synchronous with the earth’s rotation, then w = 2 tt/(3600 * 24)
rad/sec.
(a)
Is the state x = [ x x y y ] T observable?
(a)
Is the state x = [ x x y y ] T observable?
(b)
Choose x = [ x x y y ]7 a s the state vector and y as the measurement, and design a full-order
observer with poles placed at s = - 2 cj- 3 cj, and -3oj ± 3ojj.
Figure Diagram of a station-keeping satellite in orbit
Step-by-step solution
step 1 of 2
A)
There is not enough information to answer this question.
Step 2 of 2
0
1 0
3ar‘
B)
0
0 k
0
0
0
0
0
-2 o
0
z = [o
0 1
0
2 ib
1 ^3
0 1^4
o]Jsr
1^=-44.5W ,
1^=-51.5W-
4 = llo .
;,= 5 3 o ’
0
+
0
1
Problem 7.46PP
The linearized equations of motion of the simple pendulum in Fig. are
6 + w2d = u.
(a)
Write the equations of motion in state-space form.
(b)
Design an estimator (observer) that reconstructs the state of the pendulum given
measurements of &. Assume w = 5 rad/sec, and pick the estimator roots to be at s = -10 ± 10j.
(c)
Write the transfer function of the estimator between the measured value of & and the
estimated value of d.
6Stiffiaiea vaiue dr d.
(d)
Design a controller (that is, determine the state feedback gain K) so that the roots of the
closed-loop characteristic equation are at s = - 4 ± 4/.
Figure Pendulum diagram
1
Step-by-step solution
Step 1 of 4
A)
We have
m
'
:■
y=
\]x
0
y = 0 T]X
step 2 of 4
B)
d e t(£ fl-i? + i« ) = 0
£ r^ + /jS '+ ii)’ ( ; i - l ) = 0
Step 3 of 4
C)
We use the estimator equation
i= F ^ + a u + L l^ y - m )
= ( F - L H ) X + O u + Iiy
M
= [l
0 ] ( S ^ i i' + £ « ) - ‘ £
- 7 (S '-20/7)
~ S ’ + 20S + 200
Step 4 of 4
D)
d e t(S l-J i'+ O £ r) = 0
s ’ +£:jSf+oci‘ + i : , = o
K ,= l,K ^ = S
Problem 7.47PP
An error analysis of an inertial navigator leads to the set of normalized state equations
‘i I][s]^l]'
liH l
where
x1 = east—velocity emor,
x2 = platform tilt about the north axis.
x3 = north—gyro drift.
u = gyro drift rate of change.
Design a reduced-order estimator with y = x1 as the measurement, and place the observer-error
poles at -0.1 and -0.1. Be sure to provide all the relevant estimator equations.
Design a reduced-order estimator witn y = x l as tne measurement, and place the observer-error
poles at -0.1 and -0.1. Be sure to provide all the relevant estimator equations.
Step-by-step solution
step 1 of 1
Partitioning the system matrices yields
0
+ 0
1
0
=
1
0
y=
-1
0
0
1 0
0 ^1 fo
1 + 0
0 ^3 [l
o]Jsr
d e t(^ Thus
A = - 0 . 2 a n d ij = - 0 . 0 1
Problem 7.48PP
A certain process has the transfer function
G(i) = .
.
(a) Find A. B, and C for this system in observer canonicai fonn.
(b) If u = -K x, compute K so that the closed-loop control poles are located at s = - 2 ± 2j.
(c) Compute L so that the estimator error poles are located at s = -10±10y.
(d) Give the transfer function of the resulting controller [for example, using Eq.].
(d) Give the transfer function of the resulting controller [for example, using Eq.].
(e) What are the gain and phase margins of the controller and the given open-loop system?
Eq.
V(j)
D .(j) “ 7 ^ - - “ <** - A + B K + L O - 'L .
Step-by-step solution
step 1 of 6
(a)
Consider the transfer function
-(1)
V -V
The order of the transfer function is n » 2
Consider the transfer function is represented as
£7( 5)
+
By inspection,
^ sl;<^
--4
Consider the general observable canonical form in tenns of the transfer function coefficients is
•
0
0 0
0
^=[ 0
-a.
1 0 0
=
0
- 0 ,- 1
1
*+i - < v A
-fl, _
0 0 1 ]
Consider the general observable canonical form of {1) in terms of the transfer function
coefficients is
r i,i
ro
I]
fo i
UJ-[4 oJ-U"
y = 1 0 ]*
A=
0
1
4
0
0 ■
, c = [l 0]
-4
Step 2 of 6
(b)
Consider u - -K x
The closed-loop characteristic equation is given by
a , ( j ) = (^ + 2 + 2 y ) ( 5 + 2 - 2 y )
o ;p(5 )
= j*+ 4 s+8
(2)
The closed-loop characteristic equation is also determined by det ( 5 I - A + BK) s Q
detl
(■[;
" ( [ ; ;]*U a i ^
[ [L-4+4A,
-4 + 4 * .
xJ+4A
+ 4 *,,] ]
®
* ’ + 4 * , j - 4 + 4 * | . 0 ...... (3)
Compare equations (2) and (3).
*. = 3
* ,-1
Thus, |K = [3 1]|
Step 3 of 6 ^
(c)
Consider the equation to determine the estimator roots with estimator-emor poles at s = - 2 ± 2 j
a ,{s )-0
a , { s ) = {s + l 0 + l 0 j ) { s + l 0 - l 0 j )
a ,( » ) = » ’ + 2 0 s+ 2 0 0 ...... (4)
Step 4 of 6
The estimator roots are also determined by a ,( » ) = d e t ( j I - A + L C )
det 1
"([.i
4 :
:i="
4 ::i. -.'I)* t + / | j + / , - 4 . 0 ...... (5)
Compare equations (4) and (5)
/,= 2 0
/ , b 204
Thus.
Step 5 of 6 ^
(d)
Consider the equation to determine the transfer function of the resulting controller.
2 > ,( x ) — K ( x l - A + B K 4 L C ) ' L
‘■ • o - c g
-•w -i’ g
•^[212
, ,
:])■'&]
, + 4]
-2 6 4 J-6 9 2
j* + 2 4 s + 2 9 2
Thus, the transfer function of the compensator is D , ( i) =
-2 6 4 J -6 9 2
5^ + 24^ + 292
Step 6 of 6
(e)
Consider the MATLAB code to determine the phase and gain margin of the controller using
nyquist plot.
H=tf([-264-133924].[1 24 292])
nyquist{H)
The output after executing the MATLAB code is shown in Figure 1.
NyiliiistDiagnB
From Figure 1. the phase margin is —137® and gain margin is —5 3 ^ dB
Thus, the phase margin is |-137®| and gain margin is 1-53.2 d s l •
Problem 7.49PP
The linearized longitudinal motion of a helicopter near hover (see Fig) can be modeled by the
normalized third-order system
' -0 .4
I
0
0
-OJM
0
-14 9J -0.02
Figure Helicopter
Suppose our sensor measures the horizontal velocity u as the output; that is, y = u.
(a)
Find the open-loop pole locations.
(b)
Is the system controllable?
(c)
Find the feedback gain that places the poles of the system a ts = -1±1) and s = -2.
(d)
Design a full-order estimator for the system, and place the estimator poles at -8 and
-4 ± 4 V J y .
(e)
Design a reduced-order estimator with both poles at -4. What are the advantages and
disadvantages of the reduced-order estimator compared with the full-order case?
(f)
Compute the compensator transfer function using the control gain and the full-order estimator
designed in part (d), and plot Its frequency response using Matlab. Draw a Bode plot for the
closed-loop design, and indicate the corresponding gain and phase margins.
(g)
Repeat part (f) with the reduced-order estimator.
(h)
Draw the SRL and select roots for a control law that will give a control bandwidth matching
the design of part (c), and select roots for a full-order estimator that will result in an estimator
error bandwidth comparable to the design of part (d). Draw the corresponding Bode plot and
compare the pole placement and SRL designs with respect to bandwidth, stability margins, step
response, and control effort for a unit-step rotor-angle input. Use Matlab for the computations.
S te p -b y -s te p s o lu tio n
step 1 of 5
(A)
The open loop poles are the eigen values o f F
d e t ( ^ n - F ’) = 0
S = -0.6 5 6 5 and
0.1183±^ 0.3CT8
Step 2 of 5
(B)
For the helicopter
r a n k { q = ra iik [0 PO P '0 \ = 2
Step 3 of 5
(C)
i : = [0.4706
1.0 0.0627]
d e t ( 6 n - f + G « :) = 0
Step 4 of 5
(D)
A = [44.7097 18.8130 15.5800f
Step 5 of 5
d e t ( 6 n - if „ + £ 7 r , ) = 0
£= [14.2510 0.9542f
Problem 7.50PP
Suppose a DC drive motor with motor current u is connected to the wheels of a cart in order to
control the movement of an inverted pendulum mounted on the cart. The linearized and
normalized equations of motion corresponding to this system can be put in the form
e =e + v + u.
v = 6 - V - u,
where
6 = angle of the pendulum,
V = velocity of the cart.
(a)
We wish to control 6 by feedback to u of the fonn
u = - K - \e -K 2 d - K 3 v .
Find the feedback gains so that the resulting closed-loop poles are located at —1»—1
u = - K - \e -K 2 d - K 3 v .
Find the feedback gains so that the resulting closed-loop poles are located at —1, —1
(b)
Assume that 6 and v are measured. Construct an estimator for d and & of the form
t = A l-(-L (y -C t).
where x = [0 d\T and y = d. Treat both v and u as known. Select L so that the estimator poles are
it -2. and -2.
(c)
Give the transfer function of the controller, and draw the Bode plot of the closed-loop system,
indicating the corresponding gain and phase margins.
(d)
Using Matlab, plot the response of the system to an initial condition on d. and give a physical
explanation for the initial motion of the cart.
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 7.51 PP
Consider the control of
Y(s)
C(*) = i
'£/(«)
10
4 ( « + l) ‘
(a)
Let y = x1 and x1 = x2, and write state equations for the system.
(b)
Find /C1 and K2 so that u = -K ^ x ‘\ -K2x2 yields closed-loop poles with a natural frequency
ojn = 3 and a damping ratio ^ = 0.5.
(c)
Design a state estimator that yields estimator emor poles with o;n1 = 15 and ^1 = 0.5.
(d)
What is the transfer function of the controller obtained by combining parts (a) through (c)?
(e)
Sketch the root locus of the resulting closed-loop system as plant gain (nominally 10) is
varied.
Step-by-step solution
step 1 of 4
The state equation are
:a=[:
;.= [1
o]x
step 2 of 4
(B)
jr = [0.9 0.2]
Step 3 of 4
(Q
£ = [14
21 i f
Step 4 of 4
(D)
The transfer function for the controller is
D c { S )= - { S j - F + a K + L H y ''L
-(54.85+202.5)
Problem 7.52PP
Unstable dynamic equations of motion of the form
x = x + u.
arise in situations where the motion of an upside-down pendulum (such as a rocket) must be
controlled.
(a)
Let u = -K x (position feedback alone), and sketch the root locus with respect to the scalar
gain K.
(b)
Consider a lead compensator of the form
Select a and K so that the system will display a rise time of about 2 sec and no more than 25%
Select a and K so that the system will display a rise time of about 2 sec and no more than 25%
overshoot. Sketch the root locus with respect to K.
(c)
Sketch the Bode plot (both magnitude and phase) of the uncompensated plant.
(d)
Sketch the Bode plot of the compensated design, and estimate the phase margin. Design
state feedback so that the closed-loop poles are at the same locations as those of the design in
part (b).
(e)
Design an estimator for x and x using the measurement of x = y, and select the observer gain
L so that the equation for x has characteristic roots with a damping ratio ^ = 0.5 and a natural
frequency o)n = 8.
(e)
Draw a block diagram of your combined estimator and control law, and indicate where x and
appear. Draw a Bode plot for the closed-loop system, and compare the resulting bandwidth and
stability margins with those obtained using the design of part (b).
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 7.53PP
A simplified model for the control of a flexible robotic arm is shown in Fig., where
k/M = 900 rad/sec2,
y = output, the mass position,
u = input, the position of the end of the spring.
(a)
Write the equations of motion in state-space form.
(b)
Design an estimator with roots at s = -100 ± 100/.
(c)
Could both state-variables of the system be estimated if only a measurement of y was
available?
(c)
Could both state-variables of the system be estimated if only a measurement of y was
available?
(d)
Design a full-state feedback controller with roots at s = -20 ± 20/.
(e)
Would it be reasonable to design a control law for the system with roots at s = -200 ± 200/.
State your reasons.
(f)
Write equations for the compensator, including a command input for y. Draw a Bode plot for
the closed-loop system and give the gain and phase margins for the design.
Figure Simple robotic ann
Step-by-step solution
step 1 of 12
Refer to the Figure 7.98 from the text book.
It is clear that the value of — is 900 ra ^sfc * -
(a)
The equations of motion in state space form are as follows.
Consider that x ^ m y and Xj ■ y
Thus, the equations are,
X ,= X i
X, ss—
k
k
X + — tt
Thus, the state matrix representation is as follows.
0
=
1
O'
--i 0 fc *
. M
y= 1 0]jt
(1)
M.
Step 2 of 12
(b)
It is clear that roots are —100±100yUse the roots and calculate the equation.
( i + 1 0 0 + 1 0 0 y)(j + 1 0 0 -1 0 0 y )= (i+ 1 0 0 )^ + (1 0 0 )'
= « ’ + 10‘ +200s + 10‘
= « ‘ + 200«+(20000)
It is clear that *'+ 2 0 0 « + (2 0 0 0 0 ) represents the characteristic equation.
Calculate the characteristic equation for the robotic arm using the matrices shown In equation (1).
The formula to calculate the characteristic equation using the state space fonn is as follows.
d e t [ jI- ( F - L H ) ] = 0
It is clear from the equation (1) that,
0
1'
F*
0
L A/
J
<7*
0]
Step 3 of 12
Determine the characteristic equation of the system.
det s I- ( F - L H ) ] = 0
0
1"
det
=0
E :i- |i ;i-K :i
det
M
1
e :i
det
— - ij
0
^
\ .M
-L ,
.
1
det
=
0
=
0
K :}
s+L,
det
*
-1
,
^ (» + A ) + [- ^ + A !]= o
Step 4 of 12
Compare the obtained characteristic equation with * '+ 2 0 0 « + (2 0 0 0 0 )
It is clear that,
£ ,= 2 0 0 ...... (2)
(3)
• ^ + £ ,= 2 0 0 0 0
Substitute the value of ^k :In equation (2) and (3) and calculate
and
-
£ ,= 2 0 0
900+1, = 20000
l2 » 1 9 1 0 0
200
Thus, the estimator is designed
19100
Step 5 of 12
(c)
system.
The observability matrix is
Calculate h F -
I' " i [ J » i ] - i " 'I
Substitute the
tt value of H p In observability matrix,
si
[ ;:] It is clear from the value of the observability matrix that the system is observable.
Thus, both the state variables of the system can be estimated using a measurement of *
Step 6 of 12
(d)
It is clear that roots are —2 0 ± 20yUse the roots and calculate the equation.
(i+20+20y)(i+20- 20y)=(i+20)“+(20)'
= 5*+400+40j +400
= 5 *+ 4 0 5 + (8 0 0 )
It is clear that 5^+4Q $+(800) represents the characteristic equation.
Calculate the characteristic equation for the robotic arm using the matrices shown In equation (1).
The formula to calculate the characteristic equation using the state space fonn to design the
feedback controller is as follows.
d e t [ jI- ( K - G K ) ] = 0
It is clear from the equation (1) that,
0
F*
r
A
L w
0
J
0
<?=
»ndH = [l
t
0]
IM I
Step 7 of 12
Determine the characteristic equation of the system.
d e t [ r i- ( K - G K ) ] = 0
■0 1
det
[::}
±
M
0
o'
-
k
M .
■0 1'
det
E :]■
±
E :i-
det
0
0■
—(
i s " * * ')
E :]■
det
=0
-1
det
± ( i . a :.)
s.
=
± ( x ,)
0
(4)
Step 8 of 12
Compare the obtained characteristic equation with
+ 4 0 5+ (800)-
It is clear that,
kKy
M
b 40
^
i,9 0 0 j
= 0.0444
^ ( l + i f , ) = 800
/:,=
800
900
-
0.111
Thus, the feedback controller is designed [[-0.111
0.044]| -
Step 9 of 12
(e)
It is clear that roots are -200 ± 200 J ■
Use the roots and calculate the equation.
( j + 200 + 2 0 0 y ) ( i+200 - 200y) = ( i + 2 0 0 )'+ (2 0 0 )'
-
+ 40000 + 400s+40000
= s ' + 400s+(80000)
Compare the equation s ' + 400s+(80000) with equation (4) and obtain
and JC, -
kK,
M
l- = 40
A : j= 4 0 o f ;^ l
'
I.9 0 0 J
= 0.444
-^ (l+ iC ,) = 80000
M
^ _ 80000 ,
900
*87.88
It is clear from the values of
a:,
andATj that large control levels are required so it is not desirable
to design a system with the roots -200 ±200^Thus, not desirable to design a control law for the system with roots -200± 200y •
Step 10 of 12
The state space form generalized form to calculate the transfer function of the compensator is as
follows.
x = (F-GK-LH)x+L>
w= - K x
Write the following code in MATLAB and obtain the transfer function of the compensator.
» F=[ 0 1:-900 0];
» G=[0;900]:
» H=[1 0];
» L=[200;19100]:
» K=[-0.111 0.044]:
» syms s
» l=[1 0; 0 1];
» D=-K*inv{(s*l-F+G*K+L*H))*L
D=
(222*(5*s + 198))/{5*(10*s''2 + 2396*s + 278201)) + 8862049/(5*{10‘ s'^2 + 2396*s + 278201)) (8404*(s + 200))/{10*s'‘2 + 2396*s + 278201)
» simplify(D)
ans =
-(8182*s -100401 )/(10*s'^2 + 2396*s + 278201)
Step 11 of 12
Draw the bode plot for the obtained transfer function of the compensator.
Bode D ig ra m
Step 12 of 12
Thus, bode plot is obtained for the compensator and the gain margin and phase margins are
1-10.7 dB and
respectiveiy.
Problem 7.54PP
The linearized difTerential equations governing the fluid-flow dynamics for the two cascaded tanks
in Fig. are
5/j1 + aC/j1 = 6u,
6h2 + a6h2 = a6h'\,
where
5/j1 = deviation of depth in tank 1 from the nominal level.
6h2 = deviation of depth in tank 2 from the nominal level,
6u = deviation in fluid in flow rate to tank 1 (control).
Figure Coupled tanks
(a)
Level Controller for Two Cascaded Tanks: Using state feedback of the form
6u=-K■\6h^ -K25h2,
choose values of K^ and K2 that will place the closed-loop eigenvalues at
s = -2a(1 ± j).
(b)
Level Estimator for Two Cascaded Tanks: Suppose that only the deviation in the level of tank
2 is measured (that is, y = 5h2). Using this measurement, design an estimator that will give
continuous, smooth estimates of the deviation in levels of tank 1 and tank 2, with estimator error
poles at -8a(1 ± j).
(c)
Estimator/Controller for Two Cascaded Tanks: Sketch a block diagram (showing individual
integrators) of the closed-loop system obtained by combining the estimator of part (b) with the
controller of part (a).
(d)
Using Matlab, compute and plot the response at y to an initial offset in Sh^. Assume a = 1 for
the plot.
Step-by-step solution
step 1 of 2
(A)
r ^ r + e + i: ,
ldet|
t
[
-6
k
. 1
S+6j
= S’ + ( 2 6 + i : i ) i'+ 6 “ + 6 ( ^ : , + i^ ) = 0
= S=+46S+86==0
Gives /Ti = 26 and
= 56
Step 2 of 2
=
+ ( 2 6 + /,)£ T + 6 ( /i+ i,)+ 6 ” = 0
= (S + S 6 + S 6 j)(S + S 6 -S e j)
= J ’ +16&S'+1286’ = 0
Gives ly =1136 and i, = 146
Problem 7.55PP
The lateral motions of a ship that is 100 m long, moving at a constant velocity of 10 m/sec, are
described by
^1
I+1-0
r -0.0895 -0.286 0 "I T ^ 1
[ /;
I= I
-0.0439 -0.272
0 II
["
14.
0.0145 "I
®I22
where
jS = side slip angle(deg),
(fj = heading angle(deg),
6 = rudder angle{deg),
r = yaw rate (see Fig. 1).
Figure 1 View of ship from above
r = yaw rate (see Fig. 1).
Figure 1 View of ship from above
(a)
Determine the transfer function from 5to tp and the characteristic roots of the uncontrolled
ship.
(b)
Using complete state feedback of the form
6 = - K - \ p - K 2 r-K 3 {iiJ - ijjd).
where ijjd is the desired heading, determine vaiues of K^, K2, and K3 that wiii place the closedloop roots at s = -0.2,-0.2 ± 0.2).
(c)
Design a state estimator based on the measurement of ijj (obtained from a gyrocompass, for
example). Place the roots of the estimator error equation at s = -0.8 and -0.8 ± 0.8).
(d)
Give the state equations and transfer function for the compensator Dc(s) in Fig. 2, and plot it
frequency response.
(e)
Draw the Bode plot for the closed-loop system, and compute the corresponding gain and
phase margins.
(f)
Compute the feed-fonvard gains for a reference input, and plot the step response of the
system to a change in heading of 5°.
Figure 2 Ship control block diagram for Problem
S te p -b y -s te p s o lu tio n
step 1 of 3
The state space representation of the lateral motions of a ship is.
-0.0895
'fi
r = -0.0439
0
-0.286
O ir^
-0.272
1
0 r + -0.0122 S
oJ[»r
0
■0.0145
The matrix A and B ^re,
-0.0895
-0.286
O'
-0.0439
-0.272
0
0
I
0.0145
Bs
-
0.0122
0
Step 2 of 3
(a)
As it is required to find the transfer function from g to
the output equation is,
B
Y = [0
0
1]
V
V f ']
The matrix C is,
C » [0
0
1]
The transfer function of the system is,
G ( j) = C [ i I - A ] 'B
> + 0.0895
0.286
O'
0.0439
J + 0.272
0
0
-1
s
=[0 0 1]
0.0145
-
0.0122
0
Determine the inverse matrix, [ j j —A ]"' ■
_
1
« [ ( « + 0.0895)(i + 0.272) -0 .0439 x 0.286J
■ j(j+0.272)
-0.0439J
-0.0439
-0.286J
s ( f + 0.0895)
» +0.0895
0
0
( i + 0.0895)(j + 0.272) - (0.0439)(0.286)
i ( i ’ + 0 .3 6 l5 s + 0.0118)
■ j(i+0.272)
-0 .0439i
-0.0439
-0.286*
*(* +0.0895)
* + 0.0895
0
0
*’ + 0.36l5* + 0 .0 ll8
Step 3 of 3
Determine the matrix, G (*)
“ *(*’ + 0.3615*+0.01l8)^®
0 1]
■*(* + 0.272)
-0.0439*
-0.0439
■0.0145'
-0.286*
*(* + 0.0895)
* + 0.0895
-0.0122
0
0
*’ + 0.3615* + 0 .0 ll8
0
0.0145* (*+ 0.272)+0.000536
*(*’ + 0.36I5* + 0.01l8)
[0 0
1] -0.0122*(*+0.0895)-0.004147*
0
As the transfer function is 0, it is not possible to determine the roots of the system.
Also, it is not possible to determine the state feedback, or to design a state estimator, or
determine the state equations and to draw the Bode plot and step response of the system.
Hence, subparts (b), (c), (d), (e) and (f) cannot be solved.
Problem 7.56PP
Asmentioned in footnote 11 in Section 7.9.2, a reasonable approach for selecting the feedfon/vard gain in Eq. is to choose N such that when r and y are both unchanging, the DC gain from
r to i/is the negative of the DC gain from yto u. Derive a formula for W based on this selection
rule. Show that if the plant is Type 1, this choice is the same as that given by Eq.
Footnote 11
A reasonable alternative is to select N such that, when r and y are both unchanging, the DC gain
from rto u is the negative o f the DC gain from y to u. The consequences o f this choice are that
our controller can be structured as a combination o f error control and generalized derivative
control, and if the system is capable o f Type 1 behavior, that capability will be realized.
Eq.
1
N = ‘
C (A - B K )- lB [ l - K (A - LC )->{B - M )J
N=-
1
C (A - B K )-»B (1 - K (A - L C ) - > ( B - M )J
S t e p - b y - s t e p s o lu t io n
step 1 of 19
(a)
Step 2 of 19
Refer figure 7.48 and equations 7.185a and 7.185b in the text book.
Step 3 of 19
Consider the equation of the system for the feed forward gains of the controller.
x - ( A - B K - L C ) i + L ^ + M r .......(1)
u— Kx + Pr
(2)
Step 4 of 19
Assign the values and calculate the DC gain from y to u.
X*X p
r =0
step 5 of 19
Substitute the values in equation (1) and rearrange the equation.
0 = ( A - B K - L C ) x „ + L y ,+ 0
- ( A - B K - L C ) x ,- L ;- ,
M l
(A -B K -L C )
*
X ,— L y ( A - B K - L C ) ‘
(3)
Step 6 of 19
Substitute the values in equation (2) and rearrange equation.
« j= - K X j
» , = K ( A - B K - L C ) ''L ; > .
(4)
step 7 of 19
Assign the values and calculate the DC gain from rto u.
X*Xo
y=0
Step 8 of 19
Substitute the values in equation (1) and rearrange the equation.
0 = ( A - B K - L C ) X, + A / r , + 0
-(A -B K -L C )x „ -A /ji
•
(A -B K -L C )
X, —
( A - B K - LC) ‘
(5)
Step 9 of 19
Substitute the values in equation (2) and rearrange equation.
« , = - K x , + Mr,
= K A /r , ( A - B K - L C ) " '+
- [ k A /(A -B K -L C )" ' + w ]r,
« , = [ l C A / ( A - B K - L C ) '' + ^ ] r , .......(6)
Step 10 of 19
For this controller consider the DC gain from y to u is equal to the DC gain from rto u.
Step 11 of 19
Equate equation (4) and equation (6).
[ k ( a - B K - LC)"* L yo] = [ k A /( A - B K - L C )“' +
Step 12 of 19
Rearrange and find ; y .
^ = - K [ ( A - B K - L C )" ' ( L + A f) ]
Step 13 of 19 ^
Consider the general closed loop system block diagram.
Step 14 of 19
Figure 1
step 15 of 19 ^
Take the value of p as the foot note. Now the DC gain is unity for the type-1 system.
Step 16 of 19
For Type-1 system, write the equation for general closed loop system block diagram.
G ( x ) = ie ( x )
Consider the formula to calculate the DC gain of the closed loop system.
« (i)
' l-G (j)J > ,
Where,
is the negative feedback path
£) is the loop gain.
Step 17 of 19
Step 18 of 19
Step 19 of 19
Apply limit to the equation.
g (°)
« (0 )
' l-G (O )i),
[ v G ( 0 ) s I (unity feedback)]
^ ,(0 )
« (0 )
D ,( 0 )
(6)
Refer equation 7.202 in the textbook. This equation helps to achieve unity DC gain in the
controller design.
Similarly in equation (6), if the gain and the feedback are same, the system will give unity DC
gain in the controller.
Mathematically, if ND ,
y(o)
value of closed loop system —
will be unity.
« (0 )
Therefore, both the equations will give unity feedback controller design in the type-1 system.
Hence the condition is proved.
Problem 7.57PP
Assume that the linearized and time-scaled equation of motion for the ballbearing levitation
device is x - x = w + w. Here w is a constant bias due to the power amplifier. Introduce integral
error control, and select three control gains K = [K^ K2 K3]so that the closed-loop poles are at
-1 and -1 ± j and the steady-state error to w and to a (step) position command will be zero. Let y
= X and
the reference input ^ ^
be a constant. Draw a block diagram of your design
showing the locations of the feedback gains Ki. Assume that both x and x can be measured. Plot
the response of the closedloop system to a step command input and the response to a step
change in the bias input. Verify that the system is Type 1. Use Matlab (Simulink) software to
simulate the system responses.
Step-by-step solution
Step-by-step solution
step 1 of 2
The equations o f motion are given by
x -x = u+W
0 = 0
A realization of these equation is
»i[:]
=y-r
z = Xi
X
'0
K=
X
1 O'
0
0
O'
1
0
_0 1 0
1_
Step 2 of 2
The design ofthe state feedback vector, K
For closed loop poles o f 5 = -1 , - l ± j t
d e t(5 1 -^'^ + G S ,^ ) = 0
When
K,
/:,]= [2
5
3]
The closed loop system is given by
■-1"
Z = {F ^ -a ^ K )Z + 0 ^ (S > + I
y ^ M ,z
For the closed loop system we have
0
1
0
0
0
1
1 -^ j
-^ 3
This immediately gives Z z—0
AndZz—y —0
Problem 7.58PP
Consider a system with state matrices
^=[
(a)
0
- 1 ] ’
»=[!]•
Use feedback of the form u(t) = -K x{t) + Nr(t), where A/ is a nonzero scalar, to move the pole
to -3 ± 3).
(b)
Choose N so that if r is a constant, the system has zero steady-state error; that is. y (^) = r.
(c)
Show that if A changes to A+ 5A. where 5A is an arbitrary 2*2 matrix, then your choice of N
in part{b) will no longer make yC“ ) = r. Therefore, the system is not robust under changes to the
Ru<;tpm n n r a m p t p r s in A
(c)
Show that if A changes to A+ 5A. where 5A is an arbitrary 2*2 matrix, then your choice of N
in part{b) will no longer make yC“ ) = r. Therefore, the system is not robust under changes to the
system parameters in A.
(d)
The system steady-state error performance can be made robust by augmenting the system
with an integrator and using unity feedback—that is, by setting x l = r - y, where x l is the state of
the integrator. To see this, first use state feedback of the form u = -K x - K^xl so that the poles of
the augmented system are at -3. —2
(e)
Show that the resulting system will yield y (^) = r no matter how the matrices A and B are
changed, as long as the closed-loop system remains stable.
(f)
For part (d). use Matlab (Simulink) software to plot the time response of the system to a
constant input. Draw Bode plots of the controller, as well as the sensitivity function (S) and the
complementary sensitivity function (7).
S te p -b y -s te p s o lu tio n
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 7.59PP
Consider a servomechanism for following the data track on a computer-disk memory system.
Because of various unavoidable mechanical imperfections, the data track is not exactly a
centered circle, and thus the radial servo must follow a sinusoidal input of radian frequency ojO
(the spin rate of the disk). The state matrices for a linearized model of such a system are
* - [ ;
A ]-
- [ ; ] ■
>'■
The sinusoidal reference input satisfies y — —o j^r.
(a)
Let (uO = 1, and place the poles of the error system for an internal model design at
ac(s) = (s + 2 ± J2){s + 1 ± y i)
and the pole of the reduced-order estimator at
ae(s) = (s + 6).
and the pole of the reduced-order estimator at
ae(s) = (s + 6).
(b)
Draw a block diagram of the system, and clearly show the presence of the oscillator with
frequency ojO (the internal model) in the controller. Also verify the presence of the blocking zeros
at ±j(jjO.
(c)
Use Matlab (SImulink) software to plot the time response of the system to a sinusoidal input a
frequency ojO = 1.
(d)
Draw a Bode plot to show how this system will respond to sinusoidal inputs at frequencies
different from but near wO.
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 7.60PP
Compute the controller transfer function [from Y(s) to tyfsj] in Example. What is the prominent
feature of the controller that allows tracking and disturbance rejection?
Example Steady-State Tracking and Disturbance Rejection o f Motor Speed by Extended
Estimator
Construct an estimattN' to ccxitFol the state and cancel a constant bias at the
output and track a constant reference in the rootof speed system described
x
= - 3 i + h.
(7.256a)
y = x + w,
(7.256b)
w = 0»
(7.256c)
f = 0.
(7.256d)
Place the control pole at 5 = —5 and the tw o extended estimator poles at
5 = -1 5 .
Solutioii. To besin. w e destitn die control law by iiniorins the equivalent
Place the control pole at s = —5 and the tw o extended estimattMr poles
s = -1 5 .
Solution. To begin, w e design die ctmtiol law by ignoring the equivalrat
disturbance. Rather, we notice by inspectitMi that a gain c f —2 will move
the single pole from —3 to the desired —5, Therefore, K = 2. The sys­
tem augmented with equivalent external input p, which replaces the actual
disturbance w and the le fn ra c e r , is ^ v e n 1^
p=0,
i = - 2 x + u + pt
-4 r
-5
0 02 04 a s 04 1.0 U 1.4 14 1.8 2
Time (fee)
W>
Figure 7.72
Motor Speed system with extended esQmaton (a) block diagram; (b) command step response and
(fistufbMce step response
The extended estimator equatims are
^ = / i( e - x ) ,
1 = -3S-I-« + p + fe(e - i).
The estimator error gain is found to be L ^
characteristic equadon
* •[1
*+
[ 225
27
from the
+
A block diagram of the system is given in Fig. 7 .7 ^ ^ , and the step responses
to input at the command r (applied at t = 0 sec) and at the disturbance w
(applied at / w 0.5 sec) are shewn in Rg. 7.72(bX
Step-by-step solution
step 1 of 1
The related equation
p = /i ( c - x )
x = -3 x + p + 4 + i3
u = -K x -p
p l^ r o
jJ
-I,
[o
][p
- 3 - iT - i.J
« = [-!
- 4 ?
X
The controller transfer function is
r{S)
S{S+3+K+l,)
- 2 7 9 ( £ f + 4 .03 23 )
5 (5 + 3 2 )
Problem 7.61 PP
Consider the pendulum problem with control torque Tc and disturbance torque T&.
&-+ 40 = Tc + Td.
(Here
= 4.) Assume that there is a potentiometer at the pin that measures the output angle 0,
but with a constant unknown bias b. Thus the measurement equation is y = 0 + b.
(a)
Take the “augmented” state vector to be
where w is the input-equivalent bias. Write the system equations in state-space form. Give values
for the matrices A, B, and C.
for the matrices A, B, and C.
(b)
=
Using state-variable methods, show that the characteristic equation of the model is s(s2
4)
0.
(c)
Show that w is observable if we assume that y = 0, and write the estimator equations for
Pick estimator gains [ I I 12 /^Tto place all the roots of the estimator error characteristic equation
a t -10.
(d)
Using full-state feedback of the estimated (controllable) state-variables, derive a control law t
place the closed-loop poles at -2 ± j2.
(e)
Draw a block diagram of the complete closed-loop system (estimator, plant, and controller)
using integrator blocks.
(f)
Introduce the estimated bias into the control so as to yield zero steady-state error to the outpu
bias b. Demonstrate the performance of your design by plotting the response of the system to a
step change in b; that is, b changes from 0 to some constant value.
S te p -b y -s te p s o lu tio n
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 7.62PP
Consider the servomechanism problem where we wish to track a ramp reference signal. The
plant and the desired model equations are
y= [ 1 o ]i
*“ = [ 0
0 ]* ■ •
y« = [ 1 0 ] i« .
Design a model-following control law and demonstrate its tracking performance. Place the
closed-loop poles at s = - 2 ± j2.
Step-by-step solution
step 1 of 5
Write the general state space equation of the system.
x = A x + B u ...... (1)
Consider the state space equation of the system.
*■[!
Write the state description matrix from equations (1) and (2).
*■[: -J..
(4)
Write the general output equation of the system.
y= C x ...... (5)
Consider the output equation of the system.
y = [l 0]*
(6)
Write the state description matrix from equations (5) and (6).
C = [l
0]
(7)
D -[0 ]
(8)
Consider the matrix equation
•*. = A
,* . + B
^
in the system.
..........( 9 )
Consider the value of matrix jfc^in the system.
"-=[o
Write the state description matrix from equations (9) and (10).
*-=[o 3
“-= [i]
Consider the output equation of the system.
y . - C . x . + D . ...... (13)
Consider the output equation of the system.
y .= [l 0 ]* .
(14)
Write the state description matrix from equations (13) and (14).
C .
= [l
0]
(15)
D . = [ 0 ] ...... (16)
Consider the value of closed loop poles.
s = - 2 ± J 2 ...... (17)
Step 2 of 5 ^
Write the MATLAB program for design the control law and plot the tracking performance of the
system from equations (3). (4). (7). (8), (11). (12). (15). (16) and (17).
cic
Am=[0 1;0 0];
Bm=[0;1]:
Cm=[1 0];
Dm=[0]:
A=[0 1;0-1];
B=[0;1]:
C=[1 0]:
D=[0]:
pc=[-2+sqrt(-1 )*2;-2-sqrt(-1 )*2];
K=place(A.B.pc)
[N.m]=size(B):
[p.N]=size(C);
J=zeros(p.m);
[x.x]=size(Am);
aa=[kron(eye(x).A)-kron(Am'.eye(N)) kron(eye(x),B); kron(eye(x).C) kron(eye(x).D)];
bb=[zeros(x*N.1);Cm(:)]:
xx=aa\bb;
M=reshape(xx(1 :N*x).N,x)
AAm=[Am zeros(2,2);B*(N+K*M) A-B*K];
BBm=[Bm;zeros(2,1)];
CCm=[zeros(1.2)C];
DDm=[0]:
sysm^ss(AAm.BBm.CCm.DDm):
t=0:.01:5;
hold on;
[ymf,t]=impulse(sysmf,t):
plot(t.ymf)
hold on;
plot(t.t.'-’)
xlabel('Time (sec)');
ylabel('r.y');
title(Tracking performance demonstration');
nicegrid;
Step 3 of 5
The output of the MATLAB program is given below.
K=
8.0000
3.0000
M=
10
01
Step 4 of 5 ^
The tracking performance from MATLAB output is given below.
Figure 1
Step 5 of 5 ^
Hence, the model control law is designed and its tracking performance is shown in [Figure 1
Problem 7.63PP
Suppose we wish the closed-loop system to behave like a desired model, called the implicit
model
Z = A/77Z.
We may minimize a modified LQR performance index
j f “ [ (J - A ^ / q , » - A .y ) +
a.
equivalent to the standard one with the addition of a cross­
Show that this performance Index Is >
weighting term between the control sand the state of the form
<x>
where
"
J 1
0 = (CA - A«C)’'Q i (CA - A .C ),
S=
(CA - A«C),
f i = R + B’'C’’Q,CB.
Step-by-step solution
step 1 of 1
Consider the general state variable form of vector equations.
x = A i + B i / .......(1)
y = C r ...... (2)
> = C i ...... (3)
Consider the general form of LQR design.
J = J (* ’'Qx + u’'R « ) *
(4)
0
Consider the modified form of LQR performance index.
J = J ( ( y - A . y f Q , (y - A .y ) +
(5)
Substitute equations (2), (3) In equation (5).
( 6)
J = J ^ (C x - A ^ C x ) Q j (C x - A ^ C x ) +
Substitute equation (1) in equation (6).
Rearrange the equation.
[ x ^ ( C A - A .C ) ’ -Q , ( C A - A . C ) x ] +
y = J [x ^ ( C A -A .c y Q, CBH+i(’ (C B y Q, C B (C A -A .C )x ]+ dt
[ii’' ( C B y -Q , -C B b + ii’'R ii]
Here, in term 2, both the equations are in scalar quantities and are equal. Rewrite the equation.
[ x ^ ( C A - A .C ) ’ -Q , - ( C A - A . C ) x ] +
y = [ [2 « 4 (C B y Q , C B (C A - A .C ) x ]+
(7)
[ii’' ( C B ^ -Q , •C B u + ii ’'R b ]
Now substitute the following vectors in equation (7).
Q -[(C A -A .C y Q, (CA-A.C)]
S -
Q , C B ( C A - A ,C )
R - R + B ^ C ’ Q ,C B
Rewrite equation (7).
y = J(x ’^Qx+2a^Sx+i/^Ru)<*..... (8)
Equation (8) is similar to standard form of LQR equation after adding the cross-weighting term
between the control and state form.
Therefore, it is proved that LQR performance index equation is equivalent to the standard
equation after adding the cross- weighting term In the equation.
Problem 7.64PP
Explicit Model-Following: Suppose in the LQR problem, we wish the closedloop system to
behave as close as possible to a system of the form
which represents the model of desirable dynamics. We may choose a performance index of the
form
00
7=
(a)
^ j(J r - i)’ 'Q |C y -l) + ll’ ’K«j<it.
Show that this performance index can be converted to the standard one by augmenting the
states of the plant and the model and again choose the augmented state vector, f = [xT zJ]T and
write down the system equations to show that
write down the system equations to show that
+
J= j
0
where
r C ^Q iC
' ' " L - Q iC
-C ^ Q , ]
Qi
J-
(b)
Which state variables of the system are uncontrollable? Is this result surprising?
(c)
The optimal control is of the form
H = - K ,X -K 2 Z .
which means that the model’s equations must be implemented as part of the control law.
Suppose we now drive the model as follows
i = A **+
where up may be the pilot input in an aircraft system. Show that
= - C ( j I - A + B K i)-*K 2 (*I-A *)-*B p .
Ooxd-loop ityHamics Ftet^ntwti dynamics
This indicates that the feedfonward dynamics may be used to improve the transient response of
the system.
(d)
What are the transmission zeros of the overall system?
(e)
What is a possible disadvantage of this scheme compared to the standard LQR, that is, with
no explicit model?
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 7.65PP
Consider the system with the transfer function e-TsG(s), where
C(s)
=
^
s ( i + l ) ( i + 2)
The Smith compensator for this system is given by
Piot the frequency response of the compensator for 7 = 5 and Dc(s) = 1. and draw a Bode piot
that shows the gain and phase margins of the system.24
Step-by-step solution
There is no solution to this problem yet.
G et help from a C hegg subject expert.
ASK AN EXPERT
Problem 8.01 PP
The z-transform of a discrete-time filter h(k) at a 1 Hertz sample rate is
H& =
(a)
l + (l/2 );-'
[1-(1/2)J->U1 + (1/3)Z-']'
Let u(k) and y(k) be the discrete input and output of this filter. Find a difference equation
reiating u(k) andy(k).
(b)
Find the natural frequency and damping coefficient of the filter’s poles.
(c)
Is the filter stable?
Step-by-step solution
step 1 of 5
(a).
As u (k) and y (k) are the di screte input and output of the filter,
tile transfer function
'■'I
Y (a )_
1+
TJ(z)
'+1
U(z)
Y ( z ) - lz - ‘Y ( z ) 4 a '" Y ( a ) = U ( z ) + ^ z ‘U (z )
0
Z
0
Taking inverse z-transform of both tiie sides, we get
3 , ( k ) = . ly ( k . l ) . l y ( l C . 2 ) = u ( k ) 4 - l4 k - l )
Step 2 of 5
Sampling period T=lsec
to convert H (z) to H (s ) putting
2+sT
z = ------2-sT
_ 2+s
2-s
Step 3 of 5
3(6+s)(2+s)
“ (2+3s)(S+2s)
_3(s^+8s+12)
~ 6s^+28s+16
3(6+s)(2+s)
(2+3s) (8+2s)
_ 3 ( s ’+8s+12)
6s^+28s+16
3 (s“+8s+12)
1 (s’+8s+12)
Step 4 of 5
Thus the natural frequency
L«1.6329rad/sec
A nd the Damping ratio is given by
14
2^ cd^= - j
3x2 1.6329
|t= 14289|
Step 5 of 5
Poles of the filter are
2
8=—
3
and :
2
As botii the poles are in left half plane, tiie filter is stable!
Problem 8.02PP
Use the z-transform to solve the difference equation
y (t) - 3y(* - 1) + 2y(* - 2) = 2«(* - 1 ) - 2»(* - 2),
where
K(t) =
*> o .
ib <0.
0.
y ( i) = 0 ,
*< 0 .
Step-by-step solution
step 1 of 5
Step 1 of 5
Consider the following data;
Difference equation,
;> (* ) -3 ;- (t- l) + 2 y (* - 2 ) = 2 i(( * -l) -2 » ( * - 2 )
it ^ O
fc< 0
> (k ) = 0
k<0
Step 2 of 5
Apply the z-transform to the difference equation.
y (z )-3 z-'y {z)+ 2 z-^ y { i) = 2 !-'U {z)-2 z-^U (i)
( l- 3 z - ' + 2 r-’ );>(z) = 2(z-' - z - ’l / ) ( r )
y (z )
2 ( z - '- r ^ )
U (z)
l-3 z -'+ 2 z -*
A 1.
1 - 24 - + 24
2 (^ -0
z '- 3 z + 2
_
2
2 (£ -1 )
( z - l ) ( z - 2)
...... (1)
‘ z-2
Step 3 of 5
We know that. » (
Apply z-transfonn to this condition.
( z - l)
Substitute -------T
t / f z ) in the equation (1).
' >
( z - l)
y (^ )
2
z-2
Z
( z - l) '
n o
“ ( z - 2) [ ( . - l ) '
2z
( z - 2) ( z - l ) '
Apply partial fractions to above equation.
A
B
C
Z + - + - . ----- - T ...... (2)
( z - 2 ) ( z - l ) ' ” z - 2 2 - 1 ( 2 - 1)*
2
.< ( 2 - l) '+ g ( 2 - 2 ) ( 2 - l) + C (2 -2 )
( 2 - 2 ) ( 2 - 1 ) '“
(z -2 )(2 -1 )'
2 ^ A { z - \ ) '* B ( z - 2 ) ( z - \ ) + C { z - 2 )
(3)
Step 4 of 5
Put 2 = 2 in equation (3), to obtain the constant j i .
2 = . 4 ( 2 - l) * + 5 ( 2 - 2 ) ( 2 - l ) + C ( 2 - 2 )
2^A
A=2
Put z — 1 in equation (3), to obtain the constant C ■
2 = .< (l-l) ' + a ( l - 2 ) ( l- l ) + C ( l - 2 )
2— C
C = -2
z . 2 for x ^nd _2 for C in the equation (3).
Now substitute
2 = ( 2) ( 0 - l ) ' + B ( 0 - 2 ) ( 0 - l ) + ( - 2 ) ( 0 - 2 )
2+2B +4=2
-I
B ^ -2
Step 5 of 5
Now substitute 4 for ^ . 4 for ^ and « 2 for C 'n the equation (2).
, V
2z
2z
2z
Now apply the inverse z-transfonn above equation.
y(*) =2(2‘ -l-* )
Therefore, the solved difference equation using z-transform is |y ( k ) = 2 ( 2* - l - k ) |
Problem 8.03PP
The one-sided z-transform is defined as
00
F ® = !]/( * » ■ * ■
0
(a)
Show that the one-sided transform of f(k+ ^) is Z{ f(k + 1)} = zF(z) - z f ^ ).
(b)
Use the one-sided transform to solve for the transforms of the Fibonacci numbers generated
by the difference equation u(k+2) = u(k+^)+ u(k). Let u(0) = u(^) =
[Hint: You will need to find a
general expression for the transform of f(k + 2) in terms of the transform of f[k).]
(c)
Compute the pole locations of the transform of the Fibonacci numbers.
(c) Compute the pole locations of the transform of the Fibonacci numbers.
(d) Compute the inverse transform of the Fibonacci numbers.
(e) Show that, if u(k) represents the kth Fibonacci number, then the ratio u(k + 1)/u(k) will
approach
• This is the golden ratio valued so highly by the Greeks.
Step-by-step solution
step 1 of 5
w
Find the z-transform o f / (^+ 1 )
z { /{ k + i) ] = ± / { k + i) z - ^
Z { / ( * + l ) } = 2 / ( . / ) z ^ " ‘. where i + l = J
Z [ / { k + \ ) } = z Z / O ) z - ' - : ^ f ( 0)
0
Thus, we g et |z(/(A + l)) =zF(s) -
^ ( 0 )|
Step 2 of 5
0 >)
We know that,
a ( i+ 2 ) - a ( i+ l) - a ( jt ) = 0
■We have Z [ / ( i + 2)] = I V (z) - 2 V (0) - JeT (1)
Obtain the z transform and simplify further.
z ^ U (z ) - z^u (0) - za (1) - [ z U (z) - za (0 )] - U { z ) = 0
{ z ^ - z - l)C /(^ ) =
z)a(O ) + z a (l)
Step 3 of 5
Since. a (0 ) = a ( l ) a n d a (l) = l ,
W e g e ty ( 2 ) = - j^
Thus, we get
U (2 ) =
z^- z -1
(c)
Find the poles ofthe system U (z) =
- z - 1
2
z^ - z - 1 = 0
1 ± V5
s 1.618, and ot^ = -0 .618
Thus, the poles are
= 1.618, and
-0 .6 1 8 |
Step 4 of 5
We know that U (z) = - j ---------^
We can rewrite this as Z/ (z) = ----- ^
^
.
z z
By the long division rule, we get the quotient as l-fz “*-Fz"® +z~^ +...
Hence, a ( * ) = U , 2,3,5.............
Step 5 of 5
Apply the principle o f partial fraction to U (z) =
----------.
z^ —z —1
U (z) =
( 1 - 0 0 - ^ " )
a ^ -O i
E7(z)= ^-------------I=r
T
^ ^
1 - q z '^
.
O j- c ii
----- T
l------------- O jZ ‘ ^
Finding the invers e z transform of ^ ( z ) :
u [k ) =
- o f+ —S — of
O i- O i
O i- C ^
Thus, we get
(«)
S i n c e , I < 1, for large jbthe second term is s 0 , and the ratio o f a (^+ 1 ) to
u {k )is
Problem 8.04PP
Prove the seven properties of the s-plane-to-z-plane mapping listed in Section 8.2.3.
Step-by-step solution
step 1 of 10 ^
A. z-^lane p ole at Z^Zj
at the wwnplwg
o f equivaleiit s i l a n e pole
The equivalent characteristics in tfie z-plane are related to s i l a n e b y the
Where T is the sanq)le period
The equivalent characteristics in d ie z-plane are related to s-plane b y the
Where T is d ie sanqile period
The follow ing is the m^>ping betwe en the s-plane and z-plane b y z —^
Imagioafy part o f s ila n e
Imaginary part o f Z-plaoe
step 2 of 10 ^
(1)
The stability boundary in s^^ane is given by.
5 = 0 *+jeo F<»all a> between [-<x>,oo]
The m q ^ in g in z-plane
stability gives rise to.
= e '( c o 8 « ( r + j ^ a i r )
I f c r = 0 .f lM ii« " = l
The magnitude o f z gives
[z| = ^oo8^(fi;ff')+ sin ^(ffiir)
=Ji
=1
H ence d ie poles located o n the unit circle in d ie z-plane are equivalent to the pole
location on d ie imaginaty axis in the s il a n e
Thus the unit d i c k s in z-plane rqnesent d ie stability boundary.
Step 3 of 10 ^
(2)
h i d ie sm all victnity around z = 4-1 is essentially identical to d ie v id n ity around
s = 0 in d ie s-plane.
z - ^
l= e ^
This resuh is true udien s = 0 or the sampling tim e T = 0 . as d ie sanqiling tune
cannot be considered as zero w e crmsider die v id n ity around d ie s-plane a s zero.
Hence, d ie vicinity around z = + l is identical to vidnftyaround.f = 0 .
Step 4 of 10
(3 )
The stability boundary in s^^ane is given by.
s = 0 ‘+ j& F<xall <9 between [ - oo. go]
The m q ^ in g in z-plane
stability gives rise to.
z=eT^
= ^ ( c x 3 B d r + ysmtsET)
Since, e ' i s a crmstant value, die z-plane locations g ive reqxm se information
normalized to d ie sample rale, radier dian to tim e as in s-plane
Step 5 of 10 ^
(4 )
Consider a value o f
in d ie s-plane and substitute inth erelatian o f sa n d
z-plane w e have
- . M
= (c o s n '+ y s in ;r )
=-i+yo
From d ie above expression, w e can infer diat the negative real z-axis always
represents s fiequeiKty cd' ^
in s-plane.
Step 6 of 10
(5 )
N o w considering different vertical points in d ie s-plane w e can prove diatd iese
vertical poiiits map w idiin d ie unit d i c k o f d w z-plane.
A t z = 0 , d ie m ^iping o f z-plane is,
ml
Step 7 of 10
A t z s ± y — .d ie m ^ ip in g o fz -iila iie is ,
= e
- - -
= {C 0 8 X ± J ^ X )
=-l
Thus, form d ie above pdn ts; w e can infer diat, vertical lines in the le d h a lf o f the
s^dane m ^ into circles within d ie unit ctccle o f z-plane.
Step 8 of 10 ^
(6 )
Tn
tK w iy n n ta l ling-g t i a w I w m d r a w n d i a t T n y rg a w it a r o n g t a n t t f n a g tn a f y
part s = ja>
z=e^^
s ( gos<0 ^ + y s tii< 9 r)
(
2js
= cosm— + j8tn<9T
)
L
These lines are mapped onto to the radial lines in z-planes as the phase in the
m qiping o f z-plane is.
Hence, form die above points w e can infer diat, d ie horizontal points in the aplane
radial lines in the z-^lane.
Step 9 of 10
(7)
Considering N yquist f r e q u e n t
and decreasing the value o f O ' from 0 to OD.
A t point
The m q ^ in g in z-plane ccHie^Kinding to this point is.
=
If
( coB^ + j s in x ’)
-> o o , d ie mapping in z-t>lane w ould
z= 0
Step 10 of 10
Atpoiiil s = - a „ - J ^
The m ^iping in z-plane cane^m iiding to this point is,
z= e
*
(coBJT—7 s m x ’)
0*0 ->G0 , d ie mapping in z - f lane w ould b ^
z=0
I f G)g is in c ie a s e d ,d ie p o k w illb e sh ifie d a lo n g d ie p o s itiv e r e a I a x is in d ie z fd a iie to w a r d sth e p o in tz = l. I f
is decreased, the pole w ill b e shifted aknig the
z-plane towards die p oin tz = 0 .
I f d ie sampling rate is small, each period w ill be sanipled less disn tw ice die
average, the sanqiling rate is
or greater than Nyquist frequency diere w in
be over lap.
Problem 8.05PP
A unity feedback system has an open-loop transfer function given by
250
The following lag compensator added in series with the plant yields a phase margin of 50°;
s/1.25 + 1
D c (s ) =
50s+ l
Using the matched pole-zero approximation, determine an equivalent digital realization of this
compensator.
Step-by-step solution
Step-by-step solution
step 1 of 1
The
compensator transfer function is given by
” r-l
D (z>, = i2 5 _
50s-l
s-1.25
i," 50 ;
The MPZ iq>proximation is
1. z-c’ '
At low frequency
ency
1 (-1.25)
HI
or
K,
1.25x50
I.
=K,
1-e®
Thus the digital compensator transfer function is
I
,
7
M 7
^
1-e®
|z .e “ ” |
Problem 8.06PP
The following transfer function is a lead network designed to add about 60° of phase at o;1 = 3
rad/sec:
H(.s) =
s+l
O .ls + l
(a) Assume a sampling period of T= 0.25 sec, and compute and plot in the z-plane the poie and
zero iocations of the digital implementations o1 H(s) obtained using (1) Justin’s method and (2)
pole-zero mapping. For each case, compute the amount of phase lead provided by the network
at z 1 = ej(jj1 T.
(b) Using a log-log scale for the frequency rangeo; = 0.1 tocu =100 rad/sec, plot the magnitude
Bode plots for each of the equivalent digital systems you found in part (a), and compare with
H(s). {Hint Magnitude Bode plots are given by \H(z)\ = \H(eja)T)\.)
(b) Using a log-log scale for the frequency ranges = 0.1 tocu =100 rad/sec, plot the magnitude
Bode plots for each of the equivalent digital systems you found in part (a), and compare with
H(s). {Hint Magnitude Bode plots are given by \H(z)\ = \H(eja)T)\.)
Step-by-step solution
Step 1 of 7
Given that
s+l
O.ls+l
and T=0.25sec
Step 2 of 7
(1)
Tustin's Method
2
3
” 0.25 [ l + z ‘
Step 3 of 7
Therefore,
# )* '
H (z)^
....
..(1)
'■ ' 1.8+0.2z‘
Thus the poles of H (z) are
1.8-+O.2z'^=0
z ‘= H
0.2
2.=-0.111
and zero is: 9-7z’^=0
2^=1=0.778
Step 4 of 7
Step 5 of 7
(2)
Matched pole zero method
H ,( s ) = .
s+l
Step 6 of 7 ^
Let the digital i^proximation of
(s) be Hp (z ) , then
H.
here T ^ .2 5 sc c
d
is given by
0+1
^
o+ioj
=lx
1-e-'"
1
K.
: 4.1497
T h « e fo « H ( z ) = K . l^
vAere, o ^ ’** and P=e**
-O T7Q••1^
H(z)=4.1497
/
(l-0.082z‘^)
4.1497
(z-0.778)
(z-0.082)
Step 7 of 7
lm(z)
Problem 8.07PP
The following transfer function is a lag network designed to introduce a gain attenuation of
10(-20 db) at w = 3 rad/sec:
H{s) =
(a)
lOs+l
100s+ 1
Assume a sampling period of T= 0.25 sec, and compute and plot in the z-plane the poie and
zero iocations of the digital implementations o1 H(s) obtained using (1) Justin’s method and (2)
pole-zero mapping. For each case, compute the amount of gain attenuation provided by the
network at z1 = eyojl T .
(b)
For each of the equivalent digital systems in part (a), plot the Bode magnitude curves over th
frequency range w = 0.01 to 10 rad/sec.
(b)
For each of the equivalent digital systems in part (a), plot the Bode magnitude curves over th
frequency range w = 0.01 to 10 rad/sec.
Step-by-step solution
step 1 of 2
w
Let
H {s y -
lOff+l
lOOs+1
|/ / p ® ) | = ® = 3 = 0 .1 0 0 1 (-2 i^ )
Tustin's method
/ / ( z ) = 0.10112.-
-0.97531
+ 0.99750
Step 2 of 2
(b)
All there are es sentially the same and indistinguishable on the plot because the
range o f intere st is below the half s an^ le frequency
Problem 8.08PP
For the system shown in Fig, find values for K, TD, and 7/ so that the closed-loop poles satisfy ^
> 0.5 and wn > 1 rad/sec. Discretize the PID controller using
(a)
Tustin’s method
(b)
The matched pole-zero method
Use Matlab to simulate the step response of each of these digital implementations for sample
times of 7 = 1.0.1. and 0.01 sec.
Figure Control system
Step-by-step solution
step 1 of 10
Write the plant transfer function.
1
<?w =-
5 ( 5 + 1)
Continuous PID controller:
Evaluate the characteristic equation.
l + D ( j) G ( i) = 0
l + Xf--------ii._L_=o
j(j+ l)
* ’ + ( l+ A T o ) s ’ + « j + — . 0
Step 2 of 10
Consider the required specifications.
®.>i—
>0.5
sec
Select the dominant closed loop poles to exceed the specifications.
Consider s = - 0 .8 ± J .
Calculate the values of
^ .
= 1 .2 $ 2 !t
sec
Compare » = - 0 .8 ± y with
(ta , = 0 . i
^
1.28
*0.625
Therefore, the required specifications are met with s ~ -O .S ±J.
step 3 of 10
Evaluate the characteristic equation at » = - 0 .8 ± y
^1.888 - 0.36(1 + ATj,) - 0.8A:+ + ( 0 . 9 2 -1.6(1 + CTj ) + 8C)y = 0
Equate the real and imaginary parts to zero.
1.888 - 0.36(1+ATo ) - 0 .8 8 :+ — = 0
2|
1.528-(0.367„+ 0 . 8 ) 8 : + ^ = 0
‘t
1.528 = ^ 0 . 3 6 r „ + 0 . 8 - ^ j 8 : ...... (1)
o .9 2 - i.6 ( i+ 8 :r „ ) + 8 := o
0.92 + 8:
1.6
1+8T„
(2)
Arbitrarily choose the value, 7} ■ 10.
Solve equations (1) and (2) by substituting 7} s 10.
Therefore,
a:
= 1.817,71, = 0.3912
Therefore, the transfer function of continuous PID controller is.
D (5 ) = 1 .8 1 7 ^ 1 + 0 .3 9 1 2 s + ^ j
Step 4 of 10
(a)
Discretize the PID controller using Tustin’s method.
Substitute —[-i—
for every occurrence of s in Z)(5).
r ( ^ i+ 2 " 'J
'''
= 1.817 1+ 0'■
. 3 9 1 2 [|( ) ) h
10
T
( . l+ z " 'J
l- z " '
1 . 8 1 7 ( l- z - " ) + l : ^ ( l - 2 z - '+ z - " ) + 0 . 0 5 r ( l + 2 z - ' + z-*)
l-z -'
C (r) =
. 1 :^ 6 .
1 . 8 1 7 + i^ ^ + 0 . 0 5 7 - + f - ^ ® ^ ^ ^
T
1 . 1 7 - jz - '+ ^ i; ^ + 0 . 0 5 r - 1 . 8 1 7 j r
+ 0.1
Step 5 of 10
Calculate the expression D ( z ) at r = i , r = o . i , r = o .o i sec.
3.2886 - 2.7432Z-' - 0 .3 4 5 4 z ‘'
l- z - "
16.042 - 28.422z-' +12.404z-“
143.97 - 284.32z-' + 140.345Z-*
1 -Z -"
Step 6 of 10
(b)
Observe the PID controller transfer function
G ( z ) = 8 :( i+ v + ^ )
There is one more zero than pole.
Therefore, for the matched pole zero approximation add a pole at 2 s -1 .
Map poles and zeros according to the relation z —
(z + l)(z -l)
There is no DC gain for this transfer function D ( z ) .Therefore, match the gain of D ( z ) at
5 = y<v^(closed loop natural frequency).
(z + l)(z -l)
N -1—
4< = |cos(2ffl,r)-(e~^+e"^)co s(iD ;,7’)+ e '^ * * ^ j{c o s (2 ffl|,r)-l}4
jsin (2fii',r)
+ « "^ )s in
sin (2fl),r)
B = {c o s ( 2 a i;r )- (e “^ + e “^ ) c o s ( fi> ,r ) + e '^ * * ^ } s in ( 2 ffl,r ) +
{s m (2 < » ,r)-(c “^ + < “^ )s m ((» ,7 ')J {c o 8 (2 a > ^ )-l}
Step 7 of 10
Equate the DC gains.
D ( z ) = K , ^ -------------^
--------------- >-
Calculati
Calculate the expression Z )(z) at r = l, r = 0.1, 7 = 0.01 sec.
3.2886 - 2.74325-' -0.34542-*
l-z -'
16.042 - 28.422z-' + 12.404z''
l-z
143.97 - 284.32z-' +140.345Z-'
1 -z ^
Step 8 of 10
Sketch the step response for each of the digital implementations at 7 = 1, 7 = 0.1, 7 = 0.01 sec
using MATLAB.
Write a MATLAB code for step response at 7 = 1sec •
num=[3.2886 -2.7432 -0.3454];den=[1 0 -1];sys=tf(num,den,1)
sysd=feedback(sys,1)
step(sysd)
Step response for j* = 1 is shown in Figure 1.
Step 9 of 10
Write a MATLAB code for step response at 7 = 0.1 sec •
num=[16.042 -28.422 12.404];den=[1 0 -1];sys=tf(num,den,0.1)sysd=feedback(sys,1)step(sysd)
Step response for 7 = 0.1 >s shown in Figure 2.
Figure 2
Step 10 of 10
Write a MATLAB code for step response at 7 = 0.01 sec.
num=[143.97-284.32 140.345];den=[1 0
-1];sys=tf(num,den,0.01)sysd=feedback(sys,1)step{sysd)
Step response for 7 = 0.01 is shown in Figure 3.
S t ^ Response
Figures
Problem 8.09PP
Consider the system configuration shown in Fig. where.
40(j + 2)
G(i) = ( i+ 1 0 ) ( P - 1 .4 ) ‘
(a)
Find the transfer functionGfzJ for r = 1 assuming the system is preceded by a ZOH.
(b)
Use Matlab to draw the root locus of the system with respect to K.
(c)
What is the range of K for which the closed-loop system is stable? Compare your results with
the case in which an analog controller is used (that is, where the sampling switch is always
closed). Which system has a larger allowable value of K7
the case in which an analog controller is used (that is, where the sampling switch is always
closed). Which system has a larger allowable value of K7
(d)
Use Matlab to compute the step response of both the continuous and discrete systems with K
chosen to yield a damping factor of ^ = 0.5 for the continuous case.
Figure Control system
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 8.1 OPP
Single-Axis Satellite Attitude Control: Satellites often require attitude control for proper orientation
of antennas and sensors with respect to Earth. Figure 2 shows a communication satellite with a
three-axis attitude-control system. To gain Insight into the three-axis problem, we often consider
one axis at a time. Figure 1 depicts this case, where motion is allowed only about an axis
perpendicular to the page. The equations of motion of the system are given by
I6 = Mc + M
d
,
where
/ = moment of inertia of the satellite about its mass center,
MC = control torque applied by the thrusters,
MD = disturbance torques.
Figure 1 Satellite control schematic
MD = disturbance torques.
Figure 1 Satellite control schematic
6 = angle of the satellite axis with respect to an inertial reference with no angular acceleration.
We normalize the equations of motion by defining
u =
M e
—
,
Wd =
M d
— ,
and obtain
e = u+WdTaking the Laplace transfonn yields
1
^ [H ( s ) + W d ( s ) l,
e {s ) =
which, without disturbance, becomes
In the discrete case in which u is applied through a ZOH, we can use the methods described In
this chapter to obtain the discrete transfer function
r
«(z)
2
z+l
1
L f e - D ^ J '
(a)
Sketch the root locus of this system by hand, assuming proportional control.
(b)
Draw the root locus using Matlab to verify the hand sketch.
(c)
Add discrete velocity feedback to your controller so that the dominant poles correspond to ^
0.5 and wn = ZnAOT.
(d)
What is the feedback gain if T= 1 sec? If 7 = 2 sec?
(e)
Plot the closed-loop step response and the associated control time history for 7 = 1 sec.
Figure 2 Communications satellite Source: Courtesy Space Systems/Loral (SSL)
Step-by-step solution
step ^ of A ^
(a)
The loci braches depart vertically &om z s 1
Step 2 of 4
y-» I- \ T^(^+l)
step 3 of 4
(c)
\ k(z~ 0.63)
D ( z ) = - ^ -------- ^
(z+0.44)
z*0.44±0.44y,-0.113
Step 4 of 4
(d)
k=
0.692
1.383 for
0.3458 for
={
r= ls e c
7 - 2 sec
Problem 8.11 PP
It is possible to suspend a mass of magnetic material by means of an electromagnet whose
current is controlled by the position of the mass (Woodson and Melcher, 1968). The schematic of
a possible setup is shown in Fig.1, and a photo of a working system at Stanford University is
shown in Fig. 2. The equations of motion are
where the force on the ball due to the electromagnet is given by f(x, I). At equilibrium the magnet
force balances the gravity force. Suppose we let 10 represent the current at equilibrium. If we
write 1 = 10 + /. where / represents a deviation from the nominal current, 10. expand f about x = 0
and / = 10. and neglect higher-order terms, we obtain the linearized equation
Reasonable values for the constants in Eq. (8.54) are m = 0.02 kg, lc\ = 20 N/m, and K2 = 0.4
N/A.
Reasonable values for the constants in Eq. (8.54) are m = 0.02 kg, IC\ = 20 N/m. and K2 = 0.4
N/A.
(a) Compute the transfer function from / to x, and draw the (continuous) root locus for the simple
feedback / = -Kx.
(b)
Assume that the input is passed through a ZOH, and let the sampling period be 0.02 sec.
Compute the transfer function of the equivalent discrete-time plant.
(c)
Design a digital control for the magnetic levitation device so that the closed-loop system
meets the following specifications: fr< 0.1 sec. te< 0.4 sec, and overshoots 20%.
(d)
Plot a root locus with respect to /(I for your design, and discuss the possibility of using your
closed-loop system to balance balls of various masses.
(e)
Plot the step response of your design to an initial disturbance displacement on the ball, and
show both X and the control current /. If the sensor can measure x only over a range of ±1/4 cm
and the amplifier can provide a current of only 1 A, what is the maximum displacement possible
for control, neglecting the nonlinear terms in f(x. 1)7
Figure 1 Schematic of magnetic levitation device for Problem 8.11
o
Figure 2 Magnetic ball levitator used in the laboratory Source: Photo courtesy o f Gene Franklin
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 8 .1 2PP
Repeat Problem in Chapter 5 by constructing discrete root loci and performing the designs
directly in the z-plane. Assume that the output y is sampled, the input u is passed through a ZOH
as it enters the plant, and the sample rate is 15 Hz.
Problem
A servomechanism position control has the plant transfer function
10
f( j+ l) ( f+ l( ^
G(i) = -
You are to design a series compensation transfer function Dc(s) in the unity
feedback configuration to meet the following closed-loop specifications:
• Theresponse to a reference stepinput is to have no more than 16% overshoot.
• Theresponse to a reference stepinput is to have a rise time of no more than 0.4 sec.
• Theresponse to a reference stepinput is to have no more than 16% overshoot.
• Theresponse to a reference stepinput is to have a rise time of no more than 0.4 sec.
• The steady-state emor to a unit ramp at the reference input must be less than 0.05.
(a) Design a lead compensation that will cause the system to meet the dynamic response
specifications, ignoring the error requirement.
(b) What is the velocity constant Kv for your design? Does it meet the emor specification?
(c)
Design a lag compensation to be used in series with the lead you have designed to cause the
system to meet the steady-state error specification.
(d)
Give the Matlab plot of the root locus of your final design.
(e)
Give the Matlab response of your final design to a reference step.
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 8.13PP
Design a digital controller for the antenna servo system shown in Figs.1 and 2 and described in
Problem. The design should provide a step response with an overshoot of less than 10% and a
rise time of less than 80 sec.
(a) What should the sample rate be?
(b) Use discrete equivalent design with the matched pole-zero method.
(c)
Use discrete design and the z-plane root locus.
Figure 1 Satellite-tracking antenna Source: Courtesy Space Systems/Loral
ve: uounesy apace aysiems/Lorai
Problem 3.36
You wish to control the elevation of the satellite-tracking antenna shown in Fig.1 and Fig.2. The
antenna and drive parts have a moment of inertia J and a damping B; these arise to some
extent from bearing and aerodynamic friction, but mostly from the back emf of the DC drive
motor. The equations of motion are
J0 + B$=Te,
where Tc is the torque from the drive motor. Assume that
J = 600,000 kg m2 B = 20,000 N m sec.
(a) Find the transfer function between the applied torque Tc and the antenna angle 6 .
(b) Suppose the applied torque is computed so that 9 tracks a reference command 9r according
to the feedback law
Tc = K ( d r -d ),
where K is the feedback gain. Find the transfer function between 0rand 6 .
(c)
What is the maximum value of K that can be used if you wish to have an overshoot Mp <
10%?
(d) What values of K will provide a rise time of less than 80 sec? (Ignore the Mp constraint.)
(e) Use Matlab to plot the step response of the antenna system tor K = 200,400,1000, and
2000. Find the overshoot and rise time of the four step responses by examining your plots. Do
the plots to confirm your calculations in parts (c) and (d)?
Figure 1 Satellite-tracking antenna Source: Courtesy Space Systems/Loral
S te p -b y -s te p s o lu tio n
Step 1 of 2
(a)
The equation o f m otion is
"Where J = 600000 kgm^
B = 20000N.m.sec
I f we define
B
1
7!
a * —* — ,« * —
/
30
B
We get
u (s )
s(30ff-l-l)
From the specifications
^ > 0 .5 4
(21^ >0.0225
Step 2 of 2
(b)
u ( s ) = * ( 5 ( f i) - 5 ( f i) )
s = -0.0167±0.0205^
The corresponding natural frequency and damping
» , = 0.0265,^'= 0.6299
Problem 8.14PP
The system
1
G(*) =
■ (i+ 0 .1 )(* + 3 )
is to be controlled with a digital controller having a sampling period of 7 = 0.1 sec. Using a zplane root locus, design compensation that will respond to a step with a rise time tr< 1 sec and
an overshoot Mp < 5%. What can be done to reduce the steady-state error?
Step-by-step solution
step 1 of 1
Step 1 of 1
Continuous plant
0 W = (s+0.1)(s+3)
’
,
___ __
■
z+0.9019
^ ( z - 0 .7 4 0 8 ) ( z -0 .9 9 )
^ Is e c
^ 1 .8 rad/sec
z = 0.8564±0.1278j
= 2.07rad/sec,^=0.70
D (z ) = i _ 5 _
^ ^ z -1
Problem 8.15PP
The discrete transfer function for pure derivative control is
D d (z )= K T D ^ -j^,
where the pole at z = 0 adds some destabilizing phase lag. Can this phase lag be removed by
using derivative control of the form
D4(0 = KTd ^ ^ ^ 1
Support your answer with the difference equation that wouid be required and discuss the
requirements to implement it.
Step-by-step solution
Step-by-step solution
step 1 of 1 '
(a)
No we can't use derivatiTe control of the form
To remove the phase lag
The difference equation corresponding to
Z)(z) = * ,7 i
2- \
a (z )
T
E{z)
e ( A + l ) - « (jt)
Problem 9.01 PP
Figure shows a simple pendulum system in which a cord is wrapped around a fixed cylinder. The
motion of the system that results is described by the differential equation
( i+ R 9 ) e + g ^ 9 + t i0 ^ = 0,
where
/= length of the cord in the vertical (down) position.
R = radius of the cylinder.
Figure Motion of cord wrapped around a fixed cylinder
(a) Write the state-variable equations for this system.
(b)
Linearize the equation around the point 9 = 0 , and show that for small values of 6 , the system
equation reduces to an equation for a simple pendulum—that is.
9 -\-(g /i) = 0.
Step-by-step solution
step 1 of 4
Sketch the given simple pendulum S3rstem in which a cord is w n ^ e d around a fixed cylinder.
Step 2 of 4
The motion of the system that results is des cribed by the differential equation
{ l+ R 0 ) e + g s m 0 + R ^ = O
Where / is the length in the vertical position,
is the radius o f the cylinder.
Step 3 of 4
w
This is a second order non linear equation in 9,
Let3c = ^5
Find Xi, and
- ( / ? ^ + g s in ^ J
1+R9
'
gsin
; + Aca
x^=9
= ^1
Thus, the state variable equations for this system are
-(/Z x j+ g s in x a )
/ + Rx2
Step 4 of 4
0 >)
Linearize the equation around the point ^ = 0.
For small values o i9 . we get (/ + R0) = I
Simplify further.
(/ + R9) s
/ s in ^ = d
^= 0
We can rewrite this as
+ g d = 0.
Thus, it is proved that ^ + • ^ ^ = 0
I
Problem 9.02PP
The circuit shown in Fig has a nonlinear conductance G such that iG = g(vG) = vG(vG - 1){vG 4). The state difTerential equations are
di
- = - . + v.
Figure Nonlinear circuit
R=l
dv
= - i + g ( ii- v ) .
where / and v are the state variables and u is the input.
(a) One equilibrium state occurs when i/ = 1. yielding /1 = v1 = 0. Find the other two pairs of v
and /that will produce equilibrium.
(b) Find the linearized model of the system about the equilibrium point u = 1. /1 = v1 = 0.
(c) Find the linearized models about the other two equilibrium points.
Step-by-step solution
step 1 of 5
Refer to Figure 9.57 in the textbook.
The current flowing through the resistor is,
'o - s C v o )
The differential equations are,
^
= -/ + g (a -v )
Where,
i and v are the state variables
If is the input
Step 2 of 5
(a)
At equilibrium state,
4 -0
- i + v= 0
( 1)
^ = 0
dt
- i + g (tt-v ) = 0
(2)
( v « - l ) ( v o - 4 ) for ig in equation (1).
Substitute
From Figure 9.7 in the text book,
V c - lf - V
Substitute i f - v for vg in equation (3).
- ( if- v ) ( if- v - l) ( if- v - 4 ) + v = 0
Substitute 1 for If
- ( l- v ) ( l- v - l) ( l- v - 4 ) + v = 0
- (l-v ) ( -v ) (- 3 - v ) + v = 0
v(v* + 2v - 2 ) = 0 ...... (4)
Solve the equation (4).
Therefore,
v = 0 ,-l± > /3
From equation (1),
- i+ v = 0
i= v
1= o , - \ ± S
Therefore, the pairs of y and i ^^at will produce equilibrium is,
v = 0 ,- l± S
i = 0 ,- l± y /3
Step 3 of 5
^if, Svy a nd Si ■
leplace u, v, and / by 1
Tom the differential equations.
Si = - S i +
...... (5)
= -S i + g (l + Su - 5 v ) ......(6)
s -S i + ^ ( l + ^ if - ^ v ) further.
Simplify the expression
= -g i +
- ^ v ) [ ( l + Su - ^ v ) - l ] [ ( l + Su - tfv ) - 4 ]
= -S i + (1 + 5if - Sv)[Su - Sv){Su -
- 3)
^ v s - A + 3 ^ if + 3 ^ v ...... (7)
Vrite the equations (5) and (7) in matrix form.
'hus, the linearized model of the system at equilibrium point is
Su
'F '-
-1
Step 4 of 5
(c)
Write the general linearized model equation.
-1
I ■
O'
[s n
-1
Su
^
dv. W * " .dv.
When u = 1,
« ( " .v ) = « ( l.v )
= v ^ + 2 v -2
Apply partial differentiation.
^
= v’ + 2 v - 2 + v(2v + 2)
dv
= 3 v*+ 4 v -2
Substitute -1 ±
for v .
^^3(-i±S)\A{-X±S)-2
- 5 t 2 -^
Step 5 of 5
It is know that,
g g (« - v)
( 8)
-g '{ u - v)
And,
( " - *')
(9)
g '{u -v )
Similarly,
^
dv
=5w 2S
And,
du
dv
m -5 ± 2 S
The linearized model about the other two equilibrium points is.
4 1 ■' i M i .
*
=
-1
m
Su
u
5 T 2 a^ J L ^ vJ ’ ' [ - 5 ± 2> 5J
Thus, the linearized model about the other two equilibrium points is.
d[Sn
■-1
1
-1
5 T 2 ,^
r^ /1
U J"
0
^If
-5 ± 2 ^3
Problem 9.03PP
Consider the circuit shown in Fig; u^ and U2 are voltage and current sources, respectively, and
R^ and R2 are nonlinear resistors with the following
characteristics;
Resistor 1: i*i = G (V |) = V j,
Resistor 2: v2 = r(i2).
Figure A nonlinear circuit
Here the function ris defined in Fig. 2.
Figure 2 Nonlinear resistance
(a) Show that the circuit equations can be written as
i | =C<*i
+
^ « j t i - jcs- H q)Suppose we have a constant voltage source of 1 volt at u^ and a constant current source of 27
amps (that is, iif = i. ^ = 27). Find the equilibrium
state
= [ •*!• ■*2* ^
for the circuit. For a particular input uo. an equilibrium state of the
system is defined to be any constant state vector whose elements satisfy the relation
i| = i2 = i3 = 0 .
Consequently, any system started in one of its equilibrium states will remain there indefinitely
until a different input is applied.
(b) Due to disturbances, the initial state (capacitance, voltages, and inductor current) is slightly
different from the equilibrium and so are the independent sources; that is,
|| ( 0 = I ^ + <U(().
i( W = A < b ) + ii( ib ) .
Do a small-signal analysis of the network about the equilibrium found in (a), displaying the
equations in the form
+fai!ct +fasi3 +giSui +gisii2.
iii
(c)
Draw the circuit diagram that corresponds to the linearized model. Give the values of the
elements.
Step-by-step solution
step 1 of 9
(a)
Refer Figure 9.58 in the textbook.
Apply KVL in the given circuit and the corresponding equation is given below,
< , = G ( » ,- * ,)
(1)
= C -^ x ,
(2)
(3)
1. = ^ ...... m
<5 )
Where,
L is the inductance,
C is the capacitance,
ti|and
represented as voltage and current source,
^and ^ is represented as loop currents.
Refer Figure 9.59 in the textbook.
Consider the following equation for the voltage v, •
V, = 1 ^ .......(6)
Step 2 of 9
Substitute equation (1) and equation (4) in equation (2).
G { a ,- x ,) - x ,+ u ^ = C ^ x ,
C S |= G ( ii | - ; i [ i ) - j ^ + i i,
(7)
Substitute equation (4) in equation (3).
C x j - x , ...... (8)
Substitute equation (6) in equation (5).
x ,- x ^ - r i^ = L ^ I ^
(9)
Substitute equation (4) in equation (9).
x ,-x ,-n c , = L — x,
L x ,= x ,- x ^ - r ( x ,)
Substitute
(10)
and
s i in equation (7), equation (8) and equation (10).
i,= C { u ,- x ,) - x ,+ u ,
i,= x ,
x ,= x , - x , - r ( x ,)
Step 3 of 9
Thus, the circuit equation is |i^ = G ( l l | - j | ) - l ^ - n t , | . | j , = 3^ | and | i , =
- g ,- r ( jl^ ) |
Step 4 of 9
Consider the following equation to form the Equilibrium state equation.
......... ( 11)
....... ( 12)
x ,= 4 -4 -r(:^ )
Substitute iij s
s
(13)
s
0, u^ = l and
* 2 7 in equation (11), equation (12) and equation
(13).
0 = G ( 1 - * ;') - ; ^ " + 2 7 .......(14)
0 = ; ^ .......(15)
0 = x , * - ; 4 - r ( 0 ) .......(16)
Consider the following equation for the G.
G = ( l- I* f
(17)
Substitute
« 0 and equation (17) in equation (14).
( l - i |') ’ ( l- x * )- 0 + 2 7 = 0
( l - * ” )’ + 27 = 0
(18)
The roots of the Equation (18) are 4 and ^ ( - I± 3 iV 3 ^ Therefore, the value of
is 4.
Step 5 of 9
From Figure 9.59 in the text book, the value of r( 0 ) is 2.
Substitute
s 4 and r ( 0 ) = 2 in equation (16).
0 = 4 -;^ -2
4
=2
Consider the foiiowing formuia for the Equiiibrium state fomi.
x *= [x *
X*
x j ] ''
(19)
step 6 of 9 '
Substitute
x‘
s
4,
s
2 and
s 0 in equation (19)
=[4 2 of
2 O f I-
Thus, the Equilibrium state form is
Step 7 of 9
(b)
Consider the following equation from Equilibrium state equation.
....... (20)
S i k j^ ji i+ S x ^ ...... (21)
& ^ = ( x | * + 5 x , ) - ( t ^ + 5 x , ) - r ( j^ + t f a ^ )
Substitute
-
(22)
U2 -2 7 < 3^ - A, 3^ - 2 and
s Q in equation (20), equation (21)and
equation (22).
= G { - 3 + S u ^-S3 c^)-S3C2 + 2 7 + S u2 ...... (23)
6 X2 » SXy
- 2 - S x 2 - { 2 + S x2)
= SX f-S X 2 -S X 2
Consider the following equation for the G.
G = ((-3 + * ,-< 5 x ,))'
Substitute
(26)
« 0 and equation (26) in equation (23).
Sx^ -{- 3 + S U f-S 3 c ^ f - 6 x2 + 2 1 +S u 2
= - 2 7 ^ x j - 63C2 + 2 7 S u ^ + 6 U 2
Thus, the resultant equation is | ^ i, = -276x^ - 6 x j+ 2 7 6 u , + 6 U21. | ^ ^ = ^ ] a n d
Step 8 of 9
(c)
Determine the values of the following elements.
->
L = \H
Refer Figure 9.59 in the text book and find
value.
/ j. = ^
"2
“ 27
Thus, the values of the elements are calculated.
Draw the linear circuit using the values of the elements.
The linear circuit model is shown in Figure 1.
Step 9 of 9
Thus, the linear circuit model is drawn and it is shown in Figure 1.
9.04PP
Problem 9.04PP
Consider the nonlinear system
jfc= —
x( 0) =l .
(a) Assume uo = 0 and solve forxo(fj.
(b) Find the linearized model about the nominal solution In part (a).
(b) Find the linearized model about the nominal solution In part (a).
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 9.05PP
Linearizing effect of feedback: We have seen that feedback can reduce the sensitivity of the
input-output transfer function with respect to changes in the plant transfer function, and reduce
the effects of a disturbance acting on the plant. In this problem we explore another beneficial
property of feedback: It can make the input-output response more linear than the open-loop
response of the plant alone. For simplicity, let us ignore all the dynamics of the plant and assume
that the plant is described by the static nonlinearity
« < 1,
y «) =
a f i,
II > 1 .
(a)Suppose we use proportional feedback
u(t) = r(t) + a(r(t)-y(t)),
where a > 0 is the feedback gain. Find an expression for y(t) as a function of r(t) for the closedloop system. {This function is called the nonlinear characteristic at the system.) Sketch the
u(tf= r(t) + a(r(t) - y(t)).
where a > 0 is the feedback gain. Find an expression for y(t) as a function of r(t) for the closedloop system. (This function is called the nonlinear characteristic ot the system.) Sketch the
nonlinear transfer characteristic for a = 0 (which is really open loop), a = 1, and a = 2 .
(b) Suppose we use integral control:
t
h( 0
= r(t) + J ( r { r ) - y ( T ) ) d r .
The closed-loop system is therefore nonlinear and dynamic. Show that if r(t) is a constant, say r,
then lim y ( /) = r . Thus, the integral control makes the steady-state transfer characteristic of
/-►oo'
the closed-loop system exactly linear. Can the closed-loop system be described by a transfer
function from rto y?
Step-by-step solution
>1 of11 ^
Given that the plant is described by the static non linearity as
u, u £ 1
u +1
-.u > \
^ (0 =
w
The given proportional feedback is « (i) = r ( i ) +
” ,y (0 ] •
Here, a ^ 0 is the feedback gain.
Step 2 of 11 ^
The nonlinear system with saturation proportional control is sketched.
Find,y for u ^ 1.
y =r+ a { r - y )
y = { l- a ) y
y= (l-ii)r
Thus, we get
y - r - u
Obtain y for u > 1
u + \
Simplify the eg ression further.
2y s \ + r + a r - £^,
(2+ a)^ = l + (l+ a )r
y
l + ( l+ a )r
= -------i --------- i—
2+a
Step 4 of 11
Sketch the response of the open-loop sjrstem v/s the closed-loop system
2h
Find the eg ressio n for,y when
3
and 2.
Thus we get
I
y\
1
2
, = - + —/
3
3
4
4
Step 6 Of 11 ^
Sketch the open-loop v/s closed loop gri^h for a s 0,1, and 2.
Open loop
closed loop lor winous a
a=^
z
y
y'
X'
z
z
(b)
Sketch the non-linear system with saturation inte^'al control
output.
)
I
V
J in p u t
step 8 of 11 ^
Given that
The closed loop system is non-linear and dynamic.
Thus, we getu = r -I-
,
where r is a constant
Findj', w hena < 1.
y = r^ \[{r-y )
R -Y
(l + ff)r = ( l + s ) 5
Thus, we get Y= R.
Step 10 of 11 >
In stable condition,y stays bounded. Hence, y
Hence, we get ^ = r +
(r - y j) d t
y^-
oo, if y ^ * r .
The e^»ression infers that y^ = r .
Find
y -^ y
y =r- y
y -^ y -r
Find,y w h e n a > 1.
We know that =j^a).
Find X and y .
y - - \ ir - y )
Step 11 of 11
From the equation we get 2y
y ^ r.
Hence, on further simplification we obtain the e^sression —^
R (s)
Therefore, in steady-state condition.^ = r.
Thus, we get lim
= lim r ( t ) .
So, we infer that lim .y (t) = r , since, r is a constant
^— ,
2s + 1
Problem 9.06PP
This problem shows that linearization does not always work. Consider the system
X = w ?,
Jt(0)
0.
(a) Find the equilibrium point and solve toxx(t).
(b) Assume that a = 1. Is the linearized model a valid representation of the system?
(c) Assume that a = -1 . Is the linearized model a valid representation of the system?
Step-by-step solution
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 9.07PP
Consider the object moving in a straight line with constant velocity shown in Fig. The only
available measurement is the range to the object. The system equations are
where
z = constant,
X = constant = vO,
r = V P T ? .
Derive a linear model for this system.
Figure Diagram of the moving object
Derive a linear model for this system.
Figure Diagram of the moving object
f2
Object
Step-by-step solution
step 1 of 3
Consider the object moving in a straight line with constant velocity as shown in the given
figure.
Sketch the figure given.
Step 2 of 3
0
1 0
0 0 0
The system equations are
0
0
0
where z = constant,
X = constant = Vq,
r =
^ .
The output o f this system can be written as
= r,
vAere r = A(x)
Find R
6 r = — (5x
dx.
Q r= H 6x
H
[ax
dv
ds\
Step 3 of 3
Thus, we get
« ■ [ ? • ;] ■
We know that
± = Fx.
Therefore, we get Sr =
-
0
-
Problem 9.08PP
Consider the third-order system shown in Fig.
(a) Sketch the root locus for this system with respect to K, showing your calculations for the
asymptote angles, departure angles, and so on.
(b) Using graphical techniques, locate carefully the point at which the locus crosses the
imaginary axis. What is the value of K at that point?
(c) Assume that, due to some unknown mechanism, the amplifier output is given by the following
saturation non-linearity (instead of by a proportional gain K):
i«i < 1.
I«l < 1.
» = {
« > 1,
1.
-1 ,
e < - l.
Qualitatively describe how you would expect the system to respond to a unit-step input.
Figure Control system
-O Y
Step-by-step solution
step 1 of 9
(a)
Refer Figure 9.61 in the textbook.
Consider the following formula for the general fonn of characteristics equation.
l + A:G(5)ff(j) = 0 ..... (1)
Substitute ( f ^ l) _ f o r G (j)and1 for f f ( j) in Equation (1).
i+ jc ii^ = o
(2)
Consider the following formula for the roots of the general form of an equation by the root locus
method.
...... (3)
D is )
Where,
The roots of iV( 5) = 0are called the zeros of the problem.
The roots of D{s) = 0 are the poles.
Step 2 of 9
Step 1:
Consider the number of poles and zeros from the characteristics equation.
Compare Equation (2) and Equation (3).
To find zeros put numerator JV(j) = 0
( j+ l) ’ =0
Thus, the two zeros are -1 and -1.
To find poles put denominator D(s) = 0 •
*•= 0
Thus, the three poles are 0 ,0 and 0.
Step 2:
Consider the formula for the number of paths
N um ber o f path s N um ber o f poles
»3
Step 3:
Consider the formula for the angle of asymptotes.
^
I8 0 * + 3 6 (r ( /- l)
■(4)
n —m
Where,
Number of poles is n
Number of zeros is m
Substitute 1 for /, 3 for n and 2 for m in equation (4).
1 8 0 °+ 3 6 0 » (1 -1 )
3 -2
= 180*
Thus, the ahgle of asymptotes is
.
Step 3 of 9
Step 4:
Consider the formula for the centre of asymptotes.
(sum o f finite p o Ie s )-(s u m o f finite zeros)
(n u m b e ro f finite poles )-(nunU >er o f finite zeros)
( O ) - ( - l- l)
■
(3 )-(2 )
= 2
Thus, the centre of asymptotes is 2.
Step 4 of 9
Step 5:
Consider the following formula for the breakaway points.
Rearrange Equation (2).
s’
j ’ + 8 :(j+ iy = o
j ’ + J C ( i '+ 2 i + l ) = 0
(5)
«' + 2 *+ l
step 5 of 9
Differentiate Equation (5) with respect to s .
^
__ d f
ds
^
< is(i* + 2 s + l j
( j ' + 2 i+ l) 3 s ’ - » ’ ( 2 j+ 2 )
( j ’ + 2s + l) ’
iifC
Consider — s Q.
ds
- [ ( i ’ + 2 i + l)3 s ’ - s ’ (2 j + 2 ) ] = 0
-s* -4 s’ -3 s’ - 0
- s ’ (s’ +4s+3) = 0 ..... (6)
The roots of the Equation (6) are 0, 0, -land -3.
Substitute the value -3 for s in the Equation (5).
(-3)’
K =-
( _ 3 )’ + 2 ( - 3 ) + 1
= 6.75
Thus, the breakaway point is [g^ = - 3 |-
Step 6 of 9
Step 6:
Procedure to draw root locus plot for the given system:
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the centroid on the real axis, and draw the asymptotes from centroid at an angle of | 3q*
• Locate the breakaway point on the real axis.
• Draw the root locus.
The root locus plot is shown in Figure 1.
Figure 1
Hence, the root locus is plotted for the given system and it is shown in Figurel.
Step 7 of 9
(b)
Consider the following formula for the locus crosses the imaginary axis.
K G ( jo ) H ( ja ) = - \
J ' (y<»)*-i-2(ya»)tli
I
0«>)'
]
* K 2 ( j o )+AT = 0
(yV»)’
-J a ?-K a^+ K 2Jo)+ K = 0
(7)
Equate the real and imaginary terms.
-K a ^ + K = 0
a=\
- a ^ + K 2a = 0
(»)
Substitute the value Ifo r uiin the Equation (8).
-l+ 2 ^ = 0
K = 0.S
Thus, the locus crosses the imaginary axis at |ttis lra d /s e c |fo r |jj^ aQ .s|.
Step 8 of 9
The graphical technique for the locus crosses the imaginary axis is shown in Figure 2.
Figure 2
!, the locus crosses the imaginary axis which is shown in Figure 2.
Step 9 of 9
(c)
If
s 0.S which corresponds to ^ = 0.000867 on the root locus, the system is conditionally
stable with saturation which can be expected stable for the small input signals. However, as the
reference input size gets larger, the equivalent gain can get smaller due to the saturation and the
system is expected to become less well damped. Therefore, the system is expected to be
unstable at some point for the large inputs.
Thus, the system to be expected conditionally stable or unstable depending upon the input
signals.
Problem 9.09PP
Consider the system with the plant transfer function
GW =
1
?T T -
We would like to use PID control of the fonri
D tW = I0 ^ 1 + ^
+ 2 i^ ,
to control this system. It is known that the system’s actuator is a saturation nonlinearity with a
slope of unity and |r/| < 10. Compare the system response for a step input of size 10 with and
without antiwindup circuit. Plot both the step response and the control effort using Simulink.
Qualitatively describe the effect of the antiwindup circuit.
Qualitatively describe the effect of the antiwindup circuit.
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 9.1 OPP
Compute the describing function for the relay with dead-zone nonlinearity shown in Fig.
Fig. relay with dead zone;
Step-by-step solution
Step-by-step solution
Step 1 of 5
Sketch the given figure.
Step 2 of 5
To find the describing fimction for the relay with dead-zone non-linearity
The standard equation is u ( 0 ~ osin(dV) and
Step 3 of 5
Find Yi.
Yi = - V 'y { t ) s m ^ a t ) d [ e x )
7T^
(« *)
7T^
sin {a t) d (a*)
Thus, we get
Yi = ^ < t o t { a t ^ )
Step 4 of 5
We know that
k
—
a
Thtis, we get
The described function is then given by D F = —
Step 5 of 5
Find the DF,
D F =^
4V
TTa
cos(<ai)
Hence, the described function is
DF
-s fl
- ntg +
sin QiX +
Problem 9.11 PP
Compute the describing function for gain with dead-zone nonlinearity shown in Fig.
Fig. gain with dead zone;
fj
IkpeXb
-k
7
Step-by-step solution
Step-by-step solution
Step 1 of 4
Sketch the given figure.
Step 2 of 4 ^
sin QiX +
The standard equation is u (t) - asin(<zv) and
This is an odd nonlinearity so that all the cosine terms are zeroes and the DF is real
( “ ) ■='('“ )
yrJO
= ^^J^^[j4sin((Zif) - l]sin(atf)
( izk)
Since,
h = asin(4:u^)
We have,
(ZUj s sm
' —'
Step 3 of 4
Simplify the e^>ression further
?r
y
The described function is then given by DF = —
Step 4 of 4
Find the DF,
DF:
DF =
2K,
Thus, the describing function is
DF =
2K^
Problem 9 .1 2PP
Compute the describing function for the preloaded spring or Coulomb plus viscous friction
nonlinearity shown in Fig.
Fig. preloaded spring, or coulomb plus viscous friction;
y
'Slope JCo
Steo-bv-steo solution
Step-by-step solution
Step 1 of 3
Sketch the given figure
slope fC,
Step 2 of 3
We know that,
2
ka
• - m
.
f W
h.
AT
step 3 of 3
Here, there is a combination o f a gain,
plus relay nonlinearity.
Y
The described function is then given by D F = — .
a
Obtain the describing function.
4N
DF=K, + ^
yra
Hence, the describing function is
4}7
D F = K f,+ — .
______ ^
Problem 9.13PP
Consider the quantizer function shown in Fig. that resembles a staircase. Find the describing
function for this nonlinearity and write a Matlab .m function to generate it.
Figure Quantizer nonlinearity
Step-by-step solution
step 1 of 7
Given figure for the quantizer nonlinearity is.
Step 2 of 7
The abscissa brealq>oints are denoted by ^
We know that b^=—
J [a sin(ZU)sin
d[cx)
Step 3 of 7
Simplify the e^>ression further.
l \ = — ( " o X sinatf d lo jt ) + — \ ’^ q x ana > td (at) + .... + —
jrJO
x siacatd(cot)
— ( C 0 S ^ + C08<l^ + .... + C O S ^ )
Here, ^ = sin
d n ( ^ ) f o r i = 1 . 2 .. . ,»
Step 4 of 7 ^
Obtain the des cribing function
(a ) = — .
a
a
0,
0 <a < —
2
K m («3) = '
J-----;---------- O
L (^2i-l V
2 »-l
2» + l
Thus, the describing function is
0,
0 <a < 2
1*
1
M l w "\
[
2a
2« - 1
2» + 1
2
2
V
^
Step 5 of 7 ^
Obtain the MATLAB code for generating the describing function.
c le a r a l l ;
c lo s e a l l ;
n a = 99;
n n = 6;
K e q = z e r o s (1^ n n * tia ) ;
f o r n « l:n n
a i = li n s p a c e ( ( 2 * n - l) / 2 , (2 * n + l)/ 2 , n a ) ;
f o r n i= l:n a
f o r k = l;n
K e q ( ( n - l) * n a + n i) * K e q ( (n -1 )* n a + n i) +
(4 / ( p i * a i ( n i ) ) ) * s q r t ( 1 - ( ( 2 * k - l ) / ( 2 * a i ( n i ) ) ) ^ 2) ;
end;
end;
end;
p l o t ( l i n s p a c e (1 / 2 , ( 2 * n n + l)/ 2 ,n a * n n ), K e q );
t i t l e ( 'D e s c r i b i n g fu n c tio n f o r q u a n tiz e r n o n l i n e a r i t y ') ;
x l a b e l ( 'a / q ');
y l a b e l ( 'K _ { e q } ' ) ;
g r id on;
Step 6 of 7
Sketch the obtained gnq>h for nonlinearity after execution o f the code.
Desofeing funcOonfor quantizer nonineatny
i
1.2
^ ■ V '1i.v
,5L
—
i \ /
P 's /
:
!
0.4 —
02
-■
0
' i■
3
4
• .' a/q •
Step 7 of 7
The descnbing function is plotted in the figure as a function o f —.
The maximum o f the DF occurs at
^
« 1.27 corresponding to —« 0.7.
a
Since, the staircase can be ^>proximated by a straight line, it is seen that the DF will in
the limit approach the slope o f the linear ^proximation.
Hence, the slope is one.
Problem 9.14PP
Derive the describing function for the ideal contactor controller shown in Fig. Is it frequency
dependent? Would it be frequency dependent if it had a time delay or hysteresis? Graphically
sketch the time histories of the output for several amplitudes of the input and determine the
describing function values for those inputs.
Figure Contactor for Problem
Step-by-step solution
Step 1 of 7
Sketch the given figure.
Step 2 of 7 ^
sin Q at +
The standard equation is u (t) - asin(dv) and
Step 3 of 7
This is an odd nonlinearity so that all the cosine terms are zeroes and the DF is real
— J|j*^[j4sin(atf)-l]sin(aif) d(ait)
Step 4 of 7
AT
On further simplification we get 2^ = — cos(<zK^)
Since
then cos(4Xi)
■ft
The described function is then given by D F = — .
Step 5 of 7
Find the D F when d < a .
DF = ^
4T
Thus, the describing function is D F =
0,
a <d
The descnbing function is not frequency dependent
Step 6 of 7
Frequency dependence will be introduc ed with the introduction o f a time delay.
Sketch the describing function for<i = 0.1 and T = 1.0.
D e s c r ib in g f u n e f e n : D F
Step 7 of 7 ^
Sketch the time histories for the several amplitudes ofthe input frequencies.
Problem 9.15PP
A contactor controller of an inertial platform is shown in Fig, where
/ = 0.1 kg • m^,
^=1.
- = 0.01 sec,
c
= 0.1 sec,
^ = 0.01 sec,
d = 10~^ m l,
r = iN - iiL
The required stabilization resolution is approximately 10-6 rad;
T=IN-m.
The required stabilization resolution is approximately 10-6 rad;
K<pm > d for (pm > 10-6 rad.
Figure Block diagram of the system
Discuss the existence, amplitude, and frequency of possible limit cycles as a function of the gain
K and the DF of the controller. Repeat the problem for a deadband with hysteresis.
Step-by-step solution
S te p i of 17
Sketch the given figure.
M o t o r and c o n tro lle r
step 2 of 17
Given that
/ = 0.1 kg.m^,
- = 10s,
B
* = 1.
c
- = O.Ols.
e
From the figure it is obtained that
7i = 0.1 s,
O.Ols,
d = 10^rad, and
r = 1 N .m .
Step 3 of 17 >*■
The limit cycles depend on the natural behavior of a closed-loop part
The describing function of the switch is
.
Step 4 of 17
The characteristic equation is iTG+1 sO .
Find — .
Be
Kh
V + i , ____ 1
+1
B c
On further simplification we get f V l
U -O lJ
O .lff+ 1
:-l.
s(10s + l)(0 .0 l5 + l) _
Step 5 of 17
Sketch the bode plot o f the obtained transfer function
BocteUngran
I
%
Step 6 of 17
Sketch the root locus ofthe transf^ function.
Hoot Locus
step 7 of 17 >*■
The conditions for limit cycle is phase equal to -180° v4iich occurs at and when the
magnitude is one.
KK
1
Thus, we get — — « ---------- rfa t at =-95 ra d /s l.
0.01
10 X 10“^ ^
^ ^
Simplify the equation further.
= 100 (rad/s)(nm sec/rad)
X X ,.s l0 0 n m
Step 8 of 17
Thus, the ea>ression for X „is obtained as K „ =
^
^
— L fl - * ^
m i \a
Here, the maximum is at iZ !
trd
Sketch the response of the non-linear system.
Step 9 of 17
Step 10 of 17 ^
The output is a constant for levels of m > 3 d .
The fi^quency ofthe limit cycle is fixed by the phase. So (P is a constant
The result is independent o f K, for large enough K to insure the K<p » d .
rrx • •
•,
This IS consistent with
r «•
constant
„
= constant for K,. = -----------, a = K<p.
Step 11 of 17 >
Sketch the time histories representing the anq^litudes.
Obtain the eg ressio n for <p.
21
<p= (5 rad/s^)(Af)^
Step 13 of 17
We know that At = —.
4
Find a .
iTT
o » = ---p
(P= 100 rad/s
Step 14 of 17
Find At using the obtained values.
*
2^
Aa
A t= —
,
3.14
A t * -----200
At = 1 .5 7 x 1 0 -“ rad/s
Thus, we get <P= 12 x 10"*rad
Step 15 of 17 ^
If the resolution o f platform pickoff should be ~ 10~^rad and short term s ensor
noise «10~ ^rad,then
» dt is satisfied.
W eknow that K x 10“* = d, a n d d = 10^ ra d .
Hence, we get X* = 10.
Since, XX„ = 100n m , w eg etX „ s lO n m .
Step 16 of 17
We know that K , = ^
Jl - ( - 1
Let us assume that
4x1
m i \ a ) ‘\
1-
\a )
ttx
• xIO'
lO '^ U J
Find —.
• = 0 .8 x l0 -® x l0
- = 8 x 1 0 "*
Step 17 of 17
For the analysis obtain a suitable value for a.
d
a=8x10" ^
0 = 0 .1 2 5 x 1 0 ^ x 1 0 "’
a s 0.125 rad
Here, we must consider the mid frequency model because the limit cycle is at
m = 100 rad /s.
at gets inte^ated in the gyro below its break fi^quency, but
circuit for a gain o f 10 andX = 10.
So, 0 = 1 2 .5 X 10"* X 100 rad
Thus, we get a = 0.125 ra d , which is same as before.
Hence, the result is verified successfully.
goes through the lead
Problem 9.16PP
Nonlinear Clegg Integrator: There have been some attempts over the years to improve upon the
linear integratorA linear integrator has the disadvantage of having a phase lag of 90° at all
frequencies. In 1958, J. C. Clegg suggested that we modify the linear integrator to reset its state,
X,
to zero whenever the input to the integrator, e, crosses zero {that is, changes sign). The Clegg
integrator has the property that it acts like a linear integrator whenever its input and output have
the same sign. Othenwise, it resets its output to zero.
The Clegg integrator can be described by
x(t) = e(t), if e(t) ^0.
x(t+) = 0, if e{f) = 0,
where the latter equation implies that the state of the integrator, x, is reset to zero immediately
after e changes sign. It can be implemented with op-amps and diodes. A potential disadvantage
of the Clegg integrator is that it may induce oscillations.
w n e t e m e la u e i e q u a u u i i im p lie s ir ia i m e s i a i e u i m e iiiie y ia iu i, x , is l e s e i lu z e i u irn m e u ia ie iy
after e changes sign. It can be implemented with op-amps and diodes. A potential disadvantage
of the Clegg integrator is that it may induce oscillations.
(a) Sketch the output of the Clegg integrator if the input is e = a sin(cuf).
(b) Prove that the DF for the Clegg integrator is
= — - A
and this amounts to a phase lag of only 38°.
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 9 .1 7PP
Compute and sketch the optimal reversal curve and optimal control for the minimal time control of
the plant
l«l < 1Use the reverse-time method and eliminate the time.
Step-by-step solution
step 1 of 7
Step 1 of 7
Given that
X2,
X2 = -X2 + U,
Use the reverse time method and eliminate the time.
X2 = - 7 ^ + tt
For u = + 1 , time reversal means that we let r = - t,a n d that changes the time of the system ai
dx
the input matrices — = - Ibc - G « .
tJT
Here,
dx^
* - Xj, ^
So, we get
- 1, —2- *
dz
- 1.
* . = dz
.
Step 2 of 7
Integrate both sides.
I 'i T S n i - l . '"
In (^ “ 1) = T +
(O) = 0 , find z.
Since
We have C i = l n ( - l ) .
l n ( ; ^ - 1) - I n ( - l) = z
l n ( l - ;i^) = r
T = ln (l - :i^)
Step 3 of 7
Simplify further.
Xj= - 1 -I-
dx^ =f-1+«*) dz
Integrate both sides to getxi.
J^Xj =J(“l +«*) dz
0
0
Xj = « * - T - 1
Eliminate T to get Xj = -
- In (1 -
.
Step 4 of 7
This is the reversal curve for u s 1, ;^ < 0.
For w = - 1 , given that ;^ = ;^ + 1.
So, we get
. = dz.
d{x^ + \)
Integrate both sides
ln (;^ + 1) = T + Cl
Step 5 of 7
(O) = 0 . find C\.
Since
In (ji^ + 1) = T -I- Cl
C, = ln (l)
Find z.
In (:(^ + 1) = T
ji^ + 1 =e*
T = ln ( ;^ + l)
Step 6 of 7
Now, integrate to findxi.
dz
- Xj,
= 1 - e*
Simplify further.
0
0
Xi = - e* + T + Ca
Step 7 of 7 ^
Thus, when Xj (0) = 0, and Cj = 1, we get Xi = - «* +
t
+1.
Eliminate z to get, x^ = In ^1 + jx^p - x^.
This is the reversal curve forn = - 1 , and
We can then write in general, for all
> 0.
Xj = sgn (;t^)ln^l + |x^p - x^ ■
Therefore, the control law is
u = - s g n [j;i + I , - sgn(jrj)ln(l + |j^D]
Problem 9.18PP
Sketch the optimal reversal curve for the minimal time control with |u| < 1 of the linear plant
1| =J(2,
i i = -2*1 - 3x2+ M.
Step-by-step solution
step 1 of 3
Given that
X i= X2,
Given that
* i= * a .
X2 = - 2 X i - 3X2
“ •
We reverse time that means
T
=-
and that changes the sign on the system and the
input matrices,
dz
dr
Here,
= - ^ , and
X2= 2^1 + 3xj - a .
We simulate the system using the MATLAB Isim function with u = +1 and store
X2 , and repeat with a = - 1 and store
and
and X2 and plot the result to obtain the optimal
reversal curve.
Step 2 of 3
Write the M AILA B program to plot the reverse optimal curve for the give system.
t=Q: . 0 1 : 1 ;
F=[0 l ; - 2 - 3 ] ;
G =[0;1] ;
H = [ l 0] ;
J=[0] ;
% U sin g t h e r e v e r s e t i n e m ethod
s y s = s s ( - F , J ) ;
%u= +l;
[ y p , t / x p ] - I s i m ( s y s ^ o n e s ( l O l r 1) r t ) ;
p l o t (x p ( : , 1) , x p ( : , 2 ) ) }
h o ld on;
% u=-l
[ y m , t , x m ] = l s i m ( s y s , - l * o n e s (101,1) , t ) ;
p l o t (xm ( : , 1 ) , xm ( : , 2 ) ) ;
g rid ;
x la b el(» x _ l» )?
y l a b e l (’x_2’ );
t i t l e ( ' O ptim al r e v e r s a l c u r v e * ) /
Step 3 of 3 ^
Sketch the optimal reverse curve for the given system.
Thus, the required reverse optimal curve is sketched.
Problem 9.19PP
Sketch the time-optimal control law for
XI =X2,
JC 2=-X l+U .
IkI < i .
and show a trajectory for x^ (0) = 3 and x2(0) = 0.
Step-by-step solution
step 1 of 5
Step 1 of 5
Given that
- Xj + a, and
We know thata = 1.
Find the Li^lace transform of the given functions.
X ,{s )= '+ 1
Step 2 of 5
Find the inverse Ls^lace transform for the obtained functions.
= - s in ( i) ,
= - c o s ( /)
= X .-1
Step 3 of 5 ^
Therefore, we get x^ = 1 - cos ( i)..
We see that (xj - l)^ +
= 1, which is a circle with center at (l, 0), x^ < 0.
Similarly for a = - 1 we get,
(t) ~ sin ( t) ,
Xj (<) = cos (i) - 1, and
(xj + 1)^ +
=1
Which is a circle with center at (-1 , 0),
> 0.
Step 4 of 5
Sketch the reverse curve for the given system.
Step 5 of 5
The trajectories for this system are circle s centered at (±1,0 ) . This is called the Bushaw
problem in optimal control literature.
Thus, the required figure is sketched.
Problem 9.20PP
Consider the thermal control system shown in Fig. The physical plant can be a room, an oven,
etc.
(a)
What is the limit-cycle period?
(b )if Tr is commanded as a slowly increasing function, sketch the output of the system. T. Show
the solution for T r“large.”
Figure Thermal system
Step-by-step solution
Step 1 of 4
Sketch the giveti figure.
Step 2 of 4 ^
w
This is a first order system so use {T,
plot. For an oven, it is piecewise linear
We have T + oT = BNsgti (e) with hysteresis with
% = 800°C,
BN
-----w 1000®C above 7* = 0 (say room temperature)
a
Thus, we get limit cycle period ^
•
Step 3 of 4
(b)
Given that a = 0.01 s“^plot vs at, A ~ 100®C, % = 0, limit cycle period is
-(T r~
sq>proximately equal to ^
a)
aP
{ boo)
^ j and gives
l,3 0 0 j
P = 100(0.058 + 0.176)
P = 23.4 8
Thus, we get P = 23.4 s .
Step 4 of 4
Sketch the ten^erature plot for the given system.
Time (sec)
giw s
as follows.
as
Problem 9.21 PP
Several systems, such as a spacecraft, a spring-mass system with resonant frequency well
below the frequency of switching, and a large motor-driven load with very small friction can be
modeled as just an inertia. For an Ideal switching curve, sketch the phase portraits of the system.
The switching function is e = 0 + roj. Assume that r = 10 sec and the control signal = 10-3 rad/
sec2.
Now sketch the results with
(a)
deadband.
(b)
deadband plus hysteresis.
(c)
deadband plus time delay T,
(c)
deadband plus time delay T,
(d)
deadband plus a constant disturbance.
Step-by-step solution
step 1 of 8
The given switching function is e ^ 0 +
It is assumed that
r = 10 s. and
1
the control signal s 10~^rad/s^.
Find
da
de
da _ u
dd
a
d a _ lOr^rad/sec*
d&
a
We know that — = u&, and $ = 500®*.
2
Thus, we get ^ « 0.05 when a = 10^
Step 2 of 8
(a)
Sketch the phase portrait o f the system with dead band.
step 3 of 8
0 >)
Obtain the phase portrait o f the system where there is deadband as well as
hysterisis.
Step 4 of 8
(c)
The phase portrait o f the system with dead band and the time delay T is sketched.
step 5 of 8
(<?
The system has a dead band as well as a constant
We know that T a = L $ .
Thus, we get A 5 -f TO = 0.
Find the change in slope L&.
L9 = -z a + Ta
L B = -(r-T )a i
Step 6 of 8
Find 9 when u tends to zero for (^ - <^ + A + TO * 0) + 7*.
9 s a.
Find the e^>ression for a and a .
We know that 9^ = - r a j + ( d - k).
a = u + D , when t, ^
9^
a = {u + D )t +Of,
Step 7 of 8
Find 9.
9 s
#3
+ D'^ — + a ^ + 9j
Find 9 when t = T.
Step 8 of 8 ^
Sketch the phase portrait o f the sjrstem with dead band and a constant disturbance.
Problem 9.22PP
Compute the amplitude of the limit cycle in the case of satellite attitude control with delay
19 = N u { t- A),
using
ii = -sgQ(r^+0).
Sketch the phase-plane trajectory of the limit cycle and time history of 0 giving the maximum
value of 0.
Step-by-step solution
Step 1 of 9
The equation for the satellite is
li = —sgn {z& +
.
Delay o f the sateUite altitude is given as
AAT
Since, the delay is A seconds, 0 must travel - j —units during the delay.
Obtain the e}q>ressions for 0 and 0 from
/ ^ = A 6 i( i- A )
Step 2 of 9
Eliminate t- ^ ^ tin the expressions u sin g s and 6^.
a = — ( S - £ |, ) + —
Simplify further to obtain T
0+T0=O
1
r=—
step 3 of 9
Sketch the phase plane trajectory o f limit cycle.
Step 4 of 9
From the geometry o f the limit cycle we obtain the e :^ e s s io n ^ j + - ^ = 0^
The geometry of the limit cycle at the point A is
a ^ - a„ = —
Thus, we g/st0j^+ T0j^ = 0
Similarly at point S we get 0^
Step 5 of 9
Solve for the expressions w h e n c e = 0^. and ^ s 0
Eliminate 0^ from the equations
^ ( 217)
Thus, we get
step 6 of 9
Now eliminate 0^ from the equation
A 1
t
„
AATl
Eliminate 0^ from the equation
0^-¥
0.
Step 7 of 9
•
L ( LN \
^ A = j[2 e A ^ - r ]
T0J^=O
After solving for 0j^ we get
From the equation 0j^ + - j - *
.
ATA A^ATf
1
we get the result
1
Simplify the equation
2/
^
a
J
/ _
Step 8 of 9
F in d |^ |.
AAT[A + 2 ( r - A ) ]
2 /(t - A )
I4l=
2 f-A f
8/
t
-A
J
1^1=14.
Step 9 of 9
Sketch the time history o f 0
6
From the figure we infer that it is non-linear oscillatioa
Problem 9.23PP
Consider the point mass pendulum with zero friction as shown in Fig. Using the method of
isoclines as a guide, sketch the phase-plane portrait of the motion. Pay particular attention to the
vicinity of 0 = rr. Indicate a trajectory corresponding to spinning of the bob around and around
rather than oscillating back and forth.
Figure Pendulum
Step-by-step solution
step 1 of 5
Sketch the given figure o f the pendulum.
Obtain the equations governing the motion o f the pendulum.
W = mg/ sin 6
m /^d = ^ / s i n d
d = ^ s in d
/
Step 2 of 5
When 0 ~ 0 , we have sin ^ ~ 5
Simplify the obtained eiq^ressionby substitution.
e = ^0
i
i
S±
step 3 of 5
The saddle point is at d s 0, 27T,....
We know that when 5 ~ zr, we get sin 5 — 6
Obtain the eg ressio n corre spending to sin ^ ~
e^-
^.
^-e
i
s = ± j.
step 4 of 5
The centre is obtained at d » ± ^ , ± 2?r,.
O btain^ and a , using the isoclines.
d&
d = —sm d
I
d$
a t= —
d$
ac= —sin5
/
The isoclines are sinusoidal curves.
Obtain the phase portrait using MATIAB
Step 5 of 5
1htMCk«vd»itlli«n(-l.4a. l
0
)*nth«conpiiaiHn«MM.
The i ^ e r and the lower portraits correspond to the whirling motion with the circular
motion o f the pendulum.
Problem 9.24PP
Draw the phase trajectory for a system
between i(0) = 0 ,x (0 ) = 0,and x(t) = 1 mm. Find the transition time ffb y graphical means from
the parabolic curve by comparing your solution with two different interval sizes and the exact
soiution.
Step-by-step solution
step 1 of 4
Step 1 of 4
Consider the following system.
Draw the phase portrait plot for the given system using MATLAB.
Write the MATLAB script for drawing the phase portrait plot.
» syms x{t)D2x=diff(x.2);Dx=diff(x.1):» dsolve('D2x=1E-6','Dx(0)=0','x(0)=0')ans
=t''2/2000000» x=0:0.000001:1 E-3;t=sqrt{200000.*x):plot(x,t)
Step 2 of 4
Draw the phase portrait plot.
Figure 1
Step 3 of 4
Write the expression for velocity.
Ax
To calculate the transition time,
by graphical integration, write the expression for change in
time using above equation.
Here.
V is the average v in a given interval of ^
Divide x into five intervals going from left to the right.
X
V
0.2x10'’ 0.4x10'’ 0.6x10-’ 0.8x10-’ IxlO-*
0.02
0.028
0.035
0.04
0.045
Step 4 of 4
Now calculate the transition time.
B A/, + Afj + Afj -I-A/4 + A/5
_ 0.2
0,2
0.2
0.2
0.2
“ 0.02 * 0.028 * 0.035 0.04 * 0.045
= 10+7.14+5.71+5+4.44
= 32.29 sec
Therefore, the transition time.
is [32.39 see].
Better approximation can be obtained by finer division for x . Othenvise, we can obtain the exact
transition time using the following formula.
{v
Time can also be found by calculating the s
v(x)"' plot.
Problem 9.25PP
Consider the system with equations of motion
9 + 9+sine = 0.
(a) What physical system does this correspond to?
(b) Draw the phase portraits for this system.
(c)
Show a specific trajectory for 0(0) = 0.5 rad and e = 0.
Step-by-step solution
Step-by-step solution
There is no solution to this problem yet.
G et help from a C hegg subject expert.
ASK AN EXPERT
Problem 9.26PP
Consider the nonlinear upright pendulum with a motor at its base as an actuator. Design c
feedback controller to stabilize this system.
Step-by-step solution
step 1 of 2
The giTen sjrstem is a non-linear p endiilum with a motor at its base as an actuator.
We know that
9 = sin ^ + a .
When a lead network is used we get
We know that
6 = sin ^ + a .
When a lead network is used we get
We have
x^=9
x^ = 9
X, =
step 2 of 2
Obtain die matrix form of the sjrstem equations.
X'
= sin Xj - 4xj - 4xj
-3x3 “ 2xi
Hence, die equation for the lineaiized system is
0
1
0
-3
0 -4
-2
0 -3
The poles of the system are at det(s’I - F ) * (s +1)*
If die system starts near the iq u i ^ position dien die system is asynqitotically stable near the
oiigia
Problem 9.27PP
Consider the system
i s —siox.
Prove that the origin is an asymptotically stable equilibrium point.
Step-by-step solution
step 1 of 2
The given system is
X = - sin X
The given system is
X = - sin X
To prove that
r ( x ) 5 - x^(3x.
The Ly^mnov function for P = 1 is
step 2 of 2
Find V ( x ) .
^ (x) = 2xx
P^(x)= -2 x s in x
We know that
P^(x) ^
X, for 0 i X ^ 1
Since, s in x ^ - ^ x , f o r 0 ^ x ^ l , w e choose Q = \
Therefore, it is evident that the origin is an asymptotically stable equilibrium point
Problem 9.28PP
A first-order nonlinear system is described by the equation jp, = -f{x), where
^x) is a continuous and differentiable nonlinear function that satisfies the following;
m = o,
f(x) > 0, for X > 0.
f(x)<0, for x<0.
Use the Lyapunov function V(x) = x2J2 to show that the system is stable near the origin (x = 0).
Step-by-step solution
Step-by-step solution
step 1 of 2
The descrytion o f the first order non line ar system is x = 2 / (x)
Where f (x) is the continuous and non linear function
Here, / ( x ) satisfies the conditions
/(0 ) = 0
/ ( x ) > 0 for
X
<0
f (x) < 0 for X > 0
The given L3ra5>nov function is V (x) = —
To prove that the system is stable near the origin.
Step 2 of 2 ^
Find V [x )
r ( x ) = XX
"Whenx > 0 and / ( x ) > 0, we getf^(x) > 0
Whenx < 0 and / ( x ) < 0 , we getP^(x) > 0
Whenx = 0 and / ( x ) = 0, we getP^(x) = 0
Here, for all x ^ 0, T < 0
Thus, by sq>plying Lys^novs stability criterion, we conclude that the system is stable.
Problem 9.29PP
Use the Lyapunov equation
A7P + PA = -Q = -I.
to find the range of K for which the system in Fig. will be stable. Compare your answer with the
stable values for K obtained using Routh’s stability criterion.
Figure
J«f)=0
-o m
Step-by-step solution
Step-by-step solution
step 1 of 6
Consider the following formula for the Lyapunov equation.
A ''P + P A = - I ...... (1)
Consider the following matrix for the positive definite matrix
■[”
]
Refer Figure 9.67 in the textbook.
Find the closed loop transfer function from the given Figure.
r(» )
G M w (» )
R(s)
1 + G ( i) « ( i)
Substitute G ( * ) =
(2 )
» ( » ) = • in equation (2).
( i+ 4 )( s - l) '
I(£ )_ j£ ± i)M L
1+ 7
( j+ 4 ){ j-l)
y (» )
K
R{s)
s‘ +3s+(K-A)
(3)
Step 2 of 6
Consider the formula for the closed loop system matrix A in controller canonical form.
*-[T
T ]
From equation (3). the highest exponent of s is 2.
Substitute n s 2 in equation (4).
*-[T ?]
Substitute
—4and oj s 3 in equation (5).
*■[?-'r>]
* - [ ? ‘ -.1
Substitute matrix A =
[?
P=
in equation (1).
:]•[: t ; V H ; :]'
;i; ;H; ; i r ‘r ]- r :-.] -
Rearrange the equation (6) and simplify the equation is given below.
2? (4-A T ) = - l
(7)
2 q -6 p = -\
(8)
p {4 -K )+ r-3 q = 0
(9)
Rearrange the equation (7).
Step 3 of 6 ^
Substitute q ^ -
2{4-K)
in equation (8).
( z p ) ] - * ' ’(4 -A T )
-1
(4 -J C )
.- 6 p = - \
-6 /? = - l
-6 p = - l +
^
{4 -K )
6 (4 -A)
Substitute q ^
i M
i
6 ( 4 - a:)
2 (4 - if )
6 (4 -A :)^
>
U (4 -A T )J
- a: ’ h-7 a : - 2 1
6 ( 4 - a:)
Step 4 of 6
Consider the two stability conditions for /> > 0 is given below.
l( 3 - A T )
„
^ = 6 (4 -A )^®
Thus, the value of K is K > 3 and IC > 4 -
X ’ -10JC’ +42Ar-72
>0
36(K-4)
(iC-4)(jr»-6iC-M8) ^
36(JS:-4)"
p r - g * = ( ^ : - 4 ) ( A '* - 6 ^ + 1 8 ) > 0 -
( 10 )
From equation (10), the stability condition is satisfied when K > 4 - since
IS^is
always positive.
Thus, the value of K \s K > 4 ^nd it is satisfied the stability condition.
step 5 of 6 ^
From equation (4), the characteristics equation is given below.
»’ + 3 s + ( A : - 4 ) = 0 ...... (11)
Apply Routh-Hurwitz criteria to equation (11).
1
5 2
K
-
4
0
5 0
3
( K
- 4
)
Figure 1
Step 6 of 6
The system is stable if the equation satisfies the following condition.
• All the terms in the first column of the Routh’s array must be positive sign.
From Figure 1, the stable condition is
Thus, the value of
is
> 4.
> 4 and it is satisfied the stability condition.
Therefore, the value of K is I a: > 4 | same for the Lyapunov equation and Routh-Array criteria
Problem 9.30PP
Consider the system
Find all values of a and
for which the input u{t) = ay(t) + fSwill achieve the goal of maintaining
the output y(t) near 1.
Step-by-step solution
step 1 of 6
Step 1 of 6
Consider the following system.
Output equation is,
The input of the system is.
a (/)= a ;> (f)+ P
( 2)
Substitute X| for
in equation (2).
a (/) = c u ,+ P
Substitute oX|
for in equation (1).
The nonlinear, closed loop system equations are,
i , = x,-t-x,{ax,+ fi)
(3)
i j = x 2 { x j+ a x , + f i)
(4)
Step 2 of 6
To find the equilibrium points of
and
for the desired output of ^
substitute X | s l , i*|s O
s O in equations (3) and (4).
From equation (3).
0 = l+ * ,( a ( l) + /9 )
I + j( ^ ( a + ^ ) = 0
I
(5)
From equation (4).
0 = J i( jC j+ a ( l) + j9 )
j^ ( * , + a + ^ ) = 0
] 4 + x ,( a + f i) = 0
Substitute _ ] for x , ( a + / J )
j^ - l= 0
jt J - 1
Step 3 of 6 ^
Case 1:
Consider the equilibrium point,
s 1 and X2 —1•
Let.
...... (6)
y i = x 2 - i ...... (7)
0 + P = - l ......(8)
Differentiate the equation (6).
Differentiate the equation (7).
Substitute
for
^ + l f o r aT| and y 2 -\-\^ox
In equation (3).
>■1 - >>1+ i + ( > ! + > )(“ (:> 'i+ > )+ ^ )
= ; ' ! + i+ ( > ’2 + i ) ( “ .*'i+ “ + /? )
= y , + \ + a y j/ t + y t( a + p ) * a y , * a + p
My, + l+ ay,y, +;>,(-l)+aj>, -1
y ,= y ,( l+ a ) - y ,+ a y ,y j
Substitute
for
(9)
^ + l f o r aT| and y 2 -\-\^ox
>■2= ( ; '2 + O U 2
in equation (4).
+0+^)
= { y i+ ^ ) ( y ,+ ^ + a y i+ a * P )
-
+ ^ 2 + ‘*>’1^2+^2 ( « + ^ ) + ;>2+ 1
^
= ; ^ + ;" 2 + « J ’i>’2 + ; ’2( - 0+^ 2 + i+ a j> , - 1
y 2 = a y ,+ y i+ a y , y i+ } i
°)
Step 4 of 6 ^
The characteristic equation of the linearized system is,
j'- ( a + 2 ) j+ ( 2 a + l) = 0
There are no values of a which produce stable roots. So it is concluding that,
s i Is an unstable equilibrium point.
Step 5 of 6
Case 2:
s 1 and X2- —I
Consider the equilibrium point,
Let.
...... (11)
y i = x i + l ...... (12)
0 + P = l ...... (13)
Differentiate the equation (11).
>1=^1
Differentiate the equation (12).
Step 6 of 6
Substitute jc^ for
^ + 1 for x^ and y 2 —\ for X2 in equation (3).
- J ’l + l + ( j ’2 - l ) ( “ U + 0 + /®)
= ; 'i+ i + ( > ’2 - i ) ( “ J’i + ‘* + ^ )
=>>,+ l+ « y , y , + y ^ { a + P ) - a y , - ( a + p )
My, + l + a y , y j + y , ( l) - a > , -1
y, = y i ( ^ - a ) * y i * a y ^ i
Substitute i |2 for
( I'l)
^ * i> lfo r JC| and
I f o r X2 in equation (4).
.^2 =(.>'2- l ) ( . ) ’2 - '+ a ( . > ’i+ > )+ t» )
=(^2 -
> )(.> ’2 - 1 + a>’i + «+> 9 )
= A - y t + 0 W 2+>’2(«+/9)->’2
= A -y 2 +W
~{a+P)
2 + ^ '2 (> )-J ’2 + > -o ';'i - •
y ,M - a y ,-y ,+ a y ,y ,+ )i
(15)
The characteristic equation of the linearized system is,
s ^ + a s + { 2 a - i) = 0
The system is stable for small signals near the equilibrium point if.
a > —and a+ fi= \
Therefore, the values of or and
are. o r > — a n d a + > 9 = l
s 1 and
Problem 9.31 PP
Consider the nonlinear autonomous system
j r
1
r
1
(a)
Find the equilibrium point(s).
(b)
Find the linearized system about each equilibrium point.
(c)
For each case in part (b), what does Lyapunov theory tell us about the stability of the
nonlinear system near the equilibrium point?
nonlinear system near the equilibrium point?
Step-by-step solution
step 1 of 7
(a)
The non-linear autonomous system is,
X ,(2 1 i-2 C ,)
* l'
•«2 =
-X ,!,
The non- linear autonomous system can be written as follows:
x ,= x ,{ x ,- x ,)
(1)
i i - j f - l ...... (2)
=
...... <3)
Now, set the values i j , i^ ,a n d
Substitute 0 for
as 0, and solve the non-linear equations.
in equation (1).
0 = 3^( a^ - * , )
J^=0
Substitute 0 for
in equation (2).
0 = :tf-l
JCj = ± 1
Substitute 0 for jc^ in equation (2).
0 = - x fy
x,= 0
The equilibrium points are
[±1, 0, o f
Therefore, the equilibrium points are |[^ ■ o ■ o rl and
»■ o n
step 2 of 7
(b)
(i)
Consider the following equilibrium point.
x = [l, 0, o f
Assume,
y i= ^
Differentiate the equations.
step 3 of 7
Recall equation (1).
X , = X j( x ,- X , )
Substitute
for
jr , for x^.and ji , for x,
^ ,+ 1 for
Recall equation (2).
Substitute
^ ■and
+1 for x^.
Recall equation (3).
ic, =-xfy
Substitute
for i , , ji , for
and y , + l for
The linearized system is,
y^^y
0 -1
0
2
0
0 y
0
0-1
0 -1 0
2 0 0 y
0 0 -1
Therefore, the linearized system is
Step 4 of 7
(ii)
Consider the following equilibrium point.
x = [- l, 0, o f
Assume,
y ,= x,+ \
y i - ’h
Differentiate the equations.
y > -^
step 5 of 7
Recall equation (1).
x ,= x ,( x ,- x ,)
Substitute
for i | , J>|-1 for x^,
for jtj,a n d y^ for Xy.
.Vi = 3 ’2[3’3 - ( j ' . - 0 ]
= y 2 y 2 - y iy i+ y i
Recall equation (2).
Substitute
for i^ ,a n d
- I f o r a:|.
= y t-2 y,
Recall equation (3).
iy^-JC^X,
Substitute y^ for jfcj, _yj for jtj, and ^ , - 1 for a:|.
—
y ,y ,+ y .
The linearized system is,
y^Fy
0
I
O'
= -2 0 0 y
0
0 1
Therefore, the linearized system about the equilibrium, point is.
0
10
y= -2 0 0 y
0 0 1
Step 6 of 7
(c)
(i)
The linearization is developed to determine the stability of the system near the equilibrium
conditions.
Calculate the characteristic equation.
|d - F | = 0
s
0
0
0 1 0
0 0 s
-
0 -1 o '
2 0 0
0 0 -1
3
I
0
-2
»
0
0
0
3+ 1
=
=0
0
3 [ s (s + 1 ) - 0 ] - 1 [ ( - 2 ) ( j + 1 ) - 0 ] + 0 = 0
3 * ( s + 1 ) + 2 ( s + 1) = 0
( 3 * + 2 ) ( j + 1) = 0
From the characteristic equation observe that the system has two poles on the ja> axis. These
poles make the system neutrally or marginally stable. Hence, the Lyapunov theory does not tell
whether the system is stable or not.
Therefore, the nonlinear terms affect the stability at the equilibrium point
[1. 0. o f
Step 7 of 7
(ii)
Calculate the characteristic equation.
| 5 l- F | = 0
s 0 0
0
0
-2
0 0 4
0
0
4
I
0
0 0
0
1
s -1
=
0
0
0
s-
j[s ( s - l) - 0 ] - ( - l) [ ( 2 ) ( j- l) - 0 ] + 0 = 0
3*( j - 1 ) + 2 ( j - 1 ) = 0
(j “+ 2 )( 3 - 1 ) = 0
From the characteristic equation, the system has two poles on the Ja> axis and one on the real
Therefore, the system at equilibrium point
- 1, 0, o f
Problem 9.32PP
Consider the circuit shown in Fig. For what diode characteristics will this system be stable?
Figure Circuit diagram
Step-by-step solution
Step-by-step solution
step 1 of 2
Refer Figure 9.68 in the textbook.
Consider the following equation for the inductance and capacitance equation.
-(1)
-(2 )
4 -
.... (3)
Where.
L is the inductance,
C is the capacitance,
is the diode current,
is the capacitance current.
Apply KCL in the given circuit and the corresponding equation is given below,
/< .H + « o -0
(4)
Substitute equation (1) and equation (3) in Equation (4).
(5)
C ^ * k + f { y ) = 0 ...
Rearrange equation (5).
c f = - 4-/(v)
dv"
d
e
/ ( ”)
c
(6 )
Step 2 of 2
Consider the following equation for the energy equation.
(7)
F = ic v “ + iiiJ
Take differentiation on both sides.
y
2 y., dv 2 , . di,
= - C v — + - L i,- j^
2 dt 2 '■ dt
wv ^ dv
di,
V = Cv— + L i , - ^
dt
'■ dt
(8 )
Substitute equation (6) in Equation (8).
V = Cv ± . M
c
= Cv
>L
c
dt
c
/ ( > ’) '
c
= - W i- ty ( v ) + v ij
r = - y f( v )
(9)
Where.
dii _ v
dt ^ L
Thus, from the equation (9), the system is stable for any positive DC diode characteristics.
Problem 9.33PP
Van der Pol’s equation: Consider the system described by the nonlinear differential equation
1 + «(1 + J i^ ) i+ x = 0
with the constant £ > 0.
(a)
Show that the equations can be put in the form [Lienard or (x.y) plane]
i =y+s
( H
(b)
Use the Lyapunov function y =
and sketch the region of stability as predicted by
this Vin the Lienard plane.
^u/ w o cj u 1C L .y a y u \lu v lu i lu u u i i y = ^
o ivciu i i u i c i c y n j i i u i s ia u i ii iy a o p i c u i o i c u u y
this Vin the Lienard plane.
(c) Plot the trajectories of part (b) and show the initiai conditions that tend to the origin. Simulate
the system in Simulink using various initial conditions on x{0) and A ( 0 ) Consider two cases,
with £ = 0.5 and £=1.0.
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 9.34PP
Duffing’s equation: Consider the system described by the nonlinear differential equation
where u = A cos{t). This equation represents the model of a hard spring where k is the spring
constant and if £ > 0, the spring gets stiffer as the displacement increases. Let k = 0.05, ^ = 1,
and A = 7.5.
(a)
Build a simulation of the system in Simulink. Show that the system response can be very
sensitive to slight perturbations on the initial conditions x ^ ), i(0 ) (the system is said to be
chaotic). Simulate the response of the system with xfO) = 3 and
4 for f = 30 sec. Repeat
the simulation for slightly perturbed initial conditions xfO) = 3.01 and 1(0)= 4.01. Compare the
two results.
fb) Consider the unforced Duffino eouation (u = 0 ). Plot the time resoonse of the svstem for xfO)
(b)
Consider the unforced Duffing equation {u = 0). Plot the time response of the system for x(0)
= 1, 10))= 1 for f = 200 sec. Draw the phase-plane plot for the system. Show that the origin is an
equilibrium point.
(c)
Now consider the forced Duffing equation {u_= 0). Find the solution to the Duffing equation
for x(0) = -1. 10))= 1 for f = 30 sec. Draw the phase-plane plot
versus x(t)) for this case.
(d)
Repeat part (c) for k = 0.25, e = 1. and A = 8.5.
(e)
Repeat part (c) for /c = 0.1, e = 1, and ^ = 11.
(f)
We can get more insight into the system by plotting H t j ) versus x(tj) at several hundred
points at 2 n periodic observation times. In other words rather than looking at the system
continuously, we “strobe” the system and plot the behavior at strobe times only. Show that unlike
the phase-plane plots in parts (c)-(e), the points fall on a well-structured plot referred to as a
Poincare section (also called a strange attractor). Plot the Poincare sections for parts (c)-(e).
Simulate the system using the initial conditions x(0) = -1 and A ( 0 ) = 1 for f = 10,000 sec in
order to plot the Poincare sections.
(g)
What can you conclude about the nature of the solution of the Duffing equation from the
results of the previous parts?
(h)
Characterize the system behavior in terms of the ranges of the system parameters k, e, and
A.
S te p -b y -s te p s o lu tio n
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 10.01PP
Of the three components of the PID controller (proportional, integral, or derivative), which one is
the most effective in reducing the error resulting from a constant disturbance? Explain.
Step-by-step solution
step 1 of 3
Integral control is he most effectiTe n reducing the error due to constant disturbances.
Step 2 of 3
Block diagram for showing integral control is the most effective means o f reducing
steady errors.
Using the above block diagram.
Y ^O {W -\-B D c)
E = R - Y = R - 0 { W + EDc)
E *-
1
«
\+ D c O
,
•
O
\-D (D
■W
= 1« , M = li SB(s) = l i p s f — !—
^
^ ^
s-»o^ ll+ £ ) p O
O
,
\+ D fP
Step 3 of 3
Writing G'(-S') =
and using a step input R { S ) = — and a step disturbance
' '
dO{s)
^ '
S
0)(^) = ^
we can show that integral control leads to zero steady state error. While
proportional and derivative control in general, do not
Bitegral control,
= U E[S)=
i-* 0
(^S) = —
S+0
dgS + n g )
If
Proportional control, D q (S) =
Derivative control
(s) = s ,
^ 0
^ 0 if i^ p ( 0 ) ^ 0
This analysis assumes that there are no pole zero cancellations between the plant G
and the con^ensator, D ,, in general, proportional or derivative control will not have
zero steady state error
Problem 10.02PP
Is there a greater chance of instability when the sensor in a feedback control system for a
mechanical structure is not collocated with the actuator? Explain.
Step-by-step solution
step 1 of 3
Step 2 of 3
PD control o f satellite: Non - collocated
Yes for con^arison, see the following two root loci which were taken from the
discussion in the text on satellite attitude control.
In Figure 2.26, the sensor and the actuator are collocated; resulting in a stable closedloop system with PD control in fig 2.5, the sensor and the actuator are not collocated
creating an unstable system with the same PD control
Step 3 of 3
Problem 10.03PP
Consider the plant Gfsj = 1/s3. Determine whether it is possible to stabilize this plant by adding
the iead compensator
D As) = K
^ ,
(a< b).
S+ D
(a)
What is the maximum phase margin of the resuiting feedback system?
(b)
Can a system with this plant, together with any number of lead compensators, be made
unconditionally stable? Explain why or why not.
Step-by-step solution
Step-by-step solution
step 1 of 3
Gain, the poles are alwa3rs in the right half plane, Ifw e try positiTe feedback one pole
departs at 0° so again one pole starts into the rig h t-h a lf-p la n e . For lo w -enough
gain the system will be unstable.
Step 2 of 3
Q‘(dS') =
has phase angle o f -270'* for all frequencies. The masimum
(i?) = k
phase lead from a compensator
S+ b
is 90° with — = oo
a
In practice a lead compensator with —= 100. Contributes phase lead of
a
i^proximately 80°, hence the closed loop system will be unstable with
F M = -1 0 ° to have P i/« 7 0 ° we nee d ; for exan^le, a double lead
compensator Dc(S)
(S + a y
vith
=
100
Step 3 of 3
b)
No diis plant cannot be made unconditionally stable **Because the Root Locus
Departure angles from the three poles at the origin are ±60°, for low enough.
Problem 10.04PP
Consider the closed-loop system shown in Fig.
(a)
What is the phase margin if /C= 70,000?
(b) What is the gain margin if /C = 70, 000?
(c)
What value of K will yield a phase margin of ~ 70*?
(d)
What value of K will yield a phase margin of ~ 0*?
(e)
Sketch the root locus with respect to K jo r the system, and determine what value of K causes
(e)
Sketch the root locus with respect to K for the system, and determine what value of K causes
the system to be on the verge of instability.
(f)
If the disturbance w is a constant and K = ^Q, 000, what is the maximum allowable value for w
if yC“ ) is to remain less than 0.1? (Assume r= 0.)
(g)
Suppose the specifications require you to allow larger values of w than the value you
obtained in part (f) but with the same error constraint [|y(°°)| 0.1]. Discuss what steps you could
take to alleviate the problem.
Figure Control system
W(s)
m
• ns)
Step-by-step solution
step 1 of 18
(a)
Step 2 of 18
Refer Figure 10.88 in the textbook.
Step 3 of 18
Derive the transfer function from figure 10.88.
g ( j+ i)
G (*)J > W = < (« + S )(« + 1 0 )(s + l0 0 )
( 1)
Substitute 70,000 in equation and rewrite equation (1).
a:( j + i )
X . G ( j) D ( j) =
i ( j + 5 ) ( i + 1 0 ) ( f + 100)
_________7 0 0 0 0 (» + l)_________
(2)
Plot the straight Bode diagram for the transfer function <?(#) •
Step 4 of 18
Consider the standard form of transfer function for a third order system.
K {\+ s M
G (*
(3)
( i+ iM ) 0 + V < » i) ( i+ V < » > )
Compare Equations (2), and (3).
Number of zeros present in the system is 1
One pole at break frequency,
Second pole at break frequency, < O js]0.
Third pole at break frequency, di^slOO
Substitute Ja> for s in Equation (2).
G (» =r
'
(4)
[ y<»(o.2y < » + i)(o .iy v » + i)(o .o iy « » + i)]
'
Find the magnitude of G { j 0 ) in dB using Equation (4) as listed in table (1).
Corner Frequency Slope
g)
Term
(S)
Change in slope
mm
20
-
-20
0
-20
-20
-20
-40
-20
-60
-1
( y ® + i)
1
(o. 2y<»+ i)
1
<» rt-1 0
(O .ly-ffl+l)
I
-1 0 0
( o .o iy o + i)
Step 5 of 18
Calculate gain A using table 1.
Assign lower frequency ^
and higher frequency ^
sec
Calculate gain (A) at
slO O O '^^sec
■
A=\G{je>)\
-2 0 Io g M
= 42.92db
Calculate gain
at
s
.
= |G (» |
= 2 0 lo g ( l+ l)
= 6.020db
Calculate gain (A) at c t fs o
.
-|s lo p e fix ) m a )^ ,to « } ^ x lo g ^ + A ^ ^
» 0 + 6.020db
s6.020db
Step 6 of 18
i)a t
.
to«>.,xlog—
1
1+A ^
+6.020db
1>
at a>=
+A.
®caj
7.959db
a -
n f l» r tto < » ,,x lo g ^ + A ,
®c4j
— l-1 3 .9 8 d b
10 J
n a » ,4 to < » ,,x lo g ^ + A _
®c4j
“ 1-13,
3.98 db
10J
+A.
® .5 j
-33.98db
Step 7 of 18
Find the phase plot for equation (4).
-90
tan"' ( J a )
tad‘ '{0.2y«))
tan''(0.1y«>)
tan"'
*
(o.oiy<») de
g
deg deg
)
0.1
-90
1
deg
5.71
deg
deg
g
1.14
0.57
0.057
-82.52
-90 45
11.30
5.71
0.57
-62.58
5
-90
78.69
45
26.56
2.86
-85.73
10
-90
84.28
63.43
45
5.72
-119.87
100
-90
89.42
87.13
84.28
45
-216.99
1000
-90
89.94
89.71
89.42
84.28
-263.47
Draw the Bode diagram with phase and mac nitude values as shown in fic ure 1.
10'
10'
tor
Frequency (ladtsec)
Figure 1
Step 8 of 18
From figure 1, calculate the value of Phase margin.
Phase margin is I80’ - 1 6 3 '= 1 T
Gain margin is 7 dB.
Therefore, the value of phase margin at AT= 70,000 is l i E
Step 9 of 18
(b)
From figure 1, calculate the value of gain margin.
Gain margin is 7 dB.
Therefore, the value of gain margin is |7 <|B|-
Step 10 of 18
(c)
Determine the value of K to yield the phase margin of 7q* .
Consider figure 1 and find the magnitude corresponding to
.
The magnitude corresponding to 70* is 15 dB.
Use the formula and find the value of K.
-2 0 \o g K ^ \S
I
^
*5
lo g A :- ^
A . 0.1778
Therefore, the value of K that yields 7q* phase margin is (AT =0.17781-
Step 11 of 18
(d)
Determine the value of K to yield the phase margin of q».
Consider figure 1 and find the magnitude corresponding to 7g».
The magnitude corresponding to 70* is -7dB.
Use the formula and find the value of K.
-201ogA: = - 7
_7_
k>gA:»
'2 0
a:
-2 .2 3 8
Therefore, the value of K that yields 7q* phase margin is |Af = 2.23^-
Step 12 of 18
(e)
Write the matlab program and draw the root locus of the transfer function.
rlocus([1 1],[1,115,1550,5000,0])
Step 13 of 18
Figure 2
Step 14 of 18
Consider the formula and calculate the range of K.
\ + K J > {s ) G {s ) = 0 ...... (5)
Substitute equation (2) in equation (5).
i+ -
r-o
s{s+S){s + 10)(i +100)
Simplify the equation.
5^1154^+I550S*+(5000+ a:)4 + a: -
o
step 15 of 18
Apply Routh-Array criteria to the equation.
53
1
1,550
115
5,000+A"
173,250-A"
52
115
- a 2 - 155,025 K +866,250,000
5 1
115
5 0
Figure 2
Step 16 of 18
From figure 2, the system moves to stable condition at Al > 0 ^nd
-155,02SA :+866,250,000,
-> 0
173,250-A :
Solve -AT* -155,025AT+866,250,000 >0and find the value of K.
a:
>160.424.72
Therefore, the range of K for the system to be stable is |0 < A <160,424.721
Step 17 of 18
(f)
Refer Figure 10.88 in the textbook. Consider J ^ * 0 write the transfer function.
>'(4)
G (4
B '( i )
l + A J 3 ( s ) .G ( i )
(6)
Take the values from figure 10.88 in the textbook.
1
<?(*)
(7)
i(i+5)(j-tl0)
A ( f F l)
KX j { s ) D { s ) = -
t(f +S)(s+10)(«+100)
(8 )
Substitute
Substitute equations (7) and (8) in equation (6) and simplify the equation.
( 1 + 100)
y (« )= -
« (« + S )(l + 1 0 )(s + 1 0 0 )+ A (s+ l)
IF (*
(9)
If q t(/) is consfant, fhen q t ( / ) s c Converting q )(/) in S-domain yields.
H 4 )= 7
. (
10 )
Consider the formula for steady state output.
F .- lta M * )
(11)
Substitute equation (9) and (10) in equation (11) and apply limits.
100c
Therefore, the maximum allowable limit a> occurs when
lOOc
step 18 of 18
(g)
Consider the equation y
100c
Assign AT = 10.000and^<0.1 and find the value of c.
c<10
Hence, by adjusting the values of ^^and c, the gain K can be increased. But, there is the
possibility for transient behavior of the system. By adding the integral controller to the circuit
reduces the steady state output error to zero.
Problem 10.05PP
Consider the system shown in Fig., which represents the attitude rate control for a certain
aircraft.
(a) Design a compensator so that the dominant poles are at -2 ± 2j.
(b) Sketch the Bode plot for your design, and select the compensation so that the crossover
frequency is at least 2 ^
rad/sec and PM > 50*.
(c) Sketch the root locus for your design, and find the velocity constant
when ojn 2 ^ 5 and
0.5.
Figure Block diagram for aircraft-attitude rate control
when w/7 2 ^ 5 and ^>1^5^
Figure Block diagram for aircraft-attitude rate control
Compensator
Hydiuulic servq
s
r
r
Aircraft
2J + 0.1
5^ + 0.1j + 4
Rateg
Step-by-step solution
step 1 of 3
b)
The Bode plot the system loop transfer functions.
3 4 (^ r+ 0 -4 ) (^r + 0-05)
D ^ [S ) 0 [S )^
S ( ^ '+ ll- 7 ) ( ^ “ + 0.15+ 4)
Is shown on the next page using li^at lab’s bode command. As the plot shows
®, = 3 and PAf = 67-3".
Step 2 of 3
The velocity constant is most easily found from either the bode plot or from
^ .
D,{s). a{s)
So we have
- 4^ =117®
'^0
PHI=.
angle(polyval (n,s)
pi
- pt
po^yval{d,s)
With section of z s 0- 4 , we get P s 11- 7 so that our lead design is
^
s+n i
To find the compensator gain, k, we can utilize the magnitude criterion at the
desired dominant closed - loop pole location we find that
| A W L = - 2 ± j i z = l = > t = 17.0
So die lead design is
D ,( S )
'
1 7 . ^ ^
s+n i
step 3 of 3
With a constant gain compensator
^ >die root locus of
D ,( S ) 0 [S ) =
+ 0-15+A)
= =
dm
Does not pass through-2 ± 2 J. Therefore we need condensation of at least
a lead network. Let
Using the angle criterion, at the closed loop pole location S '= - 2 + 2 j , we can
write an e ^ e s s io n fort the angle. Contribution from the lead network zero,
and lead network pole .
=180®=>
+134®-180®-135®-116® = -180®
Problem 10.06PP
Consider the block diagram for the servomechanism drawn in Fig. Which of the following claims
are true?
(a)
The actuator dynamics (the pole at 1000 rad/sec) must be included in an analysis to evaluate
a usable maximum gain for which the control system is stable.
(b)
The gain K must be negative for the system to be stable.
(c)
There exists a value of K for which the control system will oscillate at a frequency between 4
and 6 rad/sec.
tfW_The_system is Iin s ta h ip if IAC1 i n
(d)
The system is unstable if |/C| 10.
(e)
If K must be negative for stability, the control system cannot counteract a positive disturbance
(f)
A positive constant disturbance will speed up the load, thereby making the final value of e
negative.
(g)
With only a positive constant command input r, the error signal e must have a final value
greater than zero.
(h)
For K = -1 the closed-loop system is stable, and the disturbance results in a speed emor
whose steady-state magnitude is less than 5 rad/sec.
Figure Servomechanism
Distmbance (0.1 N*m)
S te p -b y -s te p s o lu tio n
Step 1 of 7
a)
Tme. even though it is tempting to ^proxim ate the actuator as infinitely &st
and hence, not important The Actuator pole dramatically alters the root locus
plot o f the system to be controlled.
Step 2 of 7
b)
True
Step 3 of 7
c)
True
Step 4 of 7
d)True
Step 5 of 7
e)
False
Step 6 of 7
£) True
Step 7 of 7
U False
Problem 10.07PP
A stick balancer and its corresponding control block diagram are shown in Fig. The control is a
torque applied about the pivot.
(a)
Using root-locus techniques, design a compensator Dc(s) that will place the dominant roots e
s = -5 ± 5) {corresponding to ojn = 7 rad/sec, ^ = 0.707).
(b)
Use Bode plotting techniques to design a compensator Dc(s) to meet the following
specifications;
• Steady-state 6 displacement of less than 0.001 for a constant input torque Td=
• Phase margin > 50°. and
• Closed-loop bandwidth~= 7 rad/sec.
Figure Servomechanism
• Closed-loop bandwidth~= 7 rad/sec.
Figure Servomechanism
1
(j 2 -6 4 )
Step-by-step solution
step 1 of 2
To have the cotnpens ated plant root locus go through the pole location
S = - S ± ^ ; we employ a led con^ensator
Using the angle criterion
= -1 8 0
At the closed - loop pole location S = - S + ^
we can write an expression
for the angle contribution. From the lead network zero,
and lead network
pole 4i^. We have.
To fmd the compensator gain, k, we can utilize the magnitude criterion at the
desired dominant closed - loop pole locations. We find that
\ D ^ { S ) a [ S ) \ ^ = - S ± ^ = 'i=t^ k = A7\
Therefore, we have the compensator
'
S + 451
Step 2 of 2
We need a lag network in addition to a lead network to get the required
Let
Dcz[S) =
'■
'U o + v
U - o i+ ijU o o o + ij
This compensator will meet our design specifications. The bode plot of
D X s )a (s )
.
Problem 10.08PP
Consider the standard feedback system drawn in Fig.
(a)
Suppose
G(j) = -
2500AT
s (i + 2 5 )‘
Design a lead compensator so that the phase margin of the system is more than 45°; the steadystate error due to a ramp should be less than or equal to 0.01.
(b)
Using the plant transfer function from part (a), design a lead compensator so that the
overshoot is less than 25% and the 1% settling time is less than 0.1 sec.
(c)
Suppose
(c) Suppose
G (s)-
j
(5 + 1 + 0 .1 j ) ( H - 0 . 2 s)*
and let the performance specifications now be /Cv = 100 and PM > 40°. Is the lead compensation
effective for this system? Find a lag compensator, and plot the root locus of the compensated
system.
(d)
Using G(s) from part (c), design a lag compensator such that the peak overshoot is less than
20% and/Cv=100.
(e)
Repeat part (c) using a lead-lag compensator.
(f)
Find the root locus of the compensated system in part (e), and compare your findings with
those from part (c).
Figure Block diagram of a standard feedback control system
D ^s) — * G(5)
S te p -b y -s te p s o lu tio n
step 1 of 9
^ v * 1 0 0 ^ ^ s l0 0
We have
Dlead{S) =
22 - 1+1
Now we select the zero o f the lag at last one decade lower than W^,. With d of the lag
equal to 20, we have
S
Dlog{S) = -2-2± L
0-035+1
The Lead - Lag con^ensator is
1,0.7+lJ U 2 I+ 1 J
U
035
+ 1A 2 2 1 +U
The system bode plot and step response 2 ^ ear.
Step 2 of 9
That the requires the plant gain to be equal to 100, since
k^ = U SO{S) = jt= > it= 1 0 0
Therefore
100
G {S ):
S r(l+ 0 1 5 )(l+ 0 -2 5 )
5000
5r(2T+5)(5+10)
Hence,
S
D a { S ) = 01+1
J ^
0-003+1
And the loop gain is
100 f—
O c (S )= -
/
g
Y
1
g Y
.y ^
A 0 0 0 3 + 1J U + 1J U 0 + 1J
Step 3 of 9
e)
Again the design specification o f steady - state error provides information for
tile design o f k.
The Design o fk
= Im s 0{S) = k = > k =100
The Bode plot of
100
a (^ :
1 ,5 + lJ U o + U
Step 4 of 9
Now select the zero o f D q (£f) one decade below
CD =
vtiiich implies
0-25 rad/sec
This Results in - =
= 0- 0072
2
34-7
The Lag netwoilc is thus
S
0-0072+1
And the loop gain is
o f - ^ 1
1,0-25+ lj
D ,(s y a {s ) =
s{
^
^0 0072+ i A 5 + i A 1 0 + V
Step 5 of 9
d)
We can design a lag con^ensator using root locus methods. The velocity
const system.
Step 6 of 9
b)
For ilf, =
25%, Let
t = 0 - 4 ,F o r i, < 0-1
4 -6
Let Co, « — = 46
•
01
Thus o , = 115 rad/sec
And S = - 4 6 ± jll0 5 we set the lead zero at
= 1-5 - o ^ ( ^ ) = -1 7 2
And conq^ute the pole to be at ^ = - 1284 using the angle criterioa
The Bode plot and step response show the specifications are net with an
additional gain of 20. Therefore, the con^ensator is
Dc(S) =74-65 ■
s+m
S+\2S4
Step 7 of 9
c)
TheD esignspecificationof steady state eiror provides information for
Step 8 of 9
a)
The Design specification of steady state eiror provides information for the
design ofk.
1
it, = ^
= 0 0 1 ^ J t= 1 0 0
sO(S) = 1 0 0 i = > j t = l
The Bode plot of
2500
.
.
Step 9 of 9
The Lead con^ensator is
^+1
— +1
CD
Such that o =
^+1
= -I—
- + i
p
= Z = 37 and — = /7»110. Therefore, we have
cc
S
D c (S ) =
110+1
The Bode plot o f the conq^ensated.
Problem 10.09PP
Consider the system in Fig., where
G (s) = -
s(s +
300
0.225)(J+ 4)(f + 180)■
The compensator Dc(s) is to be designed so that the closed-loop system satisfies the following
specifications;
• Zero steady-state error for step inputs.
• PM = 55% GM > 6 db.
• Gain crossover frequency is not smaller than that of the uncompensated plant.
(a)
What kind of compensation should be used and why?
(b) rie.sipn.flw.=uitabJfifCoroDeosatocJ)c''sTtoji>©<at tbe.wj'lonifinafinnc
(b)
Design a suitable compensator Dc(s) to meet the specifications.
Figure Block diagram of a standard feedback control system
ds)
D ^s)
Step-by-step solution
ste p 1 of 2
a)
Since we need 5 5 °-1 0 ° s 45° ofphase lead, a single lead network will do
the job.
Step 2 of 2
b)
From a phase lead requirement of 45° we have —m 10
a
The co n ^ ensator and the loop gain are
S
3-5+1
D ,( s ) a ( s ) =
K ill;" )
Problem 10.10PP
We have discussed three design methods: the root-locus method of Evans, the frequencyresponse method of Bode, and the state-variable pole-assignment method. Explain which of
these methods is best described by the following statements (if you feel more than one method
fits a given statement equally well, say so and explain why);
(a)
This method is the one most commonly used when the plant description must be obtained
from experimental data.
(b)
This method provides the most direct control over dynamic response characteristics such as
rise time, percent overshoot, and settling time.
(c)
This method lends itself most easily to an automated (computer) implementation.
(c)
This method lends itself most easily to an automated (computer) implementation.
(d)
This method provides the most direct control over the steady-state error constants Kp and Kv
(e)
This method is most likely to lead to the least complex controller capable of meeting the
dynamic and static accuracy specifications.
(f)
This method allows the designer to guarantee that the final design will be unconditionally
stable.
(g)
This method can be used without modification for plants that include transportation lag terms
—for example.
G(I) =
(j + 3)2'
S te p -b y -s te p s o lu tio n
step 1 of 7
a) Frequency response method is the most convenient for experimental data because the
sinusoidal steady - state records can be obtained directly in the laboratory. Either the root locus
or state variable design generally requires a separate system identification effort between the
experimental data and the construction of a model suitable for the design method.
Step 2 of 7
b) Either the root - locus or state variable pole assignment are the most direct for control over
dynamic response.
Step 3 of 7
c)
The state variable pole assignment is most easily programmed because once the
specifications are given, the design is completely algorithmic.
Step 4 of 7
d)
The Frequency response method of bode shows the error constant directly on the graph.
Step 5 of 7
e)
The Root locus or bode method will give the least complex controller. These techniques begin
with the gain alone and then add network. Compensation only as necessary to meet the
specifications. Whereas the state variable technique requires a controller of complexity
comparable to that of the plant right from the start.
Step 6 of 7
f)
Either the root locus, where by the locus is required to be entirely in the left half plane up to the
operating gain.
Step 7 of 7
G) The Frequency response technique can be used immediately for transportation lag, while the
root locus requires a small modification and the state variable design method requires an
approximation.
Problem 10.11PP
Lead and lag networks are typically employed in designs based on frequencyresponse (Bode)
methods. Assuming a Type 1 system, indicate the effect of these compensation networks on
each of the listed performance specifications. In each case, indicate the effect as “an increase,”
“substantially unchanged,” or “a decrease.” Use the second-order plantG(s) = /C/[s(s+1)] to
illustrate your conclusions.
{a)Kv,
(b)
Phase margin,
(c)
Closed-loop bandwidth.
(c)
Closed-loop bandwidth.
(d)
Percent overshoot, and
(e)
Settling time.
Step-by-step solution
Step 1 of 1 ^
Lead
Lag
k.
unchanged
increased
Phase Margin
increased
Unchanged
Closed Loop Bandwidth
Increased
Unchanged
Percent overshoot
Decreased
Unchanged
Settling Time
Decreased
Unchanged
Problem 10.12PP
Altitude Control o f a Hot-Air Balloon: American solo balloonist Steve Fossett landed in the
Australian outback aboard Spirit o f Freedom on July 3rd, 2002, becoming the first solo balloonist
to circumnavigate the globe (see Fig. 1). The equations of vertical motion for a hot-air balloon
(Fig. 2), linearized about vertical equilibrium, are
S t + —ST = Sq,
ri
T2?
+ Z =
o8 T -\-W t
6T = deviation of the hot-air temperature from the equilibrium temperature where buoyant force
equals weight.
z = altitude of the balloon,
e ^ q u 'a ^ 'w S g f iT , ^
.....
z = altitude of the balloon,
6q = deviation in the burner heating rate from the equilibrium rate
(normalized by the thermal capacity of the hot air),
w = vertical component of wind velocity,
r1,72, a = parameters of the equations.
An altitude-hold autopilot Is to be designed for a balloon whose parameters
are
r1 = 250 sec t2 = 25 sec a = 0.3 m/(sec .”C).
Figure 1 Spirit of Freedom balloon Source: Steve Holland/AP Images
Figure 2 Hot-air balloon
Only altitude is sensed, so a control law of the form
6q(s) = Dc(s)[zd(s) - z(s)],
will be used, where zd Is the desired (commanded) altitude.
(a)
Sketch a root locus of the closed-loop eigenvalues with respect to the gain K for a
proportional feedback controller, 5q = -K(z-zd). Use Routh’s criterion (or let s = jo ) and find the
roots of the characteristic polynomial) to determine the value of the gain and the associated
frequency at which the system is marginally stable.
(b)
Our intuition and the results of part (a) indicate that a relatively large amount of lead
compensation is required to produce a satisfactory autopilot. Because Steve Fossett was a
millionaire, he could afford a more complex controller implementation. Sketch a root locus of the
closed-loop eigenvalues with respect to the gain K for a double-lead compensator, 5q = Dc(s)(zd
- z). where
/«+araV
U+0.I2,;
(c)
Sketch the magnitude portions of the Bode plots (straight-line asymptotes only) for the open-
loop transfer functions of the proportional feedback and lead-compensated systems.
(d)
Select a gain K for the lead-compensated system to give a crossover frequency of 0.06
rad/sec.
(e)
Select a gain K for the lead-compensated system to give a crossover frequency of 0.06
rad/sec.
(f)
If the emor in part (e) is too large, how would you modify the compensation to give higher low-
frequency gain? (Give a qualitative answer only.)
S te p -b y -s te p s o lu tio n
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 10.13PP
Satellite-attitude control systems often use a reaction wheel to provide angular motion. The
equations of motion for such a system are
Satellite: l ^ = Tc + Ta,
W heel:
= —Tc,
Measurement: Z = ^ —
Control: Tc = -Dc(s)(Z - Zd).
where
J = moment of inertia of the wheel,
/■= wheel speed.
Tc = control torque,
Tc = control torque,
Tex = disturbance torque.
q> = angle to be controlled,
Z = measurement from the sensor,
Zd = reference angle,
/ = satellite inertia (1000 kg/m2),
a = sensor constant (1 rad/sec),
Dc(s) = compensation.
(a)
Suppose Dc(s) = KO, a constant. Draw the root locus with respect to KO for the resulting
closed-loop system.
(b)
For what range of KO is the closed-loop system stable?
(c) Add a lead network with a pole at s = -1 so that the closed-loop system has a bandwidth
wBW = 0.04 rad/sec. a damping ratio ^ = 0.5, and compensation given by
. s+z
De{s) = K i ^ ^ ^ .
Where should the zero of the lead network be located? Draw the root locus of the compensated
system, and give the value of K^ that allows the specifications to be met.
(d)
For what range of K^ is the system stable?
(e)
What is the steady-state error (the difference between Z and some reference input Zd to a
constant disturbance torque Tex for the design of part (c)?)
(f)
(g)
What is the type of this system with respect to rejection of Tex7
Draw the Bode plot asymptotes of the open-loop system, with the gain adjusted for the value
of K^ computed in part (c). Add the compensation of part (c), and compute the phase margin of
the closed-loop system. What is the type of this system with respect to rejection of Tex?
(h)
Write state equations for the open-loop system, using the state variables g>. 'q>. and Z. Select
the gains of a state-feedback controller Tc =
to locate the closed-loop poles at
j = -0 .Q 2 :t0 .0 ^ ^ .
S te p -b y -s te p s o lu tio n
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 10.14PP
Three alternative designs are sketched in Fig. for the closed-loop control of a system with the
plant transfer function Gfsj = l/sfs + 1). The signal w is the plant noise and may be analyzed as if
it were a step; the signal v is the sensor noise and may be analyzed as if it contained power to
very high frequencies.
(a)
Compute values for the parameters K \ , a, K2, KT. K3, d. and KD so that in each case
(assuming w = 0 and v = 0),
*
j2 + 4 i
+ I6 ‘
Note that in system III, a pole is to be placed at s = -4.
(b)
V
Complete the following table, expressing the last entries as A/sk to show how fast noise from
is attenuated at high frequencies:__________________________________________________
(b) Complete the following table, expressing the last entries as A/sk to show how fast noise from
V
is attenuated at high frequencies;
System Kv
-1
i- r1l t ^
1
II
III
(c)Rank the three designs according to the following characteristics (the best as “1 the poorest
as “3”):
Performance
Tracking
Plant-noise rejection
Sensor-noise rejection
Figure Alternative feedback structures
Step-by-step solution
step 1 of 2
R
^
+ aS + k^
:> A i=16. a = 4
^
A_
R
S ^ + {U k ^ )S + k 2
= ^^1 ^= 16.
m.
^
R
r
S’ + ( l+ r f ) S ‘ + ( * „ + fl( + * i) S + 4 1 ^
^
R
A nd
k j. ^ 3
< i( g + 4 )
(S + 4 )^
k j, = 3 { d - 4 )
=> A^ = 16. d = l . k j , = 9
Step 2 of 2
£ fs) = R - y = R ^
,
^+4s+U
S^+A s
S" +4S+16
e,
•R.
= U e (t) = U SB(s) = 1
k , = — ^ k , s 4 fo r all the design
, 1 1
w
I j,
- 2 . . 1
*,
= 1
W ^L ,
4
= 1
16
m -.L \
4* i
= 1
64
Problem 10.15PP
The equations of motion for a cart-stick balancer with state variables of stick angle, stick angular
velocity, and cart velocity are
±=
r
®
3133
1
0
L -31.33 0
0
0.016
-0.216
-0.649 | k.
8.649
y = [ 10 0 0 Jx.
where the output is stick angle, and the control input is voltage on the motor that drives the cart
wheels.
(a) Compute the transfer function from u to y. and determine the poles and zeros.
(b) Determine the feedback gain K necessary to move the poles of the system to the locations
-2.832 and -0.521 ± 1.068j, with wn = 4 rad/sec.
(b) Determine the feedback gain K necessary to move the poles of the system to the locations
-2.832 and -0.521 ± 1.068j, with wn = 4 rad/sec.
(c)
Determine the estimator gain L needed to place the three estimator poles at -10.
(d)
Determine the transfer function of the estimated-state-feedback compensator defined by the
gains computed in parts (b) and (c).
(e)
Suppose we use a reduced-order estimator with poles at -10 and -10. What Is the required
estimator gain?
(f)
Repeat part (d) using the reduced-order estimator.
(g)
Compute the frequency response of the two compensators.
Step-by-step solution
step 1 of 6
a)
The Transfer Function is
a{s):
-0-649 (^+0-0028)
[S-5-59) {S-5-606) [S+0-2)
Step 2 of 6
b)
With Oc=(-Sr+2-832) (S^+2 084 ±4-272y)
The feedback gains are calculated using the ACRERMAIIN’S formula
Step 3 of 6 ^
c)
The Estimator gains with ot^ (5^) = [S +10)^ with det (57 - F + LH)
Step 4 of 6
^
The conq^ensator transfer function can be obtained from
= - i( S 7 - P + a K + L H Y ' l
0-239S{S+5-60){S-30e)
~ (S+23 A ± J 2 2 \) {S - 9 -9 S )
Step 5 of 6
e)
L = [19-85983f
Step 6 of 6
0
D,[S)
2055(8’+5-58) (8’-3-69)
(S +48-2)(S-21-4)
Problem 10.16PP
A 282-ton Boeing 747 is approaching land at sea level. If we use the state given in the case
study (Section 10.3) and assume a velocity of 221 ft/sec (Mach 0.198), then the lateral-direction
perturbation equations are
' -0 .0 8 9 0
p ■
t
0.168
—
-1 .3 3
p
0
* .
-0 .9 8 9
-0 .2 1 7
0.327
0.149
■P
^ = [0
1
0
0]
0.1478
-0 .1 6 6
-0 .9 7 5
1
0.1441 '
0
0
0
■
' 0.0148 '
P '
r
-0 .151
+
0.0636
P
0
'
r
P
. ♦ .
The corresponding transfer function is
G(5) =
r ( j)
-0.151(5-1- ].0 5 ) (j+ 0 .0 3 2 8 ± 0 .4 1 4 0
s m ~ (5-H .IO 9)(s-hO .O 425)(a-h0i)646± O .731/)‘
The corresponding transfer function is
f ( j)
-O .I5 I(J -H l.0 5 )(J -H 0 .0 3 2 8 ± 0 .4 l4 fl
“ (j-H .1 0 9 )(J -h 0 .0 4 2 5 )(s -h 0 .0 6 4 6 ± 0 .7 3 y )*
(a) Draw the uncompensated root locus [for 1 + /C6(sJ] and the frequency response of the
system. What type of classical controller could be used for this system?
(b) Try a state-variable design approach by drawing a SRL for the system. Choose the closedloop poles of the system on the SRL to be ac(s) = {s+ 1.12)(s + 0.165)(s + 0.162 ± 0.681/). and
choose the estimator poles to be five times faster at ae(s) = {s + 5.58)(s + 0.825)(s + 0.812 ±
3.40/).
(c) Compute the transfer function of the SRL compensator.
(d) Discuss the robustness properties of the system with respect to parameter variations and
unmodeled dynamics.
(e) Note the similarity of this design to the one developed for different flight conditions earlier in
the chapter. What does this suggest about providing a continuous (nonlinear) control throughout
the operating envelope?
Step-by-step solution
step 1 of 4
a)
A clas sical lag network could be used to lower the resonant gain
Step 2 of 4
The Symmetric Root locus
l+ j tG ( s ) a ( - s ) = 0
Using Mat lab’s root locus command
* = [0-0308 - 2 1220 1 1 2 -0 - 034]
The estimator gains are
i = [1546-75 39-53 9 7 3 -9 8 f
Step 3 of 4
c)
The con^ ensator transfer function is given by
-38-247(5+0-94479) (5r+0-9851±yl-CT13)
D c (s ) =
(S + 6 -2 9 8 7 )(£ ’ + 0- 85187)
( S + 0- 6 0 3 1 9 ± j 3 - 1 0 )
Step 4 of 4
d)
Use Mat lab bode command
Problem 10.17PP
(Contributed by Prof. L. Swindlehurst) The feedback control system shown in Fig. is proposed as
a position control system. A key component of this system is an armature-controlled DC motor.
The input potentiometer produces a voltage Ei that is proportional to the desired shaft position; Ei
= Kpdi. Similarly, the output potentiometer produces a voltage EO that is proportional to the actual
shaft position: £0 = Kpd. Note that we have assumed that both potentiometers have the same
proportionality constant. The error signal Ei - EO drives a compensator, which in turn produces
an armature voltage that drives the motor. The motor has an armature resistance Ra, an
armature inductance La, a torque constant Kt, and a back emf constant Ke. The moment of
inertia of the motor shaft is Jm, and the rotational damping due to bearing friction is Bm. Finaiiy,
the gear ratio is A/; 1, the moment of inertia of the load is JL, and the load damping is BL.
(a) Write the differential equations that describe the operation of this feedback system.
(aj Wnife tne ainefenfiai dquaiio'nS tnar'deschOS thfe operation bt inis reeaPaciCsystem.
(b) Find the transfer function relating 00 and 6i(s) for a general compensator Dc(s).
(c)
The open-loop frequency-response data shown in Table were taken using the armature
voltage va of the motor as an input and the output potentiometer voltage EO as the output.
Assuming that the motor is linear and minimum-phase, make an estimate of the transfer function
of the motor.
G(s) =
0m(s)
where dm is the angular position of the motor shaft.
(d)
Determine a set of perfonnance specifications that are appropriate for a position control
system and will yield good performance. Design Dc(s) to meet these specifications.
(e)
Verify your design through analysis and simulation using Matlab.
Figure A servomechanism with gears on the motor shaft and potentiometer sensore
Frequency-Response Data
Frequency (rad/sec)
Frequency (rad/sec)
0.1
60.0
10.0
14.0
0.2
54.0
20.0
2.0
0.3
50.0
40.0
-10.0
0.5
46.0
60.0
-20.0
0.8
42.0
65.0
-21.0
1.0
40.0
80.0
-24.0
2.0
34.0
100.0
-30.0
3.0
30.5
200.0
-48.0
4.0
27.0
300.0
-59.0
5.0
23.0
500.0
-72.0
7.0
19.5
Step-by-step solution
step 1 of 6
Step 2 of 6
Using KIRCHHOFF’S voltage laws we can write
The torque of the motor. T is proportional to the angular speed, hence
dt
Using NEWTON Law OF Motion, we have
JA
= T -P r.
J
Where r^, fee radius of the gear is connected to the motor shaft and
fee gear connected to the ou^ut shaft.
We have
r, = M-. => e . = -N %
is the radius of
Step 3 of 6
First we will find the transfer function from ^ to Sg and then we will find the
closed loop transfer function.
N k,
[if iJ . + J ,)
Next
0
0
=
e.
1
0
4.
So
0
-a
^
-P
0
+ 0
1
.4 .
.A..
e.
K
A..
And so the closed loop transfer function is
M £1
_
k ,a {s )D ,{S )
step 4 of 6
O (S’) .
k
S(£r+5)(S+70)
step 5 of 6
1p
^
"
* — * 0-3 sec
6
Step 6 of 6
^ '
(S + 2 0 0 f
Problem 10.18PP
Design and construct a device to keep a ball centered on a freely swinging beam. An example of
such a device is shown in Fig. It uses colls surrounding permanent magnets as the actuator to
move the beam, solar cells to sense the ball position, and a hall-effect device to sense the beam
position. Research other possible actuators and sensors as part of your design effort.
Compare the quality of the control achievable for ball-position feedback only with that of multipleloop feedback of both ball and beam position.
Figure Ball-balancer design example Source: Photo courtesy o f David Powell
Step-by-step solution
step 1 of 1
STEP RESPONSE
Problem 10.19PP
Design and construct the magnetic levitation device shown in Fig. You may wish to use LEGO
components in your design.
Figure Magnetic bail levitator used in the laboratory Source: Photo courtesy o f Gene Franklin
fl
3
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 10.20PP
Design and build a Sun tracker using an Arduino board and related software.
Step-by-step solution
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 10.21PP
Run-to-Run Control: Consider the RTP system shown in Fig. We wish to heat up a
semiconductor wafer, and control the wafer surface temperature accurately using rings of
tungsten halogen lamps. The output of the system is temperature Tas a function of time: y = T(t).
The system reference input R is a desired step in temperature {700° C). and the control input is
lamp power. A pyrometer is used to measure the wafer center temperature. The model of the
system is first order, and an integral controller is used as shown in Fig. Normally, there is no
sensor bias {b = 0).
Figure RTP system
(a) Suppose the system suddenly develops a sensor bias b ^O . where b is known. What can be
done to ensure zero steady-state tracking of temperature command R. despite the presence of
the sensor bias?
(b) Now assume b = 0. In reality, we are trying to control the thickness of the oxide film grown
(Ox) on the wafer and not the temperature. At present, no sensor can measure Ox in real time.
The semiconductor process engineer must use off-line equipment (called metrology) to measure
the thickness of the oxide film grown on the wafer. The relationship between the system output
temperature and Ox is nonlinear and given by the integral of the Arrhenius equation;
4r
Oxide diickiiess = J pe~'^dt.
where ff is the process duration, and p and c are known constants. Suggest a scheme in which
the center wafer oxide thickness Ox can be controlled to a desired value (say. Ox = 5000 A) by
employing the temperature controller and the output of the metrology.
Step-by-step solution
step 1 of 2
a)
We just increase R by + b Le. replaces R by (i? +b) to cancel the sensor bias.
Step 2 of 2
■y-T
T =
5R
s a
Problem 10.22PP
Develop a nonlinear model for a tungsten halogen lamp and simulate it in Simulink.
Step-by-step solution
step 1 of 1
Let us define the normalized temperature
T
x = — andRe-w rite Eq(12) as the non linear. First order system
%
X ——A
A =
B =
—x*^)
^ ~IT
4 e o T;^
le d
■4p.opC£^„
i = 4 A 4 j: - B
1
ic d
16s o 7
The Kadioactive power
A A 4
7
=
?
Problem 10.23PP
Develop a nonlinear model for a pyrometer. Show how temperature can be deduced from the
model.
Step-by-step solution
step 1 of 2
The total black body radiation intensity , is obtained by integrating over all
frequencies or wave length.
frequencies or wave length.
<j 7-‘
Step 2 of 2
The Black body Emissive flux is given by
Where
C, = 3-7419 X 10r“ ffl/W, Cj = 14388
0
= a’ c r *
The temperature may be determined as
r = ^
Where
’
Problem 10.24PP
Repeat the RTP case study design by summing the three sensors to form a single signal to
control the average temperature. Demonstrate the performance of the linear design, and validate
the performance on the nonlinear Simulink simulation.
Step-by-step solution
step 1 of 2
A Linear model for the system was dehTed in the text as
r
=
Y= H^T + J ^
A Linear model for the system was derived in the text as
7* =
Y= H^T + J ^
Where
and
^ =L
F =
■-0-0682
0-0149
0-0000
0-0458
0-0000
-0-1181
0-4683
0-0218
-0-1008
0-5122
G= 0-5226
0-4185
Step 2 of 2
a (^ r) = ■
0 ■4344 (3-+0- 0878) (S + 0-1485)
” (S + 0 1482) ( J + 0 0527) (S +0 0863)
We may try a simple PI controller of the form
(8' + 0 0527)
Doi.S) ■
S
Problem 10.25PP
One of the steps in semiconductor wafer manufacturing during photolithography is performed by
placement of the wafer on a heated plate for a certain period of time. Laboratory experiments
have shown that the transfer function from the heater power, u, to the wafer temperature, y, is
given by
^
u(s)
= C(s) = __________ 2:5?__________ . .
(j + 0 .1 9 )(5 + 0.78)( s + 0 .00018)
(a) Sketch the 180* root locus for the uncompensated system.
(b) Using the root-locus design techniques, design a dynamic compensator, Dc(s), such that the
system meets the following time-domain specifications
i. Mp < 5%.
3IS me Toiiowing iime-aomain specincaiions
i. Mp < 5%.
ii. fr< 20 sec.
Mi. ts < 60 sec,
iv. Steady-state error to a 1*C step input command < 0.1 *C.
Draw the 180° root locus for the compensated system.
S te p -b y -s te p s o lu tio n
There is no solution to this problem yet.
G et help from a Chegg subject expert.
ASK AN EXPERT
Problem 10.26PP
Excitation-Inhibition Model from Systems Biology {Yang and Iglesias, 2005): In Dictyostelium
cells, the activation of key signaling molecules involved in chemoattractant sensing can be
modeled by the following third order linearized model. The external disturbance to the output
transfer function is:
wd)
' '
( I + o ) ( j + 1)(J + y ) '
where, w is the external disturbance signal proportional to chemoattractant concentration, and y
is the output which Is the fraction of active response regulators. Show that there is an alternate
representation of the system with the “plant” transfer function
____________ O jlf ! ) __________
j2 + ( l + o + y )« + (a + y + o K ) ’
and the “feedback regulator"
j^ + ( l + o + K)5 + (« + y + a y )
and the “feedback regulator"
It is known that a # 1 for this version of the model. Draw the feedback block diagram of the
system showing the locations of the disturbance Input and the output. What is the significance of
this particular representation of the system? What hidden system property does it reveal? Is the
disturbance rejection a robust property for this system? Assume the system parameter values
are a = 0.5 and y = 0.2, then plot the disturbance rejection response of the system for a unit step
disturbance input.
Step-by-step solution
step 1 of 5
The external disturbance to the ou^ut transfer function is given by
wM
( s + a ) ( s + l) ( s + y )
Where w is the external disturbance signal proportional to chemo attractant
concentration, and^ is the output which is the fraction of active response regulators.
It is known that a ^ Ifor this version of the model.
Step 2 of 5
Sketch the Feedback loop representatioa
Step 3 of 5 ^
Find the Lr^lace transform of the given equation
___________ 0 - “ )___________
^ (^ ) _
+ (1 + Of +
1+
W( s )
y)s+{a+ y+ay)
ay
+ (1 + Of + y)s + ( a + y + ay)
(s + a ) ( « + l) ( f f + y)
Step 4 of 5
The significance of this particular representation is that it reveals the internal model,
namely the pure integrator. Hence the system is Type I with respect to disturbance
rejection. It rejects constant disturbances in a robust &shion
FindG(s) fo ra = 0.5 and y = 0.2.
(1 - 0 .5 ) b
0 (s) =
s" + (1 + 0.5 + 0.2)s + (0.5 + 0.2 + 0.5 X 0.2)
0 ( .) =
0.5s
s’ + 1.7s + 0.8
Step 5 of 5 ^
The disturbance response is shown in the following figure.
Step Response
Time (sec)
Thus, the disturbance rejection response of the system is sketched.