O N E
Introduction
ANSWERS TO REVIEW QUESTIONS
1. Guided missiles, automatic gain control in radio receivers, satellite tracking antenna
2. Yes - power gain, remote control, parameter conversion; No - Expense, complexity
3. Motor, low pass filter, inertia supported between two bearings
4. Closed-loop systems compensate for disturbances by measuring the response, comparing it to
the input response (the desired output), and then correcting the output response.
5. Under the condition that the feedback element is other than unity
6. Actuating signal
7. Multiple subsystems can time share the controller. Any adjustments to the controller can be
implemented with simply software changes.
8. Stability, transient response, and steady-state error
9. Steady-state, transient
10. It follows a growing transient response until the steady-state response is no longer visible. The
system will either destroy itself, reach an equilibrium state because of saturation in driving
amplifiers, or hit limit stops.
11. Transient response
12. True
13. Transfer function, state-space, differential equations
14. Transfer function - the Laplace transform of the differential equation
State-space - representation of an nth order differential equation as n simultaneous first-order
differential equations
Differential equation - Modeling a system with its differential equation
SOLUTIONS TO PROBLEMS
1. Five turns yields 50 v. Therefore K =
50 volts
= 1.59
5 x 2π rad
2 Chapter 1: Introduction
2.
Desired
temperature
Voltage
difference
Temperature
difference
Actual
temperature
Fuel
flow
+
Amplifier and
valves
Thermostat
Heater
-
3.
-
Roll
angle
Roll
rate
Aileron
position
control
+
Pilot
controls
Aileron
position
Error
voltage
Input
voltage
Desired
roll
angle
Aircraft
dynamics
Integrate
Gyro
Gyro voltage
4.
Input
voltage
Desired
speed
Speed
Error
voltage
+
transducer
Motor
and
drive
system
Amplifier
Voltage
proportional
to actual speed
Dancer
position
sensor
Actual
speed
Dancer
dynamics
5.
Input
voltage
Desired
power
Transducer
Rod
position
Power
Error
voltage
+
Amplifier
-
Voltage
proportional
to actual power
Motor
and
drive
system
Sensor &
transducer
Actual
power
Reactor
Solutions to Problems 3
6.
Desired
student
population
Desired
student
rate
Population
error
+
Graduating
and
drop-out
rate
Administration
Actual
student
rate
+
Actual
student
population
Net rate
of influx
Integrate
Admissions
-
7.
Voltage
proportional
to desired
volume
Desired
volume
+
Transducer
Volume
error
Voltage
representing
actual volume
Volume
control circuit
Radio
Effective
volume
+
-
Transducer
Speed
Voltage
proportional
to speed
Actual
volume
4 Chapter 1: Introduction
8.
a.
Fluid input
Valve
Actuator
Power
amplifier
+V
Differential
amplifier
+
-
R
Desired
level
-V
+V
R
Float
-V
Tank
Drain
b.
Desired
level
Potentiometer
voltage
in
+
Amplifiers
-
Actuator
and valve
Actual
level
Flow
rate in
+
Integrate
-
Drain
Flow
rate out
voltage
out
Displacement
Potentiometer
Float
Solutions to Problems 5
9.
Desired
force
Current
+
Transducer
Displacement
Amplifier
Actual
force
Displacement
Actuator
and load
Valve
Tire
-
Load cell
10.
Commanded
blood pressure +
Actual
blood
pressure
Isoflurane
concentration
Vaporizer
Patient
-
11.
Desired
depth +
Controller
&
motor
-
Force
Feed rate
Grinder
Depth
Integrator
12.
Coil
voltage +
Desired
position
Coil
circuit
Transducer
Coil
current
Solenoid coil
& actuator
-
LVDT
13.
a. L
di
+ Ri = u(t)
dt
Force
Armature
&
spool dynamics
Depth
6 Chapter 1: Introduction
b. Assume a steady-state solution iss = B. Substituting this into the differential equation yields RB =
1,
1
R
. The characteristic equation is LM + R = 0, from which M = - . Thus, the total
R
L
1
1
solution is i(t) = Ae-(R/L)t + . Solving for the arbitrary constants, i(0) = A +
= 0. Thus, A =
R
R
1
1
1 -(R/L)t 1
−( R / L)t
. The final solution is i(t) =
-= (1 − e
e
).
R
R R
R
from which B =
c.
14.
di 1
idt + vC (0) = v(t)
+
dt C ∫
d 2i
di
b. Differentiating and substituting values,
+ 30i = 0
2 +2
dt
dt
a. Writing the loop equation, Ri + L
Writing the characteristic equation and factoring,
2
M + 2 M + 30 = M + 1 + 29 i M + 1 -
29 i .
The general form of the solution and its derivative is
-t
i = e cos
29 t A + B sin
29 t e
-t
di = - A + 29 B e -t cos 29 t - 29 A + B e- t sin
dt
v (0) 1
di
Using i(0) = 0; (0) = L
= =2
L
L
dt
i 0 = A =0
di
(0) = − A + 29 B =2
dt
2
.
Thus, A = 0 and B =
29
The solution is
29 t
Solutions to Problems 7
i=
2
29
-t
29 t
29 e sin
c.
i
t
15.
a. Assume a particular solution of
Substitute into the differential equation and obtain
Equating like coefficients,
35
From which, C = 53
10
and D = 53 .
The characteristic polynomial is
Thus, the total solution is
35
35
Solving for the arbitrary constants, x(0) = A +53 = 0. Therefore, A = - 53 . The final solution is
b. Assume a particular solution of
xp = Asin3t + Bcos3t
8 Chapter 1: Introduction
Substitute into the differential equation and obtain
(18A − B)cos(3t) − (A + 18B)sin(3t) = 5sin(3t)
Therefore, 18A – B = 0 and –(A + 18B) = 5. Solving for A and B we obtain
xp = (-1/65)sin3t + (-18/65)cos3t
The characteristic polynomial is
2
M +6 M+ 8 = M+ 4 M+ 2
Thus, the total solution is
18
1
cos 3 t sin 3 t
65
65
18
Solving for the arbitrary constants, x(0) = C + D −
=0.
65
x =C e
-4t
+ De
-2t
+ -
Also, the derivative of the solution is
dx = - 3 cos 3 t + 54 sin 3 t
-4t
-2t
- 4 C e -2 D e
dt
65
65
.
15
3
3
.
Solving for the arbitrary constants, x(0) −
− 4C − 2D = 0 , or C = − and D =
10
65
26
The final solution is
x =-
18
3 - 4 t 15 - 2 t
1
e
e +
sin 3 t cos 3 t 26
65
10
65
c. Assume a particular solution of
xp = A
Substitute into the differential equation and obtain 25A = 10, or A = 2/5.
The characteristic polynomial is
2
M + 8 M + 25 = M + 4 + 3 i M + 4 - 3 i
Thus, the total solution is
x=
2
-4t
+e
B sin 3 t + C cos 3 t
5
Solving for the arbitrary constants, x(0) = C + 2/5 = 0. Therefore, C = -2/5. Also, the derivative of the
solution is
dx
-4t
dt = 3 B -4 C cos 3 t - 4 B + 3 C sin 3 t e
Solutions to Problems 9
.
Solving for the arbitrary constants, x(0) = 3B – 4C = 0. Therefore, B = -8/15. The final solution is
x (t ) =
2
2
⎛8
− e −4t
sin(3
t ) + cos(3t )⎞
⎝ 15
⎠
5
5
16.
a. Assume a particular solution of
Substitute into the differential equation and obtain
Equating like coefficients,
1
From which, C = - 5
1
and D = - 10 .
The characteristic polynomial is
Thus, the total solution is
11
1
Solving for the arbitrary constants, x(0) = A - 5 = 2. Therefore, A =
. Also, the derivative of the
5
solution is
dx
dt
.
3
Solving for the arbitrary constants, x(0) = - A + B - 0.2 = -3. Therefore, B = − . The final solution
5
is
1
1
3
−t ⎛ 11
x (t ) = − cos(2t ) − sin(2 t ) + e
cos(
t ) − sin(t )⎞
⎝5
⎠
5
10
5
b. Assume a particular solution of
xp = Ce-2t + Dt + E
Substitute into the differential equation and obtain
10 Chapter 1: Introduction
Equating like coefficients, C = 5, D = 1, and 2D + E = 0.
From which, C = 5, D = 1, and E = - 2.
The characteristic polynomial is
Thus, the total solution is
Solving for the arbitrary constants, x(0) = A + 5 - 2 = 2 Therefore, A = -1. Also, the derivative of the
solution is
dx
= ( − A + B )e − t − Bte −t − 10e −2 t + 1
dt
.
Solving for the arbitrary constants, x(0) = B - 8 = 1. Therefore, B = 9. The final solution is
c. Assume a particular solution of
xp = Ct2 + Dt + E
Substitute into the differential equation and obtain
1
Equating like coefficients, C = 4 , D = 0, and 2C + 4E = 0.
1
1
From which, C = 4 , D = 0, and E = - 8 .
The characteristic polynomial is
Thus, the total solution is
9
1
Solving for the arbitrary constants, x(0) = A - 8 = 1 Therefore, A = 8 . Also, the derivative of the
solution is
dx
dt
.
Solving for the arbitrary constants, x(0) = 2B = 2. Therefore, B = 1. The final solution is
Solutions to Problems 11
17.
Spring
displacement
Desired
force
Input
voltage +
Input
transducer
F up
Controller
Actuator
-
Sensor
Pantograph
dynamics
F out
Spring