INTRODUCTION TO AUTOMATIC
CHAPTER 1: CONTROL SYSTEM
Control system: Subsystems and processes (or plants) assembled for the purpose of
obtaining a desired output with desired performance.
Input: stimulus
Desired response
Control system
Output: response
Actual response
Examples:
 In nature pancreas regulating our blood sugar.
 A motor speed controller regulating the speed to the desired amount.
OPEN LOOP AND CLOSED LOOP CONTROL SYSTEMS
Open-loop: the system gives an output based on pre-determined relationship between
the output and input. OR simply is not dependent on the output “no feedback”.
Example: Automatic dish washer
Input
Advantages
• Simple design
• Cheap
Control System
output
Dis-advantages
• Can not eliminate the effect of
disturbances.
• Can not be used for systems requiring
higher accuracy.
OPEN LOOP AND CLOSED LOOP CONTROL SYSTEMS
Closed-loop: the system gives an output based on not only pre-determined relationship
between the output and input but also every time checks whether the output is in
desired range. OR simply is dependent on the output “feedback”. Are self-correcting.
Examples: Walking man (eyes open), tracking systems
Control System
Input
Advantages
• Able to reject disturbances
• More accurate
Dis-advantages
• Complex.
• Expensive.
output
MATHEMATICAL
MODELLING
OF
CHAPTER 2: PHYSICAL SYSTEMS
Mathematical modelling: representation of
Electrical networks
physical systems mathematically which make
Voltage source
=i =
= ∫id
them easy for computation and simulation
Current source i =
purposes.
Mathematical modelling of
 Mechanical systems
Translational: F=ma=Bv=kx
Rotational: T=Jα=Bω=kθ
Analogy
F-V
F-I
= ∫ d =
MATHEMATICAL MODELLING OF MECHANICAL
SYSTEMS: TRANSLATIONAL
∑
B
Damper B
k
Spring k
m
=
F-Fb-Fs=ma
Where:
F=ma-Fb-Fs
F-applied force
F=ma+Bv+kx
Fb-force by damper
F=m
Mass m
+kx
Fs-force of spring
a-acceleration
F=m
F
=
Force F
+B
=
+Bv+k ∫ d
d
d
=
d
d
Or
d
=
v- velocity
x- displacement
d
MECHANICAL TRANSLATIONAL
k
k1
B1
x1….
CONT.…
m1
k2
x1….
m1
L1
L2
B12
B
x2…..
m2
F
x2…..
m2
F
MATHEMATICAL MODELLING OF MECHANICAL
SYSTEMS: ROTATIONAL
∑
k
J
= α
T-Tb-Ts= α
T
Where:
T=Jα+Tb+Ts
T-applied torque
F=Jα+Bω+kθ
Tb-torque resistance by damper
B
T=J
+B
+kθ
Ts-torque resistance by spring
α-angular acceleration
ω- angular velocity
=
Or
θ= ∬ α d = ∫ ω d
α=
θ- angular displacement
MECHANICAL ROTATIONAL
CONT.…
k1
J1
B1
J2
k12
ϴ1
T
ϴ2
J2
T
N2
k1
B2
J1
=
=
N1
B1
ELECTRICAL NETWORKS
R
L
i
C
iR
iL
R
L
i
v
v
=
v= i +
+
+
i=
+ ∫id
F-v R-B L-m
F=m
−
+
i=
-B
C
+
+Bv+k ∫ d
F-i
iC
-k
C-m
1
+
1
d +
d
d
ANALOGOUS CIRCUITS Find f-v, f-i and draw circuit
x1….
k
k1
B1
m1
k2
x1….
m1
L1
L2
B12
B
x2…..
m2
F
x2…..
m2
F
ELECTROMECHANICAL SYSTEMS
Ra
JL
La
eb= kbω
N2
Tm=ktia
ia
ea
N1
eb
Tm, θ
TRANSFER FUNCTIONS
Transfer function: a mathematical equation that relates
input of a function to the output [
{ }
].
{ }
1. The equation is derived after converting the function into Laplace domain
2. All initial conditions are zero
3. Works only for linear functions
BASIC LAPLACE TRANSFORM TABLE
Time domain
Laplace domain
u(t) step input
1
tu(t)
1
s2
!
tnu(t)
e-atu(t)
d
( )
f(t)±g(t)
sn+1
1
s+a
( )
s
snF(s)
F(s)±G(s)
FIND TRANSFER FUNCTION
For the mechanical system given above find transfer function
a. Input force to displacement 1 (x1)
b. Input force to displacement 2 (x2)
k1
B1
R
x1….
m1
L
C
k2
i
v
B12
x2…..
m2
Voltage input to the system to voltage at capacitor
F
SIMPLIFYING MULTIPLE SUBSYSTEM
BLOCK DIAGRAM REDUCTION
Forward path
G1(s)
Takeoff point
R(s)
G2(s)
Summing point
C(s)
G3(s)
Feedback path
Block
BLOCK DIAGRAM REDUCTION
Block diagram
R(s)
R(s)
Transfer function
G(s)
C(s)
G1(s)
R(s)
R(s)
±
G2(s)
G(s)
H(s)
±
( )
( )
G(s)
C(s)
G2(s)
G1(s)
CONT.…
G1(s)G2(s)
Series blocks
G1(s)±G2(s)
Parallel blocks
G( )
1 ∓ H s G( )
feedback
C(s)
C(s)
BLOCK DIAGRAM REDUCTION MOVING TAKEOFF POINT
G3
G1
G3/G1
G2
G1
G3
G1
G2
G2
G1G3
G1
G2
BLOCK DIAGRAM REDUCTION MOVING SUMMING POINT
G3/G2
G3
G1
G2
G1
G3
G1
G2
G2G3
G2
G1
G2
BLOCK DIAGRAM REDUCTION examples
H1
G6
R(s)
G1
G2
G3
G5
G4
C(s)
SIGNAL FLOW GRAPH (SFG)
i
1
a
2
b
3
j
c
4
d
5
e
7
f
k
8
g
9
h
10
m
m
• Node: a point where a path start or stop (i.e. 1, 2, 3…)
• Touching loop: loops having common node
• Path gain: value on arrow (i.e. a, b, c…)
• Non-touching loop: loops without common node.
• Loop: start and sink on same node.
• Dummy node: a path with gain equal to 1.
• Self loop:- start and sink on same node without
intermediate nodes
SIGNAL FLOW GRAPH REDUCTION
e
j
f
a
b
k
c
d
a
b
c
d
g
=
k= ef + ab + hg
d= c(ef + ab + hg)
f
k
h
d= kc
e
ℎ( + )
1−
g
h
i
SFG Masons Gain Formula(MGF)
9 =
5
=
∑
Δ
Δ
Pi-gain of the ith forward path
N- total number of forward paths
∆= 1 - ∑ loop gains + ∑ nontouching-loop gains taken two at a time +
∑ nontouching-loop gains taken three at a time –
∑ nontouching-loop gains taken four at a time …
∆i= value of ∆ after eliminating all the loops that touch its forward path.
G9
R(s)
G1
G2
G10
G3
G4
G5
H1
G6
G7
G9
H2
G8
C(s)
BD TO SFG
Convert into SFG and find gain using MGF
H1
G6
R(s)
G1
G2
G3
G5
G4
C(s)
CONTENTS
Feedback and its properties
Stability
Time response
Experimental determination of transfer function
First order system response
Controllers
Second order system response
Physical realization of controllers
Steady state error
Simulation of Mechanical Control Systems
Using SIMULINK
By: Chalachew W.
FEEDBACK AND ITS PROPERTIES
A control system mainly categorized into open and closed loop systems.
Closed loop systems use plant response(output) for determining appropriate control action from
controller. These closed loop systems are also called feedback controllers.
A controller with a feedback may affect properties like
Parameter variation: variation of parameters which are expected (i.e. working domain).
Overall gain: the magnitude in which the input is multiplied.
Disturbance: parameter variation which is unexpected.
Sensitivity: the amount of deviation on a parameter caused by deviation in another parameter.
Stability: response of a system to be in a given band (BIBO).
Time constant: time required the system response to reach 63% of final/desired output.
PARAMETER VARIATION
For a given system with input R(s), output
C(s) and parameter G(s). So transfer
function (i.e. tf=C(s)/R(s)) becomes G(s).
R(s)
G(s)
For closed loop
R(s)
G(s)
C(s)
tf =
G(s)
1 − G s H(s)
tf =
G s + ∆G(s)
1 − [G s +∆G(s)]H(s)
H(s)
C(s)
R(s)
G(s)+∆G(s)
What will be the new tf if G(s) becomes
C(s)
H(s)
G(s)+∆G(s)
tf =
For open loop
( )
∆G(s)
( )
For G(s)>>∆G(s)
tf=G(s)+∆G(s)
R(s)
+
+
G(s)+ ∆G(s)
C(s)
From this we can conclude parameter variation has less effect
on closed loop system than open loop system.
OVERALL GAIN
Open loop
tf=G(s)
Closed loop
Generally closed loop system has smaller
over all gain.
DISTURBANCE
Disturbance on open loop system
Disturbance on a closed system may
occur on
D(s)
R(s)
( )
=
( )
G1(s)
G2(s)
G2(s)
C(s)
a.
b.
c.
Forward path
Output or
Feedback path
To study the effect of disturbance we
remove the input and examine the
response caused by the disturbance
DISTURBANCE ON FORWARD PATH
D(s)
D(s)
R(s)
G1(s)
G2(s)
G2(s)
C(s)
C(s)
G1(s)
H(s)
H(s)
Making the input zero
C(s)
G2(s)
=
(s) 1 − G1(s)G2(s)H(s)
DISTURBANCE ON OUTPUT
D(s)
D(s)
R(s)
G1(s)
C(s)
C(s)
G1(s)
H(s)
H(s)
Making the input zero
C(s)
1
=
(s) 1 − G1(s)H(s)
DISTURBANCE ON FEEDBACK PATH
R(s)
R(s)
G(s)
H1(s)
G(s)
C(s)
H2(s)
H1(s)
H2(s)
D(s)
C(s)
H1(s)G(s)
=
(s) 1 − H1(s)H2(s)G(s)
From these equations we can conclude that closed system less affected by disturbance.
C(s)
SENSITIVITY
How much is the impact of variation in
parameter ‘b’ on parameter ‘a’.
=
⋅
( )
a=
Effect of variation of forward path gain
on overall transfer function on open loop.
a=G(s)
( )
=
=
b=G(s)
=
Effect of variation of forward path gain
on overall transfer function on closed
loop.
⋅ =
( )
( )
⋅
( )
=1x1=1
( )
(
[
)
; b=G(s)
]⋅
( )
[
]
( )
So, closed loop systems are less sensitive than open
loop systems.
STABILITY AND TIME CONSTANT
In general, if properly designed closed loop systems can improve both stability and time
constant.
For stable open loop system making it closed loop may affect the system negatively (i.e.
may become unstable.)
TIME RESPONSE
Transient/natural response,
and die out as time goes. lim
→
Steady state /forced response,
transient response die out.
Total response C(t)=
: response of the system as the input is applied
=0
: response of a system which remain after
+
Order: Order (i.e. highest degree) of differential equations(denominator after cancelling
common factors in the numerator) describing the system.
For system equation represented with numerator and denominator zeros of numerator
is called zeros of the system and zeros of denominator is called poles of the system.
FIRST ORDER SYSTEM RESPONSE
Time constant: time in which the response
reaches 63% of the final value.
Let G(s) first order system with tf G(s)=
and R(s)= (i.e. unit step input)
Output C(s)=R(s)G(s)=
.
The maximum possible value of c(t) is 1
which is the final value and 63% becomes
0.63.
1-e
e
C(s)= .
= +
=>A=1, B=-1;
C(s)= −
c(t)=1-e
=0.63
= 0.37
e
=
- t=-0.994
taking inverse Laplace transform
t≈ 1/
0.37
OPEN VS CLOSED LOOP TIME CONSTANT
R(s)
C(s)
Comparing
& =
k
C(s)=R(s).
C(s)= .
C(s)= −
t=
=
;
;
b=
;
−
=
(
)
;
> 1,
ℎ
.
FIRST ORDER SYSTEM RESPONSE
Rise time ( ): time taken for the system
to reach 10%-90% of the final output.
1-e
=0.9
=
1-e
1-e
.
=
=0.1
=
=
Settling time ( ): time taken the output
to reach and stay ±2% of final output for
the first time.
.
−
=
.
−
.
=
.
=0.98
SECOND ORDER SYSTEM RESPONSE
General second order system


- natural frequency
– damping constant
Find natural frequency and damping constant of
a.
b.
( )
, if
( )
f t = 27 ( ), M = 3, fv =6, K=27
G(s)=s2+4.2s+36
CLASSIFICATION OF 2 ND ORDER SYSTEM USING DAMPING RATIO
=
1−
SECOND-ORDER UNDER DAMPED RESPONSE SPECIFICATIONS
SECOND-ORDER SYSTEM RESPONSE SPECIFICATIONS
UNDERDAMPED
lim
→
STEADY STATE ERROR
e ∞ =lim
→
=
∞ = lim
→
R(s)
=lim 1+G(s)H(s)
→
Steady state error: error of a system as time
approaches to infinity.
Determine order and type of G(s) then find steady state error for
system with
E(s)=R(s)-H(s)C(s)
a.
1
G(s)=s+2, H(s)=0
b.
1
G(s)=s+2, H(s)=1
c.
1
G(s)=s(s+2) , H(s)=1
d.
G(s)=
C(s)=E(s)G(s)
E(s)=R(s)-E(s)G(s)H(s)
E(s)(1+G(s)H(s))=R(s)
R(s)
E(s)=1+G(s)H(s)
1
, H(s)=1
(s+2)
 For step, ramp and parabolic input
RELATIONSHIP BETWEEN INPUT, SYSTEM TYPE, STATIC
ERROR CONSTANTS AND STEADY STATE ERRORS
STABILITY OF CONTROL SYSTEM
Stability and transient response of a system affected by location of closed loop poles in
s-plane.
Routh-Hurwitz: tells us number of poles on the right half s-plane.
Root-Locus: shows graphically the movement of poles on s-plane for different
parameter k.
 As the parameter changes
 If the pole goes to the right the system is becoming unstable.
 If the pole goes to the left the system is becoming more stable.
?
ROUTH-HURWITZ STABILITY
CRITERION
How can I know a pole lie on RHP
ROUTH-HURWITZ STABILITY CRITERION
CONT.…
s4
a4
a2
a0
The necessary and sufficient condition the system to be
stable is there SHOULD NOT BE any sign change in
the first column.
s3
a3
a1
0
If there is any sign change
s2
b1
b2
0
s1
c1
0
0
 The system is unstable.
 The number of sign change shows the number of poles in the
RHP.
s0
d1
0
0
If there is zero
 The system is marginally stable.
OPEN VS CLOSED LOOP STABILITY
R(s)
C(s)=
C(s)=
Closed loop
1
1
C(s)
k
C(s)=
Open loop
;
; closed loop
; open loop
- k
- k
For >0; the system is stable
If >0; to be the system stable
k should be less than 1 (i.e. k<1);
ROUTH-HURWITZ STABILITY CRITERION
Check stability of a system given by
1.
G(s)= 2
and H(s)=0.3;
s +4s+2
2.
( )
=
( ) s3
3.
( )
=
( ) s5+3s4+3s3
+6s2+5s+3
4.
( )
=
( ) s5+2s4+3s3
+6s2+5s+3
+5s2+10s+2
CONT.…
SPECIAL CASES OF ROUTH-HURWITZ
1.
Zero in the first column of row
Check stability of the system
a.
Substitute with small number ε: check for sign change for positive
and negative ε.
b.
2.
a.
( )
=
( ) s5+2s4+3s3
+6s2+5s+3
b.
( )
=
( ) s5+7s4+6s3
+42s2+8s+56
Reverse coefficients: reverse coefficients of the polynomial.
Row of zero
 Use coefficients of auxiliary polynomial.
 Auxiliary polynomial is derivative of the polynomial just above row zero
NB: In these cases the system never become stable but it can be marginally stable.
DESIGN USING ROUTH-HURWITZ
Finding the range of ‘gain’ in which the given system can be
stable.
Find value of k in which the system
a.
G(s)=s(1+0.5s)(1+0.5s) and H(s)=0.2
b.
G(s)=s(1+0.6s)(1+0.4s) and H(s)=0.3
i.
To be stable
ii.
To be marginally stable, how much is its’ frequency of
sustained oscillation
EXPERIMENTAL DETERMINATION OF TRANSFER FUNCTION
Approximated to 1st or 2nd order system
Some parameters need to be collected using
sensors.
1st order system:
 G(s)=s+a
 Whose step response is
/
/
 C(s)=s(s+a) = s - s+a
The collected data plotted with time to
determine the transfer function.
 C(t)=a (1-
)
 C(ꝏ)=a
 If we get the time constant from experiment we can
easily determine the transfer function.
EXPERIMENTAL DETER…
CONT.…
2nd order system:
Percent overshoot and settling time used
for determining denominator or poles
of the transfer function.
NB: In both cases the system should settle to some final
value.
CONTROLLERS
We can alter system dynamics (parameters) to
Commonly used industrial controllers
meet the required responses but sometimes
are:-
this cannot be done because of several
 ON-OFF
reasons.
 Proportional
 Proportional-Derivative (PD)
At this time we need to use controllers.
 Proportional-Integral (PI)
 Proportional-Integral-Derivative (PID)
CONTROLLERS
CONT.…
Water level controller is used as an example in this topic.
The in flow amount controlled by a servo motor based on the set point. The
out flow is uncontrolled.
The deviation of the water level from set point (desired level) is error signal
which is going to feed the controller.
The controller turns the servo to open and close the tap based on the error
amount.
The angle movement is from 0⁰(fully closed), 45⁰(half open) 90⁰ (fully open).
Assumption: Negative error(0⁰)_Zero error(45⁰)_Positive error(90⁰)
ON-OFF CONTROLLER
This two position works as a switch,
Dead band is introduced to prevent chattering
 If the level is below set point, tap is fully open.
effect
 If the level is on and above set point, tap is fully
 Dead band a range of values below or above set
closed.
Solenoid operated valves and relays are ONOFF type controllers
Even though, these controllers are simple and
economical they are not suitable for complex
systems.
point in which the controller take no action.
PROPORTIONAL
The control action is proportional to the present error.
u=Kp x e
If the error is about negative the tap turns toward 0⁰.
If the error is about positive the tap turns toward 90⁰
Make the system responds faster .
The motion is continuous.
May not eliminate steady state error.
INTEGRAL
The control action is proportional to summation of the past
errors.
u=Ki ∫ e
d
U(s)= E(s)
If the error was more positive, most likely the tap remain
open.
If the error was more negative, most likely the tap remain
closed.
Eliminate steady state error.
DERIVATIVE
The control action is proportional to rate of change of
the error. Dependent on the trend of the error rather
than current condition.
u=Kd
U(s)=sKd x E(s)
If the error rate is positive, the tap should be more open.
If the error is negative, the tap should be more closed.
Derivative controller does not used alone.
Reduce oscillation so reduces settling time
PROPORTIONAL INTEGRAL (PI)
Combination of proportional and integral
controllers.
u = Kp ∗ e(t) + Ki ∫ e
d
So make the response faster without
offset.
There may be oscillation.
PROPORTIONAL DERIVATIVE (PD)
Proportional and derivative controllers
are combined.
u(t)=
∗
+
()
So the system is faster and have less
overshoot.
PROPORTIONAL INTEGRAL DERIVATIVE (PID)
Combines the three controllers and their
advantages.
PID controller is assumed to have faster
response, no steady state error, small
overshoot, less settling time and
oscillation.
u t = Kp ∗ e t + Ki e d +
d ( )
d
PHYSICAL REALIZATION OF INDUSTRIAL CONTROLLERS
USING OPAMP
Proportional
Eo
R2
=−
Ei
R1
Integral
=−
Derivative
=−
(sCR1+1)
PID
=−
(
)(
)
Physically these controllers can be tuned by using variable resistor.
Varying the value of the resistor will alter the gain value.
TUNING PID CONTROLLER ZIEGLER-NICHOLS RULE
METHOD 1
There are two tuning methods in Ziegler-Nichols tuning
rules.
u t = Kp ∗ e t + Ki ∫ e(t) d +
( )
Method 1: Used for a system exhibits s-shaped response for
step input.
Type of Controller
P
Kp
T
L
PI
Ki
Kd
0
0
0.9
T
L
T 0.3
0.9 ∗
L L
1.2
T
L
T 1
1.2 ∗
L 2L
PID
0
0.6
TUNING PID CONTROLLER ZIEGLER-NICHOLS RULE
METHOD 2
First find critical gain (Kcr): Gain at
which the system exhibits sustained
oscillation with Ki and Kd zero.
Critical period (Pcr): Period of the system
at critical gain.
Type of Controller
Kp
Ki
Kd
P
0.5Kcr
0
0
PI
0.45Kcr
PID
Kcr
Pcr
0
3.6
Kcr
Pcr
0.075Kcr*Pcr
1.2
0.6Kcr