Control system with matlab Time response analysis, Frequency response analysis , Matlab

Control Engineering with
Matlab Application
Dr.P.Anbarasan
Department of EEE
St.Joseph’s College of Engineering

Why Control ?
 Modern society have sophisticated control system
which are crucial to their successful operation.
 Reason to build control system
• Power amplification
• Remote control
• Convenience of input form
• Compensation for disturbances
Radar Antenna Robotic control Solar Panel

Control System
 A control system is an interconnection of
components forming a system configuration
that will provide a desired system response .

Open loop & Closed loop System

System Modeling
 Modeling is a process of abstraction of a real system. The
abstracted model may be logical or mathematical.
 A mathematical model consists of a collection of
equations describing the behavior of the system.
 Differential equations relating to input and output
Transfer function model
State space model
Input
of
transform
Laplace
Output
of
transform
Laplace
Function
Transfer 
conditions
initial
zero
with

State Space Analysis
 The state variable approach is a powerful
technique for the analysis and design of control
system

Modeling of mechanical system
 Mechanical Translational system
 Mass
 Spring
 Dash-pot
 Analogy
 Force Voltage
 Force Current
 Mechanical Rotational system
 Moment of Inertia
 Tortional spring
 Rotational dash-pot
 Analogy
 Torque Voltage
 Torque Current

Modeling of Train System
 
  0
1
2
2
2
2
2
2
2
2
1
1
1
2
1
2
1








x
x
k
dt
dx
B
dt
x
d
M
F
x
x
k
dt
dx
B
dt
x
d
M
Newton’s laws are used in the mathematical modeling of mechanical systems.

Modeling of Electrical Network
Components
Voltage-
Current
relation
Current-
Voltage
relation
Voltage-
Charge
relation
Resistor
Inductor
Capacitor
dt
dq
R
v 
dt
di
L
v  
 vdt
L
i
1
2
2
dt
q
d
L
v 

 idt
C
v
1
dt
dv
C
i 
C
q
v 
iR
v  R
v
i 

Modeling of Electrical Network
  V
dt
t
i
C
dt
t
di
L
t
Ri 

 
1
)
(
)
(
)
(
)
(
)
(
)
( s
V
Cs
s
I
s
LsI
s
RI 


Taking Laplace Transform
By Kirchhoff's voltage law
 
  1
2



RCs
LCs
Cs
s
V
s
I
If i(t) is considered as output

Matlab Program-Representation of
TF
10
2
5
)
( 2




s
s
s
s
G
num = [ 1 5];
den = [ 1 2 10 ];
OR
G = tf([1 5],[1 2 10]);

Matlab Program-Poles and Zeroes of TF
num = [ 2 10];
den = [ 1 2 10 ];
[z, p,K] = tf2zp(num,den) % z- zeroes
% p-poles
% k-gain
10
2
)
5
(
2
)
( 2




s
s
s
s
G

Matlab Program-TF from poles and
zeroes
z=[-1;-3];
p=[0;-2;-4];
K=4;
[num,den]=zp2tf(z,p,K);
printsys(num,den,'s')
  
  
4
2
3
1
4
)
(





s
s
s
s
s
s
G

Matlab Program-Residues of TF
6
11
6
6
3
5
2
)
(
)
(
2
2
2
3







s
s
s
s
s
s
s
R
s
C
num=[2 5 3 6]
den=[1 6 11 6]
[r p k]= residue(num,den)
Using Partial fraction,
2
1
3
2
4
1
6
)
(
)
(









s
s
s
s
R
s
C
r- A,B,C p-poles k-constant

Matlab Program-TF of a parallel
system
10
2
10
)
( 2
1



s
s
s
G
 
5
5
)
(
2


s
s
G
num1 = [0 0 10];
den1 = [ 1 2 10];
num2 = [0 5];
den2 = [1 5];
[num,den]= parallel(num1,den1,num2, den2);
printsys(num,den)

Matlab Program-TF of a feedback
system
10
2
10
)
( 2
1



s
s
s
G
 
5
5
)
(
1


s
s
H
num1 = [0 0 10];
den1 = [ 1 2 10];
num2 = [0 5];
den2 = [1 5];
[num,den]= feedback(num1,den1,num2, den2);
printsys(num,den)

A = [ 0 1 ; -5 -2 ];
B = [ 0 ; 3 ];
C = [ 1 0 ]; D = 0;
H = ss(A,B,C,D)
Matlab Program-State space
representation
   
0
0
1
3
0
2
5
1
0































D
C
DI
Cx
dt
d
x
B
A
BI
Ax
dt
dx




Constructing State space model of
DC motor
R= 2.0 % Ohms
L= 0.5 % Henrys
Km = .015 %
%torque constant
Kb = .015 % emf constant
Kf = 0.2 % Nms
J= 0.02 % kg.m^2
A = [-R/L -Kb/L;
Km/J -Kf/J]
B = [1/L; 0];
C = [0 1];
D = [0];
sys_dc = ss(A,B,C,D)
velocity
angular
voltage
Applied
vapp 
 

Converting State space model of DC
motor to Transfer function model
State Space Model
R= 2.0 % Ohms
L= 0.5 % Henrys
Km = .015 %
%torque constant
Kb = .015 % emf constant
Kf = 0.2 % Nms
J= 0.02 % kg.m^2
A = [-R/L -Kb/L;
Km/J -Kf/J]
B = [1/L; 0];
C = [0 1];
D = [0];
sys_dc = ss(A,B,C,D)
Conversion from SS model to
TF model
sys_tf = tf(sys_dc)

Time response analysis?
The variation of output with respect to time.
 To obtain the satisfactory performance of the system
 Output behavior of the system
 Stability of the system
 Accuracy of the system
For example, aircraft is manufactured, it should made
flight-worthy before it should takes off.
The various disturbances occurs externally to the aircraft
are made tested using various test signals to obtain
satisfactory performance.

Time Response Behaviour
 How the system behaves for the given input and
disturbances?
 For example If we consider a mercury in glass thermometer as
a system with an input of temperature and output of the level
of thermometer is suddenly immersed in hot water, i.e., given
a step input?
 In residential heating system, input temperature is constant.
But we cannot predict the main disturbance – outdoor
temperature.
 If we use motor as system and feedback to move a work piece
in an automatic machining operation, how will the output i.e.,
the displacement of the work piece vary with time when the
input gradually increased with the time with the aim of
gradually increasing the displacement of work piece?

Time Response Types
 Transient Response
 When the response of the
system is changed from
equilibrium it takes some
time to settle down
 Steady State Response
 The part of response that
remains constant after the
transient have died out is
steady state response

Standard Test Signals
 In most cases, the input signals to a control system are not
known prior to design of control system
 To analyse the performance of control system it is excited with
standard test signals
 These inputs are chosen because they capture many of the
possible variations that can occur in an arbitrary input signal
 Step signal (Sudden change)
 Ramp signal (Constant velocity)
 Parabolic signal (Constant acceleration)
 Impulse signal (Sudden shock)
 Sinusoidal signal

r(t) =A; t≥0
= 0; t<0
u(t) =1; t≥0
= 0; t<0
Step Signal Unit Step
Ramp Signal
r(t) =At; t≥0
= 0; t<0
Unit ramp
r(t) =t; t≥0
= 0; t<0
Parabolic Signal
Impulse Signal
δ(t) = ; t=0
ꝏ
= 0; t≠0
r(t) =At2
/2 ; t≥0
= 0 ; t<0
Unit parabolic
r(t) =t2
/2; t≥0
= 0; t<0
r(t) =A; t=0
= 0; t≠0
Unit Impulse
s
t
u
L
1
)}
(
{ 
2
1
)}
(
{
s
t
r
L 
3
1
)}
(
{
s
t
r
L 
1
)}
(
{ 
t
L 

Time Response of the system
 Transient response depends upon the system poles and
not on the type of the system
 It is sufficient to analyze the transient response using a
step input
 The steady state response depends on system dynamics
and the input quantity

System Representation
   
   
3
2
1
3
2
1 ......
R(s)
C(s)
K
T(s)
p
s
p
s
p
s
z
s
z
s
z
s
K








Transfer Function in
pole zero form
   
   
s
s
s
s
s
s
K
p
p
p
z
z
z
1
1
1
3
2
1
1
1
1
......
1
1
1
R(s)
C(s)
K
T(s)














Transfer Function in
time constant form

Order & Type
4
s
4s
s
1
3
Order
10
5s
s
1
s
2
Order
2
s
1
1
Order
2
3
2







 
 
7
s
1
2
Type
5
-
s
1
1
Type
2
s
1
0
Type
2


s
s
 The order of the system is given by the maximum power
of s in the denominator transfer functions.
 The type number is specified for loop transfer function
G(s)H(s). The number of poles lying at the origin decides
the type number of the system.

Step Response of 1st
Order System
 Consider the following 1st
order system
s
K


1
)
(s
C
)
(s
R
s
s
R
1
)
( 
 
s
s
K
s
C



1
)
(
1
)
(



s
K
s
K
s
C


• In order to find out the inverse Laplace of the above equation, we need to
break it into partial fraction expansion









1
1
)
(
s
s
K
s
C


Taking Inverse Laplace of above equation
 

/
1
)
( t
e
K
t
c 



Response of 1st
order system for Unit
Step
value)
final
of
(63.2%
constant
Time
1
)
(
)
(


 
s
K
s
R
s
C
 

/
1
)
( t
e
K
t
c 



Response of 1st
order system for different
time constant
s
K
s
R
s
C



1
)
(
)
(
Higher time constant leads to sluggish
response
Lower time constant leads to faster
response

Response of 1st
order system for
different gain
s
K
s
R
s
C



1
)
(
)
(

Impulse Response of 1st
Order System
 Consider the following 1st
order system
s
K


1
)
(s
C
)
(s
R
0 t
δ(t)
1
1

 )
(
)
( s
s
R  s
K
s
C



1
)
(


/
1
/
)
(


s
K
s
C


/
)
( t
e
K
t
c 

Taking Laplace Transform

Relation Between Step and impulse response
 The step response of the first order system is
 Differentiating c(t) with respect to t yields
  
 /
/
1
)
( t
t
Ke
K
e
K
t
c 





 

/
)
( t
Ke
K
dt
d
dt
t
dc 




/
)
( t
e
K
dt
t
dc 


Practical Determination of Transfer Function
of 1st
Order Systems
 Often it is not possible or practical to obtain a system's transfer
function analytically.
 Perhaps the system is closed, and the component parts are not
easily identifiable.
 The system's step response can lead to a representation even
though the inner construction is not known.
 With a step input, we can measure the time constant and the
steady-state value, from which the transfer function can be
calculated.
 If we can identify τ and K empirically we can obtain the
transfer function of the system.
s
K
s
R
s
C



1
)
(
)
(

Practical Determination of Transfer Function
of 1st
Order Systems
 For example, assume the unit
step response given in figure.
• From the response, we can
measure the time constant, that
is, the time for the amplitude to
reach 63% of its final value.
• Since the final value is about
0.72 the time constant is
evaluated where the curve
reaches 0.63 x 0.72 = 0.45, or
about 0.13 second.
τ=0.13s
K=0.72
• K is simply steady state value.
• Thus transfer function is
obtained as:
7
7
5
5
1
13
0
72
0
.
.
.
.
)
(
)
(




s
s
s
R
s
C

Example 1
 Find the time response for the closed transfer function
1
2
6


S
s
R
s
C
)
(
)
(
 
1
2
6


S
s
s
C )
(
  1
2
1
2
6



 s
B
s
A
S
s
s
s
R
s
R
1
)
(
,
input
step
a
is
)
(
since 
  5
0
6
6
1
2
6
.



 s
s
S
s
 
t
t
e
e
t
c 5
.
0
5
.
0
1
6
6
6
)
( 






Second Order System
 We have discussed the affect of location of poles and zeros on the
transient response of 1st
order systems.
 Compared to the simplicity of a first-order system, a second-order
system exhibits a wide range of responses that must be analyzed
and described.
 Varying a first-order system's parameter (τ, K) simply changes the
speed and offset of the response
 Whereas, changes in the parameters of a second-order system can
change the form of the response.
 A second-order system can display characteristics much like a first-
order system or, depending on component values, display damped
or pure oscillations for its transient response.

Second Order System
 A general second-order system is characterized by the
following transfer function.(E.g. Series RLC circuit, Position
Servo mechanism)
2
2
2
2 n
n
n
s
s
s
R
s
C






)
(
)
(
un-damped natural frequency of the second order system, which is the
angular frequency at which system oscillate in the absence of damping.
n

damping ratio, a dimensionless quantity describing the decay
of oscillations during transient response.


Damping and its types
 Damping is an effect created in an oscillatory system that
reduces, restricts or prevents the oscillations in the system.
 System can be classified as follows depending on damping
effect
 Overdamped system: Transients in the system exponentially
decays to steady state without any oscillations
 Critically damped system: Transients in the system
exponentially decays to steady state without any oscillations
in shortest possible time
 Underdamped system: System transient oscillate with the
amplitude of oscillation gradually decreasing to zero
 Undamped system: System keeps on oscillating at its
natural frequency without any decay in amplitude

Damping ratio on pole location
system
Overdamped
1
system
damped
Critically
1
system
mped
Underda
1
0
system
ndamped
U
0










42
Step Response of 2nd
order system
2
2
2
2
2
2
2
2
1
n
n
n
n
n
s
s
s
s
s
C














)
(
• The partial fraction expansion of above equation is given as
2
2
2
2
1
n
n
n
s
s
s
s
s
C








)
(
 2
2 n
s 

 
2
2
1 
 
n
   
2
2
2
1
2
1










n
n
n
s
s
s
s
C )
(
2
2
2
2 n
n
n
s
s
s
R
s
C






)
(
)
(
 
2
2
2
2 n
n
n
s
s
s
s
C






)
(
Step Response
s
s
R
1
)
( 
Case 1: Underdamped system

43
order underdamped System
• Above equation can be written as
   
2
2
2
1
2
1










n
n
n
s
s
s
s
C )
(
  2
2
2
1
d
n
n
s
s
s
s
C








)
(
2
1 

 
 n
d
• Where , is the frequency of transient oscillations
and is called damped natural frequency.
• The inverse Laplace transform of above equation can be obtained
easily if C(s) is written in the following form:
    2
2
2
2
1
d
n
n
d
n
n
s
s
s
s
s
C














)
(

44
    2
2
2
2
1
d
n
n
d
n
n
s
s
s
s
s
C














)
(
    2
2
2
2
2
2
1
1
1
d
n
n
d
n
n
s
s
s
s
s
C



















)
(
    2
2
2
2
2
1
1
d
n
d
d
n
n
s
s
s
s
s
C

















)
(
t
e
t
e
t
c d
t
d
t n
n



 

sin
cos
)
( 





2
1
1












 
t
t
e
t
c d
d
t
n





sin
cos
)
(
2
1
1
t
e
t
e
t
c d
t
d
t n
n



 

sin
cos
)
( 





2
1
1












 
t
t
e
t
c d
d
t
n





sin
cos
)
(
2
1
1
5
and
5
.
0
if 
 n



order System
• Here 0


t
t
c n

cos
)
( 
1
2
2
2
2 n
n
n
s
s
s
R
s
C






)
(
)
(
 
2
2
2
)
(
n
n
s
s
s
C




s
s
R
1
)
( 
5

n
If 
Case 2: Undamped system

order System
• Here 1


t
n
t n
n
te
e
t
c 

 



1
)
(
2
2
2
2
)
(
)
(
n
n
n
s
s
s
R
s
C






   2
2
2
2
2
2
)
(
n
n
n
n
n
s
s
s
s
s
s
C










s
s
R
1
)
( 
   2
1
1
)
(
n
n
n s
s
s
s
C


 




5
,
1
If 
 n


Case 3: Critically damped system

order System
2
2
2
2
)
(
)
(
n
n
n
s
s
s
R
s
C





 s
s
R
1
)
( 
• Here 1


 
2
2
2
2
)
(
n
n
n
s
s
s
s
C






  
  






























1
1
1
1
1
2
1
1
1
1
1
2
1
)
(
2
2
2
2
2
2
















n
n
n
n
n
n
n
n
n
n
s
s
s
s
C
Case 4: Over damped system

order
Over damped System
 
  





























 







 


t
n
n
n
t
n
n
n
n
n
n
n
e
e
s
t
c
1
2
2
1
2
2
2
2
1
1
1
2
1
1
1
2
1
)
(
















5
,
5
.
1
If 
 n



Response of IInd
order system for Unit Step

Response for the TF
  
2
1
1
)
(




s
s
s
s
G


Best Damping Ratio for a Control
System
 Selection of damping ratio for industrial control applications
requires a trade-off between relative stability and speed of
response.
 Many system are designed for damping ratio in the range 0.4-
0.7 ( peak overshoot of 25%)
 If allowed by rise time consideration, damping ratio close to
0.7 is the most obvious choice because it results in minimum
normalized settling time
 For navigation purpose, the transient response is not primary
performance criterion to optimize: minimum steady state error
is the major objective. Therefore damping ratio is as small as
possible (steady state error proportional to damping ratio)

Application of Damped System
 Overdamped system
 Push button water tap shut-off valves
 Automatic door closers
 Critically damped system
 Elevator mechanism
 Gun mechanism (Return to neutral position in shortest
possible time)
 Underdamped system
 All string instruments, bells are under damped to make
sound appealing
 Analog electrical and mechanical measuring instruments

Matlab Program-Step response of the
system
 To find the step response of a system
n=[25];
d=[1 4 25];
step(n,d)
title('Step response of second order system');
grid
25
4
25
)
( 2



s
s
s
G

Matlab Program-Impulse response of
the system
 To find the impulse response of a system
n=[25];
d=[1 4 25];
impulse(n,d)
title(‘Impulse response of second order system');
grid
25
4
25
)
( 2



s
s
s
G

Matlab Program-Ramp response of
the system
 To find the ramp response of a system
t=0:0.1:10
alpha=2
ramp=alpha*t % Your input signal
model=tf([25],[1 4 25]); % Your transfer function
[y,t]=lsim(model,ramp,t)
plot(t,y)
25
4
25
)
( 2



s
s
s
G

Matlab Simulink -Test inputs
Ramp input from the step
Parabolic input from the ramp

Transient Response Specifications
59
For 0< <1 and ωn > 0, the 2nd
order system’s response due to a
unit step input is as follows.
Important timing characteristics: delay time, rise time, peak time,
maximum overshoot, and settling time.


Delay Time
60
• The delay (td) time is the time required for the response to
reach half the final value the very first time.

Rise Time
• The rise time is the time required for the response to rise from 10% to
90%, 5% to 95%, or 0% to 100% of its final value.
• For underdamped second order systems, the 0% to 100% rise time is
normally used. For overdamped systems, the 10% to 90% rise time is
commonly used.For critically damped systems, the 5% to 95% is
used

Peak Time
62
• The peak time is the time required for the response to reach the
first peak of the overshoot.
62
62

Maximum Overshoot
63
The maximum overshoot is the maximum peak value of the
response curve measured from unity. If the final steady-state
value of the response differs from unity, then it is common to use
the maximum percent overshoot. It is defined by
The amount of the maximum (percent) overshoot directly
indicates the relative stability of the system.

Settling Time
64
• The settling time is the time required for the response curve to
reach and stay within a range about the final value of size
specified by absolute percentage of the final value (usually 2%
or 5%).

Time Response Specifications
n
d
t


7
.
0
1

Delay Time
Rise Time
2
2
1
1
1
tan













 



n
r
t
Peak Time
2
1 




n
p
t
Maximum Overshoot
100
%
2
1

 



e
M p
Settling Time
criterion
for
t
n
s %
2
4


criterion
for
t
n
s %
5
3



Steady State Error
 If the output of a control system at steady state does not exactly
match with the input, the system is said to have steady state error
 Any physical control system inherently suffers steady-state error in
response to certain types of inputs.
 A system may have no steady-state error to a step input, but the same
system may exhibit nonzero steady-state error to a ramp input.
 The magnitudes of the steady-state errors due to these individual
inputs are indicative of the goodness of the system.

Steady State Error
 Steady state error depends upon both input and
type of the system
 As the type number is increased, accuracy is
improved.
 However, increasing the type number aggravates
the stability problem.
 A compromise between steady-state accuracy and
relative stability is always necessary.

Steady State Error
It is a value of error signal when t tends to infinity
E(s) = Error Signal
E(s) = R(s) - C(s) .H(s)
Output signal C(s) = E(s).G(s)
Substituting C(s) in E(s)
E(s) = R(s) - E(s).G(s) H(s)
) ( ) ( 1
) (
) (
s H s G
s R
s E


Let e(t) error signal in time domain
  







 

)
(
)
(
1
)
(
)
(
)
( 1
1
s
H
s
G
s
R
L
s
E
L
t
e

Steady State Error
Let ess= steady state error
)
(t
e
Lt
e
t
ss



The final value theorem states that )
(
)
(
0
s
sF
Lt
t
f
Lt
s
t 



Steady state error,
)
(
)
(
1
)
(
)
(
)
(
0
0 s
H
s
G
s
sR
Lt
s
sE
Lt
t
e
Lt
e
s
s
t
ss









Static Error Constant
 Type-0 system will have constant steady state error when input is
step signal
Positional Error Constant
ramp signal
Velocity Error Constant
parabolic signal
Acceleration Error Constant
)
(
)
(
0
s
H
s
G
Lt
K
s
p


)
(
)
(
0
s
H
s
sG
Lt
K
s
v


)
(
)
(
2
0
s
H
s
G
s
Lt
K
s
a


   
   
3
2
1
3
2
1 ......
R(s)
C(s)
K
H(s)
G(s)
p
s
p
s
p
s
s
z
s
z
s
z
s
K N









Steady state error for Step Input
)
(
)
(
1
)
(
0 s
H
s
G
s
sR
Lt
e
s
ss



)
(
)
(
1
1
0 s
H
s
G
Lt
e
s
ss



)
(
)
(
1
1
0
s
H
s
G
Lt
e
s
ss



p
ss
K
e


1
1
)
(
)
(
0
s
H
s
G
Lt
K
s
p


Where
s
s
R
1
)
( 

To find Kp
)
(
)
(
0
s
H
s
G
Lt
K
s
p


Type-0
)......
)(
(
).......
)(
(
)
(
)
(
2
1
2
1
p
s
p
s
z
s
z
s
K
s
H
s
G





)......
)(
(
).......
)(
(
2
1
2
1
0 p
s
p
s
z
s
z
s
Lt
K
s
p






const
K
e
p
ss 


1
1
Type-1
)
(
)
(
0
s
H
s
G
Lt
K
s
p

 )......
)(
(
).......
)(
(
)
(
)
(
2
1
2
1
p
s
p
s
s
z
s
z
s
K
s
H
s
G












 )......
)(
(
).......
)(
(
2
1
2
1
0 p
s
p
s
s
z
s
z
s
Lt
K
s
p
0
1
1



p
ss
K
e

Steady state error for Ramp Input
)
(
)
(
1
)
(
0 s
H
s
G
s
sR
Lt
e
s
ss



)
(
)
(
1
0 s
H
s
sG
s
Lt
e
s
ss



)
(
)
(
1
0
s
H
s
sG
Lt
e
s
ss


v
ss
K
e
1
 )
(
)
(
0
s
H
s
sG
Lt
K
s
v


Where
2
1
)
(
s
s
R 

To find Kv
)
(
)
(
0
s
H
s
sG
Lt
K
s
v


Type-0
)......
)(
(
).......
)(
(
)
(
)
(
2
1
2
1
p
s
p
s
z
s
z
s
K
s
H
s
G





0
)......
)(
(
).......
)(
(
2
1
2
1
0






 p
s
p
s
z
s
z
s
sK
Lt
K
s
v 



0
1
1
v
ss
K
e
Type-1
)
(
)
(
0
s
H
s
sG
Lt
K
s
v


)......
)(
(
).......
)(
(
)
(
)
(
2
1
2
1
p
s
p
s
s
z
s
z
s
K
s
H
s
G





const
p
p
z
z
p
s
p
s
s
z
s
z
s
sK
Lt
K
s
v 






 ......
.
.......
.
)......
)(
(
).......
)(
(
2
1
2
1
2
1
2
1
0
const
K
e
v
ss 

1

Type-2
)
(
)
(
0
s
H
s
sG
Lt
K
s
v


)......
)(
(
).......
)(
(
)
(
)
(
2
1
2
2
1
p
s
p
s
s
z
s
z
s
K
s
H
s
G












 )......
)(
(
).......
)(
(
2
1
2
2
1
0 p
s
p
s
s
z
s
z
s
sK
Lt
K
s
p
0
1


v
ss
K
e

Steady state error for Parabolic Input
)
(
)
(
1
)
(
0 s
H
s
G
s
sR
Lt
e
s
ss



)
(
)
(
1
2
2
0 s
H
s
G
s
s
Lt
e
s
ss



)
(
)
(
1
2
0
s
H
s
G
s
Lt
e
s
ss


a
ss
K
e
1
 )
(
)
(
2
0
s
H
s
G
s
Lt
K
s
a


3
1
)
(
s
s
R 

Type-0
)
(
)
(
2
0
s
H
s
G
s
Lt
K
s
a

 0
)......
)(
(
).......
)(
(
2
1
2
1
2
0






 p
s
p
s
z
s
z
s
K
s
Lt
K
s
a 



0
1
1
a
ss
K
e
Type-1
0
)......
)(
(
).......
)(
(
2
1
2
1
2
0






 p
s
p
s
s
z
s
z
s
K
s
Lt
K
s
a
Type-2
const
p
s
p
s
s
z
s
z
s
K
s
Lt
K
s
a 





 )......
)(
(
).......
)(
(
2
1
2
2
1
2
0
const
K
e
a
ss 

1
Type-3







 )......
)(
(
).......
)(
(
2
1
3
2
1
2
0 p
s
p
s
s
z
s
z
s
K
s
Lt
K
s
a
0
1


a
ss
K
e




0
1
1
a
ss
K
e
To find Ka

Type Steady State Error
Unit Step Unit Ramp Unit Parabolic
0
1 0
2 0 0
3 0 0 0
p
K

1
1
 
v
K
1

a
K
1

Significance of Static Error Constants
 The static error constants are figures of merit of control systems.
 The higher the constants, the smaller the steady-state error. As the
steady state error is inversely proportional to static error constant.
 Increasing the gain increases the static error constant. Thus in
general increases the system gain decreases the steady state error.

1
1
)
(


s
s
G
system
o
Type
a
Consider
Matlab response of Steady State Error
for Type o system
5
.
0
;
1
1
1
1
1
0







ss
s
p
p
ss
e
s
lt
K
K
e
step
unit


ss
e
ramp
unit 

ss
e
parabolic
unit

 
1
1
)
(
1


s
s
s
G
system
Type
a
Consider
for Type 1 system
 
1
;
1
1
1
1
0






ss
s
v
v
ss
e
s
s
s
lt
K
K
e
ramp
unit
0

ss
e
step
unit 

ss
e
parabolic
unit

 
1
1
)
(
2 2


s
s
s
G
system
Type
a
Consider
for Type 2 system
 
1
;
1
1
1
1
2
2
0






ss
s
a
a
ss
e
s
s
s
lt
K
K
e
parabolic
unit
0

ss
e
step
unit 0

ss
e
ramp
unit

Dynamic Error Coefficient
 The drawback in static error coefficient is that it does not show
variation of error with time and input should be standard input.
 The dynamic error constant gives steady state error as a function
of time.
 Using this method, the steady state error can be found for any type
of input.
)
(
)
(
1
)
(
)
(
s
H
s
G
s
R
s
E


The error signal is given by
)
(
)
(
1
1
)
(
s
H
s
G
s
F


where

Dynamic Error Coefficient
The error signal is obtained by dynamic error coefficients,
.....
!
2
)
(
)
( )
(
)
(
..
2
.
1
0 


 t
r
t
r
C
C
t
r
C
t
e
)
(
)
(
1
)
(
)
(
)
(
0
0 s
H
s
G
s
sR
Lt
s
sE
Lt
t
e
Lt
e
s
s
t
ss








Steady state error
)
(
0
0 s
F
Lt
C
s
 )
(
0
1 s
F
ds
d
Lt
C
s
 )
(
2
2
0
2 s
F
ds
d
Lt
C
s

C0 , C1, C2,……are called dynamic error coefficients

Integral performance Criteria
Integral Squared Error (ISE)
Integral Absolute Error (IAE)
Integral Time-weighted Absolute Error (ITAE) 





dt
t
ITAE
dt
IAE
dt
ISE


 2
ISE integrates the square of the error over time. ISE will penalise large
errors more than smaller ones (since the square of a large error will be
much bigger).
Control systems specified to minimise ISE will tend to eliminate large
errors quickly, but will tolerate small errors persisting for a long period
of time. Often this leads to fast responses, but with considerable, low
amplitude, oscillation.
It is desirable to access the quality of control system by evaluating a
performance index that can either be calculated or measured
In the area of adaptive control, we can adjust certain parameters that
will minimize the value of performance index, also known as Cost
function

IAE integrates the absolute error over time. It doesn't add weight to
any of the errors in a systems response. It tends to produce slower
response than ISE optimal systems, but usually with less sustained
oscillation.
ITAE integrates the absolute error multiplied by the time over time.
What this does is to weight errors which exist after a long time much
more heavily than those at the start of the response.
ITAE tuning produces systems which settle much more quickly than
the other two tuning methods.
The downside of this is that ITAE tuning also produces systems with
sluggish initial response (necessary to avoid sustained oscillation).
Integral performance Criteria

Increases in Type no. (or) Adding pole at
origin reduces the Steady State Error
 
1
1
)
(
0


s
s
G
Type
 
1
1
)
(
1


s
s
s
G
Type

Addition of pole very close to imaginary
axis, system becomes oscillatory
  
1
.
0
5
.
0
10
)
(



s
s
s
G
system
a
Consider

Adding a zero increases the peak-
overshoot
25
4
25
)
( 2



s
s
s
G
 
25
4
1
25
)
( 2




s
s
s
s
G

Controller
PROPORTIONAL
PROPRTIONAL INTEGRAL
PROPORTIONAL DERIVATIVE
PROPORTIONAL INTEGRAL DERIVATIVE

Proportional Controller
 It produces an output signal which is proportional to error signal
 Its transfer function is represented by Kp
 It amplifies the error signal and increase the loop gain of the system
• Steady state tracking accuracy
• Disturbance signal rejection
• Relative stability
Drawback
• Produces constant steady state error
• Decreases the sensitivity of the system

−
actuating
error signal e(t)
reference
input r(t)
error
detector
feed back
signal b(t)
Controller
output u(t)
Kp

PI Controller
 The transfer function of
PI controller







 

s
T
s
T
K
i
i
p
1
 
n
n
s
s
s
G


2
)
(
2


     



t
0
i
p
p
dt
t
e
T
K
t
e
K
t
u
 
  









s
T
1
1
K
s
E
s
U
i
p
Let open loop TF is given by
 
s
E(s)
T
K
s
E
K
U(s)
i
p
p 


 
n
n
s
s
s
G


2
)
(
2


 
 
n
n
i
i
p
n
n
i
i
p
s
s
s
T
s
T
K
s
s
s
T
s
T
K
s
G
s
G
s
R
s
C




2
1
1
2
1
)
(
1
)
(
)
(
)
(
2
2








 









 



  )
1
(
2
)
1
(
2
2
2
s
T
K
s
T
s
s
T
K
i
n
p
n
i
i
n
p








2
2
2
3
2
2
)
1
(
n
p
i
n
p
n
i
i
i
n
p
K
s
T
K
T
s
T
s
s
T
K










2
2
2
3
2
2
)
1
(
n
p
i
n
p
n
i
i
i
n
p
K
s
T
K
T
s
T
s
s
T
K









2
2
2
3
2
2
)
1
(
)
/
(
n
i
p
n
p
n
i
n
i
p
T
K
s
K
s
s
s
T
T
K









There is a increase in order by one and introduces zero in the
system
The increase in order of the system results in less stable
The type number of the open loop system increases by
one ,this will reduces the steady state error
Increase in zero increases the peak overshoot

PD Controller
     
dt
t
de
T
K
t
e
K
t
u d
p
p


 
 
 
s
T
1
K
s
E
s
U
d
p 

 
n
n
s
s
s
G


2
)
(
2


sE(s)
T
K
E(s)
K
U(s) d
p
p 


 
n
n
s
s
s
G


2
)
(
2


 
 
 
 
n
n
d
p
n
n
d
p
s
s
s
T
K
s
s
s
T
K
s
G
s
G
s
R
s
C




2
1
1
2
1
)
(
1
)
(
)
(
)
(
2
2








  s
T
K
K
s
s
s
T
K
s
T
K
s
s
s
T
K
d
n
p
n
p
n
d
n
p
d
n
p
n
d
n
p
2
2
2
2
2
2
2
)
1
(
)
1
(
2
)
1
(

















2
2
2
2
)
2
(
)
1
(
n
p
d
n
p
n
d
n
p
K
s
T
K
s
s
T
K









Increase in zero and damping ratio
Increase in zero increases the peak
overshoot
But Increase in damping ratio
reduces the peak overshoot

PID Controller
       
dt
t
de
T
K
dt
t
e
T
K
t
e
K
t
u d
p
t
o
i
p
p




 
  









 s
T
s
T
1
1
K
s
E
s
U
d
i
p
t
0
u(t)
proportional only
PD control action
PID control action
Proportional controller stabilizes the gain but produces a steady state error
The integral controller eliminates the steady state error
The derivative controller reduces the overshoot of the response

With out controller
System response using PD controller
Using PD Controller
Overshoot is very much reduced but steady state error is present
  feedback
unity
with
s
s
s
G
system
a
Consider
4
2
10
)
( 2




System response using PI controller
  feedback
unity
with
s
s
s
G
system
a
Consider
4
2
10
)
( 2



With out controller Using PI Controller
Steady error is reduced using PI controller but overshoot is present

System response using PID controller
  feedback
unity
with
s
s
s
G
system
a
Consider
4
2
10
)
( 2



With out controller Using PID Controller
Bothe Steady error and overshoot is reduced using PID controller

Effect of Increasing Kp , Ki and Kd
Parameter Rise Time Overshoot Settling Time Steady
State
Error
Kp Decreases Increases Small Change Decreases
Ki Decreases Increases Increases Eliminate
Kd Small
Change
Decreases Decreases None

Trial and Error Method
 Set integral and derivative terms to zero first and then increase the
proportional gain until the output of the control loop oscillates at a
constant rate. This increase of proportional gain should be in such
that response the system becomes faster provided it should not make
system unstable.
 Once the P-response is fast enough, set the integral term, so that the
oscillations will be gradually reduced. Change this I-value until the
steady state error is reduced, but it may increase overshoot.
 Once P and I parameters have been set to a desired values with
minimal steady state error, increase the derivative gain until the
system reacts quickly to its set point. Increasing derivative term
decreases the overshoot of the controller response.

Ziegler Nichols Tuning Technique
First Method

Ziegler Nichols Tuning Technique
 It is very similar to the trial and error
method where integral and derivative
terms are set to the zero, i.e., making Ti
infinity and Td zero.
 Increase the proportional gain such that
the output exhibits sustained oscillations.
If the system does not produce sustained
oscillations then this method cannot be
applied. The gain at which sustained
oscillations produced is called as critical
gain (Kcr).
 Once the sustain oscillations are
produced, set the values of Ti and Td as
per the given table for P, PI and PID
controllers based on critical gain and
critical period.
Second Method

Stability Analysis
 A system is stable if its output is bounded for any bounded
input

Location of roots & its stability
Roots on left half of s plane -Stable
Roots on right half of s plane -Unstable

Single pair of roots on imaginary axis-Marginally Stable
Repeated roots on imaginary axis
One or more non repeated roots on imaginary
axis
Unstable

Single pole at origin - Stable
Double pole at origin - Unstable

Fuel Cell based Converter Topologies
Conventional
Boost
Converter
Interleaved
Converter
Sepic
Converter

Fuel Cell based Converter Topologies
Characteristics Conventional Interleaved SEPIC
Peak Overshoot 27V 2.4V 19.7
Peak Time 0.025secs 0.018secs 0.057secs
Settling Time 0.27secs 0.105secs 0.24secs
Ripple Voltage 2.41V 2.5V 6.7V

PI Controller for Buck converter to
regulate the voltage

FREQUENCY RESPONSE ANALYSIS
 It is the steady state response of a system when the input of the
system is sinusoidal signal
In TF T(s), s is replaced by jω T(jω) is called sinusoidal TF

Frequency Response Plots
 Bode Plot
 Polar Plot
 Nyquist plot
 Nichols Plot
 M and N circles
 Nichols Chart

Matlab Program-Bode Plot
n=[10 5];
d=[1 4.25 1];
bode(tf(n,d));
grid on;
[gm,Pm,wpc,wgc]=margin(tf(n,d))

Gain & Phase margin
GM = -12
PM=180+φgc
PM=180-22=158

Bode Plot for state space system
A=[0 1;-25 -4];
B=[0;25];
C=[1 0];
D=[0];
bode(A,B,C,D);

Matlab Program-Root Locus
num=[48];
den = [ 1 6 8 0];
rlocus(num,den)
grid

Reference Books
S.No Title of the Book Author Publisher
1. Control Systems, Principles
and Design
M. Gopal, Tata McGraw Hill
2. Control System Engineering S.K.Bhattacharya Pearson
3. Control System Engineering
Norman S Nise John wiley & Sons
4. Control System Engineering A.Nagoor Kani RPA Publications
5. Control System – Theory and
Applications
Smarajit Ghosh Pearson

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