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CONTROL SYSTEMS
Time Response Analysis
Dr. S. SUMATHI
Associate Professor
RNSIT
Bengaluru
5/31/2019 1
Time Response Analysis The time response of a system is the output response of the
system as a function of time.
The time response of a control system is usually
divided into two parts:
Transient response
Steady state response
( ) ( ) ( )t ssc t c t c t
c(t) denote the time response of a continuous data system
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Transient response
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The transient response is defined as that part of the
time response that goes to zero as time tends to
infinity.
Steady state response
The steady state response is the part of the total
response that remains after the transient has died out.
( ) 0ttLt c t
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First step is to obtain a mathematical model of the system
Transfer function is an Important mathematical model for a
LTI system
For any specific input signal, a complete time response
can then be obtained - Laplace transform inversion of c(s)
If the input signal is such that ,we can not find Laplace
transformable, then time response is obtained through
convolution Integral.
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Steady state behavior of the system can be obtained from
c(t) expression, with time tending to infinity.
In case of simple deterministic signals steady state
response can be obtained by the use of Final value
theorem
Time response analysis is normally carried out after
stability analysis through indirect tests
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INPUT SIGNALS Inputs to the system are not known ahead of time.
Input signals to some of the systems are random in nature
The actual signals which severely stain the a control
system are
a sudden shock
a sudden change
a constant velocity
a constant acceleration
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Hence the system dynamic behavior is analyzed under
the application of standard test signals
Standard test signals
Impulse signal (sudden shock)
Step signal (sudden change)
Ramp signal (constant velocity)
Parabolic signal (constant acceleration)
Nature of transient response depends on the system
poles and not on the type of input.
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STEP SIGNAL
The step is a signal whose value changes from one level
(usually zero) to another level A in zero time.
The mathematical representation of the step function is
r(t) = Au(t)
where u(t) = 1; t>0
= 0; t<0
u(t) is called the unit step function.
In the Laplace transform, U(s)=A/s
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RAMP SIGNAL
The ramp is a signal which starts at a value of zero and
increases linearly with time.
Mathematically,
r(t) = At; t>0
= 0; t<0
In the Laplace transform form,
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PARABOLIC SIGNAL
The parabolic function represents a signal that is represented as
In Laplace transform form
; t > 0
= 0 ; t<0
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IMPULSE SIGNAL
A unit impulse is defined as a signal which has zero value
everywhere except at t=0 , where its magnitude is infinite.
It is generally called the δ function and has the following
property.
( ) 0; 0t t
( ) 1t dt
( ) 1s
Where
tends to zero.
In the Laplace transform
( ) 1s
0 t
δ(t)
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IMPULSE RESPONSE
Impulse signal is derivative of step signal.
The impulse response of a system with transfer function
( ) ( ) ( )C s G s R s
( )G s
1( ) ( ) ( )c t L G s g t
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(As R(s)=1)
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Thus, the impulse response of a system, indicated by
( )g t
which is the inverse Laplace transform of its transfer
function.
This is also referred to as weighting function of the
system.
The weighting function of a system can be used to find the
system’s responses to any input r(t) by means of
convolution integral
0
( ) ( ) ( )
t
c t g t r d 5/31/2019 13
dt
• Impulse ẟ (t)
• Step u(t)
• Ramp r(t)
• Parabolic p(t)
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For linear systems transient response is obtained for
one of the standard test signals,
normally for a step input .
As the system response depends on the system poles
and not upon the type of input
Steady state response is then examined for step signal
as well as other test signals.
For frequency response analysis of systems a
sinusoidal signal with variable frequency is used.
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TIME RESPONSE OF FIRST ORDER SYSTEMS
The transfer function is given by
( ) 1
( ) 1
C s
R s Ts
+
_
R(s) C(s)
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UNIT STEP RESPONSE OF FIRST ORDER SYSTEMS
( ) 1/R s s
1 1( ) .
1C s
s Ts
( ) 1
( ) 1
C s
R s Ts
Therefore output response is given by
By applying partial fraction expansion to C(s) we have
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UNIT STEP RESPONSE OF FIRST ORDER SYSTEMS cont..
Ttetutc /)()(
By taking Laplace inverse C(s) we have
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T is called the TIME CONSTANT
The initial slope of the output curve is given by
Time constant is indicative of how fast the system tends to
reach the final value.
A large time constant corresponds to a sluggish system
A small time constant corresponds to a fast response.
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The error response of the system is given by
( ) ( ) ( )tTe t r t c t e
The steady state error is given by
( ) 0sst
e Lt e t
Thus the first order system tracks the unit step
input with zero steady state error.
ERROR RESPONSE
20
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21
RAMP RESPONSE OF FIRST ORDER SYSTEMS
The transfer function of a first order system is given by
( ) 1
( ) 1
C s
R s Ts
For a ramp input
Therefore
By applying partial fraction expansion to C(s) we have
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22
By taking Laplace inverse C(s) we have
Let us differentiate the ramp response
This is same as step response
g (t) ie impulse response
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TIME RESPONSE OF SECOND ORDER SYSTEMS
A general second-order
system is shown below
It is a Type-1 second
order system
n
Un-damped natural frequency of oscillations
Damping factor
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TIME RESPONSE OF SECOND ORDER SYSTEMS
Its closed loop transfer function is given by
22
2
2 nn
n
sssR
sC
)(
)(
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• The time response of any system is characterized by the
roots of the denominator polynomial q(s) .
• The denominator polynomial q(s) is therefore called the
characteristic polynomial
• q(s)=0 is called the characteristic equation
• Roots of the characteristic equation are same as the poles
of the system
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Hence the characteristic equation of the system
under consideration is given by
2 22 0n ns s
Two roots of the characteristic equation or poles of
the system are
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According the value of ξ , a second-order system can be
classified into one of the four categories
Overdamped ( ξ >1) - the system poles are real and distinct .
-a -b -c σ
jω
The response rises slowly and reaches the final value
without any oscillations
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Critically damped (ξ = 1) - the system has two real but
equal poles
The two poles are
-ωn -b -c
σ
jω
The response rises slowly and reaches the final value
without any oscillations
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Underdamped (0 < ξ <1) - the system has a pair of
complex conjugate poles
-ξωn σ
jω
The transient response is oscillatory.
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Undamped (ξ = 0) -the system has two
imaginary poles.
-a -b -c
σ
jω
The system response is oscillatory.
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Most control systems are designed as under damped systems
to have fast response
Step Response of an under damped system 0 1
22
2
2 nn
n
sssR
sC
)(
)(( ) 1/R s s
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2 2
21
( 2 )
n
n n
s
s s s
Solving for A, B and C we can find that A=1 B= -1 and
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222222 2
21
nnnn
n
ss
s
ssC
)(
2221
21
nn
n
s
s
ssC )(
22
21
dn
n
s
s
ssC
)(
• Where , is the frequency of transient
oscillations and is called damped natural frequency.
21d n
2222
1
dn
n
dn
n
ss
s
ssC
)(
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2 2 2 2
1.
( ) ( )
n n d
n d d n d
s
s s s
2 2 2 22
1.
( ) ( )1
n d
n d n d
s
s s s
Taking the inverse Laplace transform of the above equation
2( ) 1 cos sin
1
n
n
tt
d d
ec t e t t
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2
21 1 cos sin
1
nt
d d
et t
2
1 sin cos cos sin1
nt
d d
et t
21 sin( )
1
nt
d
et
22 1
2
11 sin( 1 tan )
1
nt
n
et
ξ
ϴ
1
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0 time t 0
1
c(t)
Step Response of an under damped system C(t)
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Step Response of an un damped system C(t)
The step response of un damped system can be obtained
from that of Under damped system by substituting ξ =0
22 1
2
11 sin( 1 tan )
1
nt
n
et
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Step Response of an critically damped system
For step input R(s)=1/s, therefore
By applying partial fraction expansion
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The step response is characterized by following
performance indices
i) Delay time
ii) Rise time
iii) Peak time
iv) Maximum overshoot
v) Settling time
Time Response Specifications
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i) Delay time ii) Rise time iii) Peak time
ii) Maximum overshoot v) Settling time
Time Response Specifications
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DELAY TIME The delay time is the time
required for the response to reach
50% of the final value for the very
first time.
rt
RISE TIME The rise time is the time
required for the response to rise
from 0 to 100% of the final value
for under damped system and
from 10% to 90% of the final
value for over damped systems.
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MAXIMUM OVERSHOOT
The peak or 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.
Maximum percent overshoot = ( ) ( )
100%( )
pc t cX
c
PEAK TIME
The peak time is the time required for the response to reach
the first peak of the overshoot.
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SETTLING TIME
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%).
STEADY STATE ERROR
It indicates the error between the actual output and the
desired output as t tends to infinity.
[ ( ) ( )]sst
e Lt r t c t
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i) Delay time ii) Rise time iii) Peak time
ii) Maximum overshoot v) Settling time
Time Response Specifications
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EXPRESSION FOR TIME RESPONSE SPECIFICATIONS
RISE TIME
The output of a second order under damped system excited
by a unit step input is given by
2( ) 1 sin( )
1
nt
d
ec t t
Rise time is defined as the time taken by the output to rise
from 0 to 100% of the final value
rt t ( ) 1rc t Therefore at
,
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21 1 sin( )
1
n rt
d r
et
2
sin( ) 01
n rt
d r
et
20
1
n rte
sin( ) 0 sind rt
d rt d rt
21
2
1tan
1r
d n
t
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PEAK TIME
The output of a second order under-damped system
excited by a unit step input is given by
2( ) 1 sin( )
1
nt
d
ec t t
Peak time is defined as the time at which the
maximum value of magnitude occurs
Therefore, at pt t slope of c(t) must be zero
2 2
( )| cos( ). sin( ) ( ) | 0
1 1
n n
p p
t t
t t d d d n t t
dc t e et t
dt
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2sin( ) 1 cos( ) 0n d p n d pt t
2sin( ) 1 cos( ) 0d p d pt t
cos sin( ) sin cos( ) 0d p d pt t
sin( ) 0 sind pt d pt
21p
d n
t
ξ
ϴ
1
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PEAK OVERSHOOT
2( ) 1 sin( )
1
nt
d
ec t t
The peak overshoot is the difference between the peak value
and the reference input. Therefore,
2( ) 1 1 sin( ) 1
1
nt
p p d p
eM c t t
2sin( )
1
n pt
d p
et
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21
2sin
1
n
n
d
d
e
21
2sin
1
e
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SETTLING TIME
2( ) 1 sin( )
1
nt
d
ec t t
4 44s
n
t T
Assuming ξ to be small
3 33s
n
t T
2% criterion
5% criterion
The equations for obtaining are valid only for the standard second order system
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STEADY STATE ERROR
2( ) 1 sin( )
1
nt
d
ec t t
2( ) [1 ( )] sin( ) 0
1
nt
ss dt t t
ee Lt e t Lt c t Lt t
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Numericals 1)The closed loop transfer function of certain second order unity
feedback system is given below. Determine the type of damping in the system
2
( ) 8
( ) 3 8
C s
R s s s
Solution Comparing the given transfer functions with the standard form of the
transfer function of a second order system
2
2 2 2
( ) 8
( ) 3 8 2
n
n n
C s
R s s s s s
2 8n 8 2.82n 2 3n 3 3
0.532 2 2.82n
1 Hence it is an under damped system
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Numericals 2)The closed loop transfer function of certain second order unity
feedback system is given below. Determine the type of damping in the system
Solution Comparing the given transfer functions with the standard form of the
transfer function of a second order system
2
( ) 2
( ) 4
C s
R s s
2
2 2 2
( ) 2
( ) 4 2
n
n n
C s
R s s s s
2 0n 2 4n 2n
0
0
Hence it is an undamped system
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Numericals 3) Measurements conducted on a servomechanism show the
system response to be
when subjected to step input of magnitude 2. Determine the undamped
natural frequency and damping ratio.
Solution
Taking Laplace transform of the above equation we get
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For step input
Therefore
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Comparing with standard second order system equation
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4) Obtain the response of a unity feedback system whose open
loop transfer function is for a unit step input
3( )
( 4)G s
s s
Solution
( ) ( )
( ) 1 ( )
C s G s
R s G s
3
( 4)
31
( 4)
s s
s s
2
3 3
4 3 ( 1)( 3)s s s s
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For a unit-step input, r(t)=1. Therefore 1
( )R ss
3( )
( 1)( 3)C s
s s s
3 11 2 2
1 3s s s
Taking the inverse Laplace transform, the response is
33 1( ) 1
2 2
t tc t e e
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5) What is the response of the system for a unit-step input with
Solution
The closed loop transfer function of the system is
2
10
( ) 10( 3)
10( ) 4 101 (0.1 1)
( 3)
C s s s
R s s ss
s s
For a unit step input 1
( )R ss
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2
1 10( )
4 10C s
s s s
2 4 10
A Bs C
s s s
2
1 4
4 10
s
s s s
2 2 22
1 2 2 6.
6 ( 2) ( 6)( 2) ( 6)
s
s ss
Taking the inverse Laplace transform, the response is
2 22( ) 1 cos 6 sin 6
6
t tc t e t e t
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6)The open loop transfer function of a unity feedback system is 4
( )( 1)
G ss s
Determine the nature of response of the closed loop system for a unit step input.
Also determine the rise time, peak time, peak overshoot and settling time
Solution
The closed loop transfer function is 2
4
( ) 4( 1)
4( ) 41
( 1)
C s s s
R s s s
s s
Comparing it with the standard form of the closed loop transfer function
of a second order system
2
2 2 2
( ) 4
( ) 4 2
n
n n
C s
R s s s s s
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2 4n 2n 2 1n 1 1
0.252 2 2n
21d n 22 1 0.25 1.936 rad/s
2 21 11 1 0.25
tan tan 1.3100.25
The rise time 3.141 1.310
0.9451.936
r
d
t s
rad/s
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The peak time in sec 3.141
1.6221.936
p
d
t s
21
pM e
The peak overshoot = 0.4326
Therefore, percentage of peak overshoot is 100% 43.26%pM
The settling time for 5% error is 3 3
60.25 2
s
n
t
sec
The settling time for 2% error is
4 48
0.25 2s
n
t
sec
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7) A unity feedback system is characterized by an open-loop transfer
function
( )( 10)
KG s
s s
Determine the gain K so that the system will have a damping ratio of 0.5.
For this value of K, determine the settling time, peak overshoot and time
to peak overshoot for a unit step input
Solution The closed loop transfer function of the given feedback system is
2
( ) ( ) ( 10)
( ) 1 ( ) 101
( 10)
K
C s G s Ks s
KR s G s s s K
s s
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Comparing it with the standard form of the transfer function
of the second order system, we have
2
2 2 2
( )
( ) 10 2
n
n n
C s K
R s s s K s s
2
n K n K 2 10n
2 0.5 10n 10n
2 210 100nK
So the gain K=100 so that the system will have a damping ratio of 0.5
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The settling time for 5% criterion is
4 40.8
0.5 10s
n
t s
The settling time for 2% criterion is
3 30.6
0.5 10s
n
t s
The peak overshoot is
% 100 0.163 100 16.3%p pM M
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The peak time
21p
d n
t
2
3.140.363
10 1 0.5s
The time to peak overshoot is 0.363pt s
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8) A second order system is represented by a
transfer function given below
0
2
( ) 1
( )
Q s
T s Js fs K
where
0 ( )Q s the proportional to output and T is is the torque
input. A step input 10 N-m is applied to the system and the
test results are given below
Peak overshoot
pM
pM =6% Peak time pt =1s
The steady state output of the system is 0.5 radian
Determine the values of J, K and f.
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Given, the input torque T is a step of 10 N-m. Therefore 10
( )T ss
0 2 2
( ) 10( )
( )
T sQ s
Js fs K s Js fs K
00
( )sLt sQ s
The steady state value of output =
20
100.5 .
( )sLt s
s Js fs K
100.5
K 20K
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Given the peak overshoot 6% 0.06pM
210.06 e
0.667
Given that peak time 1pt s
21
1d n
4.21n rad/s
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Comparing the given equation with the standard form of
characteristic equation of a second-order system
2 2 22 0n n
f Ks s s s
J J
2
n
K
J 2
2 2
201.128
4.21n
KJ kg m
2 n
f
J
2 2 0.667 4.21 1.128 6.34 / /nf J N m rad s
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Steady State Error of Unity Feedback Systems
The closed-loop transfer function is
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For Unity feedback system
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Static Position Error Constant (Kp)
The steady-state error of the system for a unit-step input is
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The static position error constant Kp is defined by
The steady state error for step input is
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Static Velocity Error Constant (K v)
• The steady-state error of the system for a unit-ramp
input is
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• The static velocity error constant K v is defined by
• Thus, the steady-state error for a ramp input is
given by
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Static Acceleration Error Constant (Ka)
• The steady-state error of the system for parabolic
input is
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• The static acceleration error constant Ka is defined
by
• Thus, the steady-state error for parabolic input is
given by
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For N=0 there is no pole at origin and the system is
referred as TYPE 0 system.
If N=1 the system is referred as TYPE 1 system etc.
The highest degree of the characteristic polynomial gives the order of the system
TYPE and ORDER of a system
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• For a Type 0 system
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• For a Type 0 system
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• For a Type 1 system
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• For a Type 1 system
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• For a Type 2 system
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For a Type 2 system
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STEADY STATE ERROR
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1) For a closed loop system whose open loop transfer function
Find the steady state error when the input is
Solution
For
Steady state error Ess=
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For
=5
Steady state error Ess =
For
= 0
Steady state error Ess =
Total steady state error = 0 + 0.4 + =
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2)The block diagram of a unity feedback control system
with
inner output derivative feedback is shown below.
i)Calculate the steady state error for unit ramp input.
ii) choose the value of K such that the unit step response of
the system has no over shoot and yet it is as fast as
possible
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Solution
The forward transfer function with inner loop
Steady state error for unit ramp input 1/Kv
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Closed loop transfer function is
For no overshoot
For fast response
K=0.2
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Controllers
Control systems are designed to meet three time
response specifications
Steady state accuracy
Peak overshoot to step input (damping factor)
Settling time
It can be proved that to meet these specifications a
second order system needs to modified.
This modification is termed as compensation
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In the proportional control algorithm, the controller output is
proportional to the error signal, which is the difference between
the reference signal and the feedback signal
If the input error variable, e(t) the output of the controller p(t) is
p(t) = Kp*e(t)
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Therefore it increases the forward gain of the system
Hence it increases the natural undamped frequency
but decreases damping ratio
Therefore the
steady state accuracy improves, but the
transient response becomes more oscillatory
ξ
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Proportional plus Derivative Control
The controller output is equal to proportional plus
derivative of the error signal
Kd*de(t)/dt Kp*e(t) +
Taking Laplace transform of above equation gives
The effect of Increasing the coefficient of s term in
equation, which increases the damping of the system
Hence the transient performance improves
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The effect of Increasing the coefficient of s term in equation,
which increases the damping of the system
Hence the transient performance improves
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Ki*∫e(t)dt
Proportional plus Integral controller Control
The controller output is equal to proportional plus
integral of the error signal Kp*e(t) +
Taking Laplace transform of above equation gives
Integral error compensation increases the order of the system
.
If the system forward path has type-1 T.F , the Integral
compensator changes it to type 2 system.
Hence Improves the steady state accuracy
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PID Controller
In General the effects of dynamic performance caused by
PID controller is not obvious
If G(s) is second order system Introduction of controller
converts the characteristic equation to third order.
Certain values of Kp, Kd,Ki may cause Instability in the
system
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In General the effects of increasing parameters is:
Parameter: Rise Time Overshoot Settling Time S.S.Error
Kp Decrease Increase Small Change Decrease
Ki Decrease Increase Increase Eliminate
Kd Small Change Decrease Decrease None
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