lecture 13 14-time_domain_analysis_of_1st_order_systems
TRANSCRIPT
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Feedback Control Systems (FCS)
Dr. Imtiaz Hussainemail: [email protected]
URL :http://imtiazhussainkalwar.weebly.com/
Lecture-13-14 Time Domain Analysis of 1st Order Systems
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Introduction• The first order system has only one pole.
• Where K is the D.C gain and T is the time constant of the system.
• Time constant is a measure of how quickly a 1st order system responds to a unit step input.
• D.C Gain of the system is ratio between the input signal and the steady state value of output.
1
Ts
K
sR
sC
)()(
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Introduction• For the first order system given below
13
10
ssG )(
5
3
ssG )(
151
53
s//
• D.C gain is 10 and time constant is 3 seconds.
• And for following system
• D.C Gain of the system is 3/5 and time constant is 1/5 seconds.
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Impulse Response of 1st Order System
• Consider the following 1st order system
1Ts
K)(sC)(sR
0 t
δ(t)
1
1 )()( ssR
1
Ts
KsC )(
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Impulse Response of 1st Order System
• Re-arrange above equation as1
Ts
KsC )(
Ts
TKsC
//
)(1
TteT
Ktc /)(
• In order to represent the response of the system in time domain we need to compute inverse Laplace transform of the above equation.
atAeas
AL
1
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Impulse Response of 1st Order SystemTte
T
Ktc /)( • If K=3 and T=2s then
0 2 4 6 8 100
0.5
1
1.5
Time
c(t)
K/T*exp(-t/T)
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Step Response of 1st Order System• Consider the following 1st order system
1Ts
K)(sC)(sR
ssUsR
1 )()(
1
Tss
KsC )(
1
Ts
KT
s
KsC )(
• In order to find out the inverse Laplace of the above equation, we need to break it into partial fraction expansion
Forced Response Natural Response
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Step Response of 1st Order System
• Taking Inverse Laplace of above equation
1
1
Ts
T
sKsC )(
TtetuKtc /)()(
• Where u(t)=1 TteKtc /)( 1
KeKtc 63201 1 .)(
• When t=T
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Step Response of 1st Order System• If K=10 and T=1.5s then TteKtc /)( 1
0 1 2 3 4 5 6 7 8 9 100
1
2
3
4
5
6
7
8
9
10
11
Time
c(t)
K*(1-exp(-t/T))
Unit Step Input
Step Response
1
10
Input
outputstatesteadyKGainCD
.
%63
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Step Response of 1st Order System• If K=10 and T=1, 3, 5, 7 TteKtc /)( 1
0 5 10 150
1
2
3
4
5
6
7
8
9
10
11
Time
c(t)
K*(1-exp(-t/T))
T=3s
T=5s
T=7s
T=1s
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Step Response of 1st order System
• System takes five time constants to reach its final value.
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Step Response of 1st Order System• If K=1, 3, 5, 10 and T=1 TteKtc /)( 1
0 5 10 150
1
2
3
4
5
6
7
8
9
10
11
Time
c(t)
K*(1-exp(-t/T))
K=1
K=3
K=5
K=10
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Relation Between Step and impulse response
• The step response of the first order system is
• Differentiating c(t) with respect to t yields
TtTt KeKeKtc //)( 1
TtKeKdt
d
dt
tdc /)(
TteT
K
dt
tdc /)( (impulse response)
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Example#1• Impulse response of a 1st order system is given below.
• Find out– Time constant T– D.C Gain K– Transfer Function – Step Response
tetc 503 .)(
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Example#1• The Laplace Transform of Impulse response of a
system is actually the transfer function of the system.• Therefore taking Laplace Transform of the impulse
response given by following equation.tetc 503 .)(
)(..
)( sSS
sC
50
31
50
3
50
3
.)()(
)()(
SsR
sC
s
sC
12
6
SsR
sC
)()(
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Example#1• Impulse response of a 1st order system is given below.
• Find out– Time constant T=2– D.C Gain K=6– Transfer Function – Step Response– Also Draw the Step response on your notebook
tetc 503 .)(
12
6
SsR
sC
)()(
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Example#1• For step response integrate impulse response
tetc 503 .)(
dtedttc t 503 .)(
Cetc ts 506 .)(
• We can find out C if initial condition is known e.g. cs(0)=0
Ce 05060 .
6Ct
s etc 5066 .)(
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Example#1• If initial Conditions are not known then partial fraction
expansion is a better choice
12
6
SsR
sC
)()(
12
6
SssC )(
1212
6
s
B
s
A
Ss
ssRsR
1)(,)( input step a is since
50
66
12
6
.
ssSs
tetc 5066 .)(
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Partial Fraction Expansion in Matlab• If you want to expand a polynomial into partial fractions use
residue command.
Y=[y1 y2 .... yn];X=[x1 x2 .... xn];[r p k]=residue(Y, X)
kps
r
ps
r
ps
r
sx
sy
n
n
2
2
1
1
)()(
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Partial Fraction Expansion in Matlab• If we want to expand following polynomial into partial fractions
86
842
ss
s
Y=[-4 8];X=[1 6 8];[r p k]=residue(Y, X)
r =[-12 8] p =[-4 -2] k = []
2
2
1
12 86
84
ps
r
ps
r
ss
s
2
8
4
12
86
842
ssss
s
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Partial Fraction Expansion in Matlab
• If you want to expand a polynomial into partial fractions use residue command.
12
6
SssC )(
Y=6;X=[2 1 0];[r p k]=residue(Y, X)
r =[ -6 6]p =[-0.5 0]k = []
ssss
6
50
6
12
6
.
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Ramp Response of 1st Order System• Consider the following 1st order system
1Ts
K)(sC)(sR
2
1
ssR )(
12
Tss
KsC )(
• The ramp response is given as
TtTeTtKtc /)(
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0 5 10 150
2
4
6
8
10
Time
c(t)
Unit Ramp Response
Unit Ramp
Ramp Response
Ramp Response of 1st Order System• If K=1 and T=1 TtTeTtKtc /)(
error
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0 5 10 150
2
4
6
8
10
Time
c(t)
Unit Ramp Response
Unit Ramp
Ramp Response
Ramp Response of 1st Order System• If K=1 and T=3 TtTeTtKtc /)(
error
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Parabolic Response of 1st Order System• Consider the following 1st order system
1Ts
K)(sC)(sR
3
1
ssR )(
13
Tss
KsC )(
• Do it yourself
Therefore,
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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.
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Practical Determination of Transfer Function of 1st Order Systems
• If we can identify T and K from laboratory testing we can obtain the transfer function of the system.
1
Ts
K
sR
sC
)()(
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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.
T=0.13s
K=0.72
• K is simply steady state value.
• Thus transfer function is obtained as:
77
55
1130
720
..
..
)()(
sssR
sC
7
5
ssR
sC
)()(
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1st Order System with a Zero
• Zero of the system lie at -1/α and pole at -1/T.
1
1
Ts
sK
sR
sC )()()(
11
Tss
sKsC
)()(
• Step response of the system would be:
1
Ts
TK
s
KsC
)()(
TteTT
KKtc /)()(
Partial Fractions
Inverse Laplace
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1st Order System with & W/O Zero (Comparison)
1
1
Ts
sK
sR
sC )()()(
TteTT
KKtc /)()( TteKtc /)( 1
1
Ts
K
sR
sC
)()(
• If T>α the shape of the step response is approximately same (with offset added by zero)
TtenT
KKtc /)()(
Tte
T
nKtc /1)(
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1st Order System with & W/O Zero• If T>α the response of the system would look like
0 5 10 156.5
7
7.5
8
8.5
9
9.5
10
Time
c(t)
Unit Step Response
13
2110
s
s
sR
sC )()()(
3323
1010 /)()( tetc offset
𝑜𝑓𝑓𝑠𝑒𝑡=𝐾 +𝐾𝑇
(𝛼−𝑇 )
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1st Order System with & W/O Zero• If T<α the response of the system would look like
151
2110
s
s
sR
sC
.)(
)()(
511251
1010 ./)(
.)( tetc
0 5 10 159
10
11
12
13
14
Time
Uni
t S
tep
Res
pons
e
Unit Step Response of 1st Order Systems with Zeros
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1st Order System with a Zero
0 5 10 156
7
8
9
10
11
12
13
14
Time
Uni
t S
tep
Res
pons
eUnit Step Response of 1st Order Systems with Zeros
T
T
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0 2 4 6 8 100
2
4
6
8
10
12
14
Time
Uni
t Ste
p R
espo
nse
Unit Step Response of 1st Order Systems
1st Order System with & W/O Zero
T
T
1st Order System Without Zero
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Home Work
• Find out the impulse, ramp and parabolic response of the system given below.
1
1
Ts
sK
sR
sC )()()(
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Example#2
• A thermometer requires 1 min to indicate 98% of the response to a step input. Assuming the thermometer to be a first-order system, find the time constant.
• If the thermometer is placed in a bath, the temperature of which is changing linearly at a rate of 10°C/min, how much error does the thermometer show?
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PZ-map and Step Response
1
Ts
K
sR
sC
)()(
-1-2-3δ
jω
1
10
ssR
sC
)()( sT 1
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PZ-map and Step Response
1
Ts
K
sR
sC
)()(
-1-2-3δ
jω
2
10
ssR
sC
)()( sT 50.
150
5
ssR
sC
.)()(
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PZ-map and Step Response
1
Ts
K
sR
sC
)()(
-1-2-3δ
jω
3
10
ssR
sC
)()( sT 330.
1330
33
ssR
sC
..
)()(
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Comparison
1
1
ssR
sC
)()(
10
1
ssR
sC
)()(
Step Response
Time (sec)
Am
plitu
de
0 1 2 3 4 5 60
0.2
0.4
0.6
0.8
1
0 0.1 0.2 0.3 0.4 0.5 0.60
0.02
0.04
0.06
0.08
0.1Step Response
Time (sec)
Am
plitu
de
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First Order System With Delays
• Following transfer function represents the 1st order system with time lag.
• Where td is the delay time.
dsteTs
K
sR
sC
1)()(
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First Order System With Delays
dsteTs
K
sR
sC
1)()(
1
Unit StepStep Response
ttd
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First Order System With Delays
sessR
sC 2
13
10
)()(
0 5 10 15
0
2
4
6
8
10
Step Response
Time (sec)
Am
plit
ude
st d 2sT 3
10K
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Examples of First Order Systems
• Armature Controlled D.C Motor (La=0)
abt
at
RKKBJs
RK
U(s)
Ω(s)
uia T
Ra La
J
B
eb
V f=constant
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Examples of First Order Systems
• Electrical System
1
1
RCssE
sE
i
o
)()(
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Examples of First Order Systems
• Mechanical System
1
1
sk
bsX
sX
i
o
)()(
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Examples of First Order Systems
• Cruise Control of vehicle
bmssU
sV
1
)()(
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END OF LECTURES-13-14
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