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Review : Transient and Steady-State Response Analysis Step response of a control system
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Modern Control SystemEKT 308
Steady-State and Stability
Quick Review1. Laplace transform2. Poles and zeros , transfer function3. Simplification complex block diagram, signal flow
diagram4. State space modeling5. Modeling physical systems6. Time response of first and second order systems
Topic to cover1. Steady State Error2. Routh Hurwitz Stability Criterion
Review : Transient and Steady-State Response Analysis
Step response of a control system
Review: Performance Measures for step response
Delay Time : Time to reach half of the final value for the first time.
Rise Time : Time to rise to the final value.
Underdamped (0% 100%)
Overdamped (10% 90%) Peak Time : Time to reach the first peak of the overshoot. Percentage overshoot,
rT
pT
dT
%100)(
)()(
%100..
yyTy
ffM
OP
p
v
vpt
)(sG
)(sH
Y(s)R(s)E(s)+
- B(s)
)()()(1
1)( sRsHsG
sE
From the diagram
Consider 1...11
1...11)(
21
21
sTsTsTssTsTsT
KsGbjbb
naiaa and
1...11
1...11)(
21
21
sTsTsTsTsTsT
sHylyy
xkxx
Use the FINAL VALUE THEOREM and define steady state error,
sse
that is given by )(lim)(lim0
ssEteest
ss
Steady State Error
Unit stepUnit step input,
ssR 1)(
From)(
)()(11)( sR
sHsGsE
Steady state error, ssHsG
ses
ss1
)()(1lim0
We define step error coefficient, )()(lim
0sHsGK
ss
Thus, the steady state error is S
ss Ke
11
By knowing the type of open-loop transfer function, )()( sHsG
We could determine step error coefficient and thus the steady state error
)()(lim0
sHsGKs
s
1...11
1...11lim21
21
0
sTsTsTs
sTsTsTKbjbb
naiaa
s
1...11
1...11
21
21
sTsTsTsTsTsT
ylyy
xkxx
For open-loop transfer function of type 0: KKs K
ess
11
sK 01
1
sseFor open-loop transfer function of type 1: ,
sK 01
1
sseFor open-loop transfer function of type 2: ,
STEADY STATE ERROR EXAMPLEA first order plant with time constant of 9 sec and dc gain of 5 is negatively feedback with unity gain, determine the steady state error for a unit step input and the final value of the output.Solution:The block diagram of the system is
519
5)()(00
s
sHsGK LimLimss
s
As we are looking for a steady state error for a step input, we need to know step error coefficient,
195s
Y(s) R(s) +
-
.
Knowing the open-loop transfer function, thensK
And steady state error of
61
511
11
S
ss Ke
Its final value is65611 ssy
Unit Ramp
ttr )( , while its Laplace form is 21)(s
sR As in the previous slide, we know that
)()()(1
1)( sRsHsG
sE
Thus, its steady state error is2
0
1)()(1
limssHsG
ses
ss
Define ramp error coefficient, )()(lim
0sHssGK
sr
Which the steady state error isr
ss Ke 1
Just like the unit step input we can conclude the steady state error for a unit ramp through the type of the open-loop transfer function of the system.
For open-loop transfer function of type 0: 0rKFor open-loop transfer function of type 1: KKr
For open-loop transfer function of type 2: rK
Example:
A missile positioning system is shown.
(i) Find its closed-loop transfer function )()(
ss
i
m
(ii) Determine its undamped natural frequency and its damping ratio if 310K
(iii) Determine the steady state error, if the input is a unit ramp.
)14.0(01.0ss
K mi
Compensatot DC motor +
-
Solution:
(a) By Mason rule, the closed-loop transfer function is
KssK
KssK
ssK
ssK
ss
i
m
025.05.2025.0
01.04.001.0
)14.0(01.0.1
)14.0(01.0.
)()(
22
KssK
KssK
025.05.2025.0
01.04.001.0
2
2
(b) If 310K
,
255.225
)()(
2
ssss
i
m
Comparing with a standard second order transfer function
22
2
2)()(
nn
n
i
m
ssK
ss
Comparing
252 nThus undamped natural frequency
5n rad.s-1
and
5.22 n
damping ratio of 25.0
(c) To determine the ramp error coefficient, we must obtain its open-loop transfer function
)14.0(01.0.)(
ss
KsGo
As it is a type 1, the system will have a finite ramp error coefficient, putting 310K
)14.0(10)(
sssGo
10)14.0(
10.lim)(lim00
ss
sssGKs
os
r
Hence steady state error of 1.01
rss Ke
Steady State Error of Feedback Control System
Stability
Stability
Stability
Routh-Hurwitz Stability CriteriaIf a polynomial is given by
0.....)( 011
1 asasasasT n
nn
nWhere,
011 ,.....,, aaaa nn are constants and ...1nNecessary condition for stability are:
All the coefficients of the polynomial are of the same sign. If not, there are poles on the right hand side of the s-plane
(i)
All the coefficient should exist accept for the
.
(ii) 0a
For the sufficient condition, we can now formed a Routh-array,
Routh’s Array
2c 3c
2h
ns
1ns
2ns
3ns
na
1na
1b
1c
2na 4na 6na
7na5na3na
4b3b2b
4c
3s
2s
1s
0s
1h
1i 2i
1j
1k
Routh Array elements
1
321
1
31
2
1
n
nnnn
n
nn
nn
aaaaa
aaaaa
b
1
541
1
51
4
2
n
nnnn
n
nn
nn
aaaaa
aaaaa
b
1
2131
1
21
31
1 bbaab
bbbaa
c nn
nn
1
3151
1
31
51
2 bbaab
bbbaa
c nn
nn
1
2121
1
21
21
1 iihhi
iiihh
j
21 ik
Routh-Hurwitz Criteria states that the number of roots of charateristic equation is the same as the numberof sign changed of the first column.
Case 1: No zero on the first column
After the Array has been tabled, all the elements on the first column are not equal to zero.
If there is no sign changed, all the poles are in the LHP. While the number of poles on the RHP is equal to number of sign change on the first column of the Routh’s array.
Example:
Consider a fourth order characteristic equation
028122)( 234 sssssD
Example
Form the Routh’s array
2 12 2
1 8
2
2
4s3s2s1s0s
41
1612
5.84
232
There are two sign change on rows 2 and 3. Hence, there two poles on the RHP (Right-half f s-palne).
Solution:
Scilab solutionCE=poly([2 1 12 8 2],'s','c');roots(CE)
ans = 0.0885283 + 2.4380372i 0.0885283 - 2.4380372i - 0.3385283 + 0.2311130i - 0.3385283 - 0.2311130i
1 3 5
2 6 3
3
3
Solution:
Form the Routh’s array
4s3s
2s1s
0s
5s
02
66 , =1
176
5.61412
64942 2
5.32
310
1
If , is a small positive number there are two sign change at row 3 and 4, and also at row 4 and 5 . Hence, there two poles on the RHP.
Case 2: Coeffiecient of the first column is zero but not the others. Change the zero element by a small positive number, . The number of pole on the RHP will depend on the number of sign change.
Example:Consider a fifth order characteristic equation 035632)( 2345 ssssssD
Scilab Solution
-->roots(CE) ans = 0.3428776 + 1.5082902i 0.3428776 - 1.5082902i - 1.6680888 - 0.5088331 + 0.7019951i - 0.5088331 - 0.7019951i
-->CE=poly([3 5 6 3 2 1],'s','c')
Formed Routh’s array
5s
4s
3s
2s
1s0s
1 6 8
7 42 56
21 56
9.3
56
07
4242
28
07
5656
84
Case 3: All the coefficients on a row are zeros.Form an auxiliary equation from the row above it and replace the coefficient of the row with the differentiated coefficient of the auxiliary equation. For this case, if there is no sign change, the characteristic equation has a pair of poles with opposite sign of real component or/and a pair of conjugate poles on the imaginary axis.
Example:
Consider this fifth order characteristic equation
05684267)( 2345 ssssssD
Form the auxillary equation on the second row: 56427)( 24 sssPDifferentiate the equation: ss
dssdP 8428)( 3 As there is no sign change,
there is a pair of conjugate poles on the axis and/or a pair of poles with opposite sign of real component. To be sure we can use Scilab
-->CE=poly([56 8 42 6 7 1],'s','c')
-->roots(CE) ans = - 7. - 8.049D-16 + 2.i - 8.049D-16 - 2.i 1.4142136i - 1.4142136i
Use of Routh Hurwitz Criteria
Main use is to determine the position of the poles, which in turns can determinethe stability of the response.
UNSTABLE STABLE
j
1 3 K
3 2
K
K
Solution: Charactristic equation is 0212 KssssExpand the equation 0233 234 Kssss
Form the Routh’s array
3s
2s
1s
0s
4s
37
329
7914
373314 KK
ExampleA closed-loop transfer function is given by Kssss
KsRsC
21)()(
2
Determine the range for K for the system to be always stable and its oscillatingfrequency before it becomes unstable.
To ensure that there is no poles on the RHP of the s-plane, there mustbe no sign change on the first column of the Routh’s Array, therefore for no signchange:Refering to row 4: 0
7914 K
which gives 914Kand row 5 0K
Hence its range 9140 K
091437 2
Oscillating frequency 32 rad.s-1.
