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Transcript of 1 Time Response. CHAPTER 3. 2 3.2 Poles and Zeros and System Response. Figure 3.1: (a) System...
1
Time Time ResponsRespons
e.e.
CHAPTER CHAPTER 33
2
3.2 Poles and Zeros and System 3.2 Poles and Zeros and System Response.Response.
ssR
sssG
sRsGsCs
KsG
1)(
52)(
where)()()(
)(
55/35/2
5)5()2()(
)()(
ss
sB
sA
ssssC
ssKsC
t
t
eKKtc
etc
21
5
)(53
52)( Figure 3.1: (a) System showing input
and output; (b) Pole-zero plot of the system;
(c) Evolution of a system response.
3
The The transfer functiontransfer function,,
For a unit step of; 1/s,For a unit step of; 1/s,
Its response,Its response,
A, B and B’ are constant. For K=1 and A, B and B’ are constant. For K=1 and =1/a then,=1/a then,
Time constantTime constant (1/a), is defined as the time for e (1/a), is defined as the time for e-at-at to to decay to 37%of its initial value decay to 37%of its initial value or or the time it takes the time it takes for step response to reach 63% of its final value. for step response to reach 63% of its final value.
3.3 First Order System.3.3 First Order System.
1)()()(
sK
sRsCsG
111)(
sB
sA
sK
ssC
teBAtc ')(
atetc 1)(
4
Step responseStep response,,
ateKKtc 21)(
)()(
assKsC
Cont’d…Cont’d…
Figure 3.2: (a) First Order Response to a Unit Step.
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The poleThe pole of the transfer function is at of the transfer function is at –a–a, the , the farther the pole from the imaginary axis, the farther the pole from the imaginary axis, the faster the transient response.faster the transient response.
Rise time (TRise time (Trr), ), the time the response to go the time the response to go from the 0.1 to 0.9 of its final value. from the 0.1 to 0.9 of its final value. TTrr=2.2/a=2.2/a..
Settling time (TSettling time (Tss)),, time range when the time range when the response to reach and stay within 2% of its response to reach and stay within 2% of its final value. Let c(t) = 0.98 then the final value. Let c(t) = 0.98 then the TTss.=4/a..=4/a.
Cont’d…Cont’d…
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3.4 Second Order System.3.4 Second Order System. The The transfer function,transfer function,
For Impulse response,For Impulse response, Where,Where,
Standard Form,Standard Form,
Where Where KK is the dc gain, is the dc gain, is the damping ratio, is the damping ratio, nn is the undammped natural is the undammped natural
frequency.frequency.
012
0
)()(
bsbsa
sRsC
2
2
1
1)(
ss
sC
22
2
2)()(
nn
n
ss
KsRsC
02 22 nnss
122,1 nns
WhereWhere
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Example of 2Example of 2ndnd order system order system responses.responses.
Figure 3.3: Second Order System, pole plots and Step Figure 3.3: Second Order System, pole plots and Step Response.Response.
Cont’d…Cont’d…
8
General 2General 2ndnd Order System. Order System.
Natural Frequency (Natural Frequency (nn),),
Damping Ratio (Damping Ratio ()),,Example 3.1Example 3.1: Find the Natural Frequency (: Find the Natural Frequency (nn) and ) and Damping Ratio (Damping Ratio ()),,
Solution:Solution:nn = 6 and = 6 and =0.35.=0.35.
)()(
)()()(
2 basssbsC
sGsRsC
n
n
n
n
nn
n
ab
ab
sssG
2
2
2)(
2
22
2
Cont’d…Cont’d…
)362.4(36)( 2
ss
sG
9
From previous page,From previous page,
Figure 3.4: Left; Plot for an underdamped 2Figure 3.4: Left; Plot for an underdamped 2ndnd Order System. Order System. Right; Step Response for 2Right; Step Response for 2ndnd Order System Damping Cases. Order System Damping Cases.
22
2
2 2)(
nn
n
ssbassbsG
)()(
)()()(
2 basssbsC
sGsRsC
)cos()( 21 teKKtc t
Cont’d…Cont’d…
1kos
10
From previous page,From previous page,
Figure 3.5: Second Order Response as a Function of Damping Figure 3.5: Second Order Response as a Function of Damping Ratio.Ratio.
Cont’d…Cont’d…
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The transfer function,The transfer function, Poles at the origin from the Poles at the origin from the unit stepunit step and and
two real poles from the two real poles from the systemsystem.. Constant force response and force Constant force response and force
response.response.
3.4.1 Over Damped Response.3.4.1 Over Damped Response. ))146.1)(854.7(
9)99(
9)( 2
sssssssC
tt eKeKKtc 146.13
854.721)(
12
Example 3.1:Example 3.1: Over Damped Response. Over Damped Response.Find the step response of the system.
Solution:Expand the partial fraction.
Take the inverse Laplace Transform.
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Under Damped transfer function,Under Damped transfer function,
When 0 < When 0 < < 1 The transfer function is,< 1 The transfer function is, The Pole position is,The Pole position is,
3.4.2 Under Damped 3.4.2 Under Damped Response.Response.
dndn
n
jsjsK
sRsC
2
)()(
21 nd
dddd
nn
n
jsK
jsK
sssG
*
22
2
2)(
)92(9)( 2
sss
sC
14
From previous page,From previous page,
Figure 3.6: Second Order Response as a Function of Damping Figure 3.6: Second Order Response as a Function of Damping Ratio.Ratio.
Cont’d…Cont’d…
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Performance Measures.Performance Measures.
21
n
pT Peak Time
Overshoot
nsT
4
Settling Time
Figure 3.7: (Top) The 2nd Order Underdamped
Response Specification. (bottom) Percent overshoot
versus damping factor
Cont’d…Cont’d…
%1001
11
%100%
21
e
cctc
M pp
16
Performance Measures.Performance Measures.
Poles positionPoles position
Figure 3.8: Lines of constant peak time, Tp , settling time, Ts , and percent overshoot, %OS Note: Ts2 < Ts1 ; Tp2 < Tp1; %OS1 < %OS2
dn
pT
21
dnsT
44
21100%
eOS
dddd
nn
n
jsK
jsK
sssG
*
22
2
2)(
Cont’d…Cont’d…
17
Pole Placement.Pole Placement.
Figure 3.9: Step responses of second-order underdamped systems as poles move: (a) with constant real part; (b) with constant imaginary part; (c) with constant damping ratio
Cont’d…Cont’d…
POd
sd TPO
sd T
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The transfer function,The transfer function,3.4.3 Critically Damped.3.4.3 Critically Damped.
22
)()(
n
n
ssRsC
19
Example 3.3:Example 3.3: Critically Damped Critically Damped Response.Response.Find the step response of the system.Solution:Expand the partial fraction.
20
Dominant Pole.Dominant Pole. The formula that describing %OS, tThe formula that describing %OS, tss, t, tp p
were were derived only for system with two derived only for system with two complex complex poles poles and and no zeros.no zeros.
A system with more than two poles or A system with more than two poles or zeros can be zeros can be approximated as a second approximated as a second order system that has just order system that has just two complex two complex dominant poles.dominant poles.
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Dominant Pole.Dominant Pole.
Figure 3.11: Component responses of a three-pole system: (a) pole plot; (b) component responses: non-dominant pole is
near dominant second-order pair (Case I), far from the pair (Case II), and at infinity (Case III).
Cont’d…Cont’d…
Cannot be represented as second order system
Will approach second order system
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Cont’d…Cont’d…Effect of adding a zero to a two-pole systemEffect of adding a zero to a two-pole system The closer is the zero to dominant poles, The closer is the zero to dominant poles, the greater its the greater its effect on transient response.effect on transient response. As the zero move away from dominant As the zero move away from dominant poles, the poles, the response approaches that of the two response approaches that of the two pole system.pole system. Starting at poles
1±j2.828, then consecutively add zeros at -3, -5, -10.
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(i) Stable system. Natural response
approaches zero. Poles in LHP.
(ii) Unstable system. Natural response
grows. Poles in RHP.
(iii) Marginally stable system.
Natural response neither grows/approaches zero.
Poles on j axis.
3.5 Stability.3.5 Stability.
Figure 3.12: Closed-loop poles and response: a. stable system;
b. unstable system
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3.6 Routh-Hurwitz Stability 3.6 Routh-Hurwitz Stability Criteria.Criteria.
What is Routh-Hurwitz Criterion (RHC)?What is Routh-Hurwitz Criterion (RHC)? Through the RHC method we can tell how Through the RHC method we can tell how
many close-loop system many close-loop system polespoles are in the left are in the left half plane, in the right half-plane and on half plane, in the right half-plane and on the jthe j-axis. We can find the number of poles -axis. We can find the number of poles in each section of the s-plane, but in each section of the s-plane, but cannotcannot find their coordinatefind their coordinate. .
The number of roots of the polynomial that The number of roots of the polynomial that are in the right half-plane is equal to the are in the right half-plane is equal to the number of changes in the first column.number of changes in the first column.
The RHC method requires The RHC method requires two stepstwo steps;;(1) Generate the data table called Routh (1) Generate the data table called Routh table.table.(2) Interpret the Routh table to tell number (2) Interpret the Routh table to tell number of close loop system poles in the left half of close loop system poles in the left half plane, in the right half-plane and on the jplane, in the right half-plane and on the j--axis. axis.
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The Close-Loop Transfer function.The Close-Loop Transfer function.
Initial layout for the Routh-Hurwitz Table.Initial layout for the Routh-Hurwitz Table.
Completed Completed Routh TableRouth Table..
Cont’d…Cont’d…
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Example 3.4:Example 3.4: Routh-Hurwitz. Routh-Hurwitz.Make a Routh table from the system shown below.
Solution:Find the equivalent close loop system.Figure (b) above.
Interpretation:There are two sign changes in the first column.1 -72 103The system is unstable, two poles exist in the right half plane.
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Example 3.5:Example 3.5: Routh-Hurwitz. Routh-Hurwitz.
The number of RHP poles = The number of The number of RHP poles = The number of SIGN CHANGES of COL 1SIGN CHANGES of COL 1
TWOTWO sign changes: sign changes: RHP Poles =2RHP Poles =2
P(s) = s3 + 10s2 + 31s + 1030
s3 1 31 0 s2 10 1030 0 s1
b1 = 11031311
= - 72 b2 = 1
0101
= 0 b3 = 1
0101
= 0
s0 c1 = 72
0721031
=
103
c2 = 7207201
= 0 c3 = 72
07201
= 0
)595.87068.1)(595.87068.1)(4136.13()(10303110)( 23
jsjsssPssssP
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Example 3.6:Example 3.6: Routh-Hurwitz. Routh-Hurwitz.
Solution:Solution:
Two sign changes: 2 RHP (UNSTABLE)Two sign changes: 2 RHP (UNSTABLE) Poles: 2 LHP and 2 RHPPoles: 2 LHP and 2 RHP
2006116200)( 234
ssss
sT
s 4 1 1 1 2 0 0
s 3 6 1 6 1 0
s 2 1 0 1 2 0 0 2 0 0
s 1 - 1 9 0 0
s 0 2 0 0 0
)54.227.1)(54.227.1)(54.227.4)(54.227.4()(2006116)( 234
jsjsjsjssPsssssP
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Example 3.7:Example 3.7: Routh-Hurwitz. Routh-Hurwitz.
Assume Assume is small POSITIVE : TWO sign is small POSITIVE : TWO sign changeschanges
Poles: 2 RHP, 3 LHPPoles: 2 RHP, 3 LHP
3563210)( 2345
sssss
sT
s 5 1 3 5
s 4 2 6 3
s 3 0 7 /2 0
s 2
76 3 0
s 1
141264942 2
0 0
s 0 3 0 0
)7.051.)(7.051.)(5.134.)(5.134.)(66.1()(35632)( 2345
jsjsjsjsssPssssssP
Solution:Solution:
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Example 3.8:Example 3.8: Routh-Hurwitz. Routh-Hurwitz.
Solution:Solution:
Assume Assume is small is small positivepositive: : TwoTwo sign sign changeschanges
Poles: 2 RHP, 3 LHPPoles: 2 RHP, 3 LHP
1232321)( 2345
sssss
sT
1232321)( 2345
sssss
sT
s 5 2 2 2
s 4 3 3 1
s 3 0 4 / 3 0
s 2
43 1 0
s 1
12931612 2
0 0
s 0 1 0 0
)5.041.)(5.041.)(89.33.)(89.33.)(33.1()(123232)( 2345
jsjsjsjsssPssssssP
31
568426710)( 2345
sssss
sT
s 5 1 6 8
s 4 7 1 4 2 6 5 6 8
s 3 0 4 1 0 1 2 3 0 0 0
s 2 3 8 0
s 1 1 / 3 0 0
s 0 8 0 0
Example 3.9:Example 3.9: Routh-Hurwitz. Routh-Hurwitz.
Solution:Solution:
NO sign changes: No RHP (STABLE)NO sign changes: No RHP (STABLE) Row of ZEROS indicate existence of Row of ZEROS indicate existence of
complex poles & Symmetric Equationscomplex poles & Symmetric Equations Poles: 1 LHP and 4 on jw axisPoles: 1 LHP and 4 on jw axis
0124)(86)(
3
24
ssds
sdPsssP
)414.1)(414.1)(2)(2)(7()(5684267)( 2345
jsjsjsjsssPssssssP
32
When a purely even or odd polynomial is a When a purely even or odd polynomial is a factor of the original polynomial.factor of the original polynomial.
Even polynomial only have roots that are Even polynomial only have roots that are symmetrical about the origin.symmetrical about the origin.
The symmetry can occur under 3 The symmetry can occur under 3 conditions:conditions:
1. The roots are symmetrical and real.2. The roots are symmetrical and imaginary.3. The roots are quadrantal.
Entire row is zeroEntire row is zero
2
2
1 1
3
3
j
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Example 3.10:Example 3.10: Routh-Hurwitz. Routh-Hurwitz.
Solution:Solution:
2 sign changes: 2 RHP (symmetric)2 sign changes: 2 RHP (symmetric) Poles: 2 RHP, 4 LHP and 2 on jPoles: 2 RHP, 4 LHP and 2 on j axis axis
128192128964824103128)( 2345678
ssssssss
sT
s 8 1 1 0 4 8 1 2 8 1 2 8
s 7 3 1 2 4 8 9 6 3 2 1 9 2 6 4
s 6 2 1 1 6 8 6 4 3 2 1 2 8 6 4
s 5 0 6 3 0 3 2 1 6 0 6 4 3 2 0 0 0
s 4 8 / 3 1 6 4 / 3 8 6 4 2 4
s 3 - 8 - 1 - 4 0 - 5
s 2 3 1 2 4 8
s 1 3
s 0 8
064326)(64328)(
35
246
sssds
sdPssssP
)73.11)(73.11)(73.11)(73.11)(2)(2)(1)(2()(128192128964824103)( 2345678
jsjsjsjsjsjssssPsssssssssP
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Main use is to determine the position of the Main use is to determine the position of the poles, which in turns can determine the poles, which in turns can determine the stability of the response.stability of the response.
Use of Routh Hurwitz Use of Routh Hurwitz CriteriaCriteria
KssssK
sRsC
21)()(
2
ExampleA closed-loop transfer function is given by
Determine the range for K for the system to be always stable and its oscillating frequency before it becomes unstable.
35
SolutionSolution:: Characteristic equation is:Characteristic equation is:
Expand the equationExpand the equation
Form the Routh’s arrayForm the Routh’s array
0212 Kssss
0233 234 Kssss
s4 1 3 Ks3 3 2s2 K
s1
s0 K
37
329
7914
373314 KK
36
SolutionSolution:: For no sign changeFor no sign change Referring to row 4Referring to row 4
which gives,which gives, and row 5,and row 5, Hence its range,Hence its range,
Oscillating frequency, Oscillating frequency,
07914 K
914K
0K
9140 K091437 2
srad32
37
Steady stateSteady state
)()()(1
1)( sRsHsG
sE
From the diagram
Consider
And
Use the final value theorem and define steady state error, ess that is given by
)(sG
)(sH
Y(s)R(s) E(s)
+ -
B(s)
1...11
1...11)(21
21
sTsTsTs
sTsTsTKsGbjbb
naiaa
1...11
1...11)(
21
21
sTsTsTsTsTsT
sHylyy
xkxx
)(lim)(lim0
ssEteest
ss
38
Unit stepUnit step Unit step input,
From
Steady state error,
We define step error coefficient,
Thus, the steady state error is
By knowing the type of open-loop transfer function,
we can know step error coefficient and thus the steady state error
ssR 1)(
)()()(1
1)( sRsHsG
sE
ssHsGse
sss
1)()(1lim
0
)()(lim0
sHsGKs
s
Sss K
e
1
1
)()( sHsG
)()(lim0
sHsGKs
s
1...11
1...11lim21
21
0
sTsTsTs
sTsTsTKbjbb
naiaa
s
1...11
1...11
21
21
sTsTsTsTsTsT
ylyy
xkxx
39
For open-loop transfer function of type 0:
For open-loop transfer function of type 1:
For open-loop transfer function of type 2:
,KK s Kess
1
1
,sK 01
1
sse
,sK 01
1
sse
Unit stepUnit step
40
Example:A 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
As we are looking for a steady state error for a step input, we need to know,
Knowing the open-loop transfer function, then
And steady state error of,
Its final value is,
519
500
s
limsHsGlimKsss
sK
61
511
11
S
ss Ke
65611 ssy
Unit stepUnit step
195s
Y(s) R(s) + -
41
Unit RampUnit Ramp
ttr )( , while its Laplace form is 21)(s
sR As in the above section, we know that
)()()(1
1)( sRsHsG
sE
Thus, its steady state error is
20
1)()(1
1lim
ssHsGse
sss
Define ramp error coefficient, ;K r )()(lim0
sHssGKs
r
Which the steady state error as
rss K
e 1
Just like for 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: 0rK
For open-loop transfer function of type 1: KKr
For open-loop transfer function of type 2: rK
42
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.
(iv) Cadangkan satu kaedah bagi menghapuskan ralat keadaan mantap untuk (iii).
)14.0(01.0ss
K mi
Compensator DC motor +
-
Unit RampUnit Ramp
43
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
Unit RampUnit Ramp
44
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 K
e
Unit RampUnit Ramp
45
Unit ParabolaUnit Parabola
Its time function 2)( ttr ,while its Laplace 32)(s
sR ,thus its steady state error is
30
21 s)s(H)s(G
slimesss
Define parabolic error coefficient, paK
)s(H)s(GslimKspa
2
0
Similarly we can determine its steady state error by knowing the type of the open-loop transfer function
For open-loop transfer function of type 0: For open-loop transfer function of type 1: For open-loop transfer function of type 2:
0paK
0paK
KK pa
46
In summary we can make a table of the steady state error for the above input
sK11
rK1
paK2
Unit step Unit ramp Unit parabolic
Type 0
Type 1 0
Type 2 0 0