Chapter5
69
Chapter 5 Frequency Response Method Concept Graphics mode Analysis 1.Introduction 2. Frequency Response of the typical elements of the linear systems 3. Bode diagram of the open loop system 4. Nyquist-criterion 5. System analysis based on the frequency response 6. Frequency response of the closed loop systems
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Transcript of Chapter5
Chapter 5 Frequency Response Method
Concept
Graphics mode
Analysis
1. Introduction
2. Frequency Response of the typical elements of the linear systems
3. Bode diagram of the open loop system
4. Nyquist-criterion
5. System analysis based on the frequency response
6. Frequency response of the closed loop systems
5.1 IntroductionThree advantages:
* Frequency response(mathematical modeling) can be obtained directly by experimental approaches.
* easy to analyze the effects of the system with sinusoidal voices.
* easy to analyze the stability of the systems with a delay element5.1.1 frequency response
For a RC circuit:ur
uc
R
C)sin( :If 0 tAur
We have the steady-state response:
)(1
1)(
1
1
)(
jURCj
jU
CjR
CjjU rrc
5.1 Introduction
Here:
ARC
jUjGjUU rccm
1)(
1
)()()(
2
01- )(tg
)()()(
RC
jUjGjU rcc
We call:1
1
)(
)()(
RCjjU
jUjG
r
c
Frequency Response(or frequency characteristic) of the electric circuit.
)()()( jUjGjU rc
We have: )sin()( ccmc tUtu
then:
1
1
)(
)()(
RCjjUr
jUcjG
Make:
5.1 Introduction
Generalize above discussion, we have: Definition : frequency response (or characteristic) —the ratio of the complex vector of the steady-state output versus sinusoid input for a linear system, that is:
)(
)()(
jR
jCjG
Here: input sinusoid the of tionrepresentavector complex the )( jR
output the of tionrepresentavector complex the )( jC
stic)characterir response(o frequency )( jG
And we name:
)( )()( )()( sticcharacteriponsephase resjRjCjG
)( )(
)()()( sticcharacteri responsemagnitude
jR
jCjGA
(amplitude ratio of the steady-state output versus sinusoid input)
(phase difference between steady-state output and sinusoid input )
5.1.2 approaches to get the frequency characteristics
1. Experimental discrimination
Measure the amplitude and phase of the steady-state output
Input a sinusoid signal to the control system
Get the amplitude ratio of the output versus input
Get the phase difference between the output and input
Are the measured data enough ？
Data processing
Change frequency
y
N
5.1.2 approaches to get the frequency characteristics
Theorem: If the transfer function is G(s), we have:
js
sGjG )()(
Proof :
22
21
)( sin
)())((
)(
)(
)()( :
s
AsRtAr(t)and
pspsps
sM
sR
sCsGassume
n
))(()())((
)(
)()()(
21
jsjs
A
pspsps
sM
sRsGsCthen
n
Where — pi is assumed to be distinct pole (i=1,2,3…n).
2. Deductive approach
In partial fraction form:
)()()()()(
)( 21
32
2
1
1 js
A
js
A
ps
K
ps
K
ps
KsC n
Here:ips
ii sRpspspsps
sMK
)()(
)())((
)(
321
)90)((
1
2
)(
2)(
)())((
)(
ojGj
js
ejGA
j
AjG
jsjsjs
AsGA
)90)((12 2
)( ojGjejGA
AA
5.1.2 approaches to get the frequency characteristicsTaking the inverse Laplace transform:
))(sin()(
2)(
)(
n
1i
)90)(()90)((
1
211
jGtjGAeK
eejGAeK
eAeAeKtc
tpi
jGtjjGtjn
i
tpi
tjtjn
i
tpi
i
oo
i
i
))(sin()(
))](sin()([lim)()(limn
1i
jGtjGA
jGtjGAeKtctc tpi
ts
ti
For the stable system all poles (-pi) have a negative real parts,
we have the steady-state output signal:
5.1.2 approaches to get the frequency characteristics
The amplitude ratio of the steady-state output cs(t) versus sinusoid input r(t):
istic charactermagnitude jR
jCG(jω
A
jGA
)(
)()
)(
The phase difference between the steady-state output and sinusoid input:
acteristicphase char
jRjCjGtjGt
)()( )()]([
Then we have :
jssG
jR
jCjG )(
)(
)()(
Compare with the sinusoid input tAtr sin)( , we have:
))(sin()()( jGtjGAtcs the steady-state output:
5.1 IntroductionExamples 5.1.1
a unity feedback control system, the open-loop transfer function:
15.0
1)(
ssG
1) Determine the steady-state response c(t) of the system.
2) Determine the steady-state error e(t) of the system.
)4520cos(4t )604sin(10)( : oottrIf
Solution:
The closed-loop transfer function is:
25.0
1
15.011
15.01
)(1
)(
)(
)()(
ss
ssG
sG
sR
sCs
1) Determine the steady-state response c(t) of the system.
5.1 IntroductionThe frequency characteristic :
25.0
1
25.0
1)(
jsj js
The magnitude and phase response :
otgj
jA
452
5.0)()(
22
1
2)5.0(
1)()(
4
1
422
So we have the steady-state response c(t) :
tt
ttc
o
oooo
4cos25)154sin(22
5
)4545cos(4t25 )45604sin(22
5)(
The output response:
o
oo
t
tjRjC
jRAjC
45454
45604)()()(
2522
20
22
5
22
10
)()()(
0
5.1 Introduction
25.0
15.0
25.0
11)(1
)(
)(1
)(
)()(
)(
)(
s
ss
s
sR
sC
sR
sCsR
sR
sE
oooo
oooo
tt
tt
jRj
jjE
jRj
jjE
4.634)454()454.63(
4.784)604()454.63(
)(25.0
15.0)(
204
5
104
5
)(25.0
15.0)(
4
4
The steady state error e(t) is:
)4.634cos(55)4.784sin(55.2)( oo ttte
The error transfer function is :
2) Determine the steady-state error e(t) of the system.
)(25.0
15.0)(
jR
j
jjE
The error frequency response:
5.1 Introduction5.1.3 Graphic expression of the frequency response
Graphic expression —— for intuition
1. Rectangular coordinates plot
Example 5.1.2 )2()2(1
10
12
10)(
12
10)( 1
2
tg
jjG
ssG
o
o
o
o
o
o
o
. .
jGjG
29849950 5
875.8224.14
538.8064.13
964.754.22
435.6347.41
4507.75.0
0100
)()(
-90o
0. 5 1 2 3 4 50
)( jG
1
5
10
)( jG
5.1.3 Graphic expression of the frequency response
2. Polar plot
Example 5.1.3 )1(
)()( )1(
)(
Tjj
KsGjG
Tss
KsG
js
The magnitude and phase response:
T)](90[)()( ; )(1
)()( 1
2
tgjG
T
KjGA o
oooo
KTKTATT
18013511790)(
025
4)(
12
10
Calculate A(ω) and for different ω: )(Re
Im
T1
T21
0
-135o
2KT
The polar plot is easily useful for investigating system stability.
-117o
54KT
5.1.3 Graphic expression of the frequency response
The shortage of the polar plot and the rectangular coordinates plot: to synchronously investigate the cases of the lower and higher frequency band is difficult.
3. Bode diagram(logarithmic plots)
Plot the frequency characteristic in a semilog coordinate:
Magnitude response — Y-coordinate in decibels: )(log20 jG
X-coordinate in logarithm of ω: logω
Phase response — Y-coordinate in radian: )( jGX-coordinate in logarithm of ω: logω
First we discuss the Bode diagram in detail with the frequency response of the typical elements.
How to enlarge the lower frequency band and shrink (shorten) the higher frequency band ？
Idea:
5.2 Frequency Response of The Typical Elements
The typical elements of the linear control systems — refer to Chapter 2.
Transfer function: KsR
sCsG
)(
)()(
Frequency response:
ojG
KjGLKjGKjG
0)()(
log20)(log20)()()(
1. Proportional element
Re
Im
K
0dB, 0o
1001010.1)(log
)( ),( L
dB log20)( KL
o0)(
Polar plot Bode diagram
5.2 Frequency response of the typical elements
2. Integrating element
Transfer function:ssR
sCsG
1
)(
)()(
Frequency response:
ojG
jGLjG
jjG
90)()(
log20)(log20)(1
)(1)(
Polar plot
Re
Im
0
Bode diagram
0dB, 0o
1001010.1)(log
)( ),( L
decdBL /20 :)(
o90)(
5.2 Frequency response of the typical elements3. Inertial element
Transfer function:1
1
)(
)()(
TssR
sCsG
1
1)(
TjjG
TT
dBTLT
jG
1 )log(20
T1 3
T1 0
)(1log20)()(1
1)( 2
2
)()( 1 Ttg
Polar plot
Re
Im 0
Bode diagram
0dB, 0o
1001010.1)(log
)( ),( L
decdB /20
T
1
o90
o45
21T:
1)(
2
sT
KsG
Klog20
1/T: break frequency
1
5.2 Frequency response of the typical elements
4. Oscillating element
Transfer function: 10 12
1
)(
)()(
22
TssTsR
sCsG
TjTjG
2)1(
1)(
22
2222 )2()(1
1)(
TT
jG
)
1
2()(
221
T
Ttg
n
n
n
T
T
TTL
)log(40
)2log(20
)1( 0
)2()1(log20)( 2222
maximum value of ：)( jG
Make:
2r
2
12
1)(M
)2
2(0 21 0))((
r
nr
jG
jGd
d
eakresonant pM
requencyresonant f
r
r
5.2 Frequency response of the typical elementsThe polar plot and the Bode diagram:
Polar plot
Re
Im
0
)(2
1 nj
Bode diagram
0dB, 0o
1001010.1)(log
)( ),( L
decdB /40
o180
o90
rMlog20
)21log(20 Tn /1
r
1 2
Systemorder -Second Optimal , 02
2
0 .2
)( .1
ceNo resonan
ystemunstable s
M
r
nr
rnr
1
5.2 Frequency response of the typical elements5. Differentiating element
Transfer function:
aldifferentiordersecondTsTs
aldifferentiorderfirstTs
aldifferentis
sG
12
1)(2
Polar plot
Re
Im
Re
Im
1
Re
Im
1
differential 1th-order differential 2th-order differential
5.2 Frequency response of the typical elements
Because of the transfer functions of the differentiating elements are the reciprocal of the transfer functions of Integrating element, Inertial element and Oscillating element respectively,
that is:
12112
111
1
2222
TssTTssT
TsTs
ss
inverse
inverse
inverse
the Bode curves of the differentiating elements are symmetrical to the logω-axis with the Bode curves of the Integrating element, Inertial element and Oscillating element respectively.
Then we have the Bode diagram of the differentiating elements:
5.2 Frequency response of the typical elements
0dB, 0o
1001010.1)(log
)( ),( L
decdBL /20 :)( o90)(
differential
1th-order differential
0dB, 0o
1001010.1)(log
)( ),( L
decdB /20o45
o90
0dB, 0o
1001010.1)(log
)( ),( L
decdB /40
o180
o90
rMlog20
)21log(20
Tn 1
2th-order differential
5.2 Frequency response of the typical elements
6. Delay element
Transfer function: sesR
sCsG
)(
)()(
)()(
0)(1)()(
jG
LjGejG j
Polar plot
Re
ImR=1
0dB, 0o
1001010.1)(log
)( ),( L
Bode diagram
5.3 Bode diagram of the open loop systems5.3.1 Plotting methods of the Bode diagram of the open loop
systems
Assume:
mentsypical elen of the ter functiothe transfsGhere
sGsGsGsG
i )( :
)...()()()( 321
We have:
...)()()()()( 321 jGjGjGjG
...log20log20log2log20)( 321 GGGoGL
That is, Bode diagram of a open loop system is the superposition of the Bode diagrams of the typical elements.
Example 5.3.1)101.0(
)1(10)()(
2
ss
ssHsG
5.3 Bode diagram of the open loop systems
G(s)H(s) could be regarded as:
Then we have:
101.0
1
s
11)(s10
)101.0(
)1(10)()(
22
sss
ssHsG
① ② ③ ④
0dB, 0o
1001010.1)(log
)( ),( L
③
④
②①
20dB, 45o
-20dB, -45o
-40dB, -90o
40dB, 90o
-80dB,-180o
-60dB.-135o
-40dB/dec
－ 20dB/dec
20dB/dec
－ 40dB/dec
－ 20dB/dec
－ 40dB/dec
5.3.2 Facility method to plot the magnitude response of the Bode diagram
Summarizing example 5.3.1, we have the facility method to plot the magnitude response of the Bode diagram:
1) Mark all break frequencies in theω-axis of the Bode diagram.
2) Determine the slope of the L(ω) of the lowest frequency band (before the first break frequency) according to the number of the integrating elements: － 20dB/dec for 1 integrating element － 40dB/dec for 2 integrating elements …
3) Continue the L(ω) of the lowest frequency band until to the first break frequency, afterwards change the the slope of the L(ω)
which should be increased 20dB/dec for the break frequency of the 1th-order differentiating element . The slope of the L(ω) should be decreased 20dB/dec for the break frequency of the Inertial element …
5.3.2 Facility method to plot the magnitude response of the Bode diagram
Plot the L(ω) of the rest break frequencies by analogy .
Example 5.3.2)101.001.0)(11.0(
)1(10)(
22
ssss
ssG
)(L
1)( 201020 logωlog
)100()01.0(40
)1.0(2020201020
log
loglogloglog
)10010()1.0(2020
201020
loglog
loglog
)101( 20
201020
log
loglog
21
11
)01.0(1
01.0
)1.0(90)(
tg
tgtgo
1046.179
1009.174
105.56
13.51
)(
o
o
o
o
The Bode diagram is shown in following figure:
5.3.2 Facility method to plot the magnitude response of the Bode diagram
0dB, 0o
1001010.1)(log
)( ),( L
20dB, 45o
-20dB, -45o
-40dB, -90o
40dB, 90o
-80dB,-180o
-60dB.-135o
-100dB,-225o
-120dB,-270o
－ 60dB/dec
－ 20dB/dec
There is a resonant peak Mr at:
7.705.021100
21
2
2
nr
dBMr 25.1154.112
12
1.25dB
r
)101.001.0)(11.0(
)1(10
)(
22
ssss
s
sG－ 20dB/dec
5.3.3 Determine the transfer function in terms of the Bode diagram1. The minimum phase system(or transfer function)
We have:
1)(
1)(
)()()()(
2
2
4321
TK
GGGG
The magnitude responses are the same. But the net phase shifts are different when ω vary from zero to infinite. It can be illustrated as following:
)()(180)( ),()()(
)(180)()( ),()()(
114
113
112
111
tgTtgjGtgTtgjG
tgTtgjGtgTtgjG
o
o
)1(
)1(
)1(
)1()(
)1(
)1()(
)1(
)1()(
43
21
Ts
sK(s)G
Ts
sKsG
Ts
sKsG
Ts
sKsG
T
Sketch the polar plot:
Compare following transfer functions:
K ,0
TK
,
K ,0T
K ,
TK
, K ,0K
0T
K ,
5.3.3 Determine the transfer function in terms of the Bode diagramThe polar plot:
Re
Im
1
)1()(1
Ts
sKsG
Re
Im
Re
Im
Re
Im
1
)1()(3
Ts
sKsG
1
)1()(4
Ts
sKsG
1
)1()(2
Ts
sKsG
It is obvious: the net phase shifts of the
G1(s) is named: the minimum phase transfer function .
G1(jω) is the minimum when ω vary from zero to infinite.
phase shift 00
phase shift － π
phase shift － π
phase shift π
5.3.3 Determine the transfer function of the minimum phase systems in terms of the magnitude response
Definition:
2. Determine the transfer function from the magnitude response of the Bode diagram .
Example 5.3.3
A transfer function is called a minimum phase transfer func- tion if its zeros and poles all lie in the left-hand s-plane.
A transfer function is called a non-minimum phase transfer function if it has any zero or pole lie in the right-hand s-plane.
Only for the minimum phase systems we can affirmatively deter- mine the relevant transfer function from the magnitude response of the Bode diagram .
2 20 200
5.3.3 Determine the transfer function in terms of the Bode diagram
0dB, 0o
1001010.1)(log
)(L－40dB/dec
－40dB/dec
－20dB/dec
).(
).()(
:
10050
1502
ss
sKsG
diagramthe Bode
from the G(s) we can get
400502020220
2
KKoL
and
).log(loglog)(
:
Example 5.3.4
0dB 1001010.1)(log
)(L
0.5 20020dB/dec － 20dB/dec
20dB
))(()(
:
11 21
sTsT
KssG
diagramBode
from the the G(s) we can get
5.3.3 Determine the transfer function in terms of the Bode diagram
0dB 1001010.1)(log
)(L
0.5 20020dB/dec － 20dB/dec
20dB
))(()(
:
11 21
sTsT
KssG
diagramBode
from the the G(s) we can get
050020202020220
202020220
202020
22002
111
50
. )log().log(loglog)(
. loglog)(
loglog)(
:
/
.
TTL
TdBωL
KKL
and
T
Example 5.3.5
0dB1001010.1
)(log
)(L－ 20dB/dec
－ 20dB/dec－ 60dB/dec
8.136 dB20 dB
)(
).()(
:
12
10102
2
TssTs
sKsG
diagramthe Bode
from the G(s) we can get
5.3.3 Determine the transfer function in terms of the Bode diagram
20136812
120
100202020
10
2
10
. .log
loglog)(
0.1T T
1
ζ
KdBKL
0dB 1001010.1
)(log
)(L－ 20dB/dec
－ 20dB/dec－ 60dB/dec
8.136 dB20 dB
1)0.04ss(0.01s
1)100(0.01sG(s)
: 2
then
For the non-minimum phase system we must combine the magnitude response and phase response together to determine the transfer function.
)(
).()(
:
12
10102
2
TssTs
sKsG
diagramthe Bode
from the G(s) we can get
5.3.3 Determine the transfer function in terms of the Bode diagram
0dB, 0o
1001010.1)(log
)( ),( L
－ 180o
－ 90o
－ 20dB/dec
)(
).()(
)(
),(
)(
).(
)(
).(
1
11010
1
11010
1
11010
1
11010
43
21
s
ssG
s
sG
s
sG
s
sG
Example 5.3.6
All satisfy the magnitude response
1)-(s
1)10(0.1s )( :have we So,
)1(
)11.0(10)( 4
sG
y.ultaneouslsponse sime phase resatisfy th
s
ssGonlyBut
5.4 The Nyquist-criterion
A method to investigate the stability of a system in terms of the open-loop frequency response.
Assume:
. ; :
))...()((
))...()(()()(
21
211
polesopen-loop pzerosopen-loop zhere
mnpspsps
zszszsKsHsG
ji
n
m
Make :
)( ))...()((
))...()((
))...()((
))...()(())...()((
))...()((
))...()((1 )()(1)(
21
21
21
21121
21
211
sof the Fzerosspspsps
ssssssK
pspsps
zszszsKpspsps
pspsps
zszszsKsHsGsF
in
nF
n
mn
n
m
5.4.1 The argument principle(Cauchy’s theorem)
Note: si→ the zeros of the F(s), also the roots of the 1+G(s)H(s)=0
5.4.1 The argument principle
Now we consider the net phase shift if s travels 360o along a closed path Γ of the s-plane in the clockwise direction shown in Fig.5.4.1 .
)(sF
Fig. 5.4.1
n
jj
n
ii pssssFBecause
11
)()()( :
0)(
: ,
2)(
: ,
i
i
i
i
ss
thenlosed by Γis not encsthe zerosIf
ss
thenΓd byis enclosesthe zerosIf
Similarly we have:
0)( :
2)( :
jj
jj
pslosed by Γis not encp
psΓd byis enclosep
Re
Im Γ
S-plane
is
iss
iss s
5.4.1 The argument principle
If Z zeros and P poles are enclosed by Γ , then:
2)()2()2(
)()()(
11
ZPPZ
pssssFn
jj
n
ii
It is obvious that path Γ can not pass through any zeros si or poles pj . Then we have the argument principle:
If a closed path Γ in the s-plane encircles Z zeros and P poles of F(s) and does not pass through any poles or zeros of F(s) , when s travels along the contour Γ in the clockwise direction, the corres- ponding F(s) locus mapped in the F(s)-plane will encircle the origin of the F(s) plane N = P-Z times in the counterclockwise direction, that is:
N = P － Z
5.4.1 The argument principle
here: N —— number of the F(s) locus encircling the origin of the F(s)-plane in the counterclockwise direction.
5.4.2 Nyquist criterion
If we choose the closed path Γ so that the Γ encircles the entire right hand of the s-plane but not pass through any zeros or poles of F(s) shown in Fig.5.4.2 .
The path Γ is called the Nyquist-path.
Re
Im S-plane
Fig. 5.4.2
P —— number of the zeros of the F(s) encircled by the path
Γ in the s-plane.
Z —— number of the poles of the F(s) encircled by the path
Γ in the s-plane.
5.4.2 Nyquist criterion
When s travels along the the Nyquist-path:
)(1)()( )()(1)(
0)()(1)()(1)(
sFjHjGjHjGsF
sHsGsHsGsF
js
js
Re
Im S-plane
Fig. 5.4.2
When ω vary from - (or 0) →+ , G(jω)H(jω) Locus mapped in the G(jω)H(jω)-plane will encircle the point (-1, j0) in the counterclockwise direction:
]0fromfor2[ Z)/(Por NZPN here: P — the number of the poles of G(s)H(s) in the right hand of the s-plane. Z — the number of the zeros of F(s) in the right hand of the s-plane.
equivalent to the point (-1, j0) of the G(jω)H(jω)-plane, we have another statement of the argument principle:
Because the origin of the F(s)-plane is
5.4.2 Nyquist-criterion
If the systems are stable, should be Z = 0, then we have: The sufficient and necessary condition of the stability of the linear systems is : When ω vary from - (or 0) →+ , the G(jω)H(jω) Locus mapped in the G(jω)H(jω)-plane will encircle the point (-1, j0) as P (or P/2) times in the counterclockwise direction.
Here: P — the number of the poles of G(s)H(s) in the right hand of the s-plane.
——Nyquist criterion
Discussion :i) If the open loop systems are stable, that is P = 0, then:
for the stable open-loop systems, The sufficient and necessary condition of the stability of the closed-loop systems is : When ω vary from - (or 0) →+ , the G(jω)H(jω) locus mapped in the G(jω)H(jω)-plane will not encircle the point (-1, j0).
5.4.2 Nyquist-criterionii) Because that the G(jω)H(jω) locus encircles the point (-1, j0)
means that the G(jω)H(jω) locus traverse the left real axis of the point (-1, j0) , we make:
The sufficient and necessary condition of the stability of the linear systems is : When ω vary from - (or 0) →+ , the number of the net “positive traversing” is P (or P/2).
G(jω)H(jω) Locus traverses the left real axis of the point (-1, j0) in the counterclockwise direction —“positive traversing”. G(jω)H(jω) Locus traverses the left real axis of the point (-1, j0) in the clockwise direction —“negative traversing”.
Then we have another statement of the Nyquist criterion:
Here: the net “positive traversing” —— the difference between the number of the “positive traversing” and the number of the “negative traversing” .
Fig.5.4.3
5.4.2 Nyquist-criterion
Example 5.4.1
The polar plots of the open loop systems are shown in Fig.5.4.3, determine whether the systems are stable.
Re
Im
(-1, j0)
0
(1) P=2
Re
Im
(-1, j0)
0
(2) P=0
Re
Im
(-1, j0)
0
(3) P=2
Re
Im
(-1, j0)
0
(4) P=0
stable stable
unstable unstable
5.4.2 Nyquist-criterion
Note: the system with the poles (or zeros) at the imaginary axis
Example 5.4.2).)((
)()(1501
10
ssssHsG
There is a pole s = 0 at the origin in this system, but the Nyquist path can not pass through any poles of G(s)H(s).
Fig. 5.4.4
We consider a semicircular detour around the pole (s = 0) repre-sented by setting 0)( jes
at the s = 0 point we have:
ojoj
oj
ojoj
oj
ojoj
oj
ee
jHjGes
ee
jHjGes
ee
jHjGes
90
90
90
0
0
0
90
90
90
11000
11000
11000
ε)()(ε
ε)()(ε
ε)()(ε
Idea:
Re
Im
00
0
r
Radius
0
Radius
js
jsjs
－1
－ 2
5.4.2 Nyquist-criterion
It is obvious that there is a phase saltation of the G(jω)H(jω) at ω=0, and the magnitude of the G(jω)H(jω) is infinite at ω=0.
In terms of above discussion , we can plot the system’s polar plot shown as Fig.5.4.5.
The closed loop system is unstable.
Fig.5.4.5
Im
Re
(-1, j0)
0
0
0
).)((
)()(
1501
10
sss
sHsG
Fig. 5.4.4
Re
Im
00
0
r
Radius
0
Radius
js
jsjs
Example 5.4.3
))()(())(()()(
221
10
41
102 jsjssssss
sHsG
Similar to the Example 5.4.2, the system’s polar plot is shown as Fig.5.4.6 .
Fig.5.4.6
Im
Re
(-1, j0)
0 2j0
2j
The closed loop system is unstable.
Determine the stability of the system applying Nyquist criterion.
Solution
5.4.3 Application of the Nyquist criterion in the Bode diagram
5.4.3 Application of the Nyquist criterion in the Bode diagram
G(jω)H(jω) locus traverses the left real axis of the point (-1, j0) in G(jω)H(jω)-plane → L(ω)≥0dB and φ(ω) = － 180o in Bode diagram (as that mentioned in 5.4.2).
We have the Nyquist criterion in the Bode diagram : The sufficient and necessary condition of the stability of the linear closed loop systems is : When ω vary from 0→+ , the number of the net “positive traversing” is P/2. Here: the net “positive traversing” —the difference between the number of the “positive traversing” and the number of the “negative traversing” in all L(ω)≥0dB ranges of the open-loop system’s Bode diagram.
“positive traversing” — φ(ω) traverses the “-180o line” from below to above in the open-loop system’s Bode diagram; “negative traversing” — φ(ω) traverses the “-180o line” from above to below.
5.4.3 Application of the Nyquist criterion in the Bode diagram
Example 5.4.4 The Bode diagram of a open-loop stable system is shown in Fig.5.4.7, determine whether the closed loop system is stable.
In terms of the Nyquist criterion in the Bode diagram:
Because the open-loop system is stable, P = 0 .
The number of the net “positive traversing” is 0 ( = P/2 = 0 ).
The closed loop system is stable .
0dB, － 180o )(log
)(L－ 20
－ 40
－ 60
－ 40－ 20
－ 40
－ 60－ 270o
－ 90o
Fig.5.4.7L(ω)
φ(ω)
Solution
5.4.4 Nyquist criterion and the relative stability (Relative stability of the control systems)
In frequency domain, the relative stability could be described by the “gain margin” and the “phase margin”.
1. Gain margin Kg
gg
g
g jHjGdBKjHjG
K
)()(log)( )()(
201
uencysover freqPhase-crosjHjGg
)()( :g 0180
2. Phase margin γc
00 180180 )()()()()( cccc jHjGjHjG
encyover frequGain-crossjHjGωc
c
1
)()( :
3. Geometrical and physical meanings of the Kg and γc
5.4.4 Nyquist criterion and the relative stability
The geometrical meanings is shown in Fig. 5.4.8.
Re－ 1
Im
γc
1/Kg
stable
Critical stability
unstable
Fig. 5.4.8
The physical signification :
Kg— amount of the open-loop gain in decibels that can be allowed to increase before the closed-loop system reaches to be unstable. For the minimum phase system: Kg>1the closed loop system is stable . γc —amount of the phase shift of G(jω)H(jω) to be allowed beforethe closed-loop system reaches to be unstable. For the minimum phase system: γc>0 the closed loop system is stable .
5.4.4 Nyquist criterion and the relative stability
The changes of the open-loop gain only alter the magnitude of G(jω)H(jω).
The changes of the time constants of G(s)H(s) only alter the phase angle of G(jω)H(jω).
Attention :
For the linear systems:
Example 5.4.5
sess
KsHsG
).()()(
110(1) Determine Kg and γc when K =1 and τ =1.
(2) Determine the maximum K and τ based on K = 1 and τ = 1.
The open loop transfer function of a control system is:
5.4.4 Nyquist criterion and the relative stability
Solution
0180 gjHjGof: In terms )()( 1
cjHjG
)()(
1.43; .
gggtg
1
1 102
11011
2
c
).( K
cc K
(dB) ..).(
)()( .173441
1011
1
431
2
Kggg
g KjHjGK
0
1
1100 271090180
cccccc tgjHjG .)()(
(2) 4414411
.KmaximumKBecauseKg
.
4702701
. 27 0 maximumBecause cc
sess
KsHsG
).()()(
110(1) Determine Kg and γc ( K =1 and τ =1)
5.4.4 Nyquist criterion and the relative stability
Re
Im
(-1, j0)
0.8
1.52
G(jω)H(jω)
Fig.5.4.9
Example 5.4.6
(1) Determine Kg
The G(jω)H(jω) polar plot of a system is shown in Fig.5.4.9.
(2) Determine the stable range of the open loop gain.
Solution
)(...
dBKg 94125180
1
)()()( : jωGHKjωHjωGAssume 0
(1) Determine Kg
(2) Determine the stable range of the open loop gain.
. c Kop gain ishe open lovalue of tal stable the criticand
000
033
0
22
0
011
0
2
1
4
5
3
2
2
11
2
3
21
51
4
51
80
K K orKKK
K
KKjGHK
andjGHK
KjGHK
andjGHK
KKjGHK
andjGHK
fIn terms o
cg
c
g
cg
c
g
cg
c
g
: gain loop open the of range stable the have weThen
)(
)(
)(
.)(
)(
.)(
:
Re
Im
(-1, j0)
0.8
1.52
G(jω)H(jω)
Fig.5.4.9
5.5 System analysis based on the frequency response5.5.1 Performance specifications in the frequency domain
ω
A(ω)
0
A(0)0.707A(0)
Fig. 5.5.1
ωr ωb
Mr
(1) Resonance frequency ωr:
0
1
rr A
d
dsatisfy
jHjG
jGjAAssume
)( :
)()(
)()()( :
(2) Resonance peak Mr :
rAMr )(
(3) Bandwidth ωb:
)0(2
2)( : AAsatisfy
bb
The general frequency response of a closed loop systems is shown in Fig. 5.5.1
1. For the closed loop systems
2. For the open loop systems
(1) Gain crossover frequency ωc:
1 cωωc jωHjωGsatisfy )()( :
For the unity feedback systems, ωc≈ ωb , because:
1)G(j )(
1)G(j
)(
)()(
jGjG
jGj
1
1
(2) Gain margin Kg:
gg
g
g jHjGdBKjHjG
K
)()(log)( ;)()(
201
0180 gjHjGsatisfiesHere )()( : g
(3) Phase margin γc: )()()( 0180 cc jHjG
1 c
jHjGsatisfiesHere
)()( : c
5.5 System analysis based on the frequency response
5.5.2 Relationship of the performance specifications between the frequency and time domain
The relationship between the frequency response and the time response of a system can be expressed by following formula:
)()()(
)()()()( :
)()()(
jRjHjG
jGjRjjChere
dejCjCLtC tj
1
2
11
But it is difficult to apply the formula .
Generally Kg and γc could be concerned with the resonance peak Mr : Kg and γc ↑ —— Mr ↓.
ωc could be concerned with the resonance frequency ωr and bandwidth ωb : ωc↑ —— ωr and ωb↓.
5.5.2 Relationship of the performance specifications between the frequency and time domain
(1) Bandwidth ωb(or Resonance frequency ωr) Rise time tr
Generally ωb(or ωr )↑—— tr ↓ because of the “time scale” theorem:
dejCjCLtc tj)()()( of terms In2
11
)/()(2
1)( :Then
)()( , :If
/
tcdejCtc
jCjC
tj 11
alike : ωc↑—— tr ↓ because of ωc≈ ωb .
For the large ωb , there are more high-frequency portions in c(t), which make the time response to be faster.
βtβω :is That So ωb(or ωr )↑—— tr ↓
(2) Resonance peak Mr overshoot σp%
Normally Mr ↑ —σp% ↑ because of the large unbalance of the frequency signals passing to c(t) .
Kg and γc ↓ —σp% ↑is alike because of Kg and γc ↓—Mr ↑.
5.5.2 Relationship of the performance specifications between the frequency and time domain
Some experiential formulas:
cr
rrp
M
MM
sin and
)..()(..% Overshoot
1
8111140160
: of valueoptimum an problem, design most For rM 5111 . rM.
00 903511
5211
512
c
cccs k
kt
sin.
sin. , time Settling
5.5.2 Relationship of the performance specifications between the frequency and time domain
(3) A(0) → Steady state error ess
00 1
0
)()(
)()(
jHjG
jGA)A(
)()( )()( :assume sHKsHsGs
KsG HvG 001
01
11
0v
KK
vK
A
HG
H)( : then
01
110 v
K
Kv
A
G
G)(
: 1H(s) system, feedback unity the For
So for the unity feedback systems: sseAA )( ,)( 010
5.5.2 Relationship of the performance specifications between the frequency and time domain
(4) Reproductive bandwidth ωM → accuracy of Reproducing r(t)
ω
A(ω)
0
A(0)0.707A(0)
Fig. 5.5.2
ωr ωb
Mr
△
ωM
Reproductive bandwidth ωM :
error greproducin allowed :
)()()(
)()(
01
AjHjG
jGA
M
)( :and
)()()()()( :assume
j
jRjjCjRjE
e
e
for a given ωM , ↓—higher accuracy of reproducing △ r(t) . for a given , △ ωM ↑ —higher accuracy of reproducing r(t) .
Demonstration
5.5.2 Relationship of the performance specifications between the frequency and time domain
ω0
Fig. 5.5.3
ωM
)( jR
M
M
tj
M
M
tje
trdejR
dejRjte
)()(
)()()(
2
2
1
)( )()( :is That tetrte
For the frequency spectrum of r(t) shown in Fig.5.5.3 .
5.5.3 Relationship of the performance specifications between the frequency and the time domain: for the typical 2th-order system
For the typical 2th-order system:
22
22
2)(
)2()(
nn
n
n
n
sss
sssG
5.5.3 Relationship of the performance specifications between the frequency and the time domain: for the typical 2th-order system
We have:
422 44221 )(nb
)2
2(0 221nr
212
1
rM
... , , % , rspn tt
g
c
nc
K
tg24
1
24
241
2
241
... , , % , rspn tt
5.5 System analysis based on the frequency response
5.5.4 “three frequency band” theorem The performance analysis of the closed loop systems according to the open loop frequency response. 1. For the low frequency band
The more negative the slope of L(ω) is , the higher the control accuracy of the systems. The bigger the magnitude of L(ω) is, the smaller the steady state error ess is.
2. For the middle frequency band
the low frequency band is mainly concerned with the control accuracy of the systems.
The middle frequency band is mainly concerned with the transient performance of the systems.
ωc↑—tr ↓; Kg and γc ↓—σp% ↑
5.5.4 “three frequency band” theorem
The slope of L(ω) in the middle frequency band should be the –20dB/dec and with a certain width .
3. For the high frequency band
The high frequency band is mainly concerned with the ability of the systems restraining the high frequency noise. The smaller the magnitude of L(ω) is, the stronger the ability of the systems restraining the high frequency noise is.
Example 5.5.1
ω0dB
Fig. 5.5.4
)()(log20)( jHjGL
Ⅰ Ⅱ
－ 40
－ 20
－ 40 Compare the performances between the system and Ⅰsystem Ⅱ
5.5 System analysis based on the frequency responseSolution :
essⅠ> essⅡ
trⅠ > trⅡ
σpⅠ% =σpⅡ%
The ability of the system Ⅰrestraining the high frequency noise is stronger than system Ⅱ Ⅱ
ω0dB
Fig. 5.5.4
)()(log20)( jHjGL
Ⅰ
－ 40
－ 20
－ 40
Example 5.5.2 For the minimum phase system, the open loop magnitude response shown as the Fig. 5.5.5. Determine the system’s parameter to make the system being the optimal second-order system and the steady-state error ess< 0.1.
L(ω)
ω
0.1
1
－ 20dB/dec
－ 40dB/dec
Fig. 5.5.5
Solution :
5.6 Frequency response of the closed loop systems
5.6.2 The constant N circles: How to obtain the phase frequency characteristic of the closed loop systems in terms of the open loop frequency response…… (refer to P496)
5.6.3 The Nichols chart: How to obtain the closed loop frequency response in terms of the open loop frequency response…… (refer to P496)
5.6 Frequency response of the closed loop systems How to obtain the closed loop frequency response in terms of the open loop frequency response.
5.6.1 The constant M circles: How to obtain the magnitude frequency response of the closed loop systems in terms of the open loop frequency response…… (refer to P495)
Chapter 5 Frequency Response Methods
