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Chapter 5 Frequency Response Method
elements of the linear systems Concept 1. Introduction 2. Frequency Response of the typical
Graphics mode 3. Bode diagram of the open loop system 4. Nyquist-criterion 5. System analysis based on the frequency
Analysis response 6. Frequency response of the closed loop systems
5.1 Introduction
Three 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 element
R
5.1.1 frequency response For a RC circuit: If :
u r
A
sin(
t
0 )
u r
C
u c
We have the steady-state response:
U c
(
j
)
1
R
j
C
1
j
C U r
(
j
)
1
j
RC
1
U r
(
j
)
Make: then:
5.1 Introduction
G
(
j
)
Uc
(
Ur
(
j
)
j
)
1
j
RC
1
U c
(
j
)
G
(
j
)
U r
(
j
) We have:
u c
(
t
)
U cm
sin(
t
c
) Here:
U cm
U c
(
j
)
G
(
j
)
U r
(
j
) We call:
1 (
RC
) 2
1
A G
(
j
)
U c
(
U r
(
j
)
j
)
c
U c
(
j
)
G
(
j
)
U r
(
j
)
tg 1 (
RC
)
0 1
j
RC
1
Frequency Response
(or frequency characteristic) of the electric circuit.
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: Here
:
R
(
C
(
j
j
) )
G
(
j
)
C R
( (
j j
) )
the complex vector representa
the complex vector representa
G
(
j
)
frequency response(o of the sinusoid input of the output characteri stic) And we name:
A
(
)
G
(
j
)
C
(
R
(
j
j
) )
magnitude response
(
characteri stic
)
(
amplitude ratio of the steady-state output versus sinusoid input
)
(
)
G
(
j
)
C
(
j
)
R
(
j
)
phase res ponse
(
characteri stic
)
(
phase difference between steady-state output and sinusoid input
)
5.1.2 approaches to get the frequency characteristics
1. Experimental discrimination Input a sinusoid signal to the control system Measure the amplitude and phase of the steady-state output Change frequency Get the amplitude ratio of the output versus input Get the phase difference between the output and input N Are the measured data enough
？ y
Data processing
5.1.2 approaches to get the frequency characteristics
2. Deductive approach Theorem: If the transfer function is G(s), we have:
G
(
j
)
G
(
s
)
s
j
Proof :
assume
:
G
(
s
)
and r(t)
C
(
s
)
R
(
s
)
(
s
A
sin
t
p
1 )(
s
R
(
s
)
M
(
s
)
p s
2 2 )
A
2 (
s
p n
) Where —
p i
is assumed to be distinct pole (i=1,2,3…n).
then C
(
s
)
G
(
s
)
R
(
s
)
(
s
p
1 )(
s
M
(
s
)
p
2 )
(
s
p n
)
(
s
j
A
)(
s
j
)
In partial fraction form:
C
(
s
)
(
s K
1
p
1 )
(
s K
2
p
2 )
(
s K n
p
3 )
(
s A
1
j
)
(
s A
2
j
)
Here:
K i
(
s
p
1 )(
s
M
(
s
)
p
2 )
(
s
p
3 )
(
s
p i
)
R
(
s
)
s
p i A
1
G
(
s
)
(
s
G
(
j
)
A
2
j
j
A
)(
s
j
)
(
s
j
)
s
j
A G
(
j
)
e j
(
G
(
j
)
90
o
) 2
A
2
A
1
A G
(
j
)
e
j
(
G
(
j
)
90
o
) 2
5.1.2 approaches to get the frequency characteristics
Taking the inverse Laplace transform:
c
(
t
)
i n
1
K i e p i t
A
1
e j
t
A
2
e
j
t
i n
1
K i e p i t
A G
(
j
)
e j
(
t
G
(
j
)
90
o
)
e
j
(
t
G
(
j
)
90
o
) 2
i n
1
K i e p i t
A G
(
j
) sin(
t
G
(
j
)) For the stable system all poles (-p
i
) have a negative real parts, we have the steady-state output signal: n
t
lim
c
(
t
)
c s
(
t
)
t
lim
[ i
1
K i e p i t
A G
(
A G
(
j
) sin(
t
G
(
j
)
j
)) sin(
t
G
(
j
))]
5.1.2 approaches to get the frequency characteristics
the steady-state output:
c s
(
t
)
A G
(
j
) sin(
t
G
(
j
)) Compare with the sinusoid input
r
(
t
)
A
sin
t
, we have: The
amplitude ratio
of the steady-state output c s (t) versus sinusoid input r(t):
A G
(
j
)
A
G(jω
)
C
(
R
(
j
j
) )
magnitude character istic
The
phase difference
between the steady-state output and sinusoid input: [
t
G
(
j
)]
t
G
(
j
)
C
(
j
)
R
(
j
) Then we have :
G
(
j
)
phase char acteristic C
(
R
(
j
)
j
)
G
(
s
)
s
j
5.1 Introduction
Examples 5.1.1
a unity feedback control system, the open-loop transfer function:
G
(
s
)
1 0 .
5
s
1
If
:
r
(
t
)
10 sin( 4
t
60
o
)
20cos(4t
45
o
) 1) Determine the steady-state response
c(t)
of the system.
2) Determine the steady-state error
e(t)
of the system.
Solution: 1) Determine the steady-state response
c(t)
of the system.
The closed-loop transfer function is:
(
s
)
C
(
s
)
R
(
s
)
G
(
s
) 1
G
(
s
)
1 0 .
5
s
1
1 0 .
5
s
1
1
1 0 .
5
s
2
5.1 Introduction
The frequency characteristic :
(
j
)
1 0 .
5
s
2
s
j
0 .
5 1
j
2 The magnitude and phase response : The output response: So we have the steady-state response
c(t)
:
c
(
t
)
5 2
5 2 2 sin( 4
t
60
o
45
o
)
5 2 sin( 4
t
15
o
)
5 2 cos 4
t
2 cos(4t
45
o
45
o
)
5.1 Introduction
2) Determine the steady-state error
e(t)
of the system.
The error transfer function is :
E
(
s
)
R
(
s
)
R
(
s
)
C
(
s
)
R
(
s
) 1
(
s
)
1
1
C
(
s
)
R
(
s
) 1 0 .
5
s
2
E
(
j
)
E
(
j
)
0 .
5
s
0 .
5
s
1
2
E
(
j
)
The error frequency response:
0 .
5 0 .
5
j
j
j
0 .
5
j
0 .
5
1
2
1
2
R
(
j
)
j
0 .
5
j
0 .
5
1
2
R
(
j
)
4
4 4
R
(
j
)
4 5 5
10
20 The steady state error
e(t)
is:
( 63 .
4
o
( 63 .
4
o
45
o
)
( 4
t
45
o
)
( 4
t
60
o
)
45
o
)
4
t
4
t
78 .
4
o
63 .
4
o e
(
t
)
2 .
5 5 sin( 4
t
78 .
4
o
)
5 5 cos( 4
t
63 .
4
o
)
5.1 Introduction
5.1.3 Graphic expression of the frequency response
Graphic expression —— for intuition 1. Rectangular coordinates plot Example 5.1.2
0 0 .
5 1 2 3 4
G
(
j
) 10 7 .
07 4 .
47 2 .
4 1 .
64 1 .
24
G
(
s
)
2
s
10
1
G
(
j
)
10
j
2
1
G
(
j
) 0
o
45
o
63 .
435
o
75 .
964
o
80 .
538
o
82 .
875
o
5 0
.
995
84
.
29
o
10 5 1 0 0. 5 1 2 3 4 5 - 90 o
G
(
j
)
G
(
j
) 10 1
( 2
) 2
tg
1 ( 2
)
5.1.3 Graphic expression of the frequency response 2. Polar plot
The polar plot is easily useful for investigating system stability.
Example 5.1.3
G
(
s
)
s
(
K Ts
1 )
G
(
j
)
G
(
s
)
s
j
The magnitude and phase response:
A
(
)
G
(
j
)
j
(
K j
T
1 )
K
1
(
T
) 2 ;
(
)
G
(
j
)
[ 90
o
tg
1 (
T)] Calculate A(ω) and
(
) for different ω:
KT
Im 2
A
(
)
(
)
0
90
o
4 1 2
T KT
117
o
5 1
KT T
2
135
o
0
180
o
4KT
1
T
5
0 1 2
T -135 o
-117 o Re
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.
Idea: How to enlarge the lower frequency band and shrink (shorten) the higher frequency band
？
3. Bode diagram(logarithmic plots) Plot the frequency characteristic in a semilog coordinate: Magnitude response — Y-coordinate in decibels: 20 log
G
(
j
) X-coordinate in logarithm of ω: logω Phase response — Y-coordinate in radian:
G
(
j
) X-coordinate in logarithm of ω: logω First we discuss the Bode diagram in detail with the frequency response of the typical elements.
5.2 Frequency Response of The Typical Elements
The typical elements of the linear control systems — refer to Chapter 2.
1. Proportional element Transfer function:
G
(
s
)
C
(
s
)
R
(
s
)
K
Frequency response:
G
(
j
)
K
G
(
j
)
K
L
(
)
20 log
G
(
j
)
20 log
K
(
)
G
(
j
)
0
o
Im
K
Polar plot
Re 0dB, 0 o
L
( ), ( )
L
(
)
20 log
K
dB
0
.
1 1 10 100
Bode diagram
( ) 0
o
(log )
5.2 Frequency response of the typical elements
2. Integrating element Transfer function:
G
(
s
)
C
(
s
)
1
Im
R
(
s
)
s
Frequency response:
G
(
j
)
1
j
G
(
j
)
1
L
(
(
)
)
20 log
G
(
G
(
j
)
j
)
90
o
L
( ), ( )
20 log
Re
L
(
) :
20
dB
/
dec
0
0dB, 0 o 0
.
1 1 10 100 (log )
(
)
90
o
Polar plot Bode diagram
5.2 Frequency response of the typical elements
3. Inertial element Transfer function:
G
(
j
)
1/T: break frequency Im
1
(
T
) 2
1
1
0 Re
G
(
s
)
L
(
)
20 log (
)
C
(
s
)
R
(
s
)
tg
1
Ts
1
1 1
(
T
) 2
G
(
j
)
j
1
T
1
0
20 3
dB
log(
T
)
1
1 1 T T
T
(
T
0dB, 0 o
)
L
( ), ( ) 0
.
1
20 log
K
1 10
G
(
s
)
K T
2
s
1 :
1
T
100 1
T
2 (log ) 45
o
20 dB /
dec
90
o
Polar plot Bode diagram
5.2 Frequency response of the typical elements
4. Oscillating element Transfer function:
G
(
j
)
G
(
L
(
)
1
1 (
2
T
2 ) 1
2
G
(
s
)
maximum value of
：
C
(
s R
(
s
) )
T
2
s
2
1 2
Ts
1 0
j
)
( 1
20 log 2
T
2 )
( 1
j
2
T
2
T
2 ) 2
( 2
T
) 2
( 2
T
) 2
(
)
tg
1 ( 1
2
T
2
T
2
0
20 40 log( log(
2
T
) )
n
(
1
T
n n
)
) Make:
d d
(
G
(
j
) )
M r r
resonant p eak
0
M r
r
n G
(
j
r
) 1
2
2
2 (0
1 1
2
1 2 2 )
5.2 Frequency response of the typical elements
The polar plot and the Bode diagram:
2
1
Im
0 1 Re
L
( ), ( )
n
r
1 /
T
20 log
M r
20 log( 1 2
)
0dB, 0 o (log ) 0
.
1 1 10 100
j
1 2
(
n
)
90
o
40 dB /
dec
Polar plot
180
o
1 .
2 .
0
r
2
2
(
r
n
n
)
r
M r
Bode diagram
unstable s
0
No resonan ystem ce
, Optimal Second order System
5.2 Frequency response of the typical elements
5. Differentiating element Transfer function:
G
(
s
)
Ts
2
Ts
1
s
2
Ts
1
differenti al first
order differenti al second
order differenti al
Im Im Im
differential
Re 1 Re 1 Re
1th-order differential 2th-order differential Polar plot
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:
s inverse
1
s Ts
1
inverse
1
Ts
1
T
2
s
2
2
Ts
1
inverse
1
T
2
s
2
2
Ts
1 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
L
( ), ( )
L
( ), ( ) 180
o
0dB, 0
L
( ) : 20
dB
/
dec
o 0
.
1 1 10
differential
100 ( ) 90
o
(log ) 90
o
0dB, 0 o 0
.
n
1 1
T
1 10
L
( ), ( )
0dB, 0 o
90
o
45
o
20
dB
/
dec
0
.
1 (log ) 1 10 100
1th-order differential
100 40
dB
/
dec
(log ) 20 log( 1 2 ) 20 log
M r
2th-order differential
5.2 Frequency response of the typical elements
6. Delay element Transfer function:
G
(
s
)
C
(
s
)
R
(
s
)
e
s G
(
j
)
e
j
Im
G
( (
j
)
)
1
G
(
L
(
)
j
)
0
L
( ), ( )
R=1
0dB, 0 o (log ) Re 0
.
1 1 10 100
Polar plot Bode diagram
5.3 Bode diagram of the open loop systems
5.3.1 Plotting methods of the Bode diagram of the open loop systems Assume: We have:
G
(
s
)
here
:
G
1 (
G i s
(
s
) )
G
2 (
s
)
G
3 (
s
)...
the transf er functio n of the t ypical ele ments
(
)
G
(
j
)
G
1 (
j
)
G
2 (
j
)
G
3 (
j
)
...
L
(
)
20 log
G
2
o
log
G
1
20 log
G
2
20 log
G
3
...
That is, Bode diagram of a open loop system is the superposition of the Bode diagrams of the typical elements.
Example 5.3.1
G
(
s
)
H
(
s
)
10 (
s
1 )
s
2 ( 0 .
01
s
1 )
5.3 Bode diagram of the open loop systems
G(s)H(s) could be regarded as:
G
(
s
)
H
(
s
)
① ② ③ ④
10 (
s
1 )
s
2 ( 0 .
01
s
1 )
10
(s
1)
1 s 2
1 0 .
01
s
1 Then we have:
L
( ), ( ) － 40dB/dec ② 40dB, 90 o 20dB/dec ① 20dB, 45 o 0dB, 0 o -20dB, -45 o 0
.
1 1 -40dB/dec -40dB, -90 o － 20dB/dec 10 100 (log ) － ④ － 40dB/dec -60dB.-135 o ③ -80dB,-180 o
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
G
(
s
)
10 (
s
1 )
s
( 0 .
1
s
1 )( 0 .
01 2
s
2
0 .
01
s
1 )
L
(
)
20
log
10 20
log
10 20
log
20
log
10 20
log
20 20
log log
10
20
logω
( 20
log
20
log
20
log
( 40
log
( 0 .
01
)
0 .
1
( 1
20
log
20
log
(
0 .
1
) )
( 10
1)
( 100
10 )
)
100 )
(
(
) )
90
o
tg
1
1
tg
1
0 .
01
51 .
3
o
56 .
5
o
( 0 .
01
174 .
9
o
179 .
6
o
tg
) 2
1 ( 0 .
1
1 10 100 104 ) The Bode diagram is shown in following figure:
5.3.2 Facility method to plot the magnitude response of the Bode diagram
G
(
s
)
10 (
s
1 )
s
( 0 .
1
s
1 )( 0 .
01 2
s
2
0 .
01
s
1 )
20dB, 45 o 0dB, 0 o 0
.
1 -20dB, -45 o
L
( ), ( ) 40dB, 90 o －
20dB/dec
1 10 -40dB, -90 o －
20dB/dec
100
r
1.25dB
(log ) －
60dB/dec
-60dB.-135 o -80dB,-180 o -100dB,-225 o -120dB,-270 o
There is a resonant peak M
r
at:
M r
2
1 1
2
1 .
154
1 .
25
dB
r
n
100 1
2
2 1
2
0 .
5 2
70 .
7
5.3.3 Determine the transfer function in terms of the Bode diagram
1. The minimum phase system(or transfer function) Compare following transfer functions:
G
1 (
s
)
G
3 (
s
)
K
(
s
(
Ts
1 )
G
2 ( 1 )
s
)
K
(
s
(
Ts
1 ) 1 )
G
4
(s)
K
(
s
(
Ts K
(
s
(
Ts
1 )
1 )
1 )
1 )
T
We have:
G
1 (
K
)
(
) 2 (
T
) 2
G
2 (
1 1 )
G
3 (
)
G
4 (
) 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: Sketch the polar plot:
G
1 (
G
3 (
j
j
) )
tg
1 (
T
)
tg
1 (
T
)
tg
1 (
),
G
2 (
tg
1 (
),
G
4 (
j j
)
)
1
tg
180
o
(
T
tg
)
180
o
1 (
T
)
tg tg
1 (
1 (
) )
5.3.3 Determine the transfer function in terms of the Bode diagram
The polar plot: Im
Im
,
K
T
0
K
Re Re
,
K
T
Im
0 ,
K
Re
0 ,
K
,
K
T G
1 (
s
)
K
(
Ts s
1 1 ) phase shift 0 0
G
2 (
s
)
K
(
s
1 )
Ts
1 phase shift
－
π
G
3 (
s
)
K
(
s
1 )
Ts
1 phase shift
－
π
G
4 (
s
) 0 ,
K K
(
Ts s
1 1 ) Im It is obvious: the net phase shifts of the
Re
,
K
phase shift π
G
1 (jω) is the minimum when ω vary from zero to infinite.
T G
1 (s) is named: the minimum phase transfer function .
5.3.3 Determine the transfer function of the minimum phase systems in terms of the magnitude response Definition: 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. Determine the transfer function from the magnitude response of the Bode diagram .
Example 5.3.3
5.3.3 Determine the transfer function in terms of the Bode diagram
we can get the G(s) from
L
( ) － 40dB/dec
the Bode diagram
:
G
(
s
)
K
(
0
.
5
s
1
)
s
2
(
0
.
005
s
1
)
and
:
0dB, 0 o 0
.
1 － 20dB/dec 1
2 20 200
10 100 (log ) － 40dB/dec
L
(
)
2
o
log
K
20
log
2 20
log(
0
.
5
)
20
Example 5.3.4
we can get Bode the G(s) diagram
:
from the
L
( ) 0dB 20dB 20dB/dec 0.5
0
.
1 1
G
(
s
)
(
T
1
s Ks
1
)(
T
2
s
1
)
0
K
10 40 － 20dB/dec 200 (log ) 100
5.3.3 Determine the transfer function in terms of the Bode diagram
Ks
1
)(
T
2
s
L
( )
we can get the G(s) from the Bode diagram
:
G
(
and
:
s
)
L
(
)
L
(
ω
)
L
(
)
(
T
1
s
1
)
0dB 20dB 0
.
1 20dB/dec 0.5
1 20
log
K
20
log
2 20
log
2 20
log
0
.
5 20
log
20
log
1
/
T
1 0
K
20
dB
20
log(
0
.
2
)
20
T
1 2 0
.
2
log(
T
2
)
200 10 － 20dB/dec 200 (log ) 100 0
T
2 0
.
05
Example 5.3.5
we can get the G(s) from
－ 20dB/dec
L
( ) 8.136 dB
the Bode diagram
:
20 dB
G
(
s
)
K
(
0
.
01
s
1
)
2
s
(
T
2
s
2
Ts
1
)
0dB 0
.
1 (log ) 1 10 － 60dB/dec 100 － 20dB/dec
5.3.3 Determine the transfer function in terms of the Bode diagram
L
( )
we can get the G(s) from
－ 20dB/dec 20 dB 8.136 dB
the Bode diagram
:
G
(
s
)
K
(
0
.
01
s
1
)
2
s
(
T
2
s
2
Ts
1
)
0dB 0
.
1 (log ) 1 10 － 60dB/dec 100 － 20dB/dec
1 T
L
(
)
10
T
20
log
K
0.1
20
log
10 20
dB
K
20
log
2
ζ
1 1 2 8
.
136 0
.
2
then
: G(s)
100(0.01s
s(0.01s
1) 0.04s
2 1) For the non-minimum phase system we must combine the magnitude response and phase response together to determine the transfer function.
5.3.3 Determine the transfer function in terms of the Bode diagram
L
( ), ( )
Example 5.3.6
－ 20dB/dec
G
1 10
(
0
(
s
.
1
s
1
)
G
2 1
)
10
(
0
.
1
s
(
s
1
)
1
)
G
3 10
( (
0
s
,
1
s
1
)
1
)
G
4
(
s
)
10
(
0
.
1
s
(
s
1
)
1
)
0dB, 0 o － 0
.
1 90 o － 180 o 1 10 100 (log )
All satisfy the magnitude response But
only G
4 (
s
)
10 ( ( 0
s
.
1
s
1
) 1 )
satisfy th e phase re sponse sim ultaneousl y.
So, we have :
G
(
s
)
10(0.1s
(s 1)
1)
5.4 The Nyquist-criterion
A method to investigate the stability of a system in terms of the open-loop frequency response.
5.4.1 The argument principle(Cauchy’s theorem) Assume:
G
(
s
)
H
(
s
)
here
:
z i
K
1 (
s
(
s
p
1
z
1 )(
s
)(
s
z p
2 2 )...( )...(
s s
z p n m
)
open-loop zeros
;
p j
)
n open-loop
m poles
.
Make :
F
(
s
)
1 (
s
G
(
s
)
H p
1 )(
s
(
s
)
p
2 ( 1
s
)...(
s K p
(
1 1
s
(
)(
s p n s
z
1 )(
s p
1 )
)(
s K
1
(
p
2
s
)...(
z p
2
s
2 )...( )...(
s s
z
1
)(
p n s
)
z p n z
2
m
) ) )...(
K
(
F s
(
s
p
1
s
1 )(
s
)(
s
p
2
s
2 )...( )...(
s s
p n s n
) )
s i
s zeros of the F
(
z s
)
m
) Note: s
i
→ the zeros of the F(s), also the roots of the 1+G(s)H(s)=0
5.4.1 The argument principle
F
(s )
o along a closed path Γ of the s-plane in the clockwise direction shown in Fig.5.4.1 .
Because If
:
F the zeros s i
(
s
)
i n
1
(
s
is enclose d by Γ s i
)
j n
1 ,
then
:
(
(
s
s i
)
2
s
s i If the zeros s i
(
s
s i
)
is not enc losed by Γ
0 ,
then
:
p j
) Im
s i
S-plane
s
s i Γ
Re Similarly we have:
p j is enclose d by Γ p j
:
is not enc losed by Γ
(
s
p j
)
:
(
s
p j
)
2
0 Fig. 5.4.1
5.4.1 The argument principle
If Z zeros and P poles are enclosed by Γ , then:
F
(
s
)
i n
1
(
s
Z
(
2
)
s i
)
j n
1
P
(
2
)
( (
P s
p j
)
Z
)
2
It is obvious that path Γ can not pass through any zeros s
i
poles p
j
. Then we have the argument principle: or 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.
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
Im
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 .
S-plane Re
The path Γ is called the Nyquist-path.
Fig. 5.4.2
5.4.2 Nyquist criterion
Im
S-plane When s travels along the the Nyquist-path:
F
(
s
)
1
G
(
s
)
H
(
s
)
s
j
1
G
(
s
)
H
(
s
)
0
F
(
s
)
s
j
1
G
(
j
)
H
(
j
)
G
(
j
)
H
(
j
)
1
F
(
s
) Re Because the origin of the F(s)-plane is Fig. 5.4.2
equivalent to the point (-1, j0) of the G(jω)H(jω)-plane, we have another statement of the argument principle: 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:
N
P
Z
[
or N
(P
Z)/
2 for
from 0
] 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.
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.
—— Nyquist criterion
Here: P — the number of the poles of G(s)H(s) in the right hand of the s-plane.
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-criterion
ii) 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: 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: 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).
Here: the net “
positive traversing
” —— the difference between the
number of the “ positive traversing ” and the number of the “ negative traversing ” .
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.
Im Im
stable stable
(
-
1
, j0
) (
-
1
, j0
) 0 Re Re (1)
P=2
0 Im
unstable
(
-
1
, j0
) 0 Re (3)
P=2
Fig.5.4.3
0 (2)
P=0
Im (
-
1
, j0
)
unstable
Re (4)
P=0
5.4.2 Nyquist-criterion
Note: the system with the poles (or zeros) at the imaginary axis Example 5.4.2
G
(
s
)
H
(
s
)
s
(
s
10 1
)(
0
.
5
s
1
) 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).
Idea: We consider a semicircular detour around the pole (s = 0) repre sented by setting
s
e j
(
0)
at the s = 0 point we have:
0
s
ε
e
j
90
o
G
(
j
0
)
H
(
j
0
)
1
ε
e
j
90
o
0
s
ε
e j
0
o
G
(
j
0
)
H
(
j
0
)
1
ε
e j
0
o
1
e j
0
o
1
e j
90
o
0
s
ε
e j
90
o
G
(
j
0
)
H
(
j
0
)
1
ε
e j
90
o
1
e
j
90
o s
j
0 －
2
－
1
0
s
j
Im
0
Radius
0
Fig. 5.4.4
Radius r
Re
s
j
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.
s
j
Im
Radius r
0 Im 0 0
Re
G
(
s
)
H
(
s
)
Re 0
Radius
0
plot shown as Fig.5.4.5.
s
(
s
10 1
)(
0
.
5
s
1
) (-1, j0)
s
j
s
j
0
Fig. 5.4.4
Fig.5.4.5
In terms of above discussion , we can plot the system’s polar
0
The closed loop system is unstable.
Example 5.4.3
G
(
s
)
H
(
s
)
10
s
(
s
1
)(
s
2 4
)
s
(
s
1
)(
s
10
j
2
)(
s
j
2
) Determine the stability of the system applying Nyquist criterion. Solution
Im
j
2
Similar to the Example 5.4.2, the system’s polar plot is shown as Fig.5.4.6 .
(-1, j0) The closed loop system is unstable.
j
2 0
5.4.3 Application of the Nyquist criterion in the Bode diagram
Fig.5.4.6
Re 0
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) =
－
180 o 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 “-180 o line” from below to above in the open-loop system’s Bode diagram; “negative traversing” — φ(ω) traverses the “-180 o 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.
Solution Because the open-loop system is stable, P = 0 .
In terms of the Nyquist
－ 90 o
criterion in the Bode diagram:
0dB , － 180 o
The number of the net “positive traversing” is 0 ( = P/2 = 0 ).
－ 270 o
L
( ) － 20 － 40 － 60 － 40 － 20 － 40 φ(ω) － 60 (log ) L(ω)
Fig.5.4.7
The closed loop system is stable .
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 K
g K g
G
(
1
j
)
H
(
j
)
g K g
(
dB
)
20
log
G
(
j
)
H
(
j
)
g
g :
G
(
j
)
H
(
j
)
g
180 0
Phase-cros sover freq uency
2. Phase margin γ
c
c
G
(
j
)
H
(
j
)
c
(
180 0
)
G
(
j
c
)
H
(
j
c
)
180 0
ω c
:
G
(
j
)
H
(
j
)
c
1
Gain-cross over frequ ency
3. Geometrical and physical meanings of the K
g
and γ
c
5.4.4 Nyquist criterion and the relative stability
The geometrical meanings is shown in Fig. 5.4.8.
1/K g
Im
The physical signification :
K g
— 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:
－
1
γ c K g >1 the closed loop system is stable .
γ c
—amount of the phase shift of G(jω)H(jω) to be allowed before
unstable stable Critical stability
Fig. 5.4.8
the closed-loop system reaches to be unstable.
For the minimum phase system:
γ c >0 the closed loop system is stable .
Re
5.4.4 Nyquist criterion and the relative stability
Attention : For the linear systems: 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).
Example 5.4.5
The open loop transfer function of a control system is:
G
(
s
)
H
(
s
)
K s
(
0
.
1
s
1
)
e
s
(1) Determine K
g
and γ
c
when K =1 and τ =1.
(2) Determine the maximum K and τ based on K = 1 and τ = 1.
5.4.4 Nyquist criterion and the relative stability
Solution
G
(
s
)
H
(
s
)
K s
(
0
.
1
s
1
)
e
s
(1) Determine K
g
and γ
c
( K =1 and τ =1)
In terms of:
G
(
j
)
H
(
j
)
g
2
K g
tg
1 0
.
1
g
g
1
G
(
1
j
g
)
H
(
j
g
)
1
(
0
.
1
)
2
K g
180 0
1.43;
c
g
1
.
43
K
1
G
(
j
)
H
(
1
(
0
.
1
c
)
2
j
)
c K K
1
c
1
.
44 3
.
17
(dB)
1 1
c
(2)
G
(
j
c
)
H
(
j
c
)
180 0 90 0
tg
1 0
.
1
c
c
c
1 1 27 0
Because K g Because
c K
1 1 1
.
44
maximum K
27 0
c
27 0
1
.
44
maximum
0
.
47
5.4.4 Nyquist criterion and the relative stability
Example 5.4.6
The G(jω)H(jω) polar plot of a system is shown in Fig.5.4.9.
(1) Determine K
g
(2) Determine the stable range of the open loop gain.
Solution 2
Im
(1) Determine K
g
1.5
( 1, j0)
K
g
1 0
.
8 1
.
25 1
.
94
(
dB
) (2) Determine the stable range of the open loop gain.
G(jω)H(jω)
0.8
Fig.5.4.9
Assume
:
G
(
jω
)
H
(
jω
)
K
0
GH
(
jω
)
and the critic al stable value of t he open lo op gain is K
c
.
Re
In terms o f
:
K
0
GH
(
j
)
g
0
.
8
and K c
1
GH
(
j
)
g K
0
GH
(
j
)
g
1
K c
1 5
K
0 4 1
.
5
and K c
2
GH
(
j
)
g K
0
GH
(
j
)
g
1
K c
2 2 3 2
and
(
2 1.5
1, j0)
G(jω)H(jω)
0.8
Fig.5.4.9
K c
3
GH
(
j
)
g
1
K c
3 1 2
K
0
Then we have the stable range of the open loop gain
K
:
2 3
K
0
K
5 4
K
0
or K
1 2
K
0 Im Re
5.5 System analysis based on the frequency response
5.5.1 Performance specifications in the frequency domain 1. For the closed loop systems The general frequency response of a closed loop systems is shown in Fig. 5.5.1
(1) Resonance frequency ω
r
:
Assume
:
A
(
)
(
j
)
1
G
(
G
(
j
j
) )
H
(
j
)
r satisfy d
:
d
A
(
)
r
0
(2) Resonance peak M
r M r
A
(
)
r
(3) Bandwidth ω
b
: :
b satisfy
:
A
(
)
b
2 2
A
( 0 )
M r
A(0) 0.707A(0) 0 A(ω) ω
r
Fig. 5.5.1
ω
b
ω
2. For the open loop systems (1) Gain crossover frequency ω
c
:
c satisfy
:
G
(
jω
)
H
(
jω
)
ω
ω c
For the unity feedback systems, ω
c
≈ ω
b
1
, because:
(
j
)
G
(
1
j
G
( )
j
)
1
G(j
)
1
G
(
j
) G(j
)
1 (2) Gain margin K
g
:
K g
G
(
1
j
)
H
(
j
)
g
;
K g
(
dB
)
20
log
G
(
j
)
H
(
j
)
g Here
g
satisfies
:
G
( (3) Phase margin γ
c
:
c Here
c
satisfies
:
G
(
j
)
H
(
G
(
j
)
H
(
j
)
g
180 0
j
j
) )
H
(
j
c
)
c
1
(
180 0
)
5.5 System analysis based on the frequency response
Generally K
g
peak M
r
: K
g
and γ
c
could be concerned with the resonance and γ
c
↑ —— M
r
↓.
ω c
could be concerned with the resonance frequency ω
r
bandwidth ω
b
: ω
c
↑ —— ω
r
and ω
b
↓.
and 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:
C
(
t
)
L
1
C
(
j
)
1 2
C
(
j
)
e j
t d
here
:
C
(
j
)
(
j
)
R
(
j
)
1
G
(
G j
j
) )
H
(
j
)
R
(
j
) But it is difficult to apply the formula .
5.5.2 Relationship of the performance specifications between the frequency and time domain (1) Bandwidth ω
b
(or Resonance frequency ω
r
)
Generally ω b
(or ω
r
Rise time t
r
)↑—— t
r
↓ because of the “time scale” theorem: In terms of
c
(
t
) If :
L
1
C
,
C
( (
j
)
j
)
C
(
2
j
1
)
C
(
j
)
e j
t d
Then :
c
(
t
)
1 2
C
(
j
)
e j
t
/
1
d
1
c
(
t
/
) That is :
βω
t β
So ω
b
(or ω
r
)↑—— t
r
↓ alike : ω
c
↑—— t
r
↓ because of ω
c
≈ ω
b .
For the large ω
b
, there are more high-frequency portions in c(t), which make the time response to be faster.
5.5.2 Relationship of the performance specifications between the frequency and time domain (2) Resonance peak M
r
overshoot σ
p %
Normally M
r
↑ —σ
p
% ↑ because of the large unbalance of the frequency signals passing to c(t) .
K g
and γ
c
↓ —σ
p
% ↑is alike because of K
g
and γ
c
↓—M
r
↑.
Some experiential formulas: Overshoot
p
%
0
.
16 0
.
4
(
M r
1
)
(
1
.
1
M r
1
.
8
)
1
and
M r
sin
c
For most design problem, an optimum value of
M r
Settling time
t s
k
c
,
k
2 1
.
5 1
sin
c
1 2
.
5 1
sin
c
1
:
.
1 35 0
M r
c
1
.
5 90 0
5.5.2 Relationship of the performance specifications between the frequency and time domain (3) A(0) → Steady state error e
ss A(
0
)
A
(
)
0 1
G
(
G
(
j
j
) )
H
(
j
)
0
assume :
G
(
s
)
K G
1
s v
then :
A
(
0
)
1
K
1
H K G K H v
G v
1 0 0
(
s
)
H
For
A
( (
s
) the
0
)
K
unity
H
H
0
(
s
) feedback
1 1
K G
K G v v
system, H(s)
1 0
1 : So for the unity feedback systems:
A
(
0
)
1
,
A
(
0
)
e ss
5.5.2 Relationship of the performance specifications between the frequency and time domain (4) Reproductive bandwidth ω
M
→ accuracy of Reproducing r(t) A(ω) Reproductive bandwidth ω
M
:
M r A
(
)
1
G
(
G
(
j
j
) )
H
(
j
)
M
: allowed reproducin g error
A
(
0
)
A(0) 0.707A(0) 0
ω M
△
ω
r
ω
b
ω Fig. 5.5.2
for a given ω
M ,
△
↓—higher accuracy of reproducing r(t) . for a given
△
, ω
M
↑ —higher accuracy of reproducing r(t) .
Demonstration assume :
E
(
j
)
and :
e
(
j
)
R
(
j
)
C
(
j
)
e
(
j
)
R
(
j
)
5.5.2 Relationship of the performance specifications between the frequency and time domain For the frequency spectrum of r(t) shown in Fig.5.5.3 .
e
(
t
)
2 2 1
M M
M M
R
(
e
(
j
j
)
e
)
R
(
j
t j
d
)
e
j
t d
r
(
t
) That is :
e
(
t
)
r
(
t
)
e
(
t
)
0
R
(
j
)
ω M
Fig. 5.5.3
ω 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:
G
(
s
)
s
(
s
2
n
2
n
)
(
s
)
s
2
n
2 2
n s
n
2
5.5.3 Relationship of the performance specifications between the frequency and the time domain: for the typical 2th-order system We have:
b
n
r
n M r
2
(
1 2 2
)
1 2 2
(0
2 4 2 4 4
2 2 )
1 1 2
,
n
p
% ,
t s
,
t r
...
c
n
c
tg
1 1 4 4 2 2 2 1 4 4 2 2
K g
,
n
p
% ,
t s
,
t r
...
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 low frequency band is mainly concerned with the control accuracy of the systems.
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 e ss is.
2. For the middle frequency band The middle frequency band is mainly concerned with the transient performance of the systems.
ω c
↑—t
r
↓; K
g
and γ
c
↓—σ
p
% ↑
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
L
(
)
20 log
G
(
j
)
H
(
j
)
－
40 Compare the performances between the system Ⅰand system Ⅱ 0dB
－
20 ω
－
40 Fig. 5.5.4
Ⅰ Ⅱ
5.5 System analysis based on the frequency response
Solution : e ssⅠ > e ssⅡ
σ
pⅠ
% =σ
pⅡ
% t
rⅠ > t rⅡ The ability of the system Ⅰ restraining the high frequency noise is stronger than system Ⅱ 0dB
L
(
)
20 log
G
(
j
)
H
(
j
)
－
40 Fig. 5.5.4
－
20 ω Ⅰ
－
40 L(ω)
－
20dB/dec Ⅱ 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 e ss < 0.1.
Solution : 0.1
Fig. 5.5.5
1
ω
－
40dB/dec
5.6 Frequency response of the closed loop systems 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) 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)