Fundamentals of Control Systems - Chapter 4: System Stability Analy

mptotes:
)1(
0)( 
3
)12()12(
1 
 l
l
ll 


1)( 
3
14
3
2


l
-
mn 

3
10
14
)1()]24()24()3(0[zero 

  jj
mn
OA
pole
 The break points:
(1) 
)1(
)208)(3( 2  ssssK  2
234
)1(
608877263  ssss
d
dK
0)1(1  sK
s ss
0
d
dKThen (rejected)



970660
05,167,32,1
j
js (rejected) 
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)208)(3( 2  sssss   .,4,3s
The root locus method – Example 3 (cont’)
 The intersections of the root locus with the imaginary axis:
(1)  (2) 0)60(4411 234  KsKsss
Substitute s=j into the equation (2):
0)60(4411 234  KjKj 

  044 24 K 
 

0
0
K


 

0)60(11 3 

K

 

322
893,5
K


0
)208)(3(
)1(1 2  sK
 

7,61
314,1
K
j (rejected) 
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 ssss the intersections are: Critical gain: 893,5js  322crK
The root locus method – Example 3 (cont’)
 The departure angle of the root locus from the pole p3
)(180   43213 
)906,1164,153(3,146180 
0
3 7.33
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The root locus method – Example 3 (cont’)
Im s
+j5,893
33.70 +j2
1 23
0
Re s
3 14
4
j2
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j5,893
The root locus method – Example 4
 Given the system below:
10)( 2sGY(s)R(s) G (s) G(s) )39(  ss
KI)(
 C+
s
KsG PC 
 For KI = 2.7, sketch the root locus of the system when 
KI=0+, note that dKP / ds=0 has 3 roots at 3,  3, 1.5.
 For KP =270, KI = 2.7, the system is stable or not?
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The root locus method – Example 4 (cont’)
S l ti o u on:
 The characteristic equation of the system:
0)()(1  sGsGC
0107.21   K
(1)0101  sKP
392   sssP
)3)(9( 2  ss
 Poles: 9p 32 jp  33 jp 
 Zeros: 01 z
1
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The root locus method – Example 4 (cont’)
 The asymptotes:
0)(l 2/)12()12(   ll
1)(l 2/
13  mn
9)0()]3()3(9[zero  jjpole
 The break points:
213
 mnOA
0dKP 



3
31
s
s

ds
(rejected)  

5.13
2
s
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The root locus has two break points at the same location 3
The root locus method – Example 4 (cont’)
 The departure angle of the root locus from the pole p2
)]arg()[arg()arg(180 321212
0
2 ppppzp 
  3
))]3(3arg())9(3[arg()03arg(1800 jjjj 



 
  90
9
90180 10 tg
0
2 169
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The root locus method – Example 4 (cont’)
 For KI =2.7 the root locus 
is located completely in 
the left-half s-plane when 
KP =0+, so the system 
is stable when KI =2.7, 
K =270P .
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Frequency domain analysis
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Crossover frequency
 Gain crossover frequency( ): is the frequency where thec
amplitude of the frequency response is 1 (or 0 dB).
1)( cM  0)( cL 
 Phase crossover frequency (): is the frequency where
phase shift of the frequency response is equal to 1800 (or
equal to  radian).
0180)(  rad)(   
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Stability margin
 Gain margin (GM):
)(
1
 MGM  )(  LGM [dB] 
Physical meaning: The gain margin is the amount of positive 
gain at the phase crossover frequency required to bring the 
system to the stability boundary. 
( ) Phase margin M
)(1800 cM 
Physical meaning: The phase margin is the amount of 
additional phase lag at the gain crossover frequency required to 
bring the system to the stability boundary
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 . 
Graphical representation of frequency response (cont’)
Bode diagram Nyquist plot 
Gain margin
Gain margin 
Phase margin
Phase margin
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Nyquist stability criterion
 Consider a unity feedback system shown below suppose,
that we know the Nyquist plot of the open loop system G(s),
the problem is to determine the stability of the closed-loop
G ( )system cl s .
Y(s)R(s)
G(s)+
 Nyquist criterion: The closed-loop system Gcl(s) is stable if
and only if the Nyquist plot of the open-loop system G(s)
encircles the critical point (1, j0) l/2 times in the
counterclockwise direction when  changes from 0 to + (l is
the number of poles of G(s) lying in the right half s plane)
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- - .
Nyquist stability criterion – Example 1
 Consider an unity negative feedback system whose open, -
loop system G(s) is stable and has the Nyquist plots below
(three cases). Analyze the stability of the closed-loop system.
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Nyquist stability criterion – Example 1
 Solution
The number of poles of G(s) lying in the right-half s-plane is 0
because G(s) is stable. Then according to the Nyquist
criterion, the closed-loop system is stable if the Nyquist plot
G(j) does not encircle the critical point (1, j0)
 Case : G(j) does not encircle (1, j0)
 the close-loop system is stable.
 Case : G(j) pass (1, j0)
 the close-loop system is at the stability
boundary;
 Case : G(j) encircles (1, j0)
th l l t i t bl
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 e c ose- oop sys em s uns a e.
Nyquist stability criterion – Example 2
 Analyze the stability of a unity negative feedback system
whose open loop transfer function is:
)(  KsG
)1)(1)(1( 321  sTsTsTs
 Solution:
 Nyquist plot: Depending 
on the values of T1, T2, 
T d K h N i4 an , t e yqu st 
plot of G(s) could be one 
of the three curves 1, 2 
or 3.
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Nyquist stability criterion – Example 2 (cont’)
Th b f l f G( ) l i i th i ht h lf l i 0e num er o po es o s y ng n e r g - a s-p ane s
because G(s) is stable. Then according to the Nyquist
criterion, the closed-loop system is stable if the Nyquist plot
G(j) does not encircle the critical point (1, j0)
 Case : G(j) does not encircle (1, j0)
 the close-loop system is stable.
 Case : G(j) pass ( 1 j0) ,
 the close-loop system is at the stability boundary;
 Case : G(j) encircles (1, j0)
 the close-loop system is unstable.
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Nyquist stability criterion – Example 3
Given an unstable open loop systems which have the Nyquist - 
plot as below. In which cases the closed-loop system is stable?
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Stable Unstable
Nyquist stability criterion – Example 3 (cont’)
Gi t bl l t hi h h th N i tven an uns a e open- oop sys ems w c ave e yqu s
plot as below. In which cases the closed-loop system is stable?
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Unstable
Nyquist stability criterion – Example 3 (cont’)
Given an unstable open-loop systems which have the Nyquist
plot as below. In which cases the closed-loop system is stable?
St bl U t bl
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a e ns a e
Nyquist stability criterion – Example 4
 Given a open-loop system which has the transfer function:
(K>0, T>0, n>2)nTs
KsG
)1(
)( 
Find the condition of K and T for the unity negative feedback
closed-loop system to be stable.
 Solution:
 Frequency response of the open-loop system:
nTj
KjG
)1(
)(  
Magnitude:  nT KM 1)( 22  
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 Phase: )()( 1  Tntg
Nyquist stability criterion – Example 4 (cont’)
N i l yqu st p ot:
 Stability condition: the Nyquist plot of G(j) does not encircle
the critical point (1,j0). According to the Nyquist plot, this
requires:
1)( M
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
Nyquist stability criterion – Example 4 (cont’)
 )()( 1 T We have:     ntg
Ttg    )(1  tgT  )(n  n


 tg   1  nT
 Then: 1)( M  1
11
2
2

  
n
tT
K


  ngT

n
tgK 


 

 12 
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n 
Bode criterion
 Consider a unity feedback system suppose that we know the,
Nyquist plot of the open loop system G(s), the problem is to
determine the stability of the closed-loop system Gcl(s).
Y(s)R(s)
 G(s)+
 Bode criterion The closed loop s stem G (s) is stable if the: - y cl
gain margin and phase margin of open-loop system G(s) are
positive.
stable is system loop-closed he 
0
0
T



M
GM
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Bode criterion – Example
 Consider a unity negative feedback system whose open-loop
system has the Bode diagram as below. Determine the gain
margin, phase margin of the open-loop system. Is the closed-
loop system stable or not?
5c
Bode diagram:
GM
L( ) 2
dBL 35 )( 
0270)(
dBGM 35
c
000 90270180 )(M
M

(C)
180


Because GM<0 and 
M<0, the closed-loop
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 C 
system is unstable.
Remark on the frequency domain analysis
 If the closed-loop system as below, the Nyquist and Bode
criteria can also be applied and in this case the open-loop
system is G(s)H(s).
G(s)+_
R(s) Y(s)E(s)
H(s)
Yfb(s)
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