In The Name of God The Most Compassionate, The Most Merciful

Linear Control Systems

2017 Shiraz University of Technology Dr. A. Rahideh Table of Contents

1. Introduction to Control Systems

2. Mathematical Modelling of Dynamic Systems

3. Steady State and Transient Response Analysis

4. Root Locus Analysis

5. Response Analysis

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2017 Shiraz University of Technology Dr. A. Rahideh Chapter 5 Analysis

5.1. Introduction

5.2. Bode Diagrams

5.3. Polar Plots (Nyquist)

5.4. Log-Magnitude vs. Plots (Nichols Plots)

5.5. Nyquist Stability Criterion

5.6. Stability Analysis 3

2017 Shiraz University of Technology Dr. A. Rahideh Introduction • By the term frequency response, we mean the steady-state response of a system to a sinusoidal input. • In frequency-response methods, we vary the frequency of the input signal over a certain range and study the resulting response. • Assume the following system, if the input is sinusoidal

x(t)  Asinw t X (s) Y(s) G(s) The steady state output is

yss (t)  AG( jw) sinw t  G( jw)

where G(jw) is called the sinusoidal .

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2017 Shiraz University of Technology Dr. A. Rahideh Introduction • Example 1: Find the steady-state output of the following system in response to x(t)  Asinwt Y(s) X (s) K K G(s)  Ts  1 Ts 1 K G( jw)  2 2 K 1T w G( jw)  jTw 1 G( jw)   tan 1 Tw

yss (t)  AG( jw) sinwt  G( jw)

AK y (t)  sinw t  tan 1 Tw ss 2 2 5 1T w

2017 Shiraz University of Technology Dr. A. Rahideh Presenting Frequency-Response Characteristics in Graphical Forms

• The sinusoidal transfer function, a complex function of the frequency w, is characterized by its magnitude and phase angle, with frequency as the parameter.

• There are three commonly used representations of sinusoidal transfer functions: G( jw) vs. w 1. Bode diagram or logarithmic plot G( jw) vs. w

2. Nyquist plot or polar plot ImG( jw) vs. ReG( jw)

3. Log-magnitude-versus-phase plot (Nichols plots) G( jw) vs. G( jw) 6

2017 Shiraz University of Technology Dr. A. Rahideh Bode Diagrams A Bode diagram consists of two graphs:

1. One is a plot of the logarithm of the magnitude of a sinusoidal transfer function;

2. The other is a plot of the phase angle; Both are plotted against the frequency on a logarithmic scale.

The standard representation of the logarithmic magnitude of G(jw) is 20 log |G(jw)|, where the base of the logarithm is 10. The unit used in this representation of the magnitude is the (dB).

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2017 Shiraz University of Technology Dr. A. Rahideh Basic Factors The basic factors that very frequently occur in an arbitrary transfer function G(jw)H(jw) are

1. Gain K

2. Integral and derivative factors ( jw)1

3. First-order factors (1 jw)1

2 1 4. Quadratic factors 1 2 ( jw wn )  ( jw wn ) 

Note that adding the logarithms of the gains corresponds to multiplying them together. 8

2017 Shiraz University of Technology Dr. A. Rahideh Basic Factors K 1. Gain G( jw)  K G( jw)  20log K dB G(s)  K G( jw)  K 0 K  0 G( jw)    180 K  0

G( jw) G( jw) G( jw) dB K 1 K 20log K

K  0 w w 0 w K 1 K  0 180

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2017 Shiraz University of Technology Dr. A. Rahideh Basic Factors 2. Integral ( jw)1 1 G( jw)  G( jw)  20logw w dB 1 1 G(s)  G( jw)  s jw G( jw)  90

G( jw) G( jw) G( jw) Both axes are dB logarithmic 10 20

1 0 0.1 1 10 w 0.1 1 10 w 0.1 1 10 w

0.1  20 90

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2017 Shiraz University of Technology Dr. A. Rahideh Basic Factors

( jw) 3. Derivative G( jw)  w G( jw)  20logw dB G(s)  s G( jw)  jw G( jw)  90

G( jw) G( jw) G( jw) dB

10 20 90

1 0 0.1 1 10 w 0.1 1 10 w 0.1 1 10 w

0.1 Both axes are  20 logarithmic

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2017 Shiraz University of Technology Dr. A. Rahideh Basic Factors

4. First-order (1 jwT )1 1 G( jw)  (1) 2 2 1 1 1T w G(s)  G( jw)  1Ts 1 jTw G( jw)   tan 1(Tw) (2)

 1  w  1  20logTw w  1 (1) G( jw)  T G( jw)  T Tw dB  1 1 0 w  T 1 w  T 

1 G( jw) 1  20log 2  3 dB G( jw) 1  dB w w T T 2 12

2017 Shiraz University of Technology Dr. A. Rahideh Basic Factors

1  1 4. First-order (1 jwT)  w  1 G( jw)  T 1 Tw  1 G(s)  1 w  T 1Ts  20logTw w  1 1 G( jw)  T G( jw)   tan (Tw) dB  1 0 w  T

G( jw) G( jw) G( jw) dB

1 0 0.1 1 T 10 w 1 10 10 1 10 10 1 0.1 1 T T 0 0.1 1 T T 1 w  3 w 2  45

0.1  20 90

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2017 Shiraz University of Technology Dr. A. Rahideh Basic Factors 5. First-order (1 jwT ) 2 2 (1) G( jw)  1T w G(s) 1Ts G( jw) 1 jTw

G( jw)  tan 1(Tw) (2)

Tw w  1 1  T 20logTw w  T (1) G( jw)   G( jw)   1 w  1 dB 1  T 0 w  T

G( jw) 1  2 G( jw) 1  20log 2  3 dB w dB w T T 14

2017 Shiraz University of Technology Dr. A. Rahideh Basic Factors

5. First-order (1 jwT ) Tw w  1 G( jw)  T  1 G(s) 1Ts 1 w  T

20logTw w  1 1 G( jw)  T G( jw)  tan (Tw) dB  1 0 w  T

G( jw) G( jw) G( jw) dB

10 20 90

2 3 1 0.1 1 10 w 0 0.1 1 10 w 1 T 10 T 1 T 10 T 45

0 1 0.1 1 T 10 w

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2017 Shiraz University of Technology Dr. A. Rahideh Basic Factors 6. First-order (1 jwT ) 2 2 (1) G( jw)  1T w G(s)  1Ts G( jw)  1 jTw

1 Tw G( jw)  tan ( 1 ) (2)

Tw w  1 1  T 20logTw w  T (1) G( jw)   G( jw)   1 w  1 dB 1  T 0 w  T

G( jw) 1  2 G( jw) 1  20log 2  3 dB w dB w T T 16

2017 Shiraz University of Technology Dr. A. Rahideh Basic Factors

6. First-order (1 jwT ) Tw w  1 G( jw)  T  1 G(s)  1Ts 1 w  T

20logTw w  1 G( jw)  tan 1( Tw ) G( jw)  T 1 dB  1 0 w  T

G( jw) G( jw) G( jw) dB

10 20 180

2 3 135 1 0.1 1 10 w 0 0.1 1 10 w 1 T 10 T 1 T 10 T 90

1 0.1 1 T 10 w

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2017 Shiraz University of Technology Dr. A. Rahideh Basic Factors 7. Second-order

w 2 1 n G(s)  G(s)  2 2 2 s  2wns wn s 2 2  s 1 wn wn 1 G( jw)  2 2  w 2   w  1   2  1  w 2   w  G( jw)   n   n   w 2  w 1 2   j2   w  wn  wn  2  G( jw)   tan 1 wn   w 2  1 2  wn  18

2017 Shiraz University of Technology Dr. A. Rahideh Basic Factors 1 7. Second-order G( jw)  2 2  w 2   w  1   2  w 2  2    n  wn   wn  G(s)  2 2 s  2wns wn  2 w  G( jw)   tan 1 wn   w 2  1 2  wn   w 2   w  wn G( jw)   wn  1 w  wn

w  40log w  wn G( jw)  wn dB  0 w  wn 19

2017 Shiraz University of Technology Dr. A. Rahideh Basic Factors 7. Second-order

2 wn G(s)  2 2 s  2w ns wn

 2 w  G( jw)   tan1 wn   w 2  1 2  wn 

w  40log w wn G( jw)  wn dB  0 w wn

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2017 Shiraz University of Technology Dr. A. Rahideh Basic Factors 7. Second-order

The Resonant Frequency wr and the Resonant Peak Value Mr The peak value of G ( j w ) occurs when the denominator, g ( w ) , minimizes 2 2  w 2   w  g(w)  1   2   2     wn   wn 

dg(w) 2 0    1  0 wr  wn 1 2 2 dw

1 1 M r  G( jw)  G( jwr )  0    max 2 1 2 2 21

2017 Shiraz University of Technology Dr. A. Rahideh Basic Factors

Corner frequency • In the first-order system of the following form, K G(s)  Ts 1

1 the corner frequency is wc  T

• In the second-order system of the following form

K G(s)  s2  2w s w 2 n n

the corner frequency is wc  wn 22

2017 Shiraz University of Technology Dr. A. Rahideh Basic Factors

Corner frequency • Example 1: 2 4 G(s)  G(s)  3 3s  2 2 s 1

1 2 the corner frequency is wc  T  3

• Example 2:

6 3 G(s)  2 G(s)  2s  2s  4 s2  s  2

the corner frequency is wc  wn  2 23

2017 Shiraz University of Technology Dr. A. Rahideh Bode Diagrams

Example: Plot the bode diagrams of the following system

1000 1000 G(s)  G( jw)  s(s  5)(s  50) jw ( jw  5)( jw  50)

4 G( jw)  w w jw ( j 5 1)( j 50 1) 1 1 w   5 wc2   50 The corner are c 1 T 1 and T2

1 1 1 G( jw)  20log 4  20log jw  20log w  20log w dB j 5 1 j 50 1 1 1 1 G( jw)  0       jw j w 1 j w 1 24 5 50

2017 Shiraz University of Technology Dr. A. Rahideh Bode Diagrams (Magnitude)

Example: Plot the bode diagrams of the following system w  1  5 w  1  50 The corner frequencies are c 1 T 1 and c2 T2

1 1 1 G( jw)  20log 4  20log jw  20log w  20log w dB j 5 1 j 501 G( jw) dB  20 dB/decade

20 20log4 50 100 1000 0 5 10 w 0.1 1  20 20log 1 jw / 501 4  40 dB/decade 20log 1 G( jw)  jw / 51 w w 20log 1 jw ( j 5 1)( j 50 1)  60 dB/decade jw

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2017 Shiraz University of Technology Dr. A. Rahideh Bode Diagrams (Phase)

Example: Plot the bode diagrams of the following system w  1  5 w  1  50 The corner frequencies are c 1 T 1 and c2 T2 1 1 1 G( jw)  0       w w jw j 5 1 j 50 1 G( jw)

1 5 10 50 100 1000 0 0.1 w

4  45  1 G( jw)   1 jw/501 jw ( j w 1)( j w 1) jw/51 5 50 1 90  jw

 270 26

2017 Shiraz University of Technology Dr. A. Rahideh Bode Diagrams Using MATLAB Example: Using MATLAB Plot the bode diagrams of the following system 1000 4 G( jw)  G(s)  w w s(s  5)(s  50) jw ( j 5 1)( j 50 1) w=logspace(-1,3,100); numT=1000; denT=[1 55 250 0]; num1=4; den1=1; num2=1; den2=[1 0]; num3=1; den3=[1/5 1]; 27

2017 Shiraz University of Technology Dr. A. Rahideh Bode Diagrams Using MATLAB Example: Using MATLAB Plot the bode diagrams of the following system 1000 4 G( jw)  G(s)  w w s(s  5)(s  50) jw ( j 5 1)( j 50 1) num4=1; den4=[1/50 1];

[mag,phase]=bode(numT,denT,w); [mag1,phase1]=bode(num1,den1,w); [mag2,phase2]=bode(num2,den2,w); [mag3,phase3]=bode(num3,den3,w); [mag4,phase4]=bode(num4,den4,w);

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2017 Shiraz University of Technology Dr. A. Rahideh Bode Diagrams Using MATLAB Example: Using MATLAB Plot the bode diagrams of the following system 1000 4 G( jw)  G(s)  w w s(s  5)(s  50) jw ( j 5 1)( j 50 1) figure(1) loglog(w,mag,'b','linewidth',3) hold on loglog(w,mag1,'r--','linewidth',2) loglog(w,mag2,'k:','linewidth',2) loglog(w,mag3,'g-.','linewidth',2) loglog(w,mag4,'m--','linewidth',2) legend('G(j\omega)','K=4','1/j\omega','1/(j\omega/5+1)','1/(j\omega /50+1)') 29

2017 Shiraz University of Technology Dr. A. Rahideh Bode Diagrams Using MATLAB Example: Using MATLAB Plot the bode diagrams of the following system 1000 4 G( jw)  G(s)  w w s(s  5)(s  50) jw ( j 5 1)( j 50 1) figure(2) semilogx(w,phase,'b','linewidth',3) hold on semilogx(w,phase1,'r--','linewidth',2) semilogx(w,phase2,'k:','linewidth',2) semilogx(w,phase3,'g-.','linewidth',2) semilogx(w,phase4,'m--','linewidth',2) legend('G(j\omega)','K=4','1/j\omega','1/(j\omega/5+1)','1/(j\omega /50+1)') 30

2017 Shiraz University of Technology Dr. A. Rahideh Bode Diagrams Using MATLAB Example: Using MATLAB Plot the bode diagrams of the following system 1000 4 G( jw)  G(s)  w w s(s  5)(s  50) jw ( j 5 1)( j 50 1)

Note that the

magnitude is in logarithmic scale.

Magnitude Magnitude in NOT in dB

31 w 2017 Shiraz University of Technology Dr. A. Rahideh Bode Diagrams Using MATLAB Example: Using MATLAB Plot the bode diagrams of the following system 1000 4 G( jw)  G(s)  w w

s(s  5)(s  50) jw ( j 5 1)( j 50 1)

Phase

32 w 2017 Shiraz University of Technology Dr. A. Rahideh Minimum-Phase Systems and Nonminimum-Phase Systems

• Transfer functions having neither poles nor zeros in the right- half s plane are minimum-phase transfer functions,

• Whereas those having poles and/or zeros in the right-half s plane are nonminimum-phase transfer functions.

• Systems with minimum-phase transfer functions are called minimum-phase systems,

• whereas those with nonminimum-phase transfer functions

are called nonminimum-phase systems.

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2017 Shiraz University of Technology Dr. A. Rahideh Transport Lag

• Transport lag, which is also called dead time, is of non- minimumphase behavior and has an excessive phase lag with no attenuation at high frequencies.

• Such transport lags normally exist in thermal, hydraulic, and pneumatic systems.

• Consider the transport lag given by G( jw)  e jwT

• The magnitude is always equal to unity, since

G( jw)  0 G( jw)  coswT  j sinwT 1 dB 34

2017 Shiraz University of Technology Dr. A. Rahideh Transport Lag

• Consider the transport lag given by G( jw)  e jwT

• The magnitude is always equal to unity, since G( jw)  0 G( jw)  coswT  j sinwT 1 dB

• The phase angle is G( jw)  wT (radians) G( jw)  57.3wT (degrees)

G( jw) G( jw) dB

0 w

w

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2017 Shiraz University of Technology Dr. A. Rahideh Bode Diagrams

• Consider the following system (T s 1)(T s 1)(T s 1) G(s)  a b m s N (T s 1)(T s 1)(T s 1) 1 2 nN • Where the system is of type N, the order of the numerator is m and the order of the denominator is n.

• The relation between the start- and end-slopes of the magnitude Bode diagrams with the system-Type and order are as follows Start slope  20N dB/decade

36 End slope  20(n  m) dB/decade

2017 Shiraz University of Technology Dr. A. Rahideh Bode Diagrams

• Consider the following minimum-phase system (T s 1)(T s 1)(T s 1) G(s)  a b m s N (T s 1)(T s 1)(T s 1) 1 2 nN • Where the system is of type N, the order of the numerator is m and the order of the denominator is n.

• The relations between the start- and end-phase of the phase Bode diagrams with the system-Type and order in minimum- phase systems are as follows Start phase  90N degrees ONLY for minimum- phase systems 37 End phase  90(n  m) degrees

2017 Shiraz University of Technology Dr. A. Rahideh Polar Plots (Nyquist)

• The polar plot of a sinusoidal transfer function G(jw) is a plot of the magnitude of G(jw) versus the phase angle of G(jw) on polar coordinates as w is varied from zero to infinity.

• Thus, the polar plot is the locus of vectors G ( j w )  G ( j w ) as w is varied from zero to infinity.

• Note that in polar plots, a positive (negative) phase angle is measured counter-clockwise (clockwise) from the positive real axis.

• The polar plot is often called the Nyquist plot.

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2017 Shiraz University of Technology Dr. A. Rahideh Polar Plots (Nyquist)

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2017 Shiraz University of Technology Dr. A. Rahideh Polar Plots (Nyquist)

Integrator: Draw the polar plot of the following transfer function

1 G(s)  s

1 1 G( jw)  G( jw)  0  j ImG( jw) jw w G( jw) 1 Re  G ( j w )   0 & ImG( jw) w   w ReG( jw)

w  0 Im   Phase is  w  0 w   Im  0 always -90 40

2017 Shiraz University of Technology Dr. A. Rahideh Polar Plots (Nyquist)

First order: Draw the polar plot of the following transfer function

1 G(s)  s 1

1 1 jw 1 w G( jw)  G( jw)  G( jw)   j jw 1 1w 2 1w 2 1w 2

1 w  0 Re 1 Re G( jw)     2  1w w   Re  0 Note that for all w, Re>0 and Im<0 w w  0 Im  0 ImG( jw)  1w 2 w   Im  0  41

2017 Shiraz University of Technology Dr. A. Rahideh Polar Plots (Nyquist)

First order

1 1 w G(s)  ReG( jw) ImG( jw) s 1 1w 2 1w 2 ImG( jw)

Re 0.5 G( jw) w 1  Im  0.5 w   0.5 w  0 ReG( jw) 1 G( jw)

w1  0.5 w 1 42

2017 Shiraz University of Technology Dr. A. Rahideh Polar Plots (Nyquist)

Second order: Draw the polar plot of the following transfer function 1 w 2 G( jw)  n 2 G(s)  2 2  w  w s  2w s w 1   j2 n n  2   wn  wn

 w 2  w 1   2  2  wn  wn G( jw)  2  j 2  w 2  w 2  w 2  w 2 1   4 2 1   4 2  2  2  2  2  wn  wn  wn  wn

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2017 Shiraz University of Technology Dr. A. Rahideh Polar Plots (Nyquist)

2 wn Second order: G(s)  2 2 s  2wns wn  w 2  1  1 w  0  w 2  Re G( jw)   n     2 ReG( jw) 0 w  wn  w 2  w 2 1   4 2 0 w    2  2   wn  wn

w 2 0 w  0 wn  1 ImG( jw)  2 ImG( jw)  2 w  wn 2 2  w  2 w  1   4 0 w    2  2   wn  wn 44

2017 Shiraz University of Technology Dr. A. Rahideh Polar Plots (Nyquist)

2 wn Second order: G(s)  2 2 s  2wns wn  w 2  1   2   wn  ReG( jw) 2  w 2  w 2 1   4 2  2  2  wn  wn

w 2 wn ImG( jw)  2  w 2  w 2 1   4 2  2  2  wn  wn 45

2017 Shiraz University of Technology Dr. A. Rahideh Polar Plots (Nyquist)

Transport lag: Draw the polar plot of the following transfer function

G(s)  eTs G( jw)  e jTw G( jw)  cos(Tw)  j sin(Tw)

ImG( jw) 3 w  2T

w  0 ReG( jw) 2 w  2T

 w  2T 46

2017 Shiraz University of Technology Dr. A. Rahideh Polar Plots (Nyquist)

Example: Draw the polar plot of the following transfer function

eks ImG( jw) G(s)  1Ts

e jkw G( jw)  w   1 jTw w  0 Re G( jw) 1   G( jw)  1T 2w 2

G( jw)  kw  tan 1 Tw

47

2017 Shiraz University of Technology Dr. A. Rahideh Polar Plots (Nyquist)

Example: Draw the polar plot of the following transfer function

1 1 G(s)  G( jw)  s(Ts 1) jw ( jTw 1) T 1 G( jw)    j 1T 2w 2 w (1T 2w 2 )

limG( jw)  T  j w0

lim G( jw)  0  j0 w

48

2017 Shiraz University of Technology Dr. A. Rahideh General Shapes of Polar Plots The polar plot of the following transfer function (1 jwT )(1 jwT ) b ( jw)m  b ( jw)m1  G( jw)  a b G( jw)  0 1 l n n1 ( jw) (1 jwT1)(1 jwT2 ) a0 ( jw)  a1( jw)  where n>m, will have the following general shapes: 1. For l=0 or type 0 systems: The starting point of the polar plot (which corresponds to w=0) is finite and is on the positive real axis. The tangent to the polar plot at w=0 is perpendicular to the real axis. The terminal point, which corresponds to w   , is at the origin, and the curve is tangent to one of the axes. 49

2017 Shiraz University of Technology Dr. A. Rahideh General Shapes of Polar Plots 2. For l=1 or type 1 systems: At w=0, the magnitude of G(jw) is infinity, and the phase angle becomes –90°.

At low frequencies, the polar plot is asymptotic to a line parallel to the negative imaginary axis.

At w   , the magnitude becomes zero, and the curve converges to the origin and is tangent to one of the axes.

50

2017 Shiraz University of Technology Dr. A. Rahideh General Shapes of Polar Plots 2. For l=2 or type 2 systems: At w=0, the magnitude of G(jw) is infinity, and the phase angle becomes –180°.

At low frequencies, the polar plot may be asymptotic to the negative real axis.

At w   , the magnitude becomes zero, and the curve converges to the origin and is tangent to one of the axes.

51

2017 Shiraz University of Technology Dr. A. Rahideh Drawing Nyquist Plots with MATLAB Consider a transfer function as

num(s) G(s)  den(s) The Nyquist plot in MATLAB is obtained using the following command:

nyquist(num,den,w)

Where num is the vector corresponding to the coefficients of the numerator, den is the vector corresponding to the coefficients of the denominator and w is the user-specified frequency vector.

52

2017 Shiraz University of Technology Dr. A. Rahideh Drawing Nyquist Plots with MATLAB Example: Consider a transfer function as

1 G(s)  2 s  0.8s 1 The Nyquist plot in MATLAB is obtained using the following command:

num=[1]; den=[1 0.8 1]; nyquist(num,den) title('Nyquist Plot of G(s) = 1/(s^2 + 0.8s + 1)')

53

2017 Shiraz University of Technology Dr. A. Rahideh Drawing Nyquist Plots with MATLAB Example: Consider a transfer function as 1 G(s)  2 s  0.8s 1

54

2017 Shiraz University of Technology Dr. A. Rahideh Log-Magnitude versus Phase Plots (Nichols Plots) • Another approach to graphically

portraying the frequency-response characteristics is to use the log- G( jw) in dB magnitude-versus-phase plot,

• which is a plot of the logarithmic G( jw) magnitude in versus the phase angle.

• In the log-magnitude-versus-phase plot, the two curves in the Bode diagram are combined into one.

55

2017 Shiraz University of Technology Dr. A. Rahideh Log-Magnitude versus Phase Plots (Nichols Plots) (a) Bode diagram; (b) polar plot; (c) log-magnitude-versus-phase plot of a second order system.

56

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion

• The Nyquist stability criterion determines the stability of a closed-loop system from its open-loop frequency response and open-loop poles. • Consider the following closed-loop transfer function C(s) G(s)  R(s) 1 G(s)H (s)

• For stability, all roots of the characteristic equation must lie in the left-half s plane. 1 G(s)H (s)  0

• The Nyquist stability criterion relates the open-loop frequency response G(jw)H(jw) to the number of of 1+G(s)H(s) that lie in the right-half s plane. 57 2017 Shiraz University of Technology Dr. A. Rahideh Conformal Mapping

• Consider the following open-loop transfer function 2 G(s)H (s)  s 1 • The characteristic equation is 2 s 1 F(s) 1 G(s)H (s) 1   0 s 1 s 1

• The function F(s) is analytic everywhere in the s plane except at its singular points. • For each point of analyticity in the s plane, there corresponds a point in the F(s) plane. • For example, if s=2+j1, then F(s) becomes 2  j11 F(2  j1)   2  j1 2  j11 58 2017 Shiraz University of Technology Dr. A. Rahideh Conformal Mapping For a given continuous closed path in the s plane, which does not go through any singular points, there corresponds a closed curve in the F(s) plane. Mapping s    jw F(s)  ReF(s) j ImF(s)

s 1 F(s)  s 1 59

2017 Shiraz University of Technology Dr. A. Rahideh Encirclement of the Origin • Suppose that representative point s traces out a contour in the s plane in the clockwise direction. 1. If the contour in the s plane encloses the pole of F(s), there is one encirclement of the origin of the F(s) plane by the locus of F(s) in the counter-clockwise direction.

60

2017 Shiraz University of Technology Dr. A. Rahideh Encirclement of the Origin • Suppose that representative point s traces out a contour in the s plane in the clockwise direction. 2. If the contour in the s plane encloses the zero of F(s), there is one encirclement of the origin of the F(s) plane by the locus of F(s) in the clockwise direction.

61

2017 Shiraz University of Technology Dr. A. Rahideh Encirclement of the Origin • Suppose that representative point s traces out a contour in the s plane in the clockwise direction. 3. If the contour in the s plane encloses both the zero and the pole or if the counter encloses neither the zero nor the pole of F(s), then there is no encirclement of the origin of the F(s) plane by the locus of F(s).

62

2017 Shiraz University of Technology Dr. A. Rahideh Encirclement of the Origin

• The direction of encirclement of the origin of the F(s) plane by the locus of F(s) depends on whether the contour in the s plane encloses a pole or a zero.

• If the contour in the s plane encloses equal numbers of poles and zeros, then the corresponding closed curve in the F(s) plane does not encircle the origin of the F(s) plane.

63

2017 Shiraz University of Technology Dr. A. Rahideh Mapping

• Let F(s) be a ratio of two polynomials in s. • Let P be the number of poles of F(s) and Z be the number of zeros of F(s) that lie inside some closed contour in the s plane, with multiplicity of poles and zeros accounted for. • Let the contour be such that it does not pass through any poles or zeros of F(s). • This closed contour in the s plane is then mapped into the F(s) plane as a closed curve. • The total number N of clockwise encirclements of the origin of the F(s) plane, as a representative point s traces out the entire contour in the clockwise direction, is equal to Z-P. The mapping just gives the difference of Z and P, N  Z  P NOT P and Z 64

2017 Shiraz University of Technology Dr. A. Rahideh Mapping

N  0 Z  P Clockwise encirclements N  Z  P N  0 Z  P Counter-clockwise encirclements

• The number P can be readily determined for F(s) =1+G(s)H(s) from the function G(s)H(s).

• Therefore Z (the number of poles of the closed-loop system lie inside some closed contour in the s plane) can be found from P and N.

65

2017 Shiraz University of Technology Dr. A. Rahideh An Important Note • Instead of mapping into F(s) =1+G(s)H(s) the mapping is performed into G (s)=G(s)H(s).

• Therefore, instead of counting the number of clockwise encirclements of the origin, the number clockwise encirclements of the -1 point is counted.

66

2017 Shiraz University of Technology Dr. A. Rahideh Procedure of Nyquist Stability Criterion

1. Form loop transfer function G(s)H(s).

2. Form a semi-circle closed contour in the right-half of s plane that

does not pass though the poles or zeros of G(s)H (s). jw s plane The direction of the semicircle is clockwise.   3. Map the contour in s plane into G (s)=G(s)H(s). 0 4. Find the number of poles of G(s)H(s) in the right-half s plane, i.e. P. 5. Count the number of clockwise encirclements of -1 point, i.e. N. 6. Find Z  N  P which is the number of closed-loop poles in the right-half s plane. 7. If Z=0 , the closed-loop system is stable. 67

2017 Shiraz University of Technology Dr. A. Rahideh Summary of Nyquist Stability Criterion

Z  N  P where Z number of zeros of 1+G(s)H(s) in the right-half s plane N number of clockwise encirclements of the –1+j0 point P number of poles of G(s)H(s) in the right-half s plane

If Z=0 , the closed-loop system is stable.

68

2017 Shiraz University of Technology Dr. A. Rahideh Some Points

If there is any poles or zeros of G(s)H(s) on the imaginary axis, the semi-circle in right-half of s plane should encircle them

jw s plane

  0

If the locus of G(jw)H(jw) passes through the –1+j0 point, then zeros of the characteristic equation, or closed-loop poles, are located on the jw axis. 69

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion

• Example: Discuss on the stability of the following system using Nyquist stability criterion

R(s) 6 C(s) + _ s 1s  2s  3

70

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion

Solution: R(s) 6 C(s) + _ s 1s  2s  3

1. Form loop transfer function G(s)H(s).

6 G(s)H (s)  s 1s  2s  3

• The poles of G(s)H(s) are s  1 s  2 s  3

• G(s)H(s) has no zero.

71

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion

Solution: R(s) 6 C(s) + _ s 1s  2s  3

2. Form a semi-circle closed contour in the right-half of s plane that does not pass though the poles or zeros of G(s)H(s). jw s plane

 E A  3  2 1 0 B

D C 72

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion

Solution: 3. Map the contour in s plane into G (s)=G(s)H(s). jw j Section AD: s  R e s plane R

6 Γ(s)  G(s)H (s)   s 1s  2s  3 E A  3  2 1 0 6 B ΓR e j  j j j R e 1R e  2R e  3 D C

6 ΓR e j    e j3 R3 e j3 73

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion jw s plane Solution: 3. Map the contour in s plane into G (s)=G(s)H(s).  E A  3  2 1 0 j Section AD: s  R e Γ R e j   e j3 B R   D C

j0 ImΓ A  A Γ   e G plane

B  B Γ   e j 2 B C  C Γ   e j C A ReΓ 1 D D  D Γ   e j3 2

74

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion

Solution: 3. Map the contour in s plane into G (s)=G(s)H(s).

Section DE: s   jw jw s plane 6 Γ(s)  G(s)H(s)  s 1s  2s  3  E A  6 3  2 1 0 Γ jw  B  jw 1 jw  2 jw  3 D C 6 Γ jw  61w 2  jw11w 2  75

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion j w s plane Solution: 3. Map the contour in s plane into G (s)=G(s)H(s).  E A  Section DE: s   jw 3  2 1 0 B 6 Γ j w  C 61w 2  jw 11w 2  D

361w 2  ReΓ jw 2 2 361w 2  w 2 11w 2 

6w11w 2  ImΓ jw 2 2 361w 2  w 2 11w 2  76

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion

Solution: 3. Map the contour in s plane into G (s)=G(s)H(s).

Section DE: s   jw 0 w  

2  361w   0.1 w  11 ReΓ jw 2 2   361w 2  w 2 11w 2  0 w 1 1 w  0 0 w    6w11w 2   0 w  11 ImΓ jw 2 2   361w 2  w 2 11w 2  0.6 w 1  0 w  0 77 2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion jw s plane Solution:

3. Map the contour in s plane into G (s)=G(s)H(s).  Section DE: s   jw E A  3  2 1 0

0 w   B

 D C  0.1 w  11 ReΓ( jw)  ImΓ 0 w 1 G plane  0.6 1 w  0 B E 0 w   C A ReΓ  1  0.1  0 w  11 D ImΓ( jw)  0.6 w 1  0 w  0 78 2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion jw Solution: s plane 3. Map the contour in s plane into G (s)=G(s)H(s).  E A  3  2 1 0 B ImΓ G plane D C 0.6

B E C A ReΓ 1  0.1 D

79

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion jw Solution: s plane 4. Find the number of poles of G(s)H(s) in the right-half s plane, i.e. P.  E A  3  2 1 0 B ImΓ G plane D C 0.6

B E C A ReΓ 1  0.1 P  0 D

80

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion jw Solution: s plane 5. Count the number of clockwise encirclements of -1 point, i.e. N.  E A  3  2 1 0 B ImΓ G plane D C 0.6

B E C A ReΓ 1  0.1 N  0 D

81

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion jw Solution: s plane 6. Find Z  N  P which is the number of closed-loop poles in the right-half s plane.  E A  3  2 1 0 B ImΓ G plane D C 0.6

B E Z  0 C A ReΓ 1  0.1 D The system is stable.

82

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion

• Example: Discuss on the stability of the unity feedback system with the following forward path transfer function using Nyquist stability criterion

s 1 G(s)  ss 1

R(s) E(s) C(s) + _ G(s)

83

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion

Solution: R(s) E(s) C(s) + _ G(s)

1. Form loop transfer function G(s)H(s).

s 1 G(s)H (s)  ss 1

• The poles of G(s)H(s) are s  0 s  1

• The zero of G(s)H(s) is s 1

84

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion

R(s) E(s) C(s) Solution: + _ G(s)

2. Form a semi-circle closed contour in the right-half of s plane that does not pass though the poles or zeros of G(s)H(s). jw s plane

D A  1 0 1 C

B 85

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion

Solution: 3. Map the contour in s plane into G (s)=G(s)H(s).

jw Section AB: s  R e j R s plane

(s 1) Γ(s)  G(s)H (s)   ss 1 D A  1 0 1 C j j R e 1 ΓR e  R e j R e j 1 B

ΓR e j   e j 86

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion jw Solution: s plane

3. Map the contour in s plane into G (s)=G(s)H(s). 

D A  j 1 0 1 Section AB: s  R e j  j C R ΓRe   e

B

j0 ImΓ A  A Γ   e G plane

B  B Γ   e j 2 B A ReΓ 1

87

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion

Solution: 3. Map the contour in s plane into G (s)=G(s)H(s).

jw Section BC: s   jw s plane

s 1  Γ(s)  G(s)H (s)  ss 1 D A  1 0 1 C  jw 1 Γ jw   jw  jw 1 B

2w  jw 2 1 Γ jw  w w 2 1 88

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion jw s plane Solution: 3. Map the contour in s plane into G (s)=G(s)H(s).  D A  1 0 1 C Section BC: s   jw

B 0 w   2w  ImΓ ReΓ jw  1 w 1 G plane 2  w w 1  2 w  0 B  1 A ReΓ 1 2 0 w   w 2 1  ImΓ jw  0 w 1 w w 2 1     w  0  89

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion

Solution: 3. Map the contour in s plane into G (s)=G(s)H(s).

jw Section CD: s   e j  0 s plane

(s 1) Γ(s)  G(s)H (s)   ss 1 D A  1 0 1 C j j  e 1 Γ e   e j  e j 1 B

j j(  )  Γ e  R e  2    0 90

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion jw s plane Solution:  3. Map the contour in s plane into G (s)=G(s)H(s). D A  1 0 1 C Section CD: s   e j j j(  )  0 Γ e  R e ImΓ B      0 2 G plane

 j3 / 2 B C  C    Γ  R e  1 2 A ReΓ D 1 2

 D  D   0 Γ  R e j

C 91

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion jw s plane Solution: 3. Map the contour in s plane into G (s)=G(s)H(s).  D A  ImΓ 1 0 1 C

B G plane

B

A 1 ReΓ D 1 2

C 92

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion Solution: jw 4. Find the number of poles of G(s)H(s) in s plane the right-half s plane, i.e. P. ImΓ  D A  1 0 1 C

G plane B

B

A 1 ReΓ D 1 2 P  0 

C 93

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion jw Solution: s plane

5. Count the number of clockwise encirclements 

of -1 point, i.e. N. D A  ImΓ 1 0 1 C

B

G plane

B

A 1 ReΓ D 1 2 N 1

C 94

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion jw Solution: s plane 6. Find Z  N  P which is the number of  closed-loop poles in the right-half s plane. D A  ImΓ 1 0 1 C

B G plane

B

A 1 ReΓ D 1 2 Z 1

 The system is unstable. C 95

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion

• Example: Using Nyquist stability criterion find the range of positive k in which the following system is stable

R(s) 6k C(s) + _ s 1s  2s  3

96

2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion Solution: If k 10 N  2 Z  2 Unstable system

If k 10 Critically stable system

If k  10 N  0 Z  0 Stable system

ImΓ jw 6k G plane s plane 10 P  0  B E A  E 3  2 1 0 C A ReΓ k B  D 10 D C

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2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion

• Example: Using Nyquist stability criterion find the range of positive k in which the following system is stable

R(s) 1 C(s) + _ s(s 1)

1 ks

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2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion

• Solution: The characteristic equation is expressed as

1 ks s2  s 1 ks (s) 1  0  0 s(s 1) s(s 1)

Divide by the parts without k ks 2 1  0 s  s 1  ks  0 s2  s 1

ks G(s)H (s)  R(s) 1 C(s) 2 + s  s 1 _ s(s 1)

It is the virtual loop transfer function. 1 ks

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2017 Shiraz University of Technology Dr. A. Rahideh Nyquist Stability Criterion

Important note: • To investigate the stability of system with a variable, e.g. k, using Nyquist stability criterion, the variable should be as a gain in the loop transfer function. N(s) G(s)H (s)  k D(s)

• If it is not the case, the virtual loop transfer function should be formed.

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2017 Shiraz University of Technology Dr. A. Rahideh Phase Margin & Gain Margin

1. Gain Margin (GM):

• Assume wp is the frequency in which GH ( jw p )  180

wp is called phase crossover frequency.

1 • The gain margin is obtained as GM  GH ( jw p )

• Or in the case of dB it is GM   GH ( jw ) dB p dB

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2017 Shiraz University of Technology Dr. A. Rahideh Phase Margin & Gain Margin

2. Phase Margin (PM):

• Assume wg is the frequency in which GH ( j w g )  1 or

GH ( jw )  0 g dB

wg is called gain crossover frequency.

• The phase margin is obtained as PM 180  GH ( jwg )

Phase and gain margins are useful in minimum-phase systems. 102

2017 Shiraz University of Technology Dr. A. Rahideh Phase Margin & Gain Margin

In a minimum-phase system to have stability both phase margin and gain margin in dB should be positive. i.e.

PM 180  GH ( jwg )  0 180  GH ( jwg )  0 and

GM   GH ( jw )  0 GM  GH ( jw ) 1 dB p dB p

Note that phase and gain margins cannot be used for stability analysis in non-minimum-phase systems.

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2017 Shiraz University of Technology Dr. A. Rahideh Phase Margin & Gain Margin in Polar Diagram

Consider a minimum-phase system with the following polar diagram

1 Im[GH( jw)] GM  1 GM  0 OA dB

1 GM PM PM 180  GH ( jwg )  0

w w  1 p Re[GH( jw)] A O w g C GH( jw ) g The system is stable.

w  0 104

2017 Shiraz University of Technology Dr. A. Rahideh Phase Margin & Gain Margin in Polar Diagram

In polar diagram of minimum-phase systems, moving from zero frequency to infinity frequency, if point -1 is located on the left side of the trajectory (from zero to infinity frequency), the system is stable. Im[GH( jw)]

1 GM PM

w w  1 p Re[GH( jw)] A O w g C GH( jw ) g This technique cannot be used in non-minimum-phase systems. w  0 105

2017 Shiraz University of Technology Dr. A. Rahideh Phase Margin & Gain Margin in Polar Diagram

Example: Consider the following polar diagram of a minimum-phase system. Discuss on the stability if 1) Point -1 is at point A Im[GH( jw)] 2) Point -1 is at point B 3) Point -1 is at point C

w  0 w  B Re[GH( jw)] A C O

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2017 Shiraz University of Technology Dr. A. Rahideh Phase Margin & Gain Margin in Polar Diagram

Solution: 1. A=-1 PM<0 unstable 2. B=-1 PM>0 & GM> 0 stable Im[GH( jw)] 3. C=-1 PM<0 & GM<0 unstable

w  0 w  B Re[GH( jw)] A C O

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2017 Shiraz University of Technology Dr. A. Rahideh Relative Stability using Phase Margin & Gain Margin

• Comparing two stable minimum-phase systems, the one having higher gain margin or in the case of equal gain margins, the one having higher phase margin is more stable. • In the following examples system I is more stable.

Im[GH( jw)] Im[GH( jw)] Case 1 Case 2

1 PM I GM II PM II w  w  1 Re[GH( jw)] 1 Re[GH( jw)] O O 1 II GM I II

I I w  0 w  0 108

2017 Shiraz University of Technology Dr. A. Rahideh Phase Margin & Gain Margin in Bode

R(s) Diagrams Example: Calculate the gain and phase margins from the following

Bode diagrams

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2017 Shiraz University of Technology Dr. A. Rahideh Phase Margin & Gain Margin in Bode Diagrams

Solution: Find the phase and gain crossover frequency (wp and wg)

wg  0.78 rad/s

w g

w p  2.2 rad/s

w p 110

2017 Shiraz University of Technology Dr. A. Rahideh Phase Margin & Gain Margin in Bode Diagrams Solution: Find the gain margin

w p  2.2 rad/s GH( jw ) p dB

GM   GH ( jw ) dB p dB  (16) 16 dB

w p

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2017 Shiraz University of Technology Dr. A. Rahideh Phase Margin & Gain Margin in Bode Diagrams Solution: Find the phase margin

wg  0.78 rad/s

w g PM 180  GH ( jwg )

180 137 GH( jwg )  43

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2017 Shiraz University of Technology Dr. A. Rahideh A Few Points on Phase Margin & Gain Margin

1. Gain margin of first- and second-order systems is infinity since Bode phase diagram never reaches -180 degrees. 2. Non-minimum-phase system with negative phase margin and/or negative gain margin MAY be stable. 3. In minimum-phase systems with several phase and/or gain margins, only one positive phase margin and one positive gain margin leads to stability. 4. In practice for good stability PM > 45 degrees and GM > 6 dB.

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2017 Shiraz University of Technology Dr. A. Rahideh