Frequency Response Analysis: A Review - CERTHlpre.cperi.certh.gr/auth/files/AUTh_Frequency...

Prof. Costas Kiparissides

Process Dynamics and Control

Subject: Frequency Response Analysis

Professor Costas Kiparissides

Department of Chemical Engineering

Aristotle University of Thessaloniki

December 16, 2014


Frequency Response Analysis:

A Review

16-12-2014

1


2 2

αωU(s)

s ω

p2 2

p2 2 2 2 2 2

Κ αωY(s)

(τs 1) (s ω )

K α τω ω τωs

s 1 / τ1 τ ω s ω s ω

(1)

(2)

Response of First-Order Systems

Let us assume a sine wave input function , U(t) = α sin(ωt), is introduced into a first order dynamic system. Its Laplace transform will be given by

Thus, for a first-order dynamic system, the Y(s) output deviation variable will be:


p t τ2 2

K αY ( t ) ( τω e s in ω t τω c o s ω t )

1 τ ω

(3)

ps 2 2

K αY ( t ) (s in ω t τω co s ω t )

1 τ ω

(4)

p cosω + q sinω = r sin(ω+φ) (5); r = (p2+q2)1/2 and φ = tan-1(p/q)

p 1s 2 2

K αY ( t ) s in (ω t φ ) ; φ ta n ( ω τ )

1 τ ω

(6)

Response to Sine Wave Function

The time response of the first-order dynamic system will be given by the inverse Laplace transform of Y(s), Eq. (2).

At steady-state

Using the following trigonometric properties:

Equation (4) is written:

16-12-2014

2


Time response of a first-order dynamic system to a sinusoidal input

tφ

Ys( t )

U ( t )

Μ ε τ α β α τ ι κ ή π ε ρ ίο δ ο ς

α '

α

t

U ( t )

Ys( t )

Transient Stage

tφ

Ys( t )

U ( t )

Μ ε τ α β α τ ι κ ή π ε ρ ίο δ ο ς

α '

α

t

U ( t )

Ys( t )

Transient Stage



% Frequency response of a first-order system % to a sinusoidal input

Kp = 17;

T = 2;

sys = tf([Kp],[T 1]);

a = 1;

w = 2;

t = [0 : 0.01 : 12*pi/w]';

u = a*sin(t*w);

[y,t] = lsim(sys,u,t);

figure(1)

plot(t,y,'k-',t,u,'k:')

grid on

xlabel(‘Time, sec')

ylabel('Y(t)')

title(['Time response of a first-order system '...

'to a sinusoidal input'])

pos = 0;

legend('Y(t)','U(t)',pos)


16-12-2014

3


p p

s iωs iω

Κ ΚG(iω ) G(s)

τs 1 τiω 1

(7)

2 2

pp

K (1 τiω ) 1 τiωG(iω ) K

(1 τiω )(1 τiω ) 1 τ ω

(8)

1/2

2 22 2

2 2 22 2

pp

K1 τ ωG(iω) ReG(iω) ImG(iω) K

1 τ ω1 τ ω

(9)

1 1Im G(iω )φ tan tan ( ωτ )

Re G(iω )

(10)

Frequency Response Analysis

Argument

Magnitude


2 2

2 2

p pK α 1 τ ω Kα(AR) G(iω)

α α 1 τ ω

Amplitude Ratio, (AR)

1 1φφ 2π(t T) tan (ωτ) tan ( ωτ)

Phase Shift, φ


16-12-2014

4


2 2

p

A R 1lo g lo g (1 τ ω )

K 2

(11)

ω τ ω τp p

A R A Rlim lo g 0 ; lim lo g lo g ( τω )

K K

(12)

o o

ωτ ωτlim φ 0 ; lim φ 90

(13)

and φ = tan-1(-1) = -45o if ω = ωc (14)

Asymptotic analysis of (AR) and φ with respect to ωτ.

*


0

0


Amplitude Ratio w.r.t. to ωτ

0.01 0.1 1 10 100

0.01

0.1

1

= ωcτ ωτ

Λ.Ε K

p

κλίση = -1slope = -1

AR

0.01 0.1 1 10 100

0.01

0.1

1

= ωcτ ωτ

Λ.Ε K

p

κλίση = -1slope = -1

AR

Bode Plot

�

16-12-2014

5


0.01 0.1 1 10 100-90

-45

0

o

o

o

ωτ

φ

Phase Shift w.r.t. to ωτ

Bode Plot


Kp = 17 ;

T = 2 ;

sys = tf([Kp],[T 1]) ;

w = logspace(log10(10^(-2)/T),log10(10^2/T)) ;

[mag,phase] = bode(sys,w) ;

AR = zeros(1,length(mag)) ;

for j = 1:1:length(mag)

AR(1,j) = mag(:,:,j) ;

end

figure(1)

loglog(w*T,AR/Kp,'k-')

ylim([0.01 10])

grid on

xlabel('\omega\tau')

ylabel(‘AR / K_p')

title(['Plotting Frequency Response'])

phi = zeros(1,length(phase)) ;

for i = 1:1:length(phase)

phi(1,i) = phase(:,:,i) ;

end

figure(2)

semilogx(w*T,phi,'k-')

grid on

xlabel('\omega\tau')

ylabel('\phi (^o)')

title(['Plotting Frequency Response'])

Bode Plots for a First-order System

16-12-2014

6


2 2 1 2pΚ (AR)(1 ω τ ) ; τ tan(φ ) ( ω )

Solution

Angular Frequencyω (rad/sec)

Amplitude RatioAR=|G(iω)|

Phase Shiftφ

5.00 0.1 ‐11.30o

4.80 0.2 ‐21.80o

3.80 0.4 ‐38.65o

2.60 0.8 ‐57.99o

2.30 1.0 ‐63.43o

1.25 2.0 ‐75.96o

0.63 4.0 ‐82.87o

0.40 6.0 ‐85.23o

Example: Parameter Estimation

Estimate the parameters (Kp and τ) of a first-order process, from the following frequency response data.


0 .1 1 1 00 .1

1

0 .5

(α )(Λ .Ε )

ω τ

(AR)(a)

0 .1 1 1 00 .1

1

0 .5

(α )(Λ .Ε )

ω τ

(AR)(a)

0.1 1 10-90

-45

0

(β )

o

o

o

ω

φ(b)

0.1 1 10-90

-45

0

(β )

o

o

o

ω

φ(b)

Example: Parameter Estimation

Amplitude Ratio w.r.t. to ωτ

Phase Shift w.r.t. to ωτ

T = 1/ωc =1/0.5 = 2 min

At ωτ →oA.R = Kp = 5

16-12-2014

7



Let us assume a sine wave input function , U(t) = α sin(ωt), is introduced into a second order dynamic system. The Y(s) response of the system will be given by

tφ

Ys( t )

U ( t )

Μ ε τ α β α τ ι κ ή π ε ρ ί ο δ ο ς

α '

α

t

U ( t )

Ys( t )

Transient Stage

tφ

Ys( t )

U ( t )

Μ ε τ α β α τ ι κ ή π ε ρ ί ο δ ο ς

α '

α

t

U ( t )

Ys( t )

Transient Stage

Response of a second order system to a sinusoidal input

Y sK

s s

A

s

p( )

2 2 2 22 1



16-12-2014

8


1/2

2 22 2

2 2 22 2

pp

K1 τ ωG(iω) ReG(iω) ImG(iω) K

1 τ ω1 τ ω

1 1Im G(iω )φ tan tan ( ωτ )

Re G(iω )


Argument (Phase Shift)

Magnitude (Amplitude Ratio)

Where G(iω) = G(s)|s=iω is a complex transfer function


Steady-state Response to a Sine Wave

p p2 2 2 2s iω

s iω

K KG(iω) G(s)

τ s 2ζτs 1 1 τ ω 2ζτiω

(15)

G(iω) is a complex function. Its magnitude will be equal to the ratio of the output response amplitude over the respective valueof the input signal while its argument is equal to the phase shift.

or2 2

p 2 2 2 2

(1 τ ω ) i(2ζτω)G(iω) K

(1 τ ω ) (2ζτω)

(16)

This part describes the steady-state response of a second order system to a sinusoidal input function. It is obtained from the system’s transfer function by substituting the Laplace variable s in the TF with iω.

p

2 2 2 2

KG(iω )

(1 τ ω ) (2ζτω )

(17)

12 2

2τζωφ tan

1 τ ω

(18)

16-12-2014

9


Thus, for a sine wave input, the steady state response of the system is described by the following equation:

ps 2 2 2 2

K αΥ (t) G(iω) α sin(ωt φ) sin(ω t φ)

(1 τ ω ) (2τζω)

(19)

Remarks:

It is clear from Eq. (19) that a sinusoidal input (U(t)=αsinωτ) produces a sinusoidal output of the same frequency as the input.

Amplitude Ratio (AR):

The phase shift of the output lags behind that of the input by φ (Eq. 18).

p

2 2 2 2

K α '(AR) G(iω)

α (1 τ ω ) (2ζτω)

(20)

S.S Response to a Sine Wave


Asymptotic analysis of (AR) and φ with respect to ωτ.

2 2 2 2

p

AR 1log log (1 ω τ ) (2ζτω)

K 2

(21)

ωτ 0 ωτp p

AR ARlim log 0 ; lim log 2 log(ωτ )

K Κ

(22)

o

o

o

If τω 0 then tanφ 2ζωτ ; φ 0

If τω 1 then tanφ ; φ 90

2ζIf 1<τω then tanφ ; φ 180

ωτ


16-12-2014

10


Figure: Frequency response plots for a second-order system. a) (AR/Kp) w.r.t. ωτ and b) φ w.r.t. ωτ

Bode Plots

0.1 1 10-180

-135

-90

-45

0

(β)

2,0

0,8

0,50,3

ζ = 0,1

o

o

o

o

o

ωτ

φ

0.1 1 100.01

0.1

1

10

(α)

2,0

0,8

0,5

0,3

ωτ

Λ.Ε.Κ

p

ζ = 0,1

p

AR

K

0.1 1 100.01

0.1

1

10

(α)

2,0

0,8

0,5

0,3

ωτ

Λ.Ε.Κ

p

ζ = 0,1

p

AR

K

(a)

(b)


The amplitude ratio plot exhibits a maximum for certain values of ζ. We can calculate this value by differentiating Eq. (20) with respect to ωτ and setting the result to zero:

Note: As ζ0, the

2 2 2 2 2 2 2 2

d 10

d(ωτ ) 1 (ω τ ) 2ω τ 4ζ ω τ

(23)

2max(ωτ) 1 2ζ ; ζ 0.707 (24)

2p m ax

AR 1

K 2ζ 1 ζ

(25)

p

A R

K


16-12-2014

11


Ch

apte

r 4



% Frequency Response of a Second-Order System to a Sinusoidal Input

Kp = 1; T = 1;

z = 0.2;

sys = tf([Kp],[T^2 2*z*T 1]);

a = 1;

w = 2;

t = [0 : 0.01 : 24*pi/w];

u = a*sin(t*w);

[y,t] = lsim(sys,u,t);

figure(1)

plot(t,y,'k-',t,u,'k:')

grid on

xlabel(['t/\tau'])

ylabel('Y(t)')

title(['Time response of a second-order system to a sinusoidal input'])pos = 0;

legend('Y(t)','U(t)',pos)

MATLAB: Response to Sine Wave

16-12-2014

12


% BODE Plots: Second-Order SystemKp = 17 ; T = 1 ; z = 0.2 ; sys = tf([Kp],[T^2 2*z*T 1]) ;w = logspace(log10(10^(-1)/T),log10(10^1/T),200) ;[mag,phase] = bode(sys,w) ; AR = zeros(1,length(mag)) ;for j = 1:1:length(mag)

AR(1,j) = mag(:,:,j) ; endfigure(1)loglog(w*T,AR/Kp,'k-')ylim([0.01 10])grid onxlabel('\omega\tau')ylabel('AR / K_p') title(['Amplitude ratio AR / K_p w.r.t. \omega\tau'])phi = zeros(1,length(phase)) ;for i = 1:1:length(phase)

phi(1,i) = phase(:,:,i) ; endfigure(2)semilogx(w*T,phi,'k-') grid onxlabel('\omega\tau') ylabel('\phi (^o)')'\upsilon \omega\tau']) title(['Phase Shift \phi w.r.t. \omega\tau'])

MATLAB: Response to Sine Wave


Response of “n” First-Order Systems

% Sinusoidal Response of “n” First-Order Systems in Series

Kd = 1 ;n = 5 ; % number of dynamic systems in seriesT = 1/n ;v1 = [T 1] ;v2 = [T 1] ;for i = 1:(n-1)

v2 = conv(v1,v2) ;endsys = tf([Kd^n],v2) ; w = logspace(log10(10^(-1)/T),log10(10^2/T),200) ;[mag,phase] = bode(sys,w) ; AR = zeros(1,length(mag));for j = 1:1:length(mag)

AR(1,j) = mag(:,:,j) ; endfigure(1)loglog(w*T,AR/(Kd^n),'k-')xlim([0.1 100]) ylim([0.01 1])grid onxlabel('\omega\tau')ylabel('AR / K_p') title(['Amplitude ratio, AR/K_p w.r.t. \omega\tau'])phi = zeros(1,length(phase));for i = 1:1:length(phase)

phi(1,i) = phase(:,:,i);endfigure(2) semilogx(w*T,phi,'k-') grid onxlabel('\omega\tau') ylabel('\phi (^o)') title(['Phase Shift, \phi w.r.t. \omega\tau'])

16-12-2014

13


Figure : Frequency Response Plots for “n” Non-Interacting First-Order Systems in Series.a) (AR/Kp) w.r.t. ωτandb) φ w.r.t. ωτ

Bode Diagram

0.1 1 10 1000.01

0.1

1

n = 1 n = 2 n = 5

ωτ

Gd(iω)

Κd

n

0.1 1 10 100

-400

-300

-200

-100

0

n = 1 n = 2 n = 5

o

o

o

o

o

ωτ

φ

AR

0.1 1 10 1000.01

0.1

1

n = 1 n = 2 n = 5

ωτ

Gd(iω)

Κd

n

0.1 1 10 100

-400

-300

-200

-100

0

n = 1 n = 2 n = 5

o

o

o

o

o

ωτ

φ

AR



and Control System Design

16-12-2014

14


Advantages:

1. Applicable to dynamic models of any order (including non-polynomials).

2. Designer can specify desired closed-loop response characteristics.

3. Information on stability and sensitivity/robustness is

provided.

Disadvantage:

The approach tends to be iterative and time-consuming

(interactive computer graphics desirable, MATLAB)

Cha

pter

13



Some facts for complex number theory:

For a complex number:

It follows that where

such that


w a bj a w b w Re( ), Im( )

a w b w cos( ), sin( )

w w w Re( ) Im( )2 2

arg( ) tanIm( )

Re( )w

w

w1

w we j

Re

Im

w

a

b

16-12-2014

15



For a general transfer function

Frequency Response summarized by

where is the modulus of G(jω) and j is the

argument of G(jω)

Note: Substitute for s=jω in the transfer function.

G sr sq s

e s z s zs p s p

sm

n( )

( )( )

( ) ( )( ) ( )

1

1

G j G j e j( ) ( )

G j( )


The response of any linear process G(s) to a sinusoidal input is a sinusoidal.

The Amplitude Ratio of the resulting signal is given by the modulus of the transfer function model expressed in the frequency domain, G(jω).

The Phase Shift is given by the argument of the transfer function model in the frequency domain.

i.e.


AR G j G j G j

G jG j

( ) Re( ( )) Im( ( ))

tanIm( ( ))Re( ( ))

2 2

1Phase Angle

16-12-2014

16



n To study frequency response, we use two types of graphical representations.

1. The Bode Plot:

Plot of AR vs. ω on log-log scale

Plot of φ vs. ω on semi-log scale

2. The Nyquist Plot:

Plot of the trace of G(jω) in the complex plane

n Plots lead to effective stability criteria and frequency-based design methods


Frequency Response: Examples

1. Pure Capacitive Process G(s)=1/s

2. Dead Time G(s) = e-ϴs

G jKj

jj

Kj( )

ARK K

, tan/1

0 2

G j e j( )

AR 1,

16-12-2014

17


Bode Plots

10-2

10-1

100

100

101

102

AR

10-2

10-1

100

-91

-90.5

-90

-89.5

-89

Frequenc y (rad/s ec )

Pha

se A

ngle

Pure Capacitive Process

ARK

2


Cha

pter

13

Bode Plots for First-Order System

16-12-2014

18


Cha

pter

13

Bode Plots for Time-Delay System


Frequency Response of Complex Systems

3. n process in series

Frequency response of G(s)

Therefore,

G s G s G sn( ) ( ) ( ) 1

G j G j G j

G j e G j e

nj

nj n

( ) ( ) ( )

( ) ( )

1

11

AR G j G j

G j G j

ii

n

ii

ni

i

n

( ) ( )

arg( ( )) arg( ( ))

1

1 1

16-12-2014

19


4. n first order processes in series

5. First order plus delay

G sKs

Ksn

n( )

1

1 1 1

ARK Kn

n

n

1

12 2 2 2

11

1

1 1

tan tan

G sK e

sp

s

( )

1

ARKp

( ), tan ( )

1

1 2 21

Frequency Response of Complex Systems


Use a Bode plot to illustrate frequency response.

Plot of log |G| vs. log and vs. log

1 2 3

1 2 3

1 2 3

1 2 3

1

2

1 2

1 2

l o g l o g l o g l o g

l o g l o g l o g

G G G G

G G G G

G G G G

G G G G

GG

G

G G G

G G G

Cha

pter

13

Bode Plots for Complex T. Functions

16-12-2014

20


Bode Plots

1 0- 4

1 0- 3

1 0- 2

1 0- 1

1 00

1 01

1 0- 4

1 0- 2

1 00

1 0- 4

1 0- 3

1 0- 2

1 0- 1

1 00

1 01

- 3 0 0

- 2 0 0

- 1 0 0

0

G s G s G s G s( ) ( ) ( ) ( ) 1 2 3

G ss

G ss

G ss1 2 3

110 1

15 1

11

( ) , ( ) , ( )

G j( )( )( )( )

tan ( ) tan ( ) tan ( )

1

1 10 1 5 1 1

10 5

2 2 2 2 2 2

1 1 1

G1

G2

G3

GAR


Bode Plots

1 0- 4

1 0- 3

1 0- 2

1 0- 1

1 00

1 01

1 0- 4

1 0- 2

1 00

1 0- 4

1 0- 3

1 0- 2

1 0- 1

1 00

1 01

- 3 0 0

- 2 0 0

- 1 0 0

0

G=Gd

GGd

G s e s( )

G j( ) , 1

G s e G s G s G sds( ) ( ) ( ) ( ) 2

1 2 3

AR

16-12-2014

21


Cha

pter

13

Example

0.55(0.5 1)( )

(20 1)(4 1)

ss eG s

s s

Bode Plots


Recall that the frequency response is characterized by

1. Amplitude Ratio (AR)

2. Phase Angle ()

For any transfer function, G(s)

Proportional Controller

( )

( )

A R G j

G j

( ) , 0C C CG s K AR K

Cha

pter

13

Frequency Response of P Controller

16-12-2014

22


The Bode plot for a PI controller is shown in next slide.

Note b = 1/I . Asymptotic slope ( 0) is -1 on log-log plot.

PI Controller

2 2

1

1 1( ) 1 1

1tan

C C CI I

I

G s K AR Ks

Cha

pter

13

Frequency Response of PI


Frequency Response of PI

AR KcI

I

11

1

2 2

1

tan ( / )

1 0- 3

1 0- 2

1 0- 1

1 00

1 01

1 00

1 01

1 02

1 03

1 0- 3

1 0- 2

1 0- 1

1 00

1 01

-1 0 0

-8 0

-6 0

-4 0

-2 0

0

AR

16-12-2014

23


Series PID Controller. The simplest version of the series

PID controller is

Series PID Controller with a Derivative Filter.

τ 1τ 1 (14-50)

τI

c c DI

sG s K s

s

τ 1 τ 1(14-51)

τ ατ 1I D

c cI D

s sG s K

s s

Cha

pter

13

Ideal PID Controller.

1( ) (1 ) (14 48)c c D

I

G s K ss

Frequency Response of PID


Frequency Response of PID

AR Kc DI

DI

11

1

2

1tan

1 0- 3

1 0- 2

1 0- 1

1 00

1 01

1 00

1 01

1 02

1 03

1 0- 3

1 0- 2

1 0- 1

1 00

1 01

-1 0 0

-5 0

0

5 0

1 0 0

AR

16-12-2014

24


Bode plots of ideal parallel PID controller and series PID controller with derivative filter (α = 1).

Ideal parallel:

Series with Derivative Filter:

10 1 4 12

10 0.4 1cs s

G ss s

12 1 4

10cG s ss

C

hapt

er 1

3Bode Plots of PID


Nyquist Plots

Plot of G(jω) in the complex plane as ω is varied

Relation to Bode plot

AR is distance of G(jω) for the origin

Phase angle, φ , is the angle from the real positive axis

Example: First order process (K=1, τ=1)

G j( )

16-12-2014

25


Nyquist Plots

Dead-time

Second Order


Nyquist Plots

Third Order

Effect of dead-time (second order process)

G ss s s

( )

1

3 3 13 2

G ss s

Gd s e s( ) , ( )

12 3 1

2

16-12-2014

26


Analyze GOL(s) = GCGVGPGM (open loop gain)

Three methods in use:

(1) Bode plot |G|, vs. (open loop F.R.) - Chapter 13

(2)Nyquist plot - polar plot of G(j)

(3)Nichols chart |G|, vs. G/(1+G) (closed loop F.R.)

Advantages:• Do not need to compute roots of characteristic equation• Can be applied to time delay systems• Can identify stability margin, i.e., how close you are to instability.

Cha

pter

13

Controller Design in Frequency Domain


Ch

apte

r 13

Sustained Oscillations in FB Control

16-12-2014

27


1. Bode Stability Criterion2. Nyquist Stability Criterion

Bode Stability Criterion:

A closed-loop system is unstable if the FR of the open-loop T.F

GOL=GCGPGVGM, has an amplitude ratio greater than one at the

critical frequency, . Otherwise, the closed-loop system is stable.

• Note: where the open-loop phase angle is –1800.

• The Bode Stability Criterion provides info on closed-loop stability from open-loop FR info.

C

value of C Cha

pter

13

Frequency Stability Criteria


Bode Stability Criterion

“A closed-loop system is unstable if the frequency of the response of the open-loop GOL has an amplitude ratio greater than one at the critical frequency. Otherwise it is stable. “

Strategy:

1. Solve for ω in

2. Calculate AR

arg( ( ))G jOL

AR G jOL ( )

16-12-2014

28


Bode Stability Criterion Check for stability:

1. Compute open-loop transfer function

2. Solve for ω in φ=- π

3. Evaluate AR at ω

4. If AR>1 then process is unstable

Find ultimate gain:

1. Compute open-loop transfer function without controller gain

2. Solve for ω in φ=- π

3. Evaluate AR at ω

4. Let KARcu 1


For proportional-only control, the ultimate gain Kcu is defined to be the largest value of Kc that results in a stable closed-loop system.

For proportional-only control, GOL= KcG and

G = GvGpGm.

AROL(ω)=Kc ARG(ω) (14-58)

where ARG denotes the amplitude ratio of G.

At the stability limit, ω = ωc, AROL(ωc) = 1 and Kc= Kcu.

1(14-59)

(ω )cuG c

KAR

Cha

pter

13

Frequency Stability Analysis

16-12-2014

29


Example 1: Bode Stability Criterion Consider the transfer function and controller

- Open-loop transfer function

- Amplitude ratio and phase shift

- At ω=1.4128, φ=-π, AR=6.746

G se

s s

s( )

( )( . )

.

51 05 1

01G s

sc ( ) ..

0 4 11

01

G se

s s sOL

s( )

( )( . ).

.

.

5

1 05 10 4 1

1

01

01

AR

5

1

1

1 0 250 4 1

1

0 01

01 051

01

2 2 2

1 1 1

..

.

. tan ( ) tan ( . ) tan.


A process has a transfer function,

And GV = 0.1, GM = 10 . If proportional control is used, determine

closed-loop stability for 3 values of Kc: 1, 4, and 20.

Solution:

The OLTF is GOL=GCGPGVGM or...

The Bode plots for the 3 values of Kc shown in Fig. 13.9.Note: the phase angle curves are identical. From the Bode diagram:

KC AROL Stable?1 0.25 Yes4 1.0 Conditionally stable20 5.0 No

3

2( )

(0.5 1)C

OL

KG s

s

3

2( )

(0.5 1)pG ss

Example 2: Bode Stability Analysis

16-12-2014

30


Bode plots for GOL = 2Kc/(0.5s + 1)3.

Cha

pter

13

Frequency Stability Analysis


Determine the closed-loop stability of the system,

Where GV = 2.0, GM = 0.25 and GC =KC . Find C from the Bode

Diagram. What is the maximum value of Kc for a stable system?

Solution:

The Bode plot for Kc= 1 is shown in Fig. 13.11.

Note that:

15

4)(

s

esG

s

p

OL

max

1.69 rad min

0.235

1 1= 4.25

0.235

C

C

COL

AR

KAR

Cha

pter

13

Example 3: Frequency Stability Analysis

16-12-2014

31


Cha

pter

13

Bode Plot for Kc = 1


Nyquist Stability Criterion

“If N is the number of times that the Nyquist plot encircles the point (-1,0) in the complex plane in the clockwise direction, and P is the number of open-loop poles of GOLthat lie in the right-half plane, then Z=N+P is the number of unstable roots of the closed-loop characteristic equation.”

Strategy

1. Substitute s=jω in GOL(s)

2. Plot GOL(jω) in the complex plane

3. Count encirclements of (-1,0) in the clockwise direction

16-12-2014

32


Nyquist Stability Criterion

Consider the transfer function

and the PI controller

G se

s s

s( )

( )( . )

.

51 05 1

01

G ssc ( ) .

.

0 4 11

01


Ultimate Gain: KCU = maximum value of |KC| that results in a stable closed-loop system when proportional-only control is used.

Ultimate Period:

KCU can be determined from the OLFR when proportional-only control is used with KC =1. Thus

Note: First and second-order systems (without time delays) do not have a KCU value if the PID controller action is correct.

2U

C

P

1for 1

C

CU COL

K KAR

Ultimate Gain and Ultimate Period

16-12-2014

33


• The gain margin (GM) and phase margin (PM) provide measures of how close a system is to a stability limit.

• Gain Margin:Let AC = AROL at = C. Then the gain margin is

defined as: GM = 1/AC

According to the Bode Stability Criterion, GM >1 stability

• Phase Margin:Let g = frequency at which AROL = 1.0 and the corresponding phase angle is g . The phase margin is defined as: PM = 180° + g

According to the Bode Stability Criterion, PM >0 stability

Cha

pter

13

Gain and Phase Margins


Cha

pter

13

Gain and Phase Margins

16-12-2014

34


Rules of Thumb:

A well-designed FB control system will have:

Closed-Loop FR Characteristics:

An analysis of CLFR provides useful information about control

system performance and robustness. Typical desired CLFR for

disturbance and setpoint changes and the corresponding step

response are shown in Appendix J (see Text).

1.7 2.0 30 45GM PM

Cha

pter

13

Design Gain and Phase Margins


Amplitude ratio plot.fig

10-2

10-1

100

10-2

10-1

100

Bode Plot : Amplitude Ratio

16-12-2014

35


10-2

10-1

100

-300

-250

-200

-150

-100

-50

0

Bode Plot : Phase Angle


Stability Considerations

n Control is about stability

n Considered exponential stability of controlled processes using: Roth criterion Direct Substitution Root Locus Bode Criterion (Restriction on phse angle) Nyquist Criterion

n Nyquist is most general but sometimes difficult to interpret

n Roots, Bode and Nyquist all in MATLAB

n MAPLE is recommended for some applications.

16-12-2014

36

Frequency Response Analysis: A Review - CERTHlpre.cperi.certh.gr/auth/files/AUTh_Frequency...

Documents

Transcript of Frequency Response Analysis: A Review - CERTHlpre.cperi.certh.gr/auth/files/AUTh_Frequency...