lecture 02

054414 PROCESS CONTROL SYSTEM DESIGNLECTURE TWO

Daniel R. Lewin, Technion1

Frequency Domain AnalysisPROCESS CONTROL SYSTEM DESIGN - (c) Daniel R. Lewin1-2

054414 Process Control System Design

LECTURE 2: FREQUENCY DOMAIN ANALYSIS

Daniel R. LewinDepartment of Chemical Engineering

Technion, Haifa, Israel


Objectives

n Draw Bode and Nyquist plots of an arbitrary SISO linear system Both by hand and using MATLAB

o Sketch the temporal response of a SISO linear system, given its Bode plot Both by hand and using MATLAB/SIMULINK

p Determine an appropriate transfer function given the Bode plot of a SISO linear system. Both by hand and using MATLAB

On completing this section, you should be able to:




Frequency Response

P(s)( )Usin t ( )Ysin t + ( ) ( )s ip s p i=

( )Amplitude ratio, AR = Y U p i= ( ){ } ( ){ }( ){ }1

Im p iPhase shift, arg p i tan

Re p i = =


Frequency Response Example Response of dead sea pond to cyclic perturbation:

( ) ( ) ( )dc 0.05c 0.02E,c 0 0;E t 3sin t , rad/hdt 12

= + = = = Solution: ( ) ( )( )

( )( )

= + = = =

= =

0.05tc t 0.0225e 0.0229 sin t 1.383p 12 0.0229 3 0.0076

12 1.383 rad (phase lag)




Frequency Response 1o System

( )( ) ( ){ } ( ){ }( ){ } ( )

p

2 2

1 1

KAR p i

1Im p i

arg p i tan tanRe p i

= = + = = =

( ) pKp ss 1

= + The ultimate response has the following characteristics:

( )( )( )

pK 1 i1 i 1 i

+ ( )pKp i

i 1 = +

( )( )p 2 2K 1 i

1 +

Example (Dead Sea Pond):( ) 0.002 0.04p s

s 0.05 20s 1= =+ +

( )( ) ( )

2

1

0.04AR p i . For 12,AR 0.00761 400

tan 20 . For 12, 1.383= = = =+

= = = 9 9


Freq. Response Plots 1o System Bode Plot

Log

(AR)

Pha

se

In the Bode magnitude plot, log (AR) is plotted against log ()

In the Bode phase plot, Phase (in rad or degrees) is plotted against log ()

( ) ( )1tan =

( )2

1p i1

= + AR(0.01) = 1.000AR(0. 10) = 0.995

AR(1.00) = 0.707AR(10.0) = 0.099

AR(100) = 0.010

(0.01) = -0.57o(0.10) = -5.71o

(1.00) = -45o(10.0) = -84.2o

(100) = -89.4o

( ) 1p ss 1

= +




Freq. Response Plots 1o System Log

(AR)

Pha

se

pp2 20

Klim K

1=+ p

p2 2

Klim K

1= +

( )1 olim tan 90

= ( )1 o

0lim tan 0

=

Asymptotes

Asymptotes join at breakpointlocated at = 1/ rad/min

Asymptotes join at break point

Bode Plot( ) 1p ss 1

= +

-1

Gradient of high frequencyasymptote is -1 on a log-log scale


Freq. Response Plots 1o System

w=logspace(-2,2,200);s=i*w;p=1./(s+1);AR=abs(p);ph=180*phase(p)/pi;subplot(2,1,1)loglog(w,AR,'-r)subplot(2,1,2)semilogx(w,ph,'-r')

Generating Bode Plots with MATLAB




Freq. Response Plots 1o System Nyquist Plot In the Nyquist plot, p(i),

which is a complex number, is plotted directly, as a locus from = 0, to = .

-0.571.0000.01

-89.40.010100-84.20.09910.0-45.00.7071.00-5.710.9950.10

(o)AR

w=logspace(-2,2,200);s=i*w;p=1./(s+1);plot(p)


Frequency Response 1o System Nyquist Plot

Bode Plot




Frequency ResponseIntegrator

( )( ) ( ){ } ( ){ }( ){ } ( )

p

1 1

KAR p i

Im p iarg p i tan tan

Re p i 2

= = = = = =

( ) pKp ss

=

The ultimate response has the following characteristics:

( )pK i

i i ( )

pKp ii 1

= +piK


Frequency ResponseIntegrator Nyquist Plot

Bode Plot




Frequency Response Interpretation

Consider two first order systems: ( ) ( )1 21 1p s and p s5s 1 s 1= =+ +These two systems are excited by a series of steps (a square wave, approximately) as shown below.

Note that p1 responds more sluggishly than p2, because its time constant is 5 times larger.

Note also that the exciting signal is only approximately a square wave...


Frequency Response InterpretationThis is because it is actually made up of a series of harmonic terms. In fact, it can be shown that a true square wave can be expressed as the infinite series:

( ) ( )( )( ) ( ) ( ) ( )i 1 1 13 5sin 2 i 1 1 t

u t sin t sin 3t sin 5t2 i 1 1

=

+ = = + + + + More accurate approximations of the true square wave are obtained by using more terms in the above expansion.

Any signal can be expressed as a Fourier expansion of the form:

( ) ( ) ( )u t sin t

= where () is the amplitude of the signal at .





( ) ( ) ( ) ( ) ( )7 1 1 13 5 13u t sin t sin 3t sin 5t sin 13t= + + + +Lets display the magnitudes of the terms in the expansion to seven terms:

on the Bode magnitude plot. This is equivalent to a step (integrator)



Equivalent to the square wave (step)






Frequency Response Lead/Lag

( ) ( )( ) ( ){ } ( ){ }( ){ } ( )( ) ( ){ } ( ){ }( ){ } ( )

2 21 2 p

11 11 1

1

21 12 2

2

AR p i p i K 1Im p i

arg p i tan tanRe p iIm p i

arg p i tan tanRe p i

= = = + = = = = = =

( ) ( )1 pp s K s 1= +


( ) ( )1 pp i K i 1 = +( ) ( )2 pp s K s 1= + ( ) ( )2 pp i K i 1 = +




Frequency Response Lead/Lag

Nyquist Plot Bode Plot

+1

Gradient of high frequencyasymptote is +1 on a log-log scale



( ) ( ) ( )( ) ( ){ } ( ){ }( ){ }

p

2 22 2

1 12 2

KAR p i

1 2Im p i 2arg p i tan tanRe p i 1

= = +

= = =

( ) p2 2 Kp s s 2 s 1= + +


( ) ( )p2 2K

p i1 i2

= +

( ) ( )( )( )( ) ( )( )2 2

p

2 2 2 2

K 1 i2p i

1 i2 1 i2

= + ( ) ( )( ) ( )

( )( ) ( )

2 2p p

2 22 22 2 2 2

K 1 K 2p i i

1 2 1 2

= + +





( )p 2 2

p22 2

Klim K

1=

+ Bode Plot

1 o2 2

2lim tan 1801

=

-2

Gradient of high frequencyasymptote is -2 on a log-log scale


Frequency Response Complex p(s)

On the Bode log AR plot, the amplitudes of each component add upadd up:

To construct asymptotes in Bode plots for:

( ) ( ) ( ) ( )( ) ( ) ( )1 2 m sp 1 2 na s 1 a s 1 a s 1

p s K eb s 1 b s 1 b s 1

+ + += + + +

( ) p 1 21 2

log p s logK log a s 1 log a s 1 log b s 1 log b s 1

= + + + + + + +

On the Bode linear phase plot scale, the phase of each component add upadd up: ( ){ } { } { }

{ } { } = = + + + +

+ +

1 2

1 1

arg p s arg a s 1 arg a s 1 arg b s 1 arg b s 1 s

For large , log AR() vs. log has an asymptotic slope of (n m)

For large , for MP p(s) [no positive zeros or delay terms], approaches asymptotic value of (/2)(n m)




Frequency Response MP/NMP MP = Minimum phase, NMP = Non-minimum phase.

RHP zeros: It is convenient to factor the RHP zero with its LHP mirror image: . The AR of this component is AR = 1, which has a phase lag of zero. In contrast, for large , pz has a phase lag of:

( ) ( ) ( )zp s zs 1 zs 1= + +

( ){ } { } { }zlim arg p s lim arg zs 1 lim arg zs 12 2

= + += =

NMP systems are those that feature phase lags greater than anticipated based on the systems AR alone. Those whose phase lag corresponds to the systems AR are MP systems. NMP components are either: (a) Right-half plane (RHP) zeros, or (b) Dead time

Dead time: The phase lag of is . The AR of this component is AR = 1, which has a phase lag of zero.

( ) sdp s e=


Class Exercise 1 - Sketch p(i) Generate a Bode plot for the following transfer function:

( ) ( )0.5s

2s 1p s e

10s 1+= +

Solution.

( )0.5s

21s 1, and e

10s 1+ +

The Bode plot is plotted by combining the contributions of the components:




Class Exercise 1 - Sketch p(i)Solution (Contd).

( ) ( )0.5s

2s 1p s e

10s 1+= +

s 1+

0.5se

( )21

10s 1+

s 1+

( )21

10s 1+

-0.2870.01

-28710.0-28.71.00-2.870.10

(e-0.5s)


Class Exercise 2 - Determine p(s)

Determine the transfer function, p(s) for the process whose Bode diagram is given above.




Class Exercise 2 - Solution

1 zero @ = 0.1

2 poles @ = 1

Kp = 1

( ) ( )( )( )

( )+ += + +2 2

10s 1 10s 1p s or s 1 s 1


Class Exercise 2 - Solution

High asymptote -270o (a) No delay

(b) NMP system

( ) ( )( )( )

( )+ += + +2 2

10s 1 10s 1p s or s 1 s 1




Summary

n Draw Bode and Nyquist plots of an arbitrary SISO linear system Compute each component separately and combine in

Bode plot

o Sketch the response of a SISO linear system, given its Bode plot Based on interpretation of frequency response

p Determine an appropriate transfer function given the Bode plot of a SISO linear system. Uses skills developed in n

On completing this section, you should be able to:

lecture 02

Documents