Topic #5 - dspace.mit.edu · 9/13/2007 · Fall 2007 16.31 5–3 Performance Issues II • Classic...

Topic #5

16.31 Feedback Control Systems

Control Design using Bode Plots

• Performance Issues

• Synthesis

• Lead/Lag examples

Cite as: Jonathan How, course materials for 16.31 Feedback Control Systems, Fall 2007. MIT OpenCourseWare(http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

Fall 2007 16.31 5–1

Bode’s Gain Phase Relationship

• Control synthesis by classical means would be very hard if we hadto consider both the magnitude and phase plots of the loop, but

that is not the case.

• Theorem: For any stable, minimum phase system G(s), �G(jω)is uniquely related to |G(jω)|.

• The relationship is that, on a log-log plot, if the slope of themagnitude plot is constant over a decade in frequency, with slope

n, then

�G(jω) ≈ 90◦n

• So in the crossover region, where L(jω) ≈ 1 if the magnitude plotis (locally):

s0 slope of 0, so no crossover possible

s−1 slope of -1, so about 90◦PM

s−2 slope of -2, so PM very small

• Basic rule of classical control design:Select Gc(s) so that the LTF crosses over with a slope of -1.

September 13, 2007


Fall 2007 16.31 5–2

Performance Issues

• Step error response

ess =1

1 +Gc(0)Gp(0)

and we can determine Gc(0)Gp(0) from the low frequency Bode

plot for a type 0 system.

• For a type 1 system, the DC gain is infinite, but define

Kv = lims→0

sGc(s)Gp(s) ⇒ ess = 1/Kv

– So can easily determine this from the low frequency slope of

the Bode plot.

September 13, 2007


Fall 2007 16.31 5–3

Performance Issues II

• Classic question: how much phase margin do we need? Timeresponse of a second order system gives:

1. Closed-loop pole damping ratio ζ ≈ PM/100, PM < 70◦

2. Closed-loop resonant peak Mr =1

2ζ√

1−ζ2≈ 12 sin(PM/2), near

ωr = ωn�

1− 2ζ2

3. Closed-loop bandwidth

ωBW = ωn

�1− 2ζ2 +

�2− 4ζ2 + 4ζ4

and ωc = ωn

��1 + 4ζ4 − 2ζ2

Figure 29: Frequency domain performance specifications.

September 13, 2007

Band widthωBW

10

10.7

Amplituderatio|y|/|r|

0.1

20

0

-3

-20

Decibels

ω,

ωcMr

rad/s

resonant peak

Figure by MIT OpenCourseWare.


Fall 2007 16.31 5–4

Figure 30: Crossover frequency for second order system

figure(1)%z=[0:.01:1]’; zs=[0:.01:1/sqrt(2)-eps]’;wbw=sqrt(1-2*z.^2+sqrt(2-4*z.^2+4*z.^4));%wc=sqrt(sqrt(1+4*z.^4)-2*z.^2);%wr=sqrt(1-2*zs.^2);%set(gcf,’DefaultLineLineWidth’,2) %plot(z,wbw,z,wc,zs,wr)%legend(’\omega_{BW}/\omega_n’,’\omega_c/\omega_n’,’\omega_r/\omega_n’) %xlabel(’\zeta’);ylabel(’\omega’)%print -dpng -r300 fresp.png

• So typically specify ωc, PM, and the error constant as the designgoals

September 13, 2007


Fall 2007 16.31 5–5

Frequency Domain Design

• Looked at the building block

Gc(s) = Kcs + z

s + p

• Question: how chooseGc(s) to modify the LTF L(s) = Gc(s)Gp(s)to get the desired bandwidth, phase margin, and error constants?

• Lead Controller (|z| < |p|)

– Zero at a lower frequency than the pole

– Gain increases with frequency (slope +1)

– Phase positive (i.e. this adds phase lead)

Figure 31: Lead: frequency domain.

September 13, 2007

ω

ω

θ

ω pz

-90o

0o

00

20

020 log |p/z|

dB

Phase



Fall 2007 16.31 5–6

Lead Mechanics

• Maximum phase added

sinφmax =1− α1 + α

where α = |z|/|p|, which implies that

α =1− sinφmax1 + sinφmax

• Frequency of the maximum phase addition is ωmax =�|z| · |p|

– Usually try to place this near ωc

• High frequency gain increase is by 1/α

• So there is a compromise between wanting to add a large amountof phase (α small) and the tendency to generate large gains at

high frequency

– So try to keep 1/α ≤ 10, which means |p| ≤ 10|z| and

φmax ≤ 60◦

• If more phase lead is needed, use multiple lead controllers

Gc(s) = kc(s + z1)

(s + p1)

(s + z2)

(s + p2)

• Select of the overall gain is problem specific.

– One approach is to force the desired crossover frequency so that

|L(jωc)| = 1

• Use Lead to add phase ⇒ increases PM ⇒ improves transientresponse.

September 13, 2007


Fall 2007 16.31 5–7

Lead Mechanics II

• Adding a lead to the LTF changes both the magnitude and phase,so it is difficult to predict the new crossover point (which is where

we should be adding the extra phase).

Figure 32: Lead example

• The process is slightly simpler if we target the lead compensatordesign at a particular desired ωc

1. Find the φmax required

– Note that φrequired = PM − (180◦ + �G(jωc))2. Put φmax at the crossover frequency, so that

ω2c = |p| · |z|3. Select Kc so that crossover is at ωc

September 13, 2007

dB

ω

ω

Phase margin,compensated

Phase margin,uncompensated

0

0o

Phase

-90o

-180o

z p



Fall 2007 16.31 5–8

Design Example

• Design a compensator for the system G(s) = 1s(s+1)– Want ωc = 10rad/sec and PM ≈ 40◦

• Note that at 10rad/sec, the slope of |G| is -2, corresponding to aplant phase of approximately 180◦

• So need to add a lead compensator⇒ adds a slope of +1 (locally),changing the LTF slope to -1, and thus increasing the phase margin


• Design steps:

��zp =

1−sinφm1+sinφm

�,

��G(jωc) ≈ −180◦

PM = 40◦

�⇒ φm = 40◦

⇒ zp = 0.22

� ω2c = z · p = 102 ⇒ z = 4.7, p = 21.4

� Pick kc so that |GcG(jωc)| = 1

September 13, 2007

|.|

|.| = 1 Line

1

|GGc|-2

-1

1010.1

z

|G|

p

New

so ωc = 10



Fall 2007 16.31 5–9

% g=1/s/(s+1)figure(1);clf;set(gcf,’DefaultLineLineWidth’,2)wc=10;PM=40*pi/180;G=tf(1,conv([1 0],[1 1]));phi_G=phase(evalfr(G,j*wc))*180/pi;phi_m=PM-(180+phi_G);zdp=(1-sin(phi_m))/(1+sin(phi_m));z=sqrt(wc^2*zdp);p=z/zdp;Gc=tf([1 z],[1 p]);k_c=1/abs(evalfr(G*Gc,j*wc));Gc=Gc*k_c;w=logspace(-2,2,300);f_G=freqresp(G,w);f_Gc=freqresp(Gc,w);f_L=freqresp(G*Gc,w);f_G=squeeze(f_G);f_Gc=squeeze(f_Gc);f_L=squeeze(f_L);loglog(w,abs(f_G),w,abs(f_Gc),w,abs(f_L))xlabel(’Freq’);ylabel(’Magnitude’);grid onlegend(’Plant G’,’Comp G_c’,’LTF L’,’Location’,’SouthWest’)print -dpng -r300 lead_examp2.png


September 13, 2007


Fall 2007 16.31 5–10

Lag Mechanics

• If pole at a lower frequency than zero:– Gain decreases at high frequency – typically scale lag up so that

HF gain is 1, and thus the low frequency gain is higher.

– Add negative phase (i.e., adds lag)

Figure 35: Lag: frequency domain Glag = kcs/z+1s/p+1

dB

0

-90o

0 p z ω

ω

ω

θm

ω m

-20

0

0o

20 log |p/z|

Phase

• Typically use a lag to add 20 logα to the low frequency gain with(hopefully) a small impact to the PM (at crossover)

– Pick the desired gain reduction at high frequency 20 log(1/α),

where α = |z|/|p|

– Pick |Glag|s=0 = kc to give the desired low frequency gainincrease for the LTF (shift the magnitude plot up)

– Heuristic: want to limit frequency of the zero (and pole) so that

there is a minimal impact of the phase lag at ωc ⇒ z = ωc/10

September 13, 2007



Fall 2007 16.31 5–11

Lag Compensation

• Assumption is that we need to modify (increase) the DC gain ofthe LTF to reduce the tracking error.

• Two ways to get desired low frequency gain:

� Using just a gain increase, which increases ωc and decrease PM� Add Lag dynamics that increase increase the gain at low fre-

quency and leave the gain near and above ωc unchanged

September 13, 2007

dB|KcGp|

|GcGp|

20 log |p/z|

Gp(jω)

Gc(jω)Gp(jω)

ω

ω

Phase margin,compensated

Phase margin,uncompensated

0p z wc

Phase

-90o

-180o

Original ωc



Fall 2007 16.31 5–12

Lag Example

Lag example with G(s) = 3/((s+1)2(s/2+1)) and right Gc(s) = 1,

left Gc(s) = (s + 0.1)/(s + 0.01), which should reduce steady state

the step error from 0.25 to 0.032

September 13, 2007


Fall 2007 16.31 6–13

Summary

• The design summary online is old, but gives an excellent summaryof the steps involved.

• Typically find that this process is highly iterative because the fi-nal performance doesn’t match the specifications (second order

assumptions)

• Practice is the only way to absorb these approaches.

September 13, 2007


Topic #5 - dspace.mit.edu · 9/13/2007 · Fall 2007 16.31 5–3 Performance Issues II • Classic...

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