C(s) Gp(s)

Two d.o.f. Control design: dual loop

C1(s) Gp(s)

C2(s)

rd

e y++

+

_ _

2 2Choose ( ) to s factors in: ( ) ( ) ( )p pC s d s C s n s

1

1

1

( ) ( ) ( )

1 ( ) ( ) 1 ( ) ( )

When d = 0,

( ) ( )

1 ( ) ( )

( ) ( )Use for closed loop step response

1 ( ) ( )

p p

p p

p

p

p

p

C s G s G sy r d

C s G s C s G s

C s G sy r

C s G s

C s G s

C s G s

-0.1

y

_

10

1s 1

s_PI*PD +0.1s

Example from last time:Two different implementations:

-0.1

y

_

10

1s 1

s_PI

PD+0.1s

Standard

Alternative(more overshoot)

sy

C(s) Gp(s)

Two d.o.f. Control design: FF+FB

C(s) Gp(s)

Cf(s)

rd

e y+

+

+

_+

+

( ( ) ( )) ( ) ( )

1 ( ) ( ) 1 ( ) ( )

1 ( ) ( ) ( )

1 ( ) ( ) 1 ( ) ( )

( ) ( )( ) , ( )

( ) ( )

( ( ) )

f p p

p p

f p p

p p

p p C Cp

p p C C

C p f p p C

p C p C p C p C

C s C s G s G sy r d

C s G s C s G s

C s G s G se r y r d

C s G s C s G s

n s n n s nG s C s

d s d d s d

d d C s n n de r d

d d n n d d n n

( ( ) )

# in ( ( ) ) defines type w.r.t. r

# in defines type w.r.t. d

When ( )=0,

type w.r.t. r = # in

=#integrato

C p f p p C

p C p C p C p C

C p f p

p C

f

C p

d d C s n n de r d

d d n n d d n n

s d d C s n

s n d

C s

s d d

rs in loop

type w.r.t. d = # in

=#integrators in controllerCs d

( ( ) )

When ( ) 0,

1type w.r.t. d = # in C(s)

type w.r.t. r set by ( ) :

Choose ( ) so that:

#s factors in ( ( ) ) = desired type

C p f p p C

p C p C p C p C

f

f

f

C p f p

d d C s n n de r d

d d n n d d n n

C s

sC s

C s

d d C s n

C(s)

Cf(s)

rd

e y+

+

+

_+

+ 1

( 2)s s

Example:

Suppose C(s) is a lead-lag controller.Without Cf, what is the system type w.r.t. r?What is the system type w.r.t. d?Can Cf affect the type w.r.t. d?Can Cf affect the type w.r.t. r?

C(s)

Cf(s)

rd

e y+

+

+

_+

+ 1

( 2)s s

Example:

Suppose C(s) is a lead-lag controller.

Q:Find a Cf to achieve 0 ess tracking for ramp input.

C(s)

Cf(s)

rd

e y+

+

+

_+

+ 1

( 2)s s

2

2

Solution:

Choose ( ) so that

( ) cancels lowest order term in .

1, +2s, choose ( ) 2s,

then: ( ) type 2, ess = 0

f

f p p

p p f

p f p ramp

C s

C s n d

n d s C s

d C s n s

Design specs: use PID to achieve 1.Ess to ramp input = 02.As fast as possible3.Overshoot <= 25%

Plant is type 1, so need PI after PD.25% Mp PM >= 45, but with PI, need PMd 55PD can contribute a maximum of about 75So, select max wgc at phase=-200

s=tf('s'); Gp=1/s/(s+1)/(s+5);figure(2); margin(Gp); grid; V=axis;Mp=25; PMd = 70 - Mp + 10; %+10 for PI laterDPM=75; PM=PMd-DPM;hold on; plot([V(1) V(2)],[PM PM]-180,'r:');[x, y]=ginput(1); wgcd=x;z_PD = wgcd/tan(DPM*pi/180); %PD controlK = 1/abs(evalfr((s+z_PD)*Gp,j*wgcd));C=K*(s+z_PD); margin(C*Gp); grid; %Bode with PDz_PI = wgcd/10; C=C*(s+z_PI)/s; %multiply PI and PDmargin(C*Gp); grid; hold off; %Bode with PIDt=linspace(0, 15/wgcd, 301);figure(3); step(C*Gp/(1+C*Gp),t); grid;

-150

-100

-50

0

50

100M

agn

itu

de

(dB

)

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

Ph

ase

(deg

)

Bode DiagramGm = -Inf dB (at 0 rad/sec) , Pm = 48.2 deg (at 3.68 rad/sec)

Frequency (rad/sec)

0 0.5 1 1.5 2 2.5 3 3.5 40

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plit

ud

e

Figure 8-8 Unit-step response curve of PID-controlled system designed by use of the Ziegler–Nichols tuning rule (second method).

Figure 8-10 Unit-step response of the system shown in Figure 8–6 with PID controller after numerical tuning of PID parameters.

Observations:

Can shift red bump toward high freq a bit to increase speed

But at higher freq, phase is lower, so PD needs to contribute more DPM

PM=48 can be reduced by about 3

s=tf('s'); Gp=1/s/(s+1)/(s+5);figure(2); margin(Gp); grid; V=axis;Mp=25; PMd = 70 - Mp + 7; %+7 for PI laterDPM=85; PM=PMd-DPM;hold on; plot([V(1) V(2)],[PM PM]-180,'r:');[x, y]=ginput(1); wgcd=x;z_PD = wgcd/tan(DPM*pi/180); %PD controlK = 1/abs(evalfr((s+z_PD)*Gp,j*wgcd));C=K*(s+z_PD); margin(C*Gp); grid; %Bode with PDz_PI = wgcd/10; C=C*(s+z_PI)/s; %multiply PI and PDmargin(C*Gp); grid; hold off; %Bode with PIDt=linspace(0, 15/wgcd, 301);figure(3); step(C*Gp/(1+C*Gp),t); grid;

-150

-100

-50

0

50

100M

agn

itu

de

(dB

)

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

-45

Ph

ase

(deg

)

Bode DiagramGm = -Inf dB (at 0 rad/sec) , Pm = 45.5 deg (at 5.02 rad/sec)

Frequency (rad/sec)

0 0.5 1 1.5 2 2.50

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plit

ud

e

C1(s)

C2(s)

rd

e y++

+

_ _

For dual loop implementation:

1

( 1)( 5)s s s

Should have C2(s) = - 5s

So C1(s) = PID + 5s

C1=C+5*s; Gp1=1/s/1/(s+6); figure(4); step(C1*Gp1/(1+C1*Gp1),t); grid;figure(5); step(C1*Gp1/(1+C1*Gp1)/s,10*t); grid; %ramp responsefigure(6); step(C1*Gp1/(1+C1*Gp1)/s/s,10*t);grid; %acc response

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plit

ud

e

Overshoot became higher, about 33%,Significantly more than 25% spec

•Increase PMd •Increase the PI divide-number

s=tf('s'); Gp=1/s/(s+1)/(s+5);figure(2); margin(Gp); grid; V=axis;Mp=25; PMd = 70 - Mp + 10; %+10 for PI laterDPM=85; PM=PMd-DPM;hold on; plot([V(1) V(2)],[PM PM]-180,'r:');[x,y]=ginput(1); wgcd=x;z_PD = wgcd /tan(DPM*pi/180); %PD controlK = 1/abs(evalfr((s+z_PD)*Gp,j*wgcd));C=K*(s+z_PD); margin(C*Gp); grid; %Bode with PDz_PI = wgcd/10; C=C*(s+z_PI)/s; %multiply PI and PDmargin(C*Gp); grid; hold off;t=linspace(0, 15/wgcd,301);figure(3); step(C*Gp/(1+C*Gp),t); grid;figure(4); C1=C+5*s; Gp1=1/s/s/(s+6); step(C1*Gp1/(1+C1*Gp1),t); grid;t=linspace(0, 150/wgcd,3001);figure(5); step(C1*Gp1/(1+C1*Gp1)/s,t); grid; %ramp responsefigure(6); step(C1*Gp1/(1+C1*Gp1)/s/s,t); grid; %acc response

0 0.5 1 1.5 2 2.5 3 3.50

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plit

ud

e

about the same as single loop

0 5 10 15 20 25 300

5

10

15

20

25

30

35

Step Response

Time (sec)

Am

plit

ud

eRamp response

1.6 1.7 1.8 1.9 2 2.1 2.2 2.3

1.5

1.6

1.7

1.8

1.9

2

2.1

2.2

Step Response

Time (sec)

Am

plit

ud

e

29.6 29.65 29.7 29.75 29.8 29.85 29.9 29.95 30 30.05 30.129.6

29.65

29.7

29.75

29.8

29.85

29.9

29.95

30

30.05

30.1

Step Response

Time (sec)

Am

plit

ud

e

Ess_ramp = 0

0 5 10 15 20 25 300

50

100

150

200

250

300

350

400

450

500

Step Response

Time (sec)

Am

plit

ud

e

3.8 3.85 3.9 3.95 4 4.05 4.1 4.15 4.2 4.256.8

7

7.2

7.4

7.6

7.8

8

8.2

8.4

8.6

Step Response

Time (sec)

Am

plit

ud

e

29.5 29.6 29.7 29.8 29.9 30 30.1 30.2 30.3 30.4

435

440

445

450

455

460

Step Response

Time (sec)

Am

plit

ud

e

Acc response

Tuning based on desired loop shape:

We notice the max phase plot bump happens at a frequency lower than wgc

Reduce K to lower wgc so that max bump closer to wgcReduce PMd since at lower wgc, phase is higher

s=tf('s'); Gp=1/s/(s+1)/(s+5);figure(2); margin(Gp); grid; V=axis;Mp=15; PMd = 70 - Mp + 10; %+10 for PI laterDPM=85; PM=PMd-DPM;hold on; plot([V(1) V(2)],[PM PM]-180,'r:');[x,y]=ginput(1); wgcd=x;z_PD = wgcd /tan(DPM*pi/180); %PD controlK = 1/abs(evalfr((s+z_PD)*Gp,j*wgcd))*0.5;C=K*(s+z_PD); margin(C*Gp); grid; %Bode with PDz_PI = wgcd/10; C=C*(s+z_PI)/s; %multiply PI and PDmargin(C*Gp); grid; hold off;t=linspace(0, 15/wgcd,301);figure(3); step(C*Gp/(1+C*Gp),t); grid;figure(4); C1=C+5*s; Gp1=1/s/s/(s+6); step(C1*Gp1/(1+C1*Gp1),t); grid;t=linspace(0, 150/wgcd,3001);figure(5); step(C1*Gp1/(1+C1*Gp1)/s,t); grid; %ramp responsefigure(6); step(C1*Gp1/(1+C1*Gp1)/s/s,t); grid; %ramp response

0 0.5 1 1.5 2 2.5 3 3.50

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plit

ud

e

figure(3); step(C*Gp/(1+C*Gp),t); grid;

0 0.5 1 1.5 2 2.5 3 3.50

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plit

ud

e

figure(4); C1=C+5*s; Gp1=1/s/s/(s+6); step(C1*Gp1/(1+C1*Gp1),t); grid;

Ess to ramp and acc still = 0

C(s)

Cf(s)

rd

e y+

+

+

_+

+ 1

( 1)( 5)s s s

FF+FB implementation of 2 dof control

Select Cf(s) = 5s

Y=(Cf+C)Gp/(1+CGp)

s=tf('s'); Gp=1/s/(s+1)/(s+5);figure(2); margin(Gp); grid; V=axis;Mp=15; PMd = 70 - Mp + 10; %+10 for PI laterDPM=85; PM=PMd-DPM;hold on; plot([V(1) V(2)],[PM PM]-180,'r:');[x,y]=ginput(1); wgcd=x;z_PD = wgcd /tan(DPM*pi/180); %PD controlK = 1/abs(evalfr((s+z_PD)*Gp,j*wgcd))*0.5;C=K*(s+z_PD); margin(C*Gp); grid; %Bode with PDz_PI = wgcd/10; C=C*(s+z_PI)/s; %multiply PI and PDmargin(C*Gp); grid; hold off;t=linspace(0, 15/wgcd,301); t1=linspace(0, 150/wgcd,3001);figure(3); step(C*Gp/(1+C*Gp),t); grid;figure(4); C1=C+5*s; Gp1=1/s/s/(s+6); step(C1*Gp1/(1+C1*Gp1),t); grid;figure(5); step(C1*Gp1/(1+C1*Gp1)/s,t1); grid; %ramp responsefigure(6); step(C1*Gp1/(1+C1*Gp1)/s/s,t1); grid; %acc responseCf=5*s; figure(7); step((Cf+C)*Gp/(1+C*Gp),t); grid;

0 0.5 1 1.5 2 2.5 3 3.50

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plit

ud

e

Cf=5*s; figure(7); step((Cf+C)*Gp/(1+C*Gp),t); grid;

PID

This is example 8-2.

Design specs:•10% overshoot in closed-loop unit step response

The book tries to design a PID controller.But with the above specs, a P-controller will do.

0 2 4 6 8 10 12 14 16 18 200

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Step Response

Time (sec)

Am

plit

ud

e

But that was “obviously not good”.

Really? Why or why not?

Now suppose we change the design specs to:•10% overshoot in closed-loop unit step response•Ess to constant input must be zero•Limit controller to simple PID•Achieve as fast response as possible.

Now it is a valid design problem.

Design analysis:•Zero ess to step requires type 1 system, but plant is only type 0 need PI•To increase speed and Mp, we will need PD•So we will do PD followed by PI to get an overall PID•The upper limit of phase boost by PID is about 80 deg•To maximize speed, need max wgc, need to find highest freq at which when we add 80 deg we still have enough PM

s=tf('s'); Gp=1.2/(0.36*s^3+1.86*s^2+2.5*s+1);figure(1); margin(Gp); hold on; grid; V=axis;Mp=10; PMd = 70 - Mp + 10; %+10 for PI laterDPM=80; PM=PMd-DPM; plot(V(1:2),[PM PM]-180,'r:');

-150

-100

-50

0

50M

agn

itu

de

(dB

)

10-2

10-1

100

101

102

-270

-180

-90

0

Ph

ase

(deg

)

Bode DiagramGm = 19.9 dB (at 2.64 rad/sec) , Pm = 126 deg (at 0.397 rad/sec)

Frequency (rad/sec)

[x,y]=ginput(1); wgcd=x;z_PD = wgcd /tan(DPM*pi/180); %PD controlK = 1/abs(evalfr((s+z_PD)*Gp,j*wgcd));C=K*(s+z_PD); margin(C*Gp); %Bode with PD

-150

-100

-50

0

50M

agn

itu

de

(dB

)

10-2

10-1

100

101

102

-270

-180

-90

0

Ph

ase

(deg

)

Bode DiagramGm = Inf dB (at Inf rad/sec) , Pm = 62.8 deg (at 3.51 rad/sec)

Frequency (rad/sec)

z_PI = wgcd/5; C=C*(s+z_PI)/s; %PI times PDmargin(C*Gp); hold off;t=linspace(0,15/wgcd,301); t1=linspace(0,150/wgcd,3001);figure(3); step(C*Gp/(1+C*Gp),t); grid;

-150

-100

-50

0

50

100M

agn

itu

de

(dB

)

10-2

10-1

100

101

102

-270

-180

-90

0

Ph

ase

(deg

)

Bode DiagramGm = Inf dB (at Inf rad/sec) , Pm = 52.3 deg (at 3.48 rad/sec)

Frequency (rad/sec)

0 0.5 1 1.5 2 2.5 3 3.5 40

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plit

ud

e

C(s)

Let us reconsider this design problem with specs: •10% overshoot in closed-loop unit step response•Ess to constant input must be zero•Achieve as fast response as possible.•But limit controller order (nc and/or dc) to 2

PID is artificially imposed by the book. But we repeatedly said we should avoid I whenever possible.

Design analysis:•For zero ess to step, if we use an inner loop with a gain of -1.2, we can increase the plant type by 1. •So don’t need PI, but use dual loop.•For Mp <= 10%, we need PM >= 60.•For high speed, need high wgc.•But plant phase goes to -270 at high freq.•Need PD or lead to boost PM.•But order is limited to 2.•Choices are: two PDs, two leads, one of each•PD is sensitive to noise, not two PDs

Let’s do the more complex: PD*lead

s=tf('s'); Gp=1.2/(0.36*s^3+1.86*s^2+2.5*s+1);figure(1); margin(Gp); hold on; grid; V=axis;Mp=10; PMd = 70 - Mp + 7; DPM_PD=80; DPM_lead=70; PM=PMd-DPM_PD-DPM_lead; plot(V(1:2),[PM PM]-180);[x,y]=ginput(1); wgcd=x;z_PD = wgcd /tan(DPM*pi/180); %PD controlalpha=(1+sin(DPM_lead*pi/180))/(1-sin(DPM_lead*pi/180));z_lead=wgcd/alpha^0.5;p_lead=wgcd*alpha^0.5;C=(s+z_PD)*(s+z_lead)/(s+p_lead);K = 1/abs(evalfr(C*Gp,j*wgcd)); C=C*K;margin(C*Gp); hold off; %Bode with PDt=linspace(0,15/wgcd,301); t1=linspace(0,150/wgcd,3001);figure(3); step(C*Gp/(1+C*Gp),t); grid;

>> CTransfer function:2520 s^2 + 3.439e004 s + 1.173e005---------------------------------- s + 219.5

-200

-150

-100

-50

0

50

100M

agn

itu

de

(dB

)

10-2

10-1

100

101

102

103

104

-270

-180

-90

0

Ph

ase

(deg

)

Bode DiagramGm = Inf dB (at Inf rad/sec) , Pm = 67.6 deg (at 38.7 rad/sec)

Frequency (rad/sec)

0 0.05 0.1 0.15 0.2 0.25 0.3 0.350

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plit

ud

e

First design achieved 10 times fasterMp is about OK, but a little over

Ess is actually not zero but very smallDual loop implementation will make ess =0

But before we do that, should reduce Mp

s=tf('s'); Gp=1.2/(0.36*s^3+1.86*s^2+2.5*s+1);figure(1); margin(Gp); hold on; grid; V=axis;Mp=10; PMd = 70 - Mp + 10; DPM_PD=80; DPM_lead=70; PM=PMd-DPM_PD-DPM_lead; plot(V(1:2),[PM PM]-180);[x,y]=ginput(1); wgcd=x;z_PD = wgcd /tan(DPM*pi/180); %PD controlalpha=(1+sin(DPM_lead*pi/180))/(1-sin(DPM_lead*pi/180));z_lead=wgcd/alpha^0.5;p_lead=wgcd*alpha^0.5;C=(s+z_PD)*(s+z_lead)/(s+p_lead);K = 1/abs(evalfr(C*Gp,j*wgcd)); C=C*K;margin(C*Gp); hold off; %Bode with PDt=linspace(0,15/wgcd,301); t1=linspace(0,150/wgcd,3001);figure(3); step(C*Gp/(1+C*Gp),t); grid;

>> CTransfer function:1254 s^2 + 1.204e004 s + 2.891e004---------------------------------- s + 154.5

-200

-150

-100

-50

0

50M

agn

itu

de

(dB

)

10-2

10-1

100

101

102

103

104

-270

-225

-180

-135

-90

-45

0

Ph

ase

(deg

)

Bode DiagramGm = Inf dB (at Inf rad/sec) , Pm = 70.8 deg (at 27.2 rad/sec)

Frequency (rad/sec)

0 0.1 0.2 0.3 0.4 0.50

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plit

ud

e

>> C2=-1.2; C1=C-C2; Gcl=C1*Gp/(1+C*Gp);figure(4); step(Gcl,t); grid

>> C1 Transfer function:1254 s^2 + 1.204e004 s + 2.91e004--------------------------------- s + 154.5

0 0.1 0.2 0.3 0.4 0.50

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plit

ud

e