controller design by r.l

Controller design by R.L.Typical setup:

sGsC

sGsC

1:TF c.l.

01:poly char. c.l. sGsC

21

21Let psps

zszsKsC

sd

snsG

01 :Then

1

1

sd

sn

ps

zsK

This is the R.L. eq.

• With no zi, pi, controller design means to pick good K for R.L.

• Those zi, pi means to pick additional poles /zeros to R.L.

Proportional control design

1. Draw R.L. with given

2. Pick a point on R.L. to be desired c.l. pole: PD

3. Compute

KKsC P

call

01sGK

01sd

snK

sd

snsG

DPGsGK

11

When to use:If R.L. of G(s) goes through

the region for desired c.l. poles

What is that region:– From design specs, get desired Mp, ts, tr, etc.– Use formulae for 2nd order system to get desired

ωn , ζ, σ, ωd

– Identify these in s-plane

Example:

When C(s) = 1, things are okay

But we want initial response speed as fast as possible; yet we can only tolerate 10% overshoot.

Sol: From the above, we need that means:

6

1

sssG

%10pM6.0

This is a cone around –Re axis with ±60° area

We also want tr to be as small as possible.i.e. : want ωn as large as possible

i.e. : want pd to be as far away from s = 0 as possible

1. Draw R.L.

2. Pick pd on R.L. in cone & | pd | max

3. 3661

ddd

pppG

K

dp pole

desired

forRegion

?333or

6

:max

jp

p

p

d

d

d

Example:

Want: , as fast as possible

Sol:

1. Draw R.L. for2.

Draw cone ±45° about –Re axis

3. Pick pd as the cross point of the ζ = 0.7 line & R.L.

4.

62

10

ssssG

%5pM

6.010631

dddd

P ppppG

K

7.0%5 pM

Controller tuning:

1. First design typically may not work

2. Identify trends of specs changes as K is increased.e.g.: as KP , pole

3. Perform closed-loop step response

4. Adjust K to improve specse.g. If MP too much, the 2. says reduce KP

Pd M&,,

PD controller design

•

• This is introducing an additional zero to the R.L. for G(s)

• Use this if the dominant pole pair branches of G(s) do not pass through the desired region

zsKsKKsC DDP

DP KKz

Design steps:

1. From specs, draw desired region for pole.Pick from region.

2. Compute

3. Select

4. Select:

dd jp dpG dd pGzpz s.t.

dd pGz tan i.e.

DP

ddD

KzKpGzp

K1

Example:

Want:

Sol:

(pd not on R.L.)

(Need a zero to attract R.L. to pd)

sec2%,5 sp tM

7.0%5 pM

24

sec2 s

s tt

22 Choose jpd 707.0,2,2 d

2.

3.

4.

4

tan dz

22222

1

jjdpG

222 jj

4

3

4

5

24

3

4122

8

222222422

11

DP

D

KzKjjj

K

ssC 28

44

3 dpG

Closed-loop step response simulation:

» ng = [1] ;

» dg = [1 2 0] ;

» nc = [Kp, Kd]

» n = conv(nc, ng) ;

» d = dg + n ;

» stepspec(n, d) ;

Tuning: for fixed z:

Q: What’s the effect of tuning z?

PD MKK

PD MK

gcgc

gc

ddnn

nn

here 1cd

Drawbacks of PD

• Not proper : deg of num > deg of den

• High frequency gain → ∞:

• High gain for noiseSaturated circuits

cannot be implemented physically

as jKK DP

Lead Controller

• Approximation to PD

• Same usefulness as PD

•

• It contributes a lead angle:

0

zpps

zsKsC

zppC dd

ppd

Lead Design:

1. Draw R.L. for G

2. From specs draw region for desired c.l. poles

3. Select pd from region

4. LetPick –z somewhere below pd on –Re axisLetSelect

dd jp dpG

121 ,zpd

2 s.t. ppp d

2tan i.e. dp

• There are many choices of z, p

• More neg. (–z) & (–p) → more close to PD & more sensitive to noise, and worse steady-state error

• But if –z is > Re(pd), pd may not dominate

dpdp

zdp pGK

1Let

ps

zsKsC :is controllerYour

Example: Lead Design

MP is fine,

but too slow.

Want: Don’t increase MP

but double the resp. speed

Sol: Original system: C(s) = 1

Since MP is a function of ζ, speed is proportional to ωn

5.022,2 nn

4224 TF c.l.

ss

Draw R.L. & desiredregion

Pick pd right at the

vertex:

(Could pick pd a little

inside the region

to allow “flex”)

5.0 new want weHence 4 new nω

322 jpd

Clearly, R.L. does not pass through pd, nor the desired region.need PD or Lead to “bend” the R.L. into region.(Note our choice may be the easiest to achieve)

Let’s do Lead:

2 ddd pppG

623

2

Pick –z to the left of pd

4,4Pick zz

3

41

dp

663then 12

2tanthen dp 8322

3

1

7

2

4

1

dpdppdp

zdpK

8

47

s

ssC

Suppose we select –z = -1 instead.

zpd 1

612

7.2tan 2 dp

1322 j

321 j

32tan 1

32tan6

5 1

The new R.L.

closed-loop

has pole pair

at pd & pd* but

the 3rd pole is

at somewhere

(-1, 0)

Which one is better?

Particular choice of z :

2221

ddddd

pABpApzpzOzp

ppd2

OpOAp dd

dp

OApBp dd bisect

s.t. B Choose

ABpOBp dd OApd

2

1

dp2

1

22

dp

1tan dz 2tan dp

322 :example prev.In jpd 32,2 d

get weprocedure, above Follow

46.5

93.273.4

s

ssC

359.0,%21 :step c.l. rp tM

repeat. , 5.2 to2 change σ

375.0,%1.16 :step c.l. rp tM