signals and systems fall 2003 lecture #21 25 november 2003 1.feedback a)root locus b)tracking...

Signals and SystemsFall 2003

Lecture #2125 November 2003

1. Feedbacka) Root Locus b) Trackingc) Disturbance Rejection d) The Inverted Pendulum

2. Introduction to the Z-Transform

The Concept of a Root Locus

• C(s),G(s) — Designed with one or more free parameters• Question: How do the closed-loop poles move as we vary these

parameters? — Root locus of 1+ C(s)G(s)H(s)

)()()(1

)()()(

sHsGsC

sHsCsQ

The “Classical” Root Locus ProblemC(s) = K— a simple linear amplifier

A Simple Example

In either case, pole is at so = -2 - K

Sketch where pole movesas |K| increases...

Becomes more stable Becomes less stable

1)( ,2

1)(

sG

ssH

22

1

2)( (a)

Ks

K

s

Ks

K

sQ2

1

21

2

1

)( (b)

Ks

s

KssQ

What Happens More Generally ?

• For simplicity, suppose there is no pole-zero cancellation in G(s)H(s)

Closed-loop poles are the solutions of

That is

— Difficult to solve explicitly for solutions given any specific value of K, unless G(s)H(s) is second-order or lower.

— Much easier to plot the root locus, the values of s that are solutions for some value of K, because:

1) It is easier to find the roots in the limiting cases for K = 0, ±∞.

2) There are rules on how to connect between these limiting points.

)()(1

)()(

sHsKG

sHsQ

0)()(1 sHsKG

KsHsG

1)()(

Rules for Plotting Root Locus

• End points— At K = 0, G(so)H(so) = ∞⇒ so are poles of the open-loop system function G(s)H(s).

— At |K| = ∞, G(so)H(so) = 0⇒so are zeros of the open-loop system function G(s)H(s). Thus:

Rule #1:A root locus starts (at K = 0) from a pole of G(s)H(s) and ends (at

|K| = ∞) at a zero of G(s)H(s).Question: What if the number of poles ≠ the number of zeros? Answer: Start or end at ±∞.

KsHsG

1)()(

Rule #2: Angle criterion of the root locus

• Thus, s0 is a pole for some positive value of K if:

In this case, s0 is a pole if K = 1/|G(s0) H(s0)|.

• Similarly s0 is a pole for some negative value of K if:

In this case, s0 is a pole if K = -1/|G(s0) H(s0)|.

：)12()()(0 00 nsHsGK

nsHsGK 2)()(0 00

Example of Root Locus.

One zero at -2, two poles at 0, -1.

)1(

2)()(

ss

ssHsG

))()(( sHsGrlocus

Tracking

In addition to stability, we may want good tracking behavior, i.e.

for at least some set of input signals.

We want C(jω)P(jω) to be large in frequency bands in which we want good tracking

ttytxte as 0)()()(

)()()(1

1)(

)()()(1

1)(

jXjPjC

jE

sXsPsC

sE

Tracking (continued)

Using the final-value theorem

Basic example: Tracking error for a step input

)()()(1

lim)(limlim00

sXsPsC

sssEte

sst

ssXtutx

1)()()( Suppose

)()()(1

lim)(lim Then 0

sXsPsC

ste

st

Disturbance Rejection

There may be other objectives in feedback controls due to unavoidable disturbances.

Clearly, sensitivities to the disturbances D1(s) and D2(s) are much reduced when the amplitude of the loop gain

)()()(1

1)(

)()(1

)()(

)()(1

1)( 21 sD

sHsCsD

sHsC

sHsX

sHsCsE

1)()( sHsC

Internal Instabilities Due to Pole-Zero Cancellation

However

)(

Unstable

)33(

2)(

)()(1

)(

)(

Stable33

1)(

)()(1

)()( )(

2)( ,

)1(

1 )(

2

2

sXsss

ssX

sHsC

sCW(s)

sXss

sXsHsC

sHsCsY

s

ssH

sssC

Inverted Pendulum

m=Mass, I=moment of inertia

balance)(Moment )(

)(cos)(

)(sin2

2

2

2

dt

tdIt

dt

txdmltmgl

1 cos,sinsmall is assuming Linearize ：

2

2

2

2 )()(

)(

dt

txdmltmgl

dt

tdI

)()(

)(

2

2

sXmglIs

mlss

sH

──Unstable！

Feedback System to Stabilize the Pendulum

• PI feedback stabilizes θ

• Subtle problem: internal instability in x(t)!–Additional PD feedback around motor / amplifier

centers the pendulum

Root Locus & the Inverted Pendulum

• Attempt #1: Negative feedback driving the motor

• Root locus of M(s)G(s)

– Remains unstable!

after K. Lundberg

)1)(1(

/

)(

)()(

2

ss

gs

sX

ssG

LL

)1()(

)()(

ss

k

sV

sXsM

M

m

Root Locus & the Inverted Pendulum

• Attempt #2: Proportional/Integral Compensator

• Root locus of K(s)M(s)G(s)

– Stable for large enough K

after K. Lundberg

ssK

k1

1)(

Root Locus & the Inverted Pendulum

• BUT – x(t) unstable:

System subject to drift...

• Solution: add PD feedback around motor and compensator:

after K. Lundberg

)()()(1

)()(

)(

)(

sGsMsK

sMsK

x

sX

C

)1)(/()1)(1(

)1)(1(122

22

2 sgkss

ssk

s KMLMK

LKM

The z-Transform

Motivation: Analogous to Laplace Transform in CT

We now do not restrict ourselves

just to z = ejω

The (Bilateral) z-Transform

][nx ][nh ][ny

nn zzHnyznx )(][ ][

LTI DTfor ionEigenfunct

convergesit assuming ][)(

n

nznhzH

]}[{][)(][ nxZznxzXnxn

n

The ROC and the Relation Between zT and DTFT

—depends only on r = |z|, just like the ROC in s-plane only depends on Re(s)

• Unit circle (r = 1) in the ROC ⇒ DTFT X(ejω) exists

jrez ||, zr

n

njn

n

njj erenxrenxreX )][()]([)(

}][{ nrnx F

n

nj rnxrez |][|at which ROC

•

Example #1

This form for PFE and inverse z- transform

That is, ROC |z| > |a|, outside a circle

This form to findpole and zero locations

sided-right- nuanx n

-n

)( nn znuazX

0

1)(n

naz

||||,i.e.,1 If 1 azaz

az

z

az

11

1

Example #2:

Same X(z) as in Ex #1, but different ROC.

sided-left-]1[ nuanx n

1

1

1 )(

n

nn

n

nn

za

znuazX

0

11 n

n

n

nn zaza

,

11

11

1

1

1

az

zza

za

za

||||.,.,1 If 1 azeiza

Rational z-Transforms

x[n] = linear combination of exponentials for n > 0 and for n < 0

— characterized (except for a gain) by its poles and zeros

)(

)(

rational is )(

zD

zNX(z)

zX

Polynomials in z