signal and system Lecture 21

8/14/2019 signal and system Lecture 21

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Signals and SystemsFall 2003

Lecture #21

25 November 2003

1. Feedback

a) Root Locus

b) Trackingc) Disturbance Rejection

d) The Inverted Pendulum

2. Introduction to the Z-Transform


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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)


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

Closed-loop

poles are

the same.


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A Simple Example

Becomes more stable Becomes less stable

Sketch where

pole movesas |K| increases...

In either case, pole is atso = -2 -K


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What Happens More Generally ?

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

Difficult to solve explicitly for solutions given anyspecific

value ofK, unless G(s)H(s) is second-order or lower.

That is

Closed-loop poles are the solutions of

Much easier to plot the root locus, the values ofs that aresolutions forsome value ofK, 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 theselimiting points.


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Rules for Plotting Root Locus

End points

AtK= 0, G(so)H(so) =

so arepoles of the open-loop system function G(s)H(s).

At |K| = , G(so)H(so) = 0 so arezeros of the open-loop system function G(s)H(s). Thus:

Rule #1:

A root locus starts (atK= 0) from apole ofG(s)H(s) and ends (at

|K| = ) at azero ofG(s)H(s).

Question: What if the number ofpoles

the number ofzeros?Answer: Start or end at .


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Rule #2: Angle criterion of the root locus

Thus, s0 is a pole for somepositive value of K if:

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

Similarlys0 is a pole for some negative value of K if:

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


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Example of Root Locus.

One zero at -2,

two poles at 0, -1.


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

for at least some set of input signals.

Tracking

+= )(

)()(1

1)( sX

sHsCsE

)()()(1

1)(

jXjHjC

jE+

=

We want to be large in frequency bands in which we

want good tracking

)()( jPjC


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Tracking (continued)

Using the final-value theorem

Basic example: Tracking error for a step input


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Disturbance Rejection

There may be otherobjectives in feedback controls due to unavoidabledisturbances.

Clearly, sensitivities to the disturbancesD1(s) andD2(s) are much

reduced when the amplitude of the loop gain


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Internal Instabilities Due to Pole-Zero Cancellation

Hw(t)

)(

33

1)(

)()(1

)()()(

2)(

)1(

1)(

Stable

2sX

ss

sX

sHsC

sHsCsY

s

ssH,

sssC

++

=

+

=

+

=

+

=

However

)(

)33(

2)(

)()(1

)()(

Unstable

2sX

sss

ssX

sHsC

sCsW

++

+=

+=


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Inverted Pendulum

Unstable!


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Root Locus & the Inverted Pendulum

Attempt #1: Negative feedback driving the motor

Remains unstable!

Root locus of M(s)G(s)

after K. Lundberg


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Root Locus & the Inverted Pendulum

BUT x(t) unstable:

System subject to drift...

Solution: add PD feedbackaround motor and

compensator:

after K. Lundberg


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The z-Transform

The (Bilateral) z-Transform

Motivation: Analogous to Laplace Transform in CT

We now do not

restrict ourselves

just toz = ej


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The ROC and the Relation Between zT and DTFT

Unit circle (r= 1) in the ROC DTFTX(ej) exists

depends only on r= |z|, just like the ROC ins-plane

only depends onRe(s)

, r = |z|


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Example #2:

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


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Rational z-Transforms

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

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

Polynomials inz

signal and system Lecture 21

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