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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