Signals and Systems
Fall 2003
Lecture #21
25 November 2003
1,Feedback
a) Root Locus
b) Tracking
c) 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)
The,Classical” Root Locus Problem
C(s) = K — a simple linear amplifier
Closed-loop
poles are
the same.
A Simple Example
Becomes more stable
Becomes less stable
Sketch where
pole moves
as |K| increases...
In either case,pole is at s
o
= -2 - K
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 any specific
value of K,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 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.
Rules for Plotting Root Locus
End points
—At K = 0,G(s
o
)H(s
o
) = ∞
s
o
are poles of the open-loop system function G(s)H(s).
—At |K| = ∞,G(s
o
)H(s
o
) = 0
s
o
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 ±∞.
Rule #2,Angle criterion of the root locus
Thus,s
0
is a pole for some positive value of K if,
In this case,s
0
is a pole if K = 1/|G(s
0
) H(s
0
)|,
Similarly s
0
is a pole for some negative value of K if,
In this case,s
0
is a pole if K = -1/|G(s
0
) H(s
0
)|.
Example of Root Locus.
One zero at -2,
two poles at 0,-1.
In addition to stability,we may want good tracking behavior,i.e.
for at least some set of input signals.
Tracking
+
= )(
)()(1
1
)( sX
sHsC
sE
)(
)()(1
1
)( ω
ωω
ω jX
jHjC
jE
+
=
We want to be large in frequency bands in which we
want good tracking
)()( ωω jPjC
Tracking (continued)
Using the final-value theorem
Basic example,Tracking error for a step input
Disturbance Rejection
There may be other objectives in feedback controls due to unavoidable
disturbances.
Clearly,sensitivities to the disturbances D
1
(s) and D
2
(s) are much
reduced when the amplitude of the loop gain
Internal Instabilities Due to Pole-Zero Cancellation
H
w(t)
)(
33
1
)(
)()(1
)()(
)(
2
)(
)1(
1
)(
Stable
2
sX
ss
sX
sHsC
sHsC
sY
s
s
sH,
ss
sC




++
=
+
=
+
=
+
=
However
)(
)33(
2
)(
)()(1
)(
)(
Unstable
2
sX
sss
s
sX
sHsC
sC
sW






++
+
=
+
=
Inverted Pendulum
— Unstable!
Feedback System to Stabilize the Pendulum
PI feedback stabilizes θ
– Additional PD feedback around motor / amplifier
centers the pendulum
Subtle problem,internal instability in x(t)!
a
Root Locus & the Inverted Pendulum
Attempt #1,Negative feedback driving the motor
– Remains unstable!
Root locus of M(s)G(s)
after K,Lundberg
Root Locus & the Inverted Pendulum
Attempt #2,Proportional/Integral Compensator
– Stable for large enough K
Root locus of K(s)M(s)G(s)
after K,Lundberg
Root Locus & the Inverted Pendulum
BUT –x(t) unstable:
System subject to drift...
Solution,add PD feedback
around motor and
compensator:
after K,Lundberg
The z-Transform
The (Bilateral) z-Transform
Motivation,Analogous to Laplace Transform in CT
We now do not
restrict ourselves
just to z = e
jω
The ROC and the Relation Between zT and DTFT
Unit circle (r = 1) in the ROC? DTFT X(e
jω
) exists
— depends only on r = |z|,just like the ROC in s-plane
only depends on Re(s)
,r = |z|
Example #1
That is,ROC |z| > |a|,
outside a circle
This form to find
pole and zero locations
This form
for PFE
and
inverse z-
transform
1
1
1
=
az az
z
=
Example #2:
Same X(z) as in Ex #1,but different ROC.
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
Polynomials in z