I
I
∀
1
dz
1
n,
x(n) =
ˆ
X (z)zn
=
ˆ
X (z)zn−1dz.
2π
iz
2πi
Here the ‘circle’ in the integral sign, H , means that the integral in the complex z-plane
is along any closed counterclockwise circle contained in RoC(x). (An integral along a
closed contour is called a contour integral.) In summary,
I
∀
1
n ∈ Z,
x(n) =
ˆ
X (z)zn−1dz,
(13.16)
2πi
where the integral is along any closed counterclockwise circle inside RoC(x).
We can similarly use the CTFT to recover any continuous-time signal x from its
Laplace transform by
σ+i∞
Z
∀
1
t ∈ R,
x(t) =
ˆ
X (s)est ds
2πi
σ−i∞
where the integral is along any vertical line (σ − i∞, σ + i∞) contained in RoC(x).
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13.9. SUMMARY
Probing Further: Differentiation and Laplace transforms
We can use the inverse Laplace transform as given in the box on page 598 to demon-
strate the differentiation property in table 13.4. Let y be defined by
∀
d
t ∈ R,
y(t) =
x(t).
dt
We can write x in terms of its Laplace transform as
Z
σ+i∞
∀
1
t ∈ R,
x(t) =
ˆ
X (s)est ds.
2πi σ−i∞
Differentiating this with respect to t is easy,
Z
σ+i∞
∀
1
t ∈ R,
d x(t) =
s ˆ
X (s)est ds.
dt
2πi σ−i∞
Consequently, y(t) = dx(t)/dt is the inverse transform of s ˆ
X (s), so
∀ s ∈ RoC(y), ˆY(s) = s ˆX(s),
where RoC(y) ⊃ RoC(x).
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13. LAPLACE AND Z TRANSFORMS
Exercises
Each problem is annotated with the letter E, T, C which stands for exercise, requires some
thought, requires some conceptualization. Problems labeled E are usually mechanical,
those labeled T require a plan of attack, those labeled C usually have more than one
defensible answer.
1. E Consider the signal x given by
∀n, x(n) = sin(ω0n)u(n).
(a) Show that the Z transform is
∀
z sin(ω0)
z ∈ RoC(x),
ˆ
X (z) =
,
z2 − 2z cos(ω0) + 1
where
RoC(x) = {z ∈ C | |z| > 1}.
(b) Where are the poles and zeros?
(c) Is x absolutely summable?
2. T Consider the signal x given by
∀ n ∈ Z, x(n) = a|n|,
where a ∈ C.
(a) Find the Z transform of x. Be sure to give the region of convergence.
(b) Where are the poles?
(c) Under what conditions is x absolutely summable?
3. E Consider a discrete-time LTI system with transfer function given by
∀
z
z ∈ {z | |z| > 0.9},
ˆ
H(z) =
.
z − 0.9
Suppose that the input x is given by
∀ n ∈ Z, x(n) = δ(n) − 0.9δ(n − 1).
Find the Z transform of the output y, including its region of convergence.
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EXERCISES
4. E Consider the exponentially modulated sinusoid y given by
∀ n ∈ Z, y(n) = a−n cos(ω0n)u(n),
where a is a real number, ω0 is a real number, and u is the unit step signal.
(a) Find the Z transform. Be sure to give the region of convergence. Hint: Use
example 13.3 and Section 13.1.6.
(b) Where are the poles?
(c) For what values of a is this signal absolutely summable?
5. T Suppose x ∈ DiscSignals satisfies
∞
∑ |x(n)r−n| < ∞,
0 < r1 < r < r2,
n=−∞
for some real numbers r1 and r2 such that r1 < r2. Show that
∞
∑ |nx(n)r−n| < ∞,
0 < r1 < r < r2.
n=−∞
Hint: Use the fact that for any ε > 0 there exists N < ∞ such that n(1 + ε)−n < 1
for all n > N.
6. T Consider a causal discrete-time LTI system where the input x and output y are
related by the difference equation
∀ n ∈ Z, y(n) + b1y(n − 1) + b2y(n − 2) = a0x(n) + a1x(n − 1) + a2x(n − 2),
where b1, b2, a0, a1, and a2 are real-valued constants.
(a) Find the transfer function.
(b) Say as much as you can about the region of convergence.
(c) Under what conditions is the system stable?
7. E This exercise verifies the time delay property of the Laplace transform. Show
that if x is a continuous-time signal, τ is a real constant, and y is given by
∀ t ∈ R, y(t) = x(t − τ),
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13. LAPLACE AND Z TRANSFORMS
then its Laplace transform is
∀ s ∈ RoC(y), ˆY(s) = e−sτ ˆX(s),
with region of convergence
RoC(y) = RoC(x).
8. E This exercise verifies the convolution property of the Laplace transform. Sup-
pose x and h have Laplace transforms ˆ
X and ˆ
H. Let y be given by
∞
Z
∀ t ∈ R, y(t) = (x ∗ h)(t) =
x(τ)h(t − τ)dτ.
−∞
Then show that the Laplace transform is
∀s ∈ RoC(y), ˆY(s) = ˆX(s) ˆH(s),
with
RoC(y) ⊃ RoC(x) ∩ RoC(h).
9. T This exercise verifies the conjugation property of the Laplace transform, and
then uses this property to demonstrate that for real-valued signals, poles and zeros
come in complex-conjugate pairs.
(a) Let x be a complex-valued continuous-time signal and y be given by
∀ t ∈ R, y(t) = [x(t)]∗.
Show that
∀ s ∈ RoC(y), ˆY(s) = [ ˆX(s∗)]∗,
where
RoC(y) = RoC(x).
(b) Use this property to show that if x is real, then complex poles and zeros occur
in complex conjugate pairs. That is, if there is a zero at s = q, then there must
be a zero at s = q∗, and if there is a pole at s = p, then there must also be a
pole at s = p∗.
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EXERCISES
10. T This exercise verifies the time scaling property of the Laplace transform. Let y
be defined by
∀ t ∈ R, y(t) = x(ct),
for some real number c. Show that
∀ s ∈ RoC(y), ˆY(s) = ˆX(s/c)/|c|,
where
RoC(y) = {s | s/c ∈ RoC(x)}.
11. E This exercise verifies the exponential scaling property of the Laplace transform.
Let y be defined by
∀ t ∈ R, y(t) = eatx(t),
for some complex number a. Show that
∀ s ∈ RoC(y), ˆY(s) = ˆX(s − a),
where
RoC(y) = {s | s − a ∈ RoC(x)}.
12. T Consider a discrete-time LTI system with impulse response
∀n, h(n) = an cos(ω0n)u(n),
for some ω0 ∈ R. Show that if the input is
∀ n ∈ Z, x(n) = eiω0nu(n),
then the output y is unbounded.
13. E Find and plot the inverse Z transform of
1
ˆ
X (z) = (z−3)3
with
(a) Roc(x) = {z ∈ C | |z| > 3}
(b) Roc(x) = {z ∈ C | |z| < 3}.
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13. LAPLACE AND Z TRANSFORMS
14. E Obtain the partial fraction expansions of the following rational polynomials. First
divide through if necessary to get a strictly proper rational polynomial.
(a)
z + 2
(z + 1)(z + 3)
(b)
(z + 2)2
(z + 1)(z + 3)
(c)
z + 2 .
z2 + 4
15. E Find the inverse Z transform x for each of the three possible regions of conver-
gence associated with
(z + 2)2
ˆ
X (z) =
.
(z + 1)(z + 3)
For which region of convergence is x causal? For which is x strictly anti-causal?
For which is x two-sided?
16. E Find the inverse Z transform x for each of the two possible regions of convergence
associated with
z + 2
ˆ
X (z) =
.
z2 + 4
17. E Consider a stable system with impulse response
h(n) = (0.5)nx(n).
Find the steady-state response to a unit step input.
18. E Let h(n) = 2nu(−n), all n, and g(n) = 0.5nu(n), for all n. Find h ∗ u and g ∗ u,
where u is the unit step.
19. E This exercise shows how we can determine the transfer function and frequency
response of an LTI system from its step response. Suppose a causal system with
step input x = u, produces the output
∀ n ∈ Z, y(n) = (1 − 0.5n)u(n).
(a) Find the transfer function (including its region of convergence).
Lee & Varaiya, Signals and Systems
605
EXERCISES
(b) If the system is stable, find its frequency response.
(c) Find the impulse response of the system.
20. E Consider an LTI system with impulse response h given by
∀ n ∈ Z, h(n) = 2nu(n).
(a) Find the transfer function, including its region of convergence.
(b) Use the transfer function to find the Z transform of the step response.
(c) Find the inverse transform of the result of part (b) to obtain the step response
in the time domain.
21. E Determine the zero-input and zero-state responses, and the transfer function for
the following. In both cases take y(−1) = y(−2) = 0 and x(n) = u(n).
(a) y(n) + y(n − 2) = x(n), n ≥ 0.
(b) y(n) + 2y(n − 1) + y(n − 2) = x(n), n ≥ 0.
22. E Determine the zero-input and the zero-state responses for the following.
(a) 5 ˙
y + 10y = 2x, y(0) = 2, x(t) = u(t).
(b) ¨
y + 5 ˙
y + 6y = −4x − 3 ˙x, y(0) = −1, ˙y(0) = 5, x(t) = e−tu(t).
(c) ¨
y + 4y = 8x, y(0) = 1, ˙y(0) = 2, x(t) = u(t).
(d) ¨
y + 2 ˙
y + 5y = ˙
x, y(0) = 2, ˙y(0) = 0, x(t) = e−tu(t).
23. E Show that the [A, b, c, d] representation in example 13.23 is correct. Then show
that the transfer function of the state-space model is the same as that of the differ-
ence equation.
24. T Consider the circuit of figure 13.5. The input is the voltage x, the output is the
capacitor voltage v. The inductor current is called i.
(a) Derive the [A, b, c, d] representation for this system using s(t) = [i(t), v(t)]T as
the state.
(b) Obtain an [F, g, h, k] representation for a discrete-time model of the same
circuit by sampling at times kT, k = 0, 1, · · · and using the approximation
˙
s(kT ) = 1/T (s((k + 1)T ) − s(kT )). (This is called a forward-Euler approx-
imation.)
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13. LAPLACE AND Z TRANSFORMS
R
+
r
x(t)
L
+
-
C
v(t)
i(t)
-
Figure 13.5: Circuit of problem 24
25. E For the matrix A in example 13.24, determine etA,t ≥ 0.
26. E For the matrix A in example 13.26, determine An, n ≥ 0.
27. T A continuous-time SISO system has [A, b, c, d] representation with
a
b
A =
−
,
b
a
in which a, b are real constants.
(a) Find the eigenvalues of A.
(b) For what values of a, b is the SISO system stable?
(c) Calculate etA,t ≥ 0.
(d) Suppose b = c = [1 0]T , and d = 0. Find the transfer function.
28. T Let A be an N × N matrix. Let p be an eigenvalue of A. An N-dimensional
(column) vector e, possibly complex-valued, is said to be an eigenvector of A cor-
responding to p if e = 0 and Ae = pe. Note that an eigenvector always exists since
det[pI − A] = 0. Find eigenvectors for each of the two eigenvalues of the matrices
29. E Let A be a square matrix with eigenvalue p and corresponding eigenvector e.
Determine the response of the following.
(a) s(k + 1) = As(k), k ≥ 0; s(0) = e.
(b) ˙
s(t) = As(t),t ≥ 0; s(0) = e.
Hint. Show that Ane = pne and etAe = ept e.
Lee & Varaiya, Signals and Systems
607
EXERCISES
30. T Verify (13.49). Hint. First show that
1
1
s
[
sI − A]−1b =
,
sN + a
···
N−1sN−1 + · · · + a0
sN−1
by multiplying both sides by [sI − A]. Then check (13.49).
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14
Composition and Feedback
Control
Contents
14.1 Cascade composition
. . . . . . . . . . . . . . . . . . . . . . . . . 611
14.1.1 Stabilization
. . . . . . . . . . . . . . . . . . . . . . . . . . 611
14.1.2 Equalization
. . . . . . . . . . . . . . . . . . . . . . . . . . 612
14.2 Parallel composition . . . . . . . . . . . . . . . . . . . . . . . . . . 618
14.2.1 Stabilization
. . . . . . . . . . . . . . . . . . . . . . . . . . 619
14.2.2 Noise cancellation . . . . . . . . . . . . . . . . . . . . . . . 620
14.3 Feedback composition . . . . . . . . . . . . . . . . . . . . . . . . . 624
14.3.1 Proportional controllers
. . . . . . . . . . . . . . . . . . . . 626
14.4 PID controllers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 637
14.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 644
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 646
A major theme of this book is that interesting systems are often compositions of simpler
systems. Systems are functions, so their composition is function composition, as dis-
cussed in section 2.1.5. However, systems are often not directly described as functions,
so function composition is not the easiest tool to use to understand the composition. We
have seen systems described as state machines, frequency responses, and transfer func-
tions. In chapter 4 we obtained the state machine of the composite system from its compo-
nent state machines. In section 8.5 we obtained the frequency response of the composite
609
14.1. CASCADE COMPOSITION
x
w
y
H 1
H 2
H = H 1 H 2
Figure 14.1: Cascade composition of two LTI systems with transfer functions H1
and H2.
system from the frequency response of its component linear time-invariant (LTI) systems.
We extend the latter study in this chapter to the composition of LTI systems described by
their transfer functions. This important extension allows us to consider unstable systems
whose impulse response has a Z or Laplace transform, but not a Fourier transform.
As before, feedback systems prove challenging. A particularly interesting issue is how to
maintain stability, and how to construct stable systems out of unstable ones. We will find
that some feedback compositions of stable systems result in unstable systems, and con-
versely, some compositions of unstable systems result in stable systems. For example, we
can stabilize the helicopter in example 12.2 using feedback, in fact we can precisely con-
trol its orientation, despite the intrinsic instability. The family of techniques for doing this
is known as feedback control. This chapter serves as an introduction to that topic. Feed-
back control can also be used to drive stable systems, in which case it serves to improve
their response. For example, feedback can result in faster or more precise responses, and
can also prevent overshoot, where a system overreacts to a command.
We will consider three styles of composition, cascade composition, parallel composi-
tion, and feedback composition. In each case, two LTI systems with transfer functions
ˆ
H1 and ˆ
H2 are combined to get a new system. The transfer functions ˆ
H1 and ˆ
H2 are
the (Z or Laplace) transforms of the respective impulse responses, h1 and h2. Much of
our discussion applies equally well whether the system is a continuous-time system or a
discrete-time system, so in many cases we leave this unspecified.
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14. COMPOSITION AND FEEDBACK CONTROL
14.1
Cascade composition
Consider the cascade composition shown in figure 14.1. The composition is the grey box, and it has transfer function
ˆ
H = ˆ
H ˆ
1H2.
Notice that because of this simple form, if we know the pole and zero locations of the
component systems, then it is easy to determine the pole and zero locations of the compo-
sition. Unless a pole of one is cancelled by a zero of the other, the poles and zeros of the
composition are simply the aggregate of the poles and zeros of the components. More-
over, any pole of ˆ
H must be a pole of either ˆ
H1 or ˆ
H2, so if ˆ
H1 and ˆ
H2 are both stable,
then so is ˆ
H.
14.1.1
Stabilization
The possibility for pole-zero cancellation suggests that cascade composition might be
used to stabilize an unstable system.
Example 14.1: Consider a discrete-time system with transfer function
∀
z
z ∈ {z | |z| > |1.1|},
ˆ
H1(z) =
.
z − 1.1
This is a proper rational polynomial with a region of convergence of the form for a
causal signal, so it must be a causal system. However, it is not stable, because the
region of convergence does not include the unit circle.
To stabilize this system, we might consider putting it in cascade with
∀
z − 1.1
z ∈ C,
ˆ
H2(z) =
.
z
This is a causal and stable system. The transfer function of the cascade composition
is
z
z − 1.1
ˆ
H(z) =
= 1 .
z − 1.1
z
The pole at z = 1.1 has been cancelled, and the resulting region of convergence
is the entire complex plane. Thus, the cascade composition is a causal and stable
system, and we can recognize from table 13.1 that the impulse response is h(n) =
δ(n).
Lee & Varaiya, Signals and Systems
611
14.1. CASCADE COMPOSITION
Stabilizing systems by cancelling their poles in a cascade composition, however, is almost
never a good idea. If the pole is not precisely cancelled, then no matter how small the
error, the resulting system is still unstable.
Example 14.2:
Suppose that in the previous example the pole location is not
known precisely, and turns out to be at z = 1.1001 instead of z = 1.1. Then the
cascade composition has transfer function
z
z − 1.1
z − 1.1
ˆ
H(z) =
=
,
z − 1.1001
z
z − 1.1001
which is unstable.
14.1.2
Equalization
While cascade compositions do not usually work well for stabilization, they do often
work well for equalization. An equalizer is a compensator that reverses distortion. The
source of the distortion, which is often called a channel, must be an LTI system, and the
equalizer is composed in cascade with it. At first sight this is easy to do. If the channel
has transfer function ˆ
H1, then the equalizer could have transfer function
ˆ
H2 = ˆ
H−1,
1
in which case the cascade composition will have transfer function
ˆ
H = ˆ
H ˆ
1H2 = 1,
which is certainly distortion-free.
Example 14.3:
Some acoustic environments for audio have resonances, where
certain frequencies are enhanced as the sound propagates through the environment.
This will typically occur if the physics of the acoustic environment results in a
transfer function with poles near the unit circle (for a discrete-time model) or near
the imaginary axis (for a continuous-time model). Suppose for example that the
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14. COMPOSITION AND FEEDBACK CONTROL
acoustic environment is well modeled by a discrete-time LTI system with transfer
function
∀
z2
z ∈ {z | |z| > 0.95},
ˆ
H1(z) =
,
(z − a)(z − a∗)
where a = 0.95ei