pulse response h ∗ g. Show that
h ∗ g ≤ h × g .
Lee & Varaiya, Signals and Systems
533
EXERCISES
h 1
g 1
x
y
+
h 2
g 2
Figure 12.7: System composition for Exercise 13.
13. E Show that the series-parallel composition of Figure 12.7 is stable if the four
component systems are stable. Let h be the impulse response of the composition.
Express h in terms of the component impulse responses and then estimate h in
terms of the norms of the components.
14. E Let x be a discrete-time signal of finite duration, i.e. x(n) = 0 for n < M and n > N
where M and N are finite integers (positive or negative). Let ˆ
X be its Z transform.
(a) Show that all its poles (if any) are at z = 0.
(b) Show that if x is causal it has N poles at z = 0.
15. T This problem relates the Z and Laplace transforms. Let x be a discrete-time signal
with Z transform ˆ
X : RoC(x) → C. Consider the continuous-time signal y related to
x by
∞
∀t ∈ R, y(t) = ∑ x(n)δ(t −nT).
n=−∞
Here T > 0 is a fixed period. So y comprises delta functions located at t = nT of
magnitude x(n).
(a) Use the sifting property and the definition (12.16) to find the Laplace trans-
form ˆ
Y of y. What is RoC(y)?
(b) Show that ˆ
Y (s) = ˆ
X (esT ), where ˆ
X (esT ) is ˆ
X (z) evaluated at s = esT .
(c) Suppose ˆ
X (z) = 1 with RoC(x) = {z | |z| > 1}. What are ˆ
Y and RoC(y)?
z−1
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13
Laplace and Z Transforms
Contents
13.1 Properties of the Z tranform . . . . . . . . . . . . . . . . . . . . . 536
13.1.1 Linearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 539
13.1.2 Delay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 542
13.1.3 Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . 543
13.1.4 Conjugation . . . . . . . . . . . . . . . . . . . . . . . . . . . 544
13.1.5 Time reversal . . . . . . . . . . . . . . . . . . . . . . . . . . 545
13.1.6 Multiplication by an exponential . . . . . . . . . . . . . . . . 546
13.1.7 Causal signals and the initial value theorem . . . . . . . . . . 546
13.2 Frequency response and pole-zero plots . . . . . . . . . . . . . . . 548
13.3 Properties of the Laplace transform . . . . . . . . . . . . . . . . . 550
13.3.1 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . 553
13.3.2 Sinusoidal signals
. . . . . . . . . . . . . . . . . . . . . . . 553
13.3.3 Differential equations . . . . . . . . . . . . . . . . . . . . . . 554
13.4 Frequency response and pole-zero plots . . . . . . . . . . . . . . . 555
13.5 The inverse transforms . . . . . . . . . . . . . . . . . . . . . . . . 557
13.5.1 Inverse Z transform . . . . . . . . . . . . . . . . . . . . . . . 557
13.5.2 Inverse Laplace transform . . . . . . . . . . . . . . . . . . . 566
13.6 Steady state response . . . . . . . . . . . . . . . . . . . . . . . . . 568
13.7 Linear difference and differential equations . . . . . . . . . . . . . 571
13.7.1 LTI differential equations . . . . . . . . . . . . . . . . . . . . 577
13.8 State-space models . . . . . . . . . . . . . . . . . . . . . . . . . . . 582
13.8.1 Continuous-time state-space models . . . . . . . . . . . . . . 588
13.9 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594
Probing Further: Derivatives of Z transforms . . . . . . . . . . . . . 595
Probing Further: Inverse transform as an integral . . . . . . . . . . . 598
Probing Further: (Continued) Inverse transform . . . . . . . . . . . . 599
535
13.1. PROPERTIES OF THE Z TRANFORM
Probing Further: Differentiation and Laplace transforms . . . . . . . 600
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 601
In the previous chapter, we defined Laplace and Z transforms to deal with signals that
are not absolutely summable and systems that are not stable. The Z transform of the
discrete-time signal x is given by
∞
∀ z ∈ RoC(x),
ˆ
X (z) = ∑ x(m)z−m,
m=−∞
where RoC(x) is the region of convergence, the region in which the sum above converges
absolutely.
The Laplace transform of the continuous-time signal x is given by
∞
Z
∀ s ∈ RoC(x),
ˆ
X (s) =
x(t)e−st dt,
−∞
where RoC(x) is again the region of convergence, the region in which the integral above
converges absolutely.
In this chapter, we explore key properties of the Z and Laplace transforms and give ex-
amples of transforms. We will also explain how, given a rational polynomial in z or s,
plus a region of convergence, one can find the corresponding time-domain function. This
inverse transform proves particularly useful, because compositions of LTI systems, stud-
ied in the next chapter, often lead to rather complicated rational polynomial descriptions
of a transfer function.
Z transforms of common signals are given in table 13.1. Properties of the Z transform are
summarized in table 13.2 and elaborated in the first section below.
13.1
Properties of the Z tranform
The Z transform has useful properties that are similar to those of the four Fourier trans-
forms. They are summarized in table 13.2 and elaborated in this section.
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13. LAPLACE AND Z TRANSFORMS
Discrete-time signal
Z transform
Roc(x) ⊂ C Reference
∀ n ∈ Z
∀ z ∈ RoC(x)
x(n) = δ(n − M)
ˆ
X (z) = z−M
C
Example
z
x(n) = u(n)
ˆ
X (z) =
{z | |z| > 1} Example
z − 1
z
x(n) = anu(n)
ˆ
X (z) =
{z | |z| > |a|} Example
z − a
1
x(n) = anu(−n)
ˆ
X (z) =
{z | |z| < |a|} Exercise 1
1 − a−1z
in Chapter
x(n) = cos(ω0n)u(n)
ˆ
X (z) =
{z | |z| > 1} Example
z2 − z cos(ω0)
z2 − 2z cos(ω0) + 1
x(n) = sin(ω0n)u(n)
ˆ
X (z) =
{z | |z| > 1} Exercise 1
z sin(ω0)
,
z2 − 2z cos(ω0) + 1
1
x(n) =
ˆ
X (z) =
{z | |z| > |a|} ( 13.13)
1
(z − a)N
(n − 1) · · · (n − N + 1)
(N − 1)!
an−N u(n − N)
1
x(n) =
ˆ
X (z) =
{z | |z| < |a|} (13.14)
(z − a)N
(−1)N (N −1−n)···(1−n)
(N − 1)!
an−N u(−n)
Table 13.1: Z transforms of key signals. The signal u is the unit step (12.13), δ is
the Kronecker delta, a is any complex constant, ω0 is any real constant, M is any
integer constant, and N > 0 is any integer constant.
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537
13.1. PROPERTIES OF THE Z TRANFORM
Time domain
Frequency
RoC
Name
∀ n ∈ Z
domain
(reference)
∀ z ∈ RoC
w(n) = ax(n) + by(n)
ˆ
W (z) =
RoC(w) ⊃
Linearity
a ˆ
X (z) + b ˆ
Y (z)
RoC(x) ∩ RoC(y)
(Section 13.1.1)
y(n) = x(n − N)
ˆ
Y (z) = z−N ˆ
X (z)
RoC(y) = RoC(x)
Delay
(Section 13.1.2)
y(n) = (x ∗ h)(n)
ˆ
Y (z) = ˆ
X (z) ˆ
H(z)
RoC(y) ⊃
Convolution
RoC(x) ∩ RoC(h)
(Section 13.1.3)
y(n) = x∗(n)
ˆ
Y (z) = [ ˆ
X (z∗)]∗
RoC(y) = RoC(x)
Conjugation
(Section 13.1.4)
y(n) = x(−n)
ˆ
Y (z) = ˆ
X (z−1)
RoC(y) =
Time reversal
{z | z−1 ∈ RoC(x)} (Section 13.1.5)
d
y(n) = nx(n)
ˆ
Y (z) = −z
ˆ
X (z)
RoC(y) = RoC(x)
Scaling by n
dz
(page 595)
y(n) = a−nx(n)
ˆ
Y (z) = ˆ
X (az)
RoC(y) =
Exponential
{z | az ∈ RoC(x)}
scaling
(Section 13.1.6)
Table 13.2: Properties of the Z transform. In this table, a, b are complex constants,
and N is an integer constant.
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13. LAPLACE AND Z TRANSFORMS
13.1.1
Linearity
Suppose x and y have Z transforms ˆ
X and ˆ
Y , that a, b are two complex constants, and that
w = ax + by.
Then the Z transform of w is
∀ z ∈ RoC(w),
ˆ
W (z) = a ˆ
X (z) + b ˆ
Y (z).
This follows immediately from the definition of the Z transform,
∞
ˆ
W (z)
=
∑ w(m)z−m
m=−∞
∞
=
∑ (ax(m) + by(m))z−m
m=−∞
=
a ˆ
X (z) + b ˆ
Y (z).
The region of convergence of w must include at least the regions of convergence of x and y,
since if x(n)r−n and y(n)r−n are absolutely summable, then certainly (ax(n) + by(n))r−n
is absolutely summable. Conceivably, however, the region of convergence may be larger.
Thus, all we can assert in general is
RoC(w) ⊃ RoC(x) ∩ RoC(y).
(13.1)
Linearity is extremely useful because it makes it easy to find the Z transform of compli-
cated signals that can be expressed a linear combination of signals with known Z trans-
forms.
Example 13.1: We can use the results of example 12.12 plus linearity to find, for
example, the Z transform of the signal x given by
∀ n ∈ Z, x(n) = δ(n) + 0.9δ(n − 4) + 0.8δ(n − 5).
This is simply
ˆ
X (z) = 1 + 0.9z−4 + 0.8z−5.
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539
13.1. PROPERTIES OF THE Z TRANFORM
We can identify the poles by writing this as a rational polynomial in z (multiply top
and bottom by z5),
z5 + 0.9z + 0.8
ˆ
X (z) =
,
z5
from which we see that there are 5 poles at z = 0. The signal is causal, so the region
of convergence is the region outside the circle passing through the pole with the
largest magnitude, or in this case,
RoC(x) = {z ∈ C | z = 0}.
Example 13.1 illustrates how to find the transfer function of any finite impulse response
(FIR) filter. It also suggests that the transfer function of an FIR filter always has a region
of convergence that includes the entire complex plane, except possibly z = 0. The region
of convergence will also not include z = ∞ if the FIR filter is not causal.
Linearity can also be used to invert a Z transform. That is, given a rational polynomial and
a region of convergence, we can find the time-domain function that has this Z transform.
The general method for doing this will be considered in the next chapter, but for certain
simple cases, we just have to recognize familiar Z transforms.
Example 13.2: Suppose we are given the Z transform
∀
z5 + 0.9z + 0.8
z ∈ {z ∈ C | z = 0},
ˆ
X (z) =
.
z5
We can immediately recognize this as the Z transform of a causal signal, because
it is a proper rational polynomial and the region of convergence includes the entire
complex plane except z = 0 (thus, it has the form of Figure 12.2(a)).
If we divide through by z5, this becomes
∀ z ∈ {z ∈ C | z = 0},
ˆ
X (z) = 1 + 0.9z−4 + 0.8z−5.
By linearity, we can see that
∀ n ∈ Z, x(n) = x1(n) + 0.9x2(n) + 0.8x3(n),
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13. LAPLACE AND Z TRANSFORMS
where x1 has Z transform 1, x2 has Z transform z−4, and x3 has Z transform z−5.
The regions of convergence for each Z transform must be at least that of x, or at
least {z ∈ C | z = 0}. From example 12.12, we recognize these Z transforms, and
hence obtain
∀ n ∈ Z, x(n) = δ(n) + 0.9δ(n − 4) + 0.8δ(n − 5).
Another application of linearity uses Euler’s relation to deal with sinusiodal signals.
Example 13.3: Consider the exponential signal x given by
∀ n ∈ Z, x(n) = anu(n),
where a is a complex constant. Its Z transform is
∞
∞
1
z
ˆ
X (z) = ∑ x(m)z−m = ∑ amz−m =
=
,
(13.2)
1 − az−1
z − a
m=−∞
m=0
where we have used the geometric series identity (12.9). This has a zero at z = 0
and a pole at z = a. The region of convergence is
∞
RoC(x) = {z ∈ C | ∑ |a|m|z|−m < ∞} = {z | |z| > |a|},
(13.3)
m=0
the region of the complex plane outside the circle that passes through the pole. A
pole-zero plot is shown in Figure 13.1(a).
We can use this result plus linearity of the Z transform to determine the Z transform
of the causal sinusoidal signal y given by
∀ n ∈ Z, y(n) = cos(ω0n)u(n).
Euler’s relation implies that
1
y(n) =
{eiω0nu(n) + e−iω0nu(n)}.
2
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541
13.1. PROPERTIES OF THE Z TRANFORM
Im z
Im z
a
ei w0
Re z
Re z
e-i w0
RoC( x)
| z|=| a|
RoC( y)
| z|=1
(a)
(b)
Figure 13.1: Pole-zero plots for the exponential signal x and the sinusoidal signal
y of example 13.3.
Using (13.2) and linearity,
1
z
z
ˆ
Y (z)
=
+
2
z − eiω0
z − e−iω0
1 2z2 − z(eiω0 + e−iω0)
=
2 (z − eiω0)(z − e−iω0)
z(z − cos(ω0))
=
.
z2 − 2z cos(ω0) + 1
This has a zero at z = 0, another zero at z = cos(ω0), and two poles, one at z = eiω0
and the other at z = e−iω0 . Both of these poles lie on the unit circle. A pole-zero
plot is shown in Figure 13.1(b), where we assume that ω0 = π/4. We know from
(13.1) and (13.3) that the region of convergence is at least the area outside the unit circle. In this case, we can conclude that it is exactly the area outside the unit circle,
because it must be bordered by the poles, and it must have the form of a region of
convergence of a causal signal.
13.1.2
Delay
For any integer N (positive or negative) and signal x, let y = DN(x) be the signal given by
∀n ∈ Z, y(n) = x(n − N).
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13. LAPLACE AND Z TRANSFORMS
Suppose x has Z transform ˆ
X with domain RoC(x). Then RoC(y) = RoC(x) and
∞
∞
∀z ∈ RoC(y), ˆY(z) = ∑ y(m)z−m = ∑ x(m−N)z−m = z−N ˆX(z).
(13.4)
m=−∞
m=−∞
Thus
If a signal is delayed by N samples, its Z transform is multiplied by z−N .
13.1.3
Convolution
Suppose x and h have Z transforms ˆ
X and ˆ
H. Let
y = x ∗ h.
Then
∀z ∈ RoC(y), ˆY(z) = ˆX(z) ˆH(z).
(13.5)
This follows from using the definition of convolution,
∞
∀n ∈ Z, y(n) = ∑ x(m)h(n−m),
m=−∞
in the definition of the Z transform,
∞
∞
∞
ˆ
Y (z)
=
∑ y(n)z−n = ∑
∑ x(m)z−mh(n − m)z−(n−m)
n=−∞
n=−∞ m=−∞
∞
∞
=
∑
∑ x(m)z−mh(l)z−l = ˆ
X (z) ˆ
H(z).
l=−∞ m=−∞
The Z transform of y converges absolutely at least at values of z where both ˆ
X and ˆ
H
converge absolutely. Thus,
RoC(y) ⊃ RoC(x) ∩ RoC(h).
This is true because the double sum above can be written as
∞
∞
∞
∑ y(n)z−n =
∑ x(m)z−m
∑ h(l)z−l .
n=−∞
m=−∞
l=−∞
Lee & Varaiya, Signals and Systems
543
13.1. PROPERTIES OF THE Z TRANFORM
This obviously converges absolutely if each of the two factors converges absolutely. Note
that the region of convergence may actually be larger than RoC(x) ∩ RoC(h). This can
occur, for example, if the product (13.5) results in zeros of ˆ
X (z) cancelling poles of ˆ
H(z)
(see Exercise 3).
If h is the impulse response of an LTI system, then its Z transform is called the transfer
function of the system. The result (13.5) tells us that the Z transform of the output is the
product of the Z transform of the input and the transfer function. The transfer function,
therefore, serves the same role as the frequency response. It converts convolution into
simple multiplication.
13.1.4
Conjugation
Suppose x is a complex-valued signal. Let y be defined by
∀ n ∈ Z, y(n) = [x(n)]∗.
Then
∀ z ∈ RoC(y), ˆY(z) = [ ˆX(z∗)]∗,
where
RoC(y) = RoC(x).
This follows because
∞
∀ z ∈ RoC(x), ˆY(z) =
∑ y(n)z−n
n=−∞
∞
=
∑ x∗(n)z−n
n=−∞
∗
∞
=
∑ x(n)(z∗)−n
n=−∞
=
[ ˆ
X (z∗)]∗.
If x happens to be a real signal, then y = x, so ˆ
Y = ˆ
X , so
ˆ
X (z) = [ ˆ
X (z∗)]∗.
The key consequence is:
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13. LAPLACE AND Z TRANSFORMS
For the Z transform of a real-valued signal, poles and zeros occur in complex-conjugate
pairs. That is, if there is a zero at z = q, then there must be a zero at z = q∗, and if there
is a pole at z = p, then there must be a pole at z = p∗.
This is because
0 = ˆ
X (q) = ( ˆ
X (q∗))∗
Similarly, if there is a pole at z = p, then there must also be a pole at z = p∗.
Example 13.4: Example 13.3 gave the Z transform of a signal of the form x(n) =
anu(n), where a is allowed to be complex, and the Z tranform of a signal of the
form y(n) = cos(ω0n)u(n), which is real-valued. The pole-zero plots are shown
in Figure 13.1. In that figure, the complex signal has a pole at z = a, and none at
z = a∗. But the real signal has a pole at z = eiω0 and a matching pole at the complex
conjugate, z = e−iω0 .
13.1.5