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m=−∞
Notice that if this series converges, then so does
∞
∑ |x(m)z−m|
m=−∞
Lee & Varaiya, Signals and Systems
509
12.2. THE Z TRANSFORM
Im z
Im z
Im z
Re z
Re z
Re z
RoC
RoC
RoC
causal or right sided
two-sided
anti-causal
(a)
(b)
(c)
Figure 12.2: Three possible structures for the region of convergence of a Z trans-
form.
for any complex number z with magnitude r. This is because
|x(m)z−m| = |x(m)(reiω)−m| = |x(m)| · |r−m| · |e−iωm| = |x(m)| · |r−m|.
Thus, the set RoC could equally well be defined to be the set of all complex numbers z
such that x(n)z−n is absolutely summable.
Notice that whether this series converges depends only on r, the magnitude of the complex
number z = reiω, and not on ω, its angle. Thus, if any point z = reiω is in the set RoC,
then all points z with the same magnitude are also in RoC. This implies that the set RoC,
a subset of C, will have circular symmetry.
The set RoC turns out to have even more structure. There are only three possible patterns,
illustrated by the shaded areas in Figure 12.2. Each figure illustrates the complex plane,
and the shaded area is a region of convergence. Each possibility has circular symmetry,
in that whether a point is in the RoC depends only on its magnitude.
Figure 12.2(a) shows the RoC of a causal signal. A discrete-time signal x is causal if
x(n) = 0 for all n < 0. The RoC is the set of complex numbers z = reiω where following
series converges:
∞
∑ |x(m)r−m|.
m=−∞
But if x is causal, then
∞
∞
∑ |x(m)r−m| = ∑ |x(m)r−m|.
m=−∞
m=0
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Lee & Varaiya, Signals and Systems
12. STABILITY
If this series converges for some given r, then it must also converge for any ˜r > r (because
for all m ≥ 0, ˜r−m < r−m. Thus, if z ∈ RoC, then the RoC must include all points in the
complex plane on the circle passing through z and every point outside that circle.
Note further that not only must the RoC include every point outside the circle, but the
series must also converge in the limit as z goes to infinity. Thus, for example, H(z) = z
cannot be the Z transform of a causal signal because its RoC cannot possibly include
infinity (H(z) is not finite there).
Figure 12.2(c) shows the RoC of an anti-causal signal. A discrete-time signal x is anti-
causal if x(n) = 0 for all n > 0. By a similar argument, if z ∈ RoC, then the RoC must
include all points in the complex plane on the circle passing through z and every point
inside that circle.
Figure 12.2(b) shows the RoC of a signal that is neither causal nor anti-causal. Such a
signal is called a two-sided signal. Such a signal can always be expressed as a sum of
a causal signal and an anti-causal signal. The RoC is the intersection of the regions of
convergence for these two components. To see this, just note that the RoC is the set of
complex numbers z = reiω where following series converges:
∞
−1
∞
∑ |x(m)r−m| = ∑ |x(m)r−m| + ∑ |x(m)r−m|.
m=−∞
m=−∞
m=0
The first sum on the right corresponds to an anti-causal signal, and the second sum on the
right to a causal signal. For this series to converge, both sums must converge. Thus, for a
two-sided signal, the RoC has a ring structure.
Example 12.7: Consider the discrete-time unit step signal u, given by
0, n < 0
u(n) =
.
(12.13)
1, n ≥ 0
The Z transform is, using geometric series identity (12.9),
∞
∞
1
z
Û (z) = ∑ u(m)z−m = ∑ z−m =
=
,
1 − z−1
z − 1
m=−∞
m=0
with domain
∞
RoC(u) = {z ∈ C | ∑ |z|−m < ∞} = {z | |z| > 1}.
m=1
Lee & Varaiya, Signals and Systems
511
12.2. THE Z TRANSFORM
This region of convergence has the structure of Figure 12.2(a), where the dashed
circle has radius one (that circle is called the unit circle). Indeed, this signal is
causal, so this structure makes sense.
Example 12.8: The signal v given by
−1, n < 0
v(n) =
,
0,
n ≥ 0
has Z transform
∞
1
∞
z
ˆ
V (z) = ∑ v(m)z−m = − ∑ z−m = −z ∑ zk =
,
z − 1
m=−∞
m=−∞
k=0
with domain
1
RoC(v) = {z ∈ C | ∑ |z|−m < ∞} = {z | |z| < 1}.
m=−∞
This region of convergence has the structure of Figure 12.2(c), where the dashed
circle is again the unit circle. Indeed, this signal is anti-causal, so this structure
makes sense.
Notice that although the Z transform Û of u and ˆ
V of v have the same algebraic form,
namely, z/(z − 1), they are different functions, because their domains are different. Thus
the Z transform of a signal comprises both the algebraic form of the Z transform as well
as its RoC.
A right-sided signal x is where for some integer N,
x(n) = 0,
∀ n < N.
Of course, if N ≥ 0, then this signal is also causal. However, if N < 0, then the signal is
two sided. Suppose N < 0. Then we can write the Z transform of x as
∞
−1
∞
∑ |x(m)r−m| = ∑ |x(m)r−m| + ∑ |x(m)r−m|.
m=−∞
m=N
m=0
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Lee & Varaiya, Signals and Systems
12. STABILITY
The left summation on the right side is finite, and each term is finite for all z ∈ C, so
therefore it converges for all z ∈ C. Thus, the region of convergence is determined entirely
by the right summation, which is the Z transform of the causal part of x. Thus, the region
of convergence of a right-sided signal has the same form as that of a causal sequence, as
shown in figure 12.2(a). (However, if the signal is not causal, the Z transform does not
converge at infinity.)
A left-sided signal x is where for some integer N,
x(n) = 0,
∀ n > N.
Of course, if N ≤ 0, then this signal is also anti-causal. However, if N > 0, then the signal
is two sided. Suppose N > 0. Then we can write the Z transform of x as
∞
0
N
∑ |x(m)r−m| = ∑ |x(m)r−m| + ∑ |x(m)r−m|.
m=−∞
m=−∞
m=1
The right summation is finite, and therefore converges for all z ∈ C except z = 0, where
the individual terms of the sum are not finite. Thus, the region of convergence is that
of the left summation, except for the point z = 0. Thus, the region of convergence of a
left-sided signal has the same form as that of an anti-causal sequence, as shown in figure
12.2(c), except that the origin (z = 0) is excluded. This, of course, is simply the structure
of 12.2(b) where the inner circle has zero radius.
Some signals have no meaningful Z transform.
Example 12.9:
The signal x with x(n) = 1, for all n, does not have a Z trans-
form. We can write x = u − v, where u and v are defined in the previous examples.
Thus, the region of convergence of x must be the intersection of the regions of con-
vergence of u and v. However, these two regions of convergence have an empty
intersection, so RoC(x) = /0.
Viewed another way, the set RoC(x) is the set of complex numbers z where
∞
∞
∑ |x(m)z−m| = ∑ |z−m| < ∞.
m=−∞
m=−∞
But there is no such complex number z.
Lee & Varaiya, Signals and Systems
513
12.2. THE Z TRANSFORM
Note that the signal x in example 12.9 is periodic with any integer period p (because
x(n + p) = x(n) for any p ∈ Z). Thus, it has a Fourier series representation. In fact, as
shown in Section 10.6.3, a periodic signal also has a Fourier transform representation, as
long as we are willing to allow Dirac delta functions in the Fourier transform. (Recall
that this means that there are values of ω where X (ω) will not be finite.) With periodic
signals, the Fourier series is by far the simplest frequency-domain tool to use. The Fourier
transform can also be used if we allow Dirac delta functions. The Z transform, however,
is more problematic, because the region of convergence is empty.
12.2.2
Stability and the Z transform
If a discrete-time signal x is absolutely summable, then it has a DTFT X that is finite for
all ω ∈ R. Moreover, the DTFT is equal to the Z transform evaluated on the unit circle,
∀ ω ∈ R, X(ω) = ˆX(z)|z=eiω = ˆX(eiω).
The complex number z = eiω has magnitude one, and therefore lies on the unit circle. Re-
call that an LTI system is stable if and only if its impulse response is absolutely summable.
Thus,
A discrete-time LTI system with impulse response h is stable if and only if the transfer
function ˆ
H, which is the Z transform of h, has a region of convergence that includes the
unit circle.
Example 12.10: Continuing example 12.6, the transfer function of the bank ac-
count system has region of convergence given by
RoC(h) = {z = reiω ∈ C | r > a},
where a > 1. Thus, the region of convergence includes only complex numbers with
magnitude greater than one, and therefore does not include the unit circle. The bank
account system is therefore not stable.
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Lee & Varaiya, Signals and Systems
12. STABILITY
12.2.3
Rational Z tranforms and poles and zeros
All of the Z transforms we have seen so far are rational polynomials in z. A rational
polynomial is simply the ratio of two finite-order polynomials. For example, the bank
account system has transfer function
1
ˆ
H(z) = z−a
(see example 12.6). The unit step of example 12.7 and its anti-causal cousin of example
12.8 have Z transforms given by
z
z
Û (z) =
, ˆV(z) =
,
z − 1
z − 1
albeit with different regions of convergence.
In practice, most Z transforms of practical interest can be written as the ratio of two finite
order polynomials in z,
A(z)
ˆ
X (z) =
.
B(z)
The order of the polynomial A or B is the power of the highest power of z. For the unit
step, the numerator polynomial is A(z) = z, a first-order polynomial, and the denominator
is B(z) = z − 1, also a first-order polynomial.
Recall from algebra that a polynomial of order N has N (possibly complex-valued) roots,
which are values of z where the polynomial evaluates to zero. The roots of the numerator
A are called the zeroes of the Z transform, and the roots of the denominator B are called
the poles of the Z transform. The term “zero” refers to the fact that the Z transform
evaluates to zero at a zero. The term “pole” suggests an infinitely high tent pole, where
the Z transform evaluates to infinity. The locations in the complex plane of the poles and
zeros turn out to yield considerable insight about a Z transform. A plot of these locations
is called a pole-zero plot. The poles are shown as crosses and the zeros as circles.
Example 12.11: The unit step of example 12.7 and its anti-causal cousin of exam-
ple 12.8 have pole-zero plots shown in Figure 12.3. In each case, the Z transform has the form
z
A(z)
=
,
z − 1
B(z)
Lee & Varaiya, Signals and Systems
515
12.2. THE Z TRANSFORM
Im z
Im z
Re z
Re z
| z|=1
RoC( u)
| z|=1
RoC( v)
Figure 12.3: Pole-zero plots for the unit step u and its anti-causal cousin v. The
regions of convergence are the shaded area in the complex plane, not including
the unit circle. Both Z tranforms, Û and ˆ
V , have one pole at z = 1 and one zero at
z = 0.
where A(z) = z and B(z) = z − 1. A(z) has only one root, at z = 0, so the Z trans-
forms each have one zero, at the origin in the complex plane. B(z) also has only
one root, at z = 1, so the Z transform has one pole, at z = 1.
These plots also show the unit circle, with a dashed line, and the regions of conver-
gence for the two examples, as shaded areas. Note that RoC(u) has the form of a
region of convergence of a causal signal, as it should, and RoC(v) has the form of a
region of convergence of an anti-causal signal, as it should (see Figure 12.2). Note
that neither RoC includes the unit circle, so if these signals were impulse responses
of LTI systems, then these systems would be unstable.
Consider a rational Z transform
A(z)
ˆ
X (z) =
.
B(z)
The denominator polynomial B evaluates to zero at a pole. That is, if there is a pole at
location z = p (a complex number), then B(p) = 0. Assuming that A(p) = 0, then ˆ
X (p) is
not finite. Thus, the region of convergence cannot include any pole p that is not cancelled
by a zero. This fact, combined with the fact that a causal signal always has a RoC of the
form of the left one in Figure 12.2, leads to the following simple stability criterion for
causal systems:
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Lee & Varaiya, Signals and Systems
12. STABILITY
Im z
2
2
Re z
| z|=1
RoC( x)
Figure 12.4: Poles and zeros with multiplicity greater than one are indicated by a
number next to the cross or circle.
A discrete-time causal system is stable if and only if all the poles of its transfer function
lie inside the unit circle.
A subtle fact about rational Z transforms is that the region of convergence is always bor-
dered by the pole locations. This is evident in Figure 12.3, where the single pole at z = 1
lies on the boundary of the two possible regions of convergence. In fact, the rational
polynomial
z
z − 1
can be associated with only three possible Z transforms, two of which have the two regions
of convergence shown in Figure 12.3, plus the one not shown where RoC = /0.
Although a polynomial of order N has N roots, these roots are not necessarily distinct.
Consider the (rather trivial) polynomial
A(z) = z2.
This has order 2, and hence two roots, but both roots are at z = 0. Consider a Z transform
given by
∀
z2
z ∈ RoC(x),
ˆ
X (z) =
.
(z − 1)2
This has two zeros at z = 0, and two poles at z = 1. We say that the zero at z = 0 has
multiplicity two. Similarly, the pole at z = 1 has multiplicity two. This multiplicity is
indicated in a pole-zero plot by a number adjacent to the pole or zero, as shown in Figure
Lee & Varaiya, Signals and Systems
517
12.2. THE Z TRANSFORM
Example 12.12: Consider a signal x that is equal to the delayed Kronecker delta
function,
∀ n ∈ Z, x(n) = δ(n − M),
where M ∈ Z is a constant. Its Z transform is easy to find using the sifting rule,
∞
∀ z ∈ RoC(x),
ˆ
X (z) = ∑ δ(m − M)z−m = z−M = 1/zM.
m=−∞
If M > 0, then this converges absolutely for any z = 0. Thus, if M > 0,
RoC(x) = {z ∈ C | z = 0}.
This Z transform has M poles at z = 0. Notice that this region of convergence,
appropriately, has the form of that of a causal signal, Figure 12.2(a), but where the
circle has radius zero.
If M < 0, then the region of convergence is the entire set C, and the Z transform
has M zeros at z = 0. This signal is anti-causal, and its RoC matches the structure
of 12.2(c), where the radius of the circle is infinite. Note that this Z transform does
not converge at infinity, which it would have to do if the signal were causal.
If M = 0, then ˆ
X (z) = 1 for all z ∈ C, so RoC = C, and there are no poles or zeros.
This is a particularly simple Z transform.
Recall that for a causal signal, the Z transform must converge as z → ∞. The region
of convergence must include everything outside some circle, including infinity.1 This
implies that for a causal signal with a rational Z transform, the Z transform must be
proper. A rational polynomial is proper when the order of the numerator is smaller than
or equal to the order of the denominator. For example, if M = −1 in the previous example,
then x(n) = δ(n + 1) and ˆ
H(z) = z, which has numerator order one and denominator order
zero. It is not proper, and indeed, it does not converge as z → ∞. Any rational polynomial
that has a denominator of higher order than the numerator will not converge as z goes to
infinity, and hence cannot be the Z transform of a causal signal.
1Some texts consider poles and zeros at infinity, in which case a causal signal cannot have a pole at
infinity.
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Lee & Varaiya, Signals and Systems
12. STABILITY
In the following chapter, table 13.1 gives many common Z tranforms, all of which are
rational polynomials. Together with the properties discussed in the that chapter, we can
find the Z transforms of many signals.
12.3
The Laplace transform
Consider a continuous-time signal x that is not absolutely integrable. Consider the scaled
signal x given by2
σ
∀ t ∈ R, x (t) = x(t)e−σt,
(12.14)
σ
for some real number σ. Often, this signal is absolutely integrable when σ is chosen
appropriately. This new signal, therefore, will have a CTFT, even if x does not.
Example 12.13: Consider the impulse response of the simplified helicopter system
described in example 12.2. The output as a function of the input is given by
t
Z
∀
1
t ∈ R,
y(t) =
x(τ)dτ.
M
0
The impulse response is found by letting the input be a Dirac delta function and
using the sifting rule to get
∀ t ∈ R, h(t) = u(t)/M,
where u is the continuous-time unit step in (12.4). This is not absolutely integrable,
so this system is not stable. However, the scaled signal
∀ t ∈ R, h (t) = h(t)e−σt
σ
2The reason that this is different from the scaling by r−n used to get the Z transform is somewhat subtle.
The two methods are essentially equivalent, if we let r = eσ. But scaling by e−σt turns out to be more
convenient for continuous-time systems, as we will see.
Lee & Varaiya, Signals and Systems
519
12.3. THE LAPLACE TRANSFORM
is absolutely integrable if σ > 0. Its CTFT is
∞
Z
∀ σ > 0,∀ω ∈ R, H (
h(t)e−σt e−iωt dt
σ ω)
=
−∞
∞
1 Z
=
e−σt e−iωt dt
M
0
∞
1 Z
=
e−(σ+iω)t dt
M
0
1
=
.
M(σ + iω)
The last step in example 12.13 uses the following useful fact from calculus,
b
Z
1
ect dt =
(ecb − eca) ,
(12.15)
c
a
for any c ∈ C and a, b ∈ R ∪ {−∞, ∞} where ecb and eca are finite.
In general, the CTFT of the scaled signal x in (12.14) is
σ
Z
∞
∀ ω ∈ R, X (
x(t)e−(σ+iω)t dt.
σ ω) =
−∞
Notice that this is a function not just of ω, but also of σ. We are only sure it is valid for
values of σ that yield an absolutely integrable signal h .
σ
Define the complex number
s = σ + iω.
Then we can write this CTFT as
∞
∀ s ∈ RoC(x),
ˆ
X (s) = R x(t)e−st dt,
(12.16)
−∞
where ˆ
X is a function called the Laplace transform of x,
ˆ
X : RoC(x) → C
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Lee & Varaiya, Signals and Systems
12. STABILITY
where RoC(x) ⊂ C is given by
RoC(x) = {s = σ + iω ∈ C | x(t)e−σt is absolutely integrable.}
(12.17)
The Laplace tranform ˆ
H of the impulse response h of an LTI system is called the transfer
function of the system, just as with discrete-time systems.
Example 12.14: Continuing example 12.13, we can recognize from the form of
H (
σ ω) that the transfer function of the helicopter system is
∀
1
s ∈ RoC(h),
ˆ
H(s) =
.
Ms