Signal3

z-Transforms
1

Deﬁnitions
• Given an input signal x[n], the two-sided z-transform is given by
∞

X(z) =

x[n]z −n .

(1)

k=−∞

• Given an input signal x[n], the one-sided z-transform is given by
∞

X(z) =

x[n]z −n .

(2)

k=0

• By default, the term z-transform means the two-sided z-transform, i.e.,
if we simply say z-transform without explicitly specifying if it is a onesided z-transform or a two-sided z-transform, then we mean a two-sided
z-transform.
• A series of the form (1) is called a Laurent series.
• Notation: The two-sided z-transform of a signal x[n] is denoted by
z{x[n]}.

2

Basic Properties of the z-transform
• If x[n] is a causal signal, then the two-sided z-transform of x[n] is equal
to the one-sided z-transform of x[n], i.e., if x[n] is a causal signal, then
the z-transform of x[n] contains only negative powers of z.
• If x[n] is of ﬁnite length and is zero for n > N2 and n < −N1 , then the
limits of summation in (1) reduce to −N1 and N2 , i.e., the two-sided
z-transform of such a signal x[n] is
N2

X(z) =

x[n]z −n .
k=−N1

1

(3)

• Linearity: z{αx1 [n] + βx2 [n]} = αz{x1 [n]} + βz{x2 [n]}
• z{x[n − n0 ]} = z −n0 z{x[n]} ... time shift by n0 is equivalent to multiplication of the z-transform by z −n0 .

3

Examples
• z{δ[n]} = 1
• z{δ[n − n0 ]} = z −n0 for any constant n0 .
• z{αn u[n]} =

1
1−αz −1

for any constant α.

• z{−αn u[−n − 1]} =

4

1
1−αz −1

for any constant α.

More Properties of the z-transform
• z{x1 [n] ∗ x2 [n]} = z{x1 [n]}z{x2 [n]} ... convolution in the time domain is equivalent to multiplication in the z domain, i.e., convolution
is mathematically equivalent to polynomial multiplication.
Proof: Let x[n] = x1 [n] ∗ x2 [n], i.e., x[n] =

∞
k=−∞

x1 [k]x2 [n − k]. Then,

∞

z{x[n]} =

x[n]z −n
n=−∞
∞

∞

x1 [k]x2 [n − k]z −n

=
n=−∞ k=−∞
∞

∞

x2 [n − k]z −n

x1 [k]

=

n=−∞

k=−∞
∞

=

∞

x1 [k]z

x2 [n − k]z −(n−k)

−k
n=−∞
∞

k=−∞
∞

x1 [k]z −k

=

x2 [n1 ]z −n1
n1 =−∞

k=−∞

2

(4)

where n1 = n − k. Hence,
∞

∞

x1 [k]z −k

z{x[n]} =

x2 [n1 ]z −n1
n1 =−∞

k=−∞

= z{x1 [n]}z{x2 [n]}.

(5)

• If z{x[n]} = X(z), then z{an x[n]} = X(a−1 z).
Proof:
∞

z{an x[n]} =

an x[n]z −n
n=−∞
∞

=

x[n](a−1 z)−n
n=−∞
−1

= X(a z).

(6)

• If z{x[n]} = X(z), then z{nx[n]} = −z dX(z) .
dz
Proof: Note that
∞

d
dX(z)
=
dz
dz

x[n]z −n

n=−∞
∞

(−n)x[n]z −n−1

=
n=−∞

∞

= −z

nx[n]z −n .

−1

(7)

n=−∞

Hence,
∞

z{nx[n]} =

nx[n]z −n = −z
n=−∞

dX(z)
.
dz

(8)

• If x[n] is a causal signal, then x[0] = limz→∞ X(z).
Proof: If x[n] is a causal signal, then x[n] = 0 for n < 0 so that X(z)
contains only negative powers of z, i.e., X(z) is of the form
X(z) = x[0] + x[1]z −1 + x[2]z −2 + x[3]z −3 + . . .

(9)

Taking z → ∞, only the ﬁrst term in (9) remains so that
lim X(z) = x[0].

z→∞

3

(10)

• If z{x[n]} = X(z), then z{x[−n]} = X(z −1 ).
Proof:
∞

z{x[−n]} =

x[−n]z −n
n=−∞
−∞

x[n1 ]z n1

=

(11)

n1 =∞

where n1 = −n. Hence,
∞

z{x[−n]} =

x[n1 ](z −1 )−n1
n1 =−∞
−1

= X(z ).

(12)

• If z{x[n]} = X(z), then z{x∗ [n]} = X ∗ (z ∗ ).
Proof: Note that
∞

X ∗ (z ∗ ) =

x[n](z ∗ )−n

∗

n=−∞
∞

=

x∗ [n]z −n

(13)

n=−∞

since (z ∗ )∗ = z. Hence,
∞

x∗ [n]z −n = X ∗ (z ∗ ).

z{x∗ [n]} =

(14)

n=−∞
1
• If z{x[n]} = X(z), then z{Re(x[n])} = 2 [X(z) + X ∗ (z ∗ )].

Proof: From the property proved above, we know that z{x∗ [n]} =
X ∗ (z ∗ ). Noting that
1
(x[n] + x∗ [n]),
(15)
Re(x[n]) =
2
the result follows.
• If z{x[n]} = X(z), then z{Im(x[n])} =

1
[X(z)
2j

− X ∗ (z ∗ )].

Proof: We know that z{x∗ [n]} = X ∗ (z ∗ ). Noting that
1
(x[n] − x∗ [n]),
Im(x[n]) =
2j
the result follows.
4

(16)

5

Region of Convergence
• The set of values of z for which the summation in the definition of the
z-transform (1) converges is called the Region of Convergence (ROC).
The ROC is a subset of the complex plane.
• The ROC of the z-transform of a finite length signal is the entire complex plane. Since the z-transform of a finite length signal has only a
finite number of terms in the summation, the z-transform converges for
all z.
• From the theory of complex analysis, it is known that the ROC of the
z-transform of any signal is a connected set and is bounded by circles
(with the centers of the circles being at the origin of the complex plane).
Hence, there are three possibilities for the ROC as shown in Figure 1:
– Interior of a circle
– Exterior of a circle
– Annular region between two circles
−n −jωn
• If z = rejω , then X(z) = ∞
e
. It can be shown that
n=−∞ x[n]r
X(z) converges if and only if
∞

|x[n]r −n | < ∞.

(17)

n=−∞

Example: If x[n] = u[n], then (17) reduces to ∞ |r−n | < ∞ which
n=0
holds if |r −1 | < 1, i.e., if |r| > 1. Hence, the ROC of the z-transform
of u[n] is 1 < |z| < ∞, i.e., the exterior of a circle with radius 1.
Example: If x[n] = αn u[n], then (17) reduces to ∞ |αn r−n | < ∞
n=0
which holds if |αr −1 | < 1, i.e., if |r| > |α|. Hence, the ROC of the
z-transform of αn u[n] is |α| < |z| < ∞, i.e., the exterior of a circle with
radius |α|.
Example: If x[n] = −αn u[−n−1], then (17) reduces to −∞ |αn r−n | <
n=−1
∞ which holds if |α−1 r| < 1, i.e., if |r| < |α|. Hence, the ROC of the
z-transform of −αn u[−n−1] is 0 ≤ |z| < |α|, i.e., the interior of a circle
with radius |α|.
5

Figure 1: Possibilities for the region of convergence: (a) interior of a circle;
(b) exterior of a circle; (c) annular region between two circles.
• From the above examples, it can be generalized that the ROC of the
z-transform of a causal signal is the exterior of a circle and the ROC
of the z-transform of an anti-causal signal (i.e., a signal which is zero
for n > 0) is the interior of a circle.
• The above statement can be generalized to right-sided and left-sided
signals. A right-sided signal is a signal which is zero for all n smaller
than some constant n1 . A left-sided signal is a signal which is zero
for all n greater than some constant n1 . The ROC of the z-transform
of a right-sided signal is the exterior of a circle while the ROC of the
z-transform of a left-sided signal is the interior of a circle (possibly not
including the origin).
• Example: Consider the signal
αn for n ≥ 0
β n for n ≤ −1.

x[n] =

(18)

The signal x[n] in (18) is a two-sided signal. If |α| < |β|, then the ROC
6

of the z-transform of the signal x[n] is the annular region |α| < |z| < |β|.
If |α| ≥ |β|, then the z-transform of x[n] does not converge for any z.
• Special Case: If X(z) is a rational function (i.e., a ratio of polynomials), then the ROC is bounded by the poles of X(z), i.e., the values of
z for which X(z) is inﬁnity.
Example: For the signal in (18), the z-transform is
∞

−∞

αn z −n +

X(z) =
n=0
∞

β n z −n
n=−1
∞

β −n z n

αn z −n +

=

n=1

n=0

1
1
+
−1
=
−1
1 − αz
1 − β −1 z
β −1 z
1
+
=
1 − αz −1 1 − β −1 z
1
1
=
−
.
−1
1 − αz
1 − βz −1

(19)

From (19), the poles of X(z) are z = α and z = β. As seen above, if
|α| < |β| the ROC of X(z) is the annular region |α| < |z| < |β|.
Example: If X(z) is a rational function, then the ROC of X(z) is
always bounded by poles. This means that the boundaries of the ROC
have poles on them. It does not mean that if X(z) has two poles, then
the ROC is the annular region between the poles. For example, consider
the signal x[n] = αn u[n] + β n u[n]. It can be shown that the ROC of
the z-transform of X(z) is max(|α|, |β|) < |z| < ∞, i.e., the exterior
of a circle with radius max(|α|, |β|). The boundary of this ROC is the
circle with radius max(|α|, |β|) which has a pole on it (the pole being
the larger of α and β). Note that the poles of X(z) are α and β.

7

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