The z Transform
Relation to the Laplace Transform
• The z transform is to DT signals and systems
what the Laplace transform is to CT signals
and systems
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Definition
The z transform can be viewed as a generalization of the
DTFT or as natural result of exciting a discrete-time
LTI system with its eigenfunction. The DTFT is
defined by
∞
1
F
jΩn
− jΩn
x[n ] =
X
j
Ω
e
d
Ω
←

→
X
j
Ω
=
x
n
e
(
)
(
)
[
]
∑
2π ∫2π
n =−∞
If a strict analogy with the Laplace transform were
made Ω would replace ω, Σ would replace σ, S would
replace s, a summation would replace the integral and
the z transform would be defined by
X( S ) =
∞
∑
n =−∞
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x[n ]e − Sn =
∞
∑
n =−∞
x[n ]e − ( Σ + jΩ ) n =
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∞
∑ (x[n]e )e
− nΣ
− jΩn
n =−∞
3
Definition
X( S ) =
∞
∑
n =−∞
x[n ]e − Sn =
∞
∑
n =−∞
x[n ]e −( Σ + jΩ ) n =
∞
∑ (x[n]e )e
− nΣ
− jΩn
n =−∞
Viewed this way the factor, e − nΣ, would be a “convergence”
factor in that same way that the factor, e − σt , was for the
Laplace transform.
The other approach to defining the z transform is to excite a
n
DT system with its eigenfunction, Az . The response would
be
y[n ] = x[n ] ∗ h[n ] = Az n ∗ h[n ] =
∞
∑
m =−∞
∞
−m
h
m
z
[
]
∑
=−∞
x[ n ] m
1
4243
h[ m ]Az ( n − m ) = {
Az n
z transform of h [ n ]
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Definition
The universally accepted definition of the z transform of a DT
function, x, is
X( z ) =
∞
∑
n =−∞
x[n ]z − n
and x and X form a “z-transform pair”,
Z
x[n ]←
→ X( z )
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Convergence
The DTFT’s of some common functions do not, in the strict
sense, converge. The DTFT of the unit sequence would be
X( jΩ) =
∞
∑
n =−∞
u[n ]e
− jΩn
∞
= ∑ e − jΩn
n=0
which does not converge. But the z transform of the unit
sequence does exist. It is
X( z ) =
∞
∑
n =−∞
∞
u[n ]z − n = ∑ z − n
n=0
and the z transform exists for values of z whose magnitudes are
greater than one. This defines a region of convergence (ROC)
for the z transform of the unit sequence, the exterior of the unit
circle in the z plane.
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Convergence
∞
The series, ∑ z − n, is a geometric series. The general formula
n=0
for the summation of a finite geometric series is
, r =1
1

n
r = 1 − r N
∑
n=0
 1 − r , r ≠ 1
N −1
This formula also applies to the infinite series above if the
magnitude of z is greater than one. In that case the z transform
of the unit sequence is
1
z
X( z ) =
=
−1 , z > 1
z −1 1− z
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Transfer Functions
If x is the excitation, h is the impulse response and y is the
system response of a discrete-time LTI system, then
Y( z ) = X( z ) H( z )
and H is called the transfer function of the system. This
is directly analogous to previous transfer functions,
Y( jω ) = X( jω ) H( jω )
Y( jΩ) = X( jΩ) H( jΩ)
Y( s ) = X( s ) H( s )
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Region of Convergence
Taking a path analogous to that used the development of the
Laplace transform, the z transform of the causal DT signal
Aα n u[n ] , α > 0
is
∞
∞
n =−∞
n=0
X( z ) = A ∑ α n u[n ]z − n = A ∑ α n z − n
α
= A∑
 
n=0 z
∞
n
and the series converges if |z| > |α|. This
defines the region of convergence as the
exterior of a circle in the z plane centered
at the origin, of radius, |α|. The z transform
is
z
X( z ) = A
, z>α
z −α
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Region of Convergence
By similar reasoning, the z transform and region of convergence
of the anti-causal signal, Aα − n u[− n ] , α > 0 are
1
A
Az −1
X( z ) =
= −1
, z<
1 − αz z − α
α
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The Unilateral z Transform
Just as it was convenient to define a unilateral Laplace transform
it is convenient for analogous reasons to define a unilateral z
transform
∞
X( z ) = ∑ x[n ]z − n
n=0
which will simply be referred to as the z transform from this
point on.
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Properties
If two causal DT signals form these transform pairs,
Z
Z
g[n ]←
→ G( z ) and h[n ]←
→ H( z )
then the following properties hold for the z transform.
Linearity
α g[n ] + β h[n ]←Z → α G( z ) + β H( z )
Time Shifting
Delay:
Z
g[ n − n0 ] ←
→ z − n0 G( z ) , n0 ≥ 0
n0 −1

Z
n0
−m 
Advance: g[ n + n0 ] ←→ z  G( z ) − ∑ g[ m ]z  , n0 > 0


m =0
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Properties
Change of Scale
 z
α g[n ]←→ G
α
Initial Value Theorem
Z
n
g[0 ] = lim G( z )
z →∞
z-Domain Differentiation
d
− n g[n ]←→ z G( z )
dz
Z
Convolution in Discrete Time
Z
g [ n ] ∗ h [ n ] ←
→ H (z)G (z)
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Properties
Differencing
(
)
Z
g [ n ] − g [ n − 1] ←
→ 1 − z −1 G ( z )
Accumulation
n
1
z
∑ g [ m ]←→ z − 1 G ( z ) = 1 − z −1 G ( z )
m=0
Z
Final Value Theorem
lim g[n ] = lim( z − 1) G( z )
n→∞
z →1
(if the limit exists)
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The Inverse z Transform
There is an inversion integral for the z transform,
1
n −1
x[n ] =
X
z
z
dz
(
)
∫
j 2π C
but doing it requires integration in the complex plane and
it is rarely used in engineering practice.
There are two other common methods,
Synthetic Division
Partial-Fraction Expansion
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Synthetic Division
Suppose it is desired to find the inverse z transform of
2
z
z3 −
2
H( z ) =
1
15 2 17
3
3
67 −2
z − z + z−
1 + z −1 +
z +L
12
36 18
4
144
Synthetically dividing
the numerator by the
denominator yields the
infinite series
3 −1 67 −2
+ z +
z +L
4
144
This will always work but
is not in closed form.
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)
15 2 17
1 3 z2
z − z + z−
z −
12
36 18
2
15
17
1
z 3 − z2 + z −
12
36 18
3 2 17
1
z − z+
4
36 18
3 2 45
51
3
− z −1
z − z+
4
48 144 72
67
zL
the answer
144
M
M
3
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Partial-Fraction Expansion
Algebraically, partial fraction expansion for finding inverse
z transforms is identical to the same method applied to
inverse Laplace transforms. For example,
1

z z−

2
H( z ) =
2 
1 
1

z−
z−
z−





3
3
4
2
This fraction is improper in z. We could synthetically
divide the numerator by the denominator once, yielding a
remainder that is proper in z as with the Laplace transform
but there is an alternate method that may be preferred in
some situations.
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Partial-Fraction Expansion
1

z z−

H( z )
2
=
2 
1 
1
z

z−
z−
z−





3
3
4
Dividing both sides by z makes the fraction proper in z and
partial fraction expansion proceeds normally.
4
9
H( z )
2
5
=
+
− 5
2
1
1
z
z−
z−
z−
3
3
4
Then
4
9
z
z
2
z
H( z ) = 5 +
− 5
1
2
1
z−
z−
z−
3
3
4
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Solving Difference Equations
The unilateral z transform is well suited to solving difference
equations with initial conditions. For example,
3
1
 1
y[n + 2 ] − y[n + 1] + y[n ] =
 4
2
2
n
, for n ≥ 0
y[0 ] = 10 and y[1] = 4
z transforming both sides,
[
]
3
1
z
z Y( z ) − y[0 ] − z y[1] − z[ Y( z ) − y[0 ] ] + Y( z ) =
1
2
2
z−
4
the initial conditions are called for systematically.
2
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Solving Difference Equations
Applying initial conditions and solving,
2 
 16
4

3
Y( z ) = z
+
+ 3 

1
1 z − 1
z−

z
−


4
2
and
n
16  1  n
1
  2
y[n ] = 
+4
+  u[n ]
 2
3
 3  4
This solution satisfies the difference equation and the initial
conditions.
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z Transform - Laplace Transform
Relationships
Let a signal, x(t), be sampled to form
x[n ] = x( nTs )
and impulse sampled to form
xδ (t ) = x(t ) fs comb( fs t )
These two signals are equivalent in the sense that their
impulse strengths are the same at corresponding times and
the correspondence between times is t = nTs .
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z Transform - Laplace Transform
Relationships
Let a DT system have the impulse response, h[n], and let a CT
system have the impulse response, hδ (t ) =
∞
∑ h[n]δ (t − nT ) .
s
n =−∞
If x[n] is applied to the
DT system and xδ (t ) is
applied to the CT system,
their responses will be
equivalent in the sense
that the impulse strengths
are the same.
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z Transform - Laplace Transform
Relationships
The transfer function of the DT system is
H( z ) =
∞
∑
n =−∞
h[n ]z − n
and the transfer function of the CT system is
Hδ ( s ) =
∞
∑
n =−∞
h[n ]e − nTs s
The equivalence between them is seen in the transformation,
Hδ ( s ) = H( z ) z→e sTs
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z Transform - Laplace Transform
Relationships
The relationship, z = e sTs, maps points in the s plane into points
in the z plane and vice versa.
Different contours in the s plane map into the same contour in
the z plane.
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z Transform - Laplace Transform
Relationships
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The Bilateral z Transform
The bilateral z transform can be used to analyze non-causal
signals and/or systems. It is defined by
X( z ) =
∞
∑
n =−∞
x[n ]z
−n
∞
= ∑ x[n ]z
−n
+
n=0
−1
∑
n =−∞
x[n ]z − n
This can be manipulated into
X( z ) = Xc ( z ) − x[0 ] + X ac ( z )
where
∞
Xc ( z ) = ∑ x[n ]z
n=0
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−n
∞
and
X ac ( z ) = ∑ x[− n ]z n
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n=0
26
The Bilateral z Transform
The bilateral z transform can be found using the unilateral
z transform by these four steps.
1. Find the unilateral z transform Xc ( z ) and its ROC.
1
2. Find the unilateral z transform, X ac   , of the discrete z
time inverse of the anti-causal part of x[n].
1
3. Make the change of variable, z → , in the result of
z
step 2 and in its ROC.
4. Add the results of steps 1 and 3 and subtract x[0] to
form X(z).
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