Digital Signal Processing
The z-Transform and Its Application to the Analysis of LTI Systems
Moslem Amiri, V´aclav Pˇrenosil
Embedded Systems Laboratory
Faculty of Informatics, Masaryk University
Brno, Czech Republic
amiri@mail.muni.cz
prenosil@fi.muni.cz
March, 2012
The Direct z-Transform
z-transform of x(n) is defined as power series:
X(z) ≡
∞
n=−∞
x(n)z−n
where z is a complex variable
For convenience
X(z) ≡ Z{x(n)}
x(n)
z
←→ X(z)
Since z-transform is an infinite power series, it exists only for those
values of z for which this series converges
Region of convergence (ROC) of X(z) is set of all values of z for
which X(z) attains a finite value
Any time we cite a z-transform, we should also indicate its ROC
ROC of a finite-duration signal is entire z-plane except possibly points
z = 0 and/or z = ∞
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The Direct z-Transform
Example
Determine z-transforms of following finite-duration signals
x(n) = {1
↑
, 2, 5, 7, 0, 1}
X(z) = 1 + 2z−1 + 5z−2 + 7z−3 + z−5
ROC: entire z-plane except z = 0
x(n) = {1, 2, 5
↑
, 7, 0, 1}
X(z) = z2 + 2z + 5 + 7z−1 + z−3
ROC: entire z-plane except z = 0 and z = ∞
x(n) = δ(n)
X(z) = 1 [i.e., δ(n)
z
←→ 1], ROC: entire z-plane
x(n) = δ(n − k), k > 0
X(z) = z−k [i.e., δ(n − k)
z
←→ z−k], ROC: entire z-plane except z = 0
x(n) = δ(n + k), k > 0
X(z) = zk [i.e., δ(n + k)
z
←→ zk], ROC: entire z-plane except z = ∞
3 / 63
The Direct z-Transform
Example
Determine z-transform of
x(n) = (1
2)nu(n)
z-transform of x(n)
X(z) = 1 +
1
2
z−1
+ (
1
2
)2
z−2
+ (
1
2
)n
z−n
+ · · ·
=
∞
n=0
(
1
2
)n
z−n
=
∞
n=0
(
1
2
z−1
)n
For |1
2z−1| < 1 or |z| > 1
2, X(z) converges to
X(z) =
1
1 − 1
2z−1
, ROC: |z| >
1
2
4 / 63
The Direct z-Transform
Expressing complex variable z in polar form
z = rejθ
where r = |z| and θ = z
X(z) =
∞
n=−∞
x(n)r−n
e−jθn
In ROC of X(z), |X(z)| < ∞
|X(z)| =
∞
n=−∞
x(n)r−n
e−jθn
≤
∞
n=−∞
|x(n)r−n
e−jθn
| =
∞
n=−∞
|x(n)r−n
|
|X(z)| ≤
−1
n=−∞
|x(n)r−n
| +
∞
n=0
x(n)
rn
≤
∞
n=1
|x(−n)rn
| +
∞
n=0
x(n)
rn
If X(z) converges in some region of complex plane, both sums must be
finite in that region
5 / 63
The Direct z-Transform
Figure 1: Region of convergence for X(z) and its corresponding causal and
anticausal components.
6 / 63
The Direct z-Transform
Example
Determine z-transform of
x(n) = αn
u(n) =
αn, n ≥ 0
0, n < 0
We have
X(z) =
∞
n=0
αn
z−n
=
∞
n=0
(αz−1
)n
x(n) = αn
u(n)
z
←→ X(z) =
1
1 − αz−1
, ROC: |z| > |α|
7 / 63
The Direct z-Transform
Example (continued)
Figure 2: The exponential signal x(n) = αn
u(n) (a), and the ROC of its
z-transform (b).
8 / 63
The Direct z-Transform
Example
Determine z-transform of
x(n) = −αn
u(−n − 1) =
0, n ≥ 0
−αn, n ≤ −1
We have
X(z) =
−1
n=−∞
(−αn
)z−n
= −
∞
l=1
(α−1
z)l
where l = −n
x(n) = −αn
u(−n − 1)
z
←→ X(z) = −
α−1z
1 − α−1z
=
1
1 − αz−1
ROC: |z| < |α|
9 / 63
The Direct z-Transform
Example (continued)
Figure 3: Anticausal signal x(n) = −αn
u(−n − 1) (a), and the ROC of its
z-transform (b).
10 / 63
The Direct z-Transform
From two preceding examples
Z{αn
u(n)} = Z{−αn
u(−n − 1)} =
1
1 − αz−1
This implies that a closed-form expression for z-transform does not
uniquely specify the signal in time domain
Ambiguity can be resolved if ROC is also specified
A signal x(n) is uniquely determined by its z-transform X(z) and region
of convergence of X(z)
ROC of a causal signal is exterior of a circle of some radius r2
ROC of an anticausal signal is interior of a circle of some radius r1
11 / 63
The Direct z-Transform
Example
Determine z-transform of
x(n) = αnu(n) + bnu(−n − 1)
We have
X(z) =
∞
n=0
αn
z−n
+
−1
n=−∞
bn
z−n
=
∞
n=0
(αz−1
)n
+
∞
l=1
(b−1
z)l
First sum converges if |z| > |α|, second sum converges if |z| < |b|
If |b| < |α|, X(z) does not exist
If |b| > |α|
X(z) =
1
1 − αz−1
−
1
1 − bz−1
=
b − α
α + b − z − αbz−1
ROC: |α| < |z| < |b|
12 / 63
The Direct z-Transform
Example (continued)
Figure 4: ROC for z-transform in the example.
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The Direct z-Transform
Table 1: Characteristic families of signals with their corresponding ROCs
14 / 63
Properties of the z-Transform
Combining several z-transforms, ROC of overall transform is, at least,
intersection of ROCs of individual transforms
Linearity
If
x1(n)
z
←→ X1(z)
and
x2(n)
z
←→ X2(z)
then
x(n) = a1x1(n) + a2x2(n)
z
←→ X(z) = a1X1(z) + a2X2(z)
for any constants a1 and a2
To find z-transform of a signal, express it as a sum of elementary signals
whose z-transforms are already known
15 / 63
Properties of the z-Transform
Example
Determine z-transform and ROC of
x(n) = [3(2n) − 4(3n)]u(n)
Defining signals x1(n) and x2(n)
x1(n) = 2nu(n) and x2(n) = 3nu(n)
x(n) = 3x1(n) − 4x2(n)
According to linearity property
X(z) = 3X1(z) − 4X2(z)
Recall that
αnu(n)
z
←→ 1
1−αz−1 , ROC: |z| > |α|
Setting α = 2 and α = 3
X1(z) = 1
1−2z−1 , ROC: |z| > 2 and X2(z) = 1
1−3z−1 , ROC: |z| > 3
Intersecting ROCs, overall transform is
X(z) = 3
1−2z−1 − 4
1−3z−1 , ROC: |z| > 3
16 / 63
Properties of the z-Transform
Time shifting
If
x(n)
z
←→ X(z)
then
x(n − k)
z
←→ z−k
X(z)
ROC of z−k
X(z) is same as that of X(z) except for z = 0 if k > 0 and
z = ∞ if k < 0
17 / 63
Properties of the z-Transform
Scaling in z-domain
If
x(n)
z
←→ X(z), ROC: r1 < |z| < r2
then
an
x(n)
z
←→ X(a−1
z), ROC: |a|r1 < |z| < |a|r2
for any constant a, real or complex
Proof
Z{an
x(n)} =
∞
n=−∞
an
x(n)z−n
=
∞
n=−∞
x(n)(a−1
z)−n
= X(a−1
z)
Since ROC of X(z) is r1 < |z| < r2, ROC of X(a−1
z) is
r1 < |a−1
z| < r2
or
|a|r1 < |z| < |a|r2
18 / 63
Properties of the z-Transform
Time reversal
If
x(n)
z
←→ X(z), ROC: r1 < |z| < r2
then
x(−n)
z
←→ X(z−1
), ROC: 1
r2
< |z| < 1
r1
Proof
Z{x(−n)} =
∞
n=−∞
x(−n)z−n
=
∞
l=−∞
x(l)(z−1
)−l
= X(z−1
)
where l = −n
ROC of X(z−1
)
r1 < |z−1
| < r2 or 1
r2
< |z| < 1
r1
19 / 63
Properties of the z-Transform
Differentiation in z-domain
If
x(n)
z
←→ X(z)
then
nx(n)
z
←→ −z dX(z)
dz
Proof
X(z) =
∞
n=−∞
x(n)z−n
differentiating both sides
dX(z)
dz
=
∞
n=−∞
x(n)(−n)z−n−1
= −z−1
∞
n=−∞
[nx(n)]z−n
= −z−1
Z{nx(n)}
Both transforms have same ROC
20 / 63
Properties of the z-Transform
Convolution of two sequences
If
x1(n)
z
←→ X1(z)
x2(n)
z
←→ X2(z)
then
x(n) = x1(n) ∗ x2(n)
z
←→ X(z) = X1(z)X2(z)
ROC of X(z) is, at least, intersection of that for X1(z) and X2(z)
Proof: convolution of x1(n) and x2(n)
x(n) =
∞
k=−∞
x1(k)x2(n − k)
z-transform of x(n)
X(z) =
∞
n=−∞
x(n)z−n
=
∞
n=−∞
∞
k=−∞
x1(k)x2(n − k) z−n
=
∞
k=−∞
x1(k)
∞
n=−∞
x2(n − k)z−n
= X2(z)
∞
k=−∞
x1(k)z−k
= X2(z)X1(z) 21 / 63
Properties of the z-Transform
Example
Compute convolution x(n) of signals
x1(n) = {1
↑
, −2, 1} and x2(n) =
1, 0 ≤ n ≤ 5
0, elsewhere
z-transforms of these signals
X1(z) = 1 − 2z−1
+ z−2
X2(z) = 1 + z−1
+ z−2
+ z−3
+ z−4
+ z−5
Convolution of two signals is equal to multiplication of their transforms
X(z) = X1(z)X2(z) = 1 − z−1 − z−6 + z−7
Hence
x(n) = {1
↑
, −1, 0, 0, 0, 0, −1, 1}
22 / 63
Properties of the z-Transform
Convolution property is one of most powerful properties of z-transform
It converts convolution of two signals (time domain) to multiplication of
their transforms
Computation of convolution of two signals using z-transform
1 Compute z-transforms of signals to be convolved
X1(z) = Z{x1(n)}
X2(z) = Z{x2(n)}
2 Multiply the two z-transforms
X(z) = X1(z)X2(z)
3 Find inverse z-transform of X(z)
x(n) = Z−1
{X(z)}
23 / 63
Properties of the z-Transform
Correlation of two sequences
If
x1(n)
z
←→ X1(z)
x2(n)
z
←→ X2(z)
then
rx1x2
(l) =
∞
n=−∞
x1(n)x2(n − l)
z
←→ Rx1x2
(z) = X1(z)X2(z−1
)
Proof
rx1x2
(l) = x1(l) ∗ x2(−l)
Using convolution and time-reversal properties
Rx1x2 (z) = Z{x1(l)}Z{x2(−l)} = X1(z)X2(z−1
)
ROC of Rx1x2
(z) is at least intersection of that for X1(z) and X2(z−1
)
24 / 63
Properties of the z-Transform
The initial value theorem
If x(n) is causal (x(n) = 0 for n < 0), then
x(0) = lim
z→∞
X(z)
Proof: since x(n) is causal
X(z) =
∞
n=0
x(n)z−n
= x(0) + x(1)z−1
+ x(2)z−2
+ · · ·
as z → ∞, z−n
→ 0 and hence X(z) = x(0)
25 / 63
Properties of the z-Transform
Table 2: Some common z-transform pairs
Signal, x(n) z-Transform, X(z) ROC
δ(n) 1 All z
u(n) 1
1−z−1 |z| > 1
anu(n) 1
1−az−1 |z| > |a|
nanu(n) az−1
(1−az−1)2 |z| > |a|
−anu(−n − 1) 1
1−az−1 |z| < |a|
−nanu(−n − 1) az−1
(1−az−1)2 |z| < |a|
(cos ω0n)u(n) 1−z−1 cos ω0
1−2z−1 cos ω0+z−2 |z| > 1
(sin ω0n)u(n) z−1 sin ω0
1−2z−1 cos ω0+z−2 |z| > 1
(an cos ω0n)u(n) 1−az−1 cos ω0
1−2az−1 cos ω0+a2z−2 |z| > |a|
(an sin ω0n)u(n) az−1 sin ω0
1−2az−1 cos ω0+a2z−2 |z| > |a|
26 / 63
Rational z-Transforms
An important family of z-transforms are those for which X(z) is a
rational function
X(z) is a ratio of two polynomials in z−1
(or z)
Some important issues of rational z-transforms are discussed here
27 / 63
Poles and Zeros
Zeros of a z-transform X(z) are values of z for which X(z) = 0
Poles of a z-transform are values of z for which X(z) = ∞
If X(z) is a rational function (and if a0 = 0 and b0 = 0)
X(z) =
B(z)
A(z)
=
b0 + b1z−1 + · · · + bMz−M
a0 + a1z−1 + · · · + aNz−N
=
M
k=0 bkz−k
N
k=0 akz−k
=
b0z−M
a0z−N
zM + (b1/b0)zM−1 + · · · + bM/b0
zN + (a1/a0)zN−1 + · · · + aN/a0
=
b0
a0
z−M+N (z − z1)(z − z2) · · · (z − zM)
(z − p1)(z − p2) · · · (z − pN)
= GzN−M
M
k=1(z − zk)
N
k=1(z − pk)
X(z) has M finite zeros at z = z1, z2, . . . , zM
N finite poles at z = p1, p2, . . . , pN
|N − M| zeros if N > M or poles if N < M at z = 0
X(z) has exactly same number of poles as zeros
28 / 63
Poles and Zeros
Example
Determine pole-zero plot for signal
x(n) = anu(n), a > 0
From Table 2
X(z) =
1
1 − az−1
=
z
z − a
, ROC: |z| > a
X(z) has one zero at z1 = 0 and one pole at p1 = a
Figure 5: Pole-zero plot for the causal exponential signal x(n) = an
u(n).
29 / 63
Poles and Zeros
Example
Determine pole-zero plot for signal
x(n) =
an, 0 ≤ n ≤ M − 1
0, elsewhere
where a > 0
z-transform of x(n)
X(z) =
M−1
n=0
an
z−n
=
M−1
n=0
(az−1
)n
=
1 − (az−1)M
1 − az−1
=
zM − aM
zM−1(z − a)
Since a > 0, zM = aM has M roots at
zk = aej2πk/M
, k = 0, 1, . . . , M − 1
30 / 63
Poles and Zeros
Example (continued)
Zero z0 = a cancels pole at z = a. Thus
X(z) =
(z − z1)(z − z2) · · · (z − zM−1)
zM−1
which has M − 1 zeros and M − 1 poles
Figure 6: Pole-zero pattern for the finite-duration signal x(n) = an
,
0 ≤ n ≤ M − 1 (a > 0), for M = 8.
31 / 63
Poles and Zeros
Example
Determine z-transform and signal corresponding to following pole-zero
plot
Figure 7: Pole-zero pattern.
32 / 63
Poles and Zeros
Example (continued)
We use
X(z) = GzN−M
M
k=1(z − zk)
N
k=1(z − pk)
There are two zeros (M = 2) at z1 = 0, z2 = r cos ω0
There are two poles (N = 2) at p1 = rejω0 , p2 = re−jω0
X(z) = G
(z − z1)(z − z2)
(z − p1)(z − p2)
= G
z(z − r cos ω0)
(z − rejω0 )(z − re−jω0 )
= G
1 − rz−1 cos ω0
1 − 2rz−1 cos ω0 + r2z−2
, ROC: |z| > r
From Table 2 we find that
x(n) = G(rn cos ω0n)u(n)
33 / 63
Poles and Zeros
z-transform X(z) is a complex function of complex variable z
|X(z)| is a real and positive function of z
Since z represents a point in complex plane, |X(z)| is a surface
z-transform
X(z) =
z−1 − z−2
1 − 1.2732z−1 + 0.81z−2
has one zero at z1 = 1 and two poles at p1, p2 = 0.9e±jπ/4
Figure 8: Graph of |X(z)| for the above z-transform.
34 / 63
Pole Location & Time-Domain Behavior for Causal Signals
Characteristic behavior of causal signals depends on whether poles of
transform are contained in region
|z| < 1
or |z| > 1
or on circle |z| = 1
Circle |z| = 1 is called unit circle
If a real signal has a z-transform with one pole, this pole has to be real
The only such signal is the real exponential
x(n) = an
u(n)
z
←→ X(z) =
1
1 − az−1
, ROC: |z| > |a|
having one zero at z1 = 0 and one pole at p1 = a on real axis
35 / 63
Pole Location & Time-Domain Behavior for Causal Signals
Figure 9: Time-domain behavior of a single-real-pole causal signal as a function
of the location of the pole with respect to the unit circle.
36 / 63
Pole Location & Time-Domain Behavior for Causal Signals
A causal real signal with a double real pole has the form
x(n) = nan
u(n)
z
←→ X(z) =
az−1
(1 − az−1)2
, ROC: |z| > |a|
In contrast to single-pole signal, a double real pole on unit circle results
in an unbounded signal
37 / 63
Pole Location & Time-Domain Behavior for Causal Signals
Figure 10: Time-domain behavior of causal signals corresponding to a double
(m = 2) real pole, as a function of the pole location.
38 / 63
Pole Location & Time-Domain Behavior for Causal Signals
Configuration of poles as a pair of complex-conjugates results in an
exponentially weighted sinusoidal signal
x(n) = (an
cos ω0n)u(n)
z
←→ X(z) =
1 − az−1 cos ω0
1 − 2az−1 cos ω0 + a2z−2
ROC: |z| > |a|
x(n) = (an
sin ω0n)u(n)
z
←→ X(z) =
az−1 sin ω0
1 − 2az−1 cos ω0 + a2z−2
ROC: |z| > |a|
Distance r of poles from origin determines envelope of sinusoidal signal
Angle ω0 with real positive axis determines relative frequency
39 / 63
Pole Location & Time-Domain Behavior for Causal Signals
Figure 11: A pair of complex-conjugate poles corresponds to causal signals with
oscillatory behavior. 40 / 63
Pole Location & Time-Domain Behavior for Causal Signals
Figure 12: Causal signal corresponding to a double pair of complex-conjugate
poles on the unit circle.
In summary
Causal real signals with simple real poles or simple complex-conjugate
pairs of poles, inside or on unit circle, are always bounded in amplitude
A signal with a pole, or a complex-conjugate pair of poles, near origin
decays more rapidly than one near (but inside) unit circle
Thus, time behavior of a signal depends strongly on location of its poles
relative to unit circle
Zeros also affect behavior of a signal but not as strongly as poles
E.g., for sinusoidal signals, presence and location of zeros affects only
their phase 41 / 63
Inversion of the z-Transform
There are three methods for evaluation of inverse z-transform
1 Direct evaluation by contour integration
2 Expansion into a series of terms, in variables z and z−1
3 Partial-fraction expansion and table lookup
Inverse z-transform by contour integration
x(n) =
1
2πj C
X(z)zn−1
dz
Integral is a contour integral over a closed path C
C encloses origin and lies within ROC of X(z)
Figure 13: Contour C. 42 / 63
The Inverse z-Transform by Power Series Expansion
Given X(z) with its ROC, expand it into a power series of form
X(z) =
∞
n=−∞
cnz−n
By uniqueness of z-transform, x(n) = cn for all n
When X(z) is rational, expansion can be performed by long division
Long division method becomes tedious when n is large
Although this method provides a direct evaluation of x(n), a closed-form
solution is not possible
Hence this method is used only for determining values of first few
samples of signal
43 / 63
The Inverse z-Transform by Power Series Expansion
Example
Determine inverse z-transform of
X(z) =
1
1 − 1.5z−1 + 0.5z−2
1 When ROC: |z| > 1
2 When ROC: |z| < 0.5
ROC: |z| > 1
Since ROC is exterior of a circle, x(n) is a causal signal. Thus we seek
negative powers of z by dividing numerator of X(z) by its denominator
X(z) =
1
1 − 3
2z−1 + 1
2z−2
= 1 +
3
2
z−1
+
7
4
z−2
+
15
8
z−3
+
31
16
z−4
+ · · ·
x(n) = {1
↑
,
3
2
,
7
4
,
15
8
,
31
16
, . . .}
44 / 63
The Inverse z-Transform by Power Series Expansion
Example (continued)
ROC: |z| < 0.5
Since ROC is interior of a circle, x(n) is anticausal. To obtain positive
powers of z, write the two polynomials in reverse order and then divide
X(z) =
1
1 − 3
2z−1 + 1
2z−2
= 2z2
+ 6z3
+ 14z4
+ 30z5
+ 62z6
+ · · ·
x(n) = {· · · 62, 30, 14, 6, 2, 0, 0
↑
}
45 / 63
The Inverse z-Transform by Partial-Fraction Expansion
In table lookup method, express X(z) as a linear combination
X(z) = α1X1(z) + α2X2(z) + · · · + αK XK (z)
X1(z), . . . , XK (z) are expressions with inverse transforms
x1(n), . . . , xK (n) available in a table of z-transform pairs
Using linearity property
x(n) = α1x1(n) + α2x2(n) + · · · + αK xK (n)
If X(z) is a rational function
X(z) =
B(z)
A(z)
=
b0 + b1z−1 + · · · + bMz−M
a0 + a1z−1 + · · · + aNz−N
dividing both numerator and denominator by a0
X(z) =
B(z)
A(z)
=
b0 + b1z−1 + · · · + bMz−M
1 + a1z−1 + · · · + aNz−N
This form of rational function is called proper if aN = 0 and M < N
46 / 63
The Inverse z-Transform by Partial-Fraction Expansion
An improper rational function (M ≥ N) can always be written as sum
of a polynomial and a proper rational function
Example
Express improper rational function
X(z) =
1 + 3z−1 + 11
6 z−2 + 1
3z−3
1 + 5
6z−1 + 1
6z−2
in terms of a polynomial and a proper function
Terms z−2 and z−3 should be eliminated from numerator
Do long division with the two polynomials written in reverse order
Stop division when order of remainder becomes z−1
X(z) = 1 + 2z−1
+
1
6z−1
1 + 5
6z−1 + 1
6z−2
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The Inverse z-Transform by Partial-Fraction Expansion
Any improper rational function (M ≥ N) can be expressed as
X(z) =
B(z)
A(z)
= c0 + c1z−1
+ · · · + cM−Nz−(M−N)
+
B1(z)
A(z)
Inverse z-transform of the polynomial can easily be found by inspection
We focus our attention on inversion of proper rational transforms
1 Perform a partial fraction expansion of proper rational function
2 Invert each of the terms
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The Inverse z-Transform by Partial-Fraction Expansion
Let X(z) be a proper rational function (aN = 0 and M < N)
X(z) =
B(z)
A(z)
=
b0 + b1z−1 + · · · + bMz−M
1 + a1z−1 + · · · + aNz−N
Eliminating negative powers of z
X(z) =
b0zN + b1zN−1 + · · · + bMzN−M
zN + a1zN−1 + · · · + aN
Since N > M, the function
X(z)
z
=
b0zN−1 + b1zN−2 + · · · + bMzN−M−1
zN + a1zN−1 + · · · + aN
is also proper
To perform a partial-fraction expansion, this function should be
expressed as a sum of simple fractions
First factor denominator polynomial into factors that contain poles
p1, p2, . . . , pN of X(z)
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The Inverse z-Transform by Partial-Fraction Expansion
Distinct poles
Suppose poles p1, p2, . . . , pN are all different. We seek expansion
X(z)
z
=
A1
z − p1
+
A2
z − p2
+ · · · +
AN
z − pN
To determine coefficients A1, A2, . . . , AN , multiply both sides by each of
terms (z − pk ), k = 1, 2, . . . , N, and evaluate resulting expressions at
corresponding pole positions, p1, p2, . . . , pN
(z − pk )X(z)
z
=
(z − pk )A1
z − p1
+ · · · + Ak + · · · +
(z − pk )AN
z − pN
Ak =
(z − pk )X(z)
z z=pk
, k = 1, 2, . . . , N
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The Inverse z-Transform by Partial-Fraction Expansion
Example
Determine partial-fraction expansion of
X(z) =
1 + z−1
1 − z−1 + 0.5z−2
Eliminate negative powers by multiplying by z2
X(z)
z
=
z + 1
z2 − z + 0.5
−→ p1 =
1
2
+ j
1
2
and p2 =
1
2
− j
1
2
p1=p2
−−−→
X(z)
z
=
z + 1
(z − p1)(z − p2)
=
A1
z − p1
+
A2
z − p2
A1 =
(z − p1)X(z)
z z=p1
=
z + 1
z − p2 z=p1
=
1
2 + j 1
2 + 1
1
2 + j 1
2 − 1
2 + j 1
2
=
1
2
− j
3
2
A2 =
(z − p2)X(z)
z z=p2
=
z + 1
z − p1 z=p2
=
1
2 − j 1
2 + 1
1
2 − j 1
2 − 1
2 − j 1
2
=
1
2
+ j
3
2
Complex-conjugate poles result in complex-conjugate coefficients 51 / 63
The Inverse z-Transform by Partial-Fraction Expansion
Multiple-order poles
If X(z) has a pole of multiplicity m (there is factor (z − pk )m
in
denominator), partial-fraction expansion must contain the terms
A1k
z − pk
+
A2k
(z − pk )2
+ · · · +
Amk
(z − pk )m
Coefficients {Aik } can be evaluated through differentiation
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The Inverse z-Transform by Partial-Fraction Expansion
Example
Determine partial-fraction expansion of
X(z) =
1
(1 + z−1)(1 − z−1)2
Expressing in terms of positive powers of z
X(z)
z
=
z2
(z + 1)(z − 1)2
X(z) has a simple pole at p1 = −1 and a double pole at p2 = p3 = 1
X(z)
z
=
z2
(z + 1)(z − 1)2
=
A1
z + 1
+
A2
z − 1
+
A3
(z − 1)2
A1 =
(z + 1)X(z)
z z=−1
=
1
4
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The Inverse z-Transform by Partial-Fraction Expansion
Example (continued)
A3 =
(z − 1)2X(z)
z z=1
=
1
2
To obtain A2
(z − 1)2X(z)
z
=
(z − 1)2
z + 1
A1 + (z − 1)A2 + A3
Differentiating both sides and evaluating at z = 1, A2 is obtained
A2 =
d
dz
(z − 1)2X(z)
z z=1
=
3
4
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The Inverse z-Transform by Partial-Fraction Expansion
Having performed partial-fraction expansion, final step in inversion is
as follows
If poles are distinct
X(z)
z
=
A1
z − p1
+
A2
z − p2
+ · · · +
AN
z − pN
X(z) = A1
1
1 − p1z−1
+ A2
1
1 − p2z−1
+ · · · + AN
1
1 − pN z−1
x(n) = Z−1
{X(z)} is obtained by inverting each term and taking the
corresponding linear combination
From table 2
Z−1 1
1 − pk z−1
=
(pk )n
u(n), if ROC:|z| > |pk | (causal)
−(pk )n
u(−n − 1), if ROC:|z| < |pk | (anticausal)
If x(n) is causal, ROC is |z| > pmax , where
pmax = max{|p1|, |p2|, . . . , |pN |}
In this case all terms in X(z) result in causal signal components
x(n) = (A1pn
1 + A2pn
2 + · · · + AN pn
N )u(n)
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The Inverse z-Transform by Partial-Fraction Expansion
If all poles are distinct but some of them are complex, and if signal
x(n) is real, complex terms can be reduced into real components
If pj is a pole, its complex conjugate p∗
j is also a pole
If x(n) is real, the polynomials in X(z) have real coefficients
If a polynomial has real coefficients, its roots are either real or occur in
complex-conjugate pairs
Their corresponding coefficients in partial-fraction expansion are also
complex-conjugates
Contribution of two complex-conjugate poles is
xk (n) = [Ak (pk )n
+ A∗
k (p∗
k )n
)]u(n)
Expressing Aj and pj in polar form
Ak = |Ak |ejαk
and pk = rk ejβk
which gives
xk (n) = |Ak |rn
k [ej(βk n+αk )
+ e−j(βk n+αk )
]u(n)
or xk (n) = 2|Ak |rn
k cos(βk n + αk )u(n)
Thus
Z−1 Ak
1 − pk z−1
+
A∗
k
1 − p∗
k z−1
= 2|Ak |rn
k cos(βk n + αk )u(n)
if ROC is |z| > |pk | = rk
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The Inverse z-Transform by Partial-Fraction Expansion
In case of multiple poles, either real or complex, inverse transform of
terms of the form A/(z − pk)n is required
In case of a double pole, from table 2
Z−1 pz−1
(1 − pz−1)2
= npn
u(n)
provided that ROC is |z| > |p|
In case of poles with higher multiplicity, multiple differentiation is used
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The Inverse z-Transform by Partial-Fraction Expansion
Example
Determine inverse z-transform of
X(z) =
1
1 − 1.5z−1 + 0.5z−2
1 If ROC: |z| > 1
2 If ROC: |z| < 0.5
3 If ROC: 0.5 < |z| < 1
Partial-fraction expansion for X(z)
X(z) =
z2
z2 − 1.5z + 0.5
p1=1
−−−−→
p2=0.5
X(z)
z
=
z
(z − 1)(z − 0.5)
=
A1
z − 1
+
A2
z − 0.5
A1 =
(z − 1)X(z)
z z=1
= 2
A2 =
(z − 0.5)X(z)
z z=0.5
= −1
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The Inverse z-Transform by Partial-Fraction Expansion
Example (continued)
X(z) =
2
1 − z−1
−
1
1 − 0.5z−1
When ROC is |z| > 1, x(n) is causal and both terms in X(z) are causal
1
1 − pkz−1
z
←→ (pk)n
u(n)
x(n) = 2(1)n
u(n) − (0.5)n
u(n) = (2 − 0.5n
)u(n)
When ROC is |z| < 0.5, x(n) is anticausal and both terms in X(z) are
anticausal
1
1 − pkz−1
z
←→ −(pk)n
u(−n − 1)
x(n) = [−2 + (0.5)n
]u(−n − 1)
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The Inverse z-Transform by Partial-Fraction Expansion
Example (continued)
When ROC is 0.5 < |z| < 1 (ring), signal x(n) is two-sided
One of the terms corresponds to a causal signal and the other to an
anticausal signal
Since the ROC is overlapping of |z| > 0.5 and |z| < 1, pole p2 = 0.5
provides causal part and pole p1 = 1 anticausal
x(n) = −2(1)n
u(−n − 1) − (0.5)n
u(n)
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The Inverse z-Transform by Partial-Fraction Expansion
Example
Determine causal signal x(n) whose z-transform is
X(z) =
1 + z−1
1 − z−1 + 0.5z−2
We have already obtained partial-fraction expansion as
X(z) =
A1
1 − p1z−1
+
A2
1 − p2z−1
−→ A1 = A∗
2 =
1
2
−j
3
2
and p1 = p∗
2 =
1
2
+j
1
2
For a pair of complex-conjugate poles (Ak = |Ak |ejαk
and pk = rk ejβk
)
Z−1 Ak
1 − pk z−1
+
A∗
k
1 − p∗
k z−1
= 2|Ak |rn
k cos(βk n + αk )u(n)
A1 = (
√
10/2)e−j71.565
and p1 = (1/
√
2)ejπ/4
x(n) =
√
10(1/
√
2)n
cos(πn/4 − 71.565◦
)u(n)
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The Inverse z-Transform by Partial-Fraction Expansion
Example
Determine causal signal x(n) having z-transform
X(z) =
1
(1 + z−1)(1 − z−1)2
We have already obtained partial-fraction expansion as
X(z) =
1
4
1
1 + z−1
+
3
4
1
1 − z−1
+
1
2
z−1
(1 − z−1)2
For causal signals
1
1 − pz−1
z
←→ (p)n
u(n) and
pz−1
(1 − pz−1)2
z
←→ npn
u(n)
x(n) =
1
4
(−1)n
u(n) +
3
4
u(n) +
1
2
nu(n) =
1
4
(−1)n
+
3
4
+
n
2
u(n)
62 / 63
References
John G. Proakis, Dimitris G. Manolakis, Digital Signal
Processing: Principles, Algorithms, and Applications, Prentice
Hall, 2006.
63 / 63