The z-transform is an extension of the DTFT in a way that is analogous to the
Laplace transform for continuous-time signals being an extension of the
Fourier transform. It allows the use of complex variable theory and is
particularly useful in analyzing and describing systems. The question of
convergence becomes still more complicated and depends on values of
z
z
used in the inverse transform which must be in the ``region of convergence"
(ROC).
The z-transform (ZT) is defined as a polynomial in the complex variable
z
z
with the discrete-time signal values as its coefficients
[1]
[2]
[3]
.
It is given by
F
(
z
)
=
∑
n
=
−
∞
∞
f
(
n
)
z
−
n
F
(
z
)
=
∑
n
=
−
∞
∞
f
(
n
)
z
−
n
(1)
and the inverse transform (IZT) is
f
(
n
)
=
1
2
π
j
∮
R
O
C
F
(
z
)
z
n
−
1
ⅆ
z
.
f
(
n
)
=
1
2
π
j
∮
R
O
C
F
(
z
)
z
n
−
1
ⅆ
z
.
(2)
The inverse transform can be derived by using the residue theorem
[
4]
[
5]
from complex variable theory to find
f
(
0
)
f
(
0
)
from
z
−
1
F
(
z
)
z
−
1
F
(
z
)
,
f
(
1
)
f
(
1
)
from
F
(
z
)
F
(
z
)
,
f
(
2
)
f
(
2
)
from
z
F
(
z
)
z
F
(
z
)
,
and in general,
f
(
n
)
f
(
n
)
from
z
n
−
1
F
(
z
)
z
n
−
1
F
(
z
)
.
Verification by substitution is more difficult than for the DFT or DTFT. Here
convergence and the interchange of order of the sum and integral is a serious
question that involves values of the complex variable
z
z
.
The complex contour integral in
(
Equation 2) must be taken
in the ROC of the z plane.
A unilateral z-transform is sometimes needed where the definition (2.23) uses
a lower limit on the transform summation of zero. This allow the
transformation to converge for some functions where the regular bilateral
transform does not, it provides a straightforward way to solve initial
condition difference equation problems, and it simplifies the question of
finding the ROC. The bilateral z-transform is used more for signal analysis
and the unilateral transform is used more for system description and analysis.
Unless stated otherwise, we will be using the bilateral z-transform.
A few examples together with the above properties will enable one to solve and
understand a wide variety of problems. These use the unit step function to
remove the negative time part of the signal. This function is defined as
u
(
n
)
=
{
1
if
n
≥
0
0
if
n
<
0
u
(
n
)
=
{
1
if
n
≥
0
0
if
n
<
0
(3)
and several bilateral z-transforms are given by
-
{
δ
(
n
)
}
=
1
{
δ
(
n
)
}
=
1
for all
z
z
.
-
{
u
(
n
)
}
=
z
z
−
1
{
u
(
n
)
}
=
z
z
−
1
for
|
z
|
>
1
|
z
|
>
1
.
-
{
u
(
n
)
a
n
}
=
z
z
−
a
{
u
(
n
)
a
n
}
=
z
z
−
a
for
|
z
|
>
|
a
|
|
z
|
>
|
a
|
.
Notice that these are similar to but not the same as a term of a partial
fraction expansion.
The z-transform can be inverted in three ways. The first two have similar
procedures with Laplace transformations and the third has no counter part.
-
The z-transform can be inverted by the defined contour integral in the ROC of
the complex
z
z
plane. This integral can be evaluated using the residue theorem
[4]
[5]
.
-
The z-transform can be inverted by expanding
1
z
F
(
z
)
1
z
F
(
z
)
in a partial fraction expansion followed by use of tables for the first or
second order terms.
-
The third method is not analytical but numerical. If
F
(
z
)
=
P
(
z
)
Q
(
z
)
F
(
z
)
=
P
(
z
)
Q
(
z
)
,
f
(
n
)
f
(
n
)
can be obtained as the coefficients of long division.
For example
z
z
−
a
=
1
+
a
z
−
1
+
a
2
z
−
2
+
⋯
z
z
−
a
=
1
+
a
z
−
1
+
a
2
z
−
2
+
⋯
(4)
which is
u
(
n
)
a
n
u
(
n
)
a
n
as used in the examples above.
We must understand the role of the ROC in the convergence and inversion of the
z-transform. We must also see the difference between the one-sided and
two-sided transform.
The FS coefficients are weights on the delta functions in a FT of the
periodically extended signal. The FT is the LT evaluated on the imaginary
axis:
s
=
j
ω
s
=
j
ω
.
The DFT values are samples of the DTFT of a finite length signal. The DTFT is
the z-transform evaluated on the unit circle in the z plane.
F
(
z
)
=
∑
n
=
−
∞
∞
x
(
n
)
z
−
n
=
{
x
(
n
)
}
F
(
z
)
=
∑
n
=
−
∞
∞
x
(
n
)
z
−
n
=
{
x
(
n
)
}
(5)
F
(
e
j
ω
)
=
∑
n
=
−
∞
∞
x
(
n
)
e
−
j
ω
n
=
ℱ
{
x
(
n
)
}
F
(
e
j
ω
)
=
∑
n
=
−
∞
∞
x
(
n
)
e
−
j
ω
n
=
ℱ
{
x
(
n
)
}
(6)
and if
x
(
n
)
x
(
n
)
is of length
N
N
F
(
e
j
2
π
N
k
)
=
∑
n
=
0
N
−
1
x
(
n
)
e
−
j
2
π
N
k
n
=
ℱ
{
x
(
n
)
}
F
(
e
j
2
π
N
k
)
=
∑
n
=
0
N
−
1
x
(
n
)
e
−
j
2
π
N
k
n
=
ℱ
{
x
(
n
)
}
(7)
It is important to be able to relate the time-domain signal
x
(
n
)
x
(
n
)
,
its spectrum
X
(
ω
)
X
(
ω
)
,
and its z-transform represented by the pole-zero locations on the z plane.