Databases Reference
In-Depth Information
12.9 Z-Transform
In the previous section we saw how to extend the Fourier series for use with sampled functions.
We can also do the same with the Fourier transform. Recall that the Fourier transform was
given by the equation
e j ω t dt
(ω) =
(
)
(68)
F
f
t
−∞
Replacing f
(
t
)
with its sampled version, we get
n =−∞ δ(
e j ω t dt
F
(ω) =
f
(
t
)
t
nT
)
(69)
−∞
n =−∞
f n e j ω nT
=
(70)
where f n
=
f
(
nT
)
. This is called the discrete time Fourier transform. The Z-transform of
the sequence
{
f n }
is a generalization of the discrete time Fourier transform and is given by
f n z n
F
(
z
) =
(71)
n
=−∞
where
e σ T + j w T
z
=
(72)
σ
equal zero, we get the original expression for the Fourier transform
of a discrete time sequence. We denote the Z-transform of a sequence by
Notice that if we let
(
) = Z [
f n ]
F
z
We can express this another way. Notice that the magnitude of z is given by
e σ T
|
z
| =
Thus, when
equals zero, the magnitude of z is one. Because z is a complex number, the
magnitude of z is equal to one on the unit circle in the complex plane. Therefore, we can say
that the Fourier transform of a sequence can be obtained by evaluating the Z-transform of the
sequence on the unit circle. Notice that the Fourier transform thus obtained will be periodic,
which we expect because we are dealing with a sampled function. Further, if we assume T
to be one,
σ
ω
π
π
.
5to0.5Hz.
This makes sense because, by the sampling theorem, if the sampling rate is one sample per
second, the highest frequency component that can be recovered is 0.5Hz.
For the Z-transform to exist—in other words, for the power series to converge—we need
to have
varies from
to
, which corresponds to a frequency range of
0
f n z n <
n
=−∞
Whether this inequality holds will depend on the sequence itself and the value of z . The values
of z for which the series converges are called the region of convergence of the Z-transform.
From our earlier discussion, we can see that for the Fourier transform of the sequence to exist,
the region of convergence should include the unit circle. Let us look at a simple example.
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