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148 Chapter 5 THE DISCRETE FOURIER TRANSFORM
5.1.3 RELATION TO THE DTFT
Since x(n) in(5.8) is of finite duration of length N, itisalso absolutely
summable. Hence its DTFT exists and is given by
X(e jω )=
N−1
∑
n=0
x(n)e −jωn =
Comparing (5.13) with (5.11), we have
Let
ω 1
△
=
2π
N
N−1
∑
n=0
˜x(n)e −jωn (5.13)
˜X(k) =X(e jω ) ∣ ∣
ω= 2π
N k (5.14)
and
ω k
△
=
2π
N k = kω 1
Then the DFS X(k) =X(e jω k
)=X(e jkω1 ), which means that the DFS is
obtained by evenly sampling the DTFT at ω 1 = 2π N
intervals. From (5.12)
and (5.14) we observe that the DFS representation gives us a sampling
mechanism in the frequency domain that, in principle, is similar to sampling
in the time domain. The interval ω 1 = 2π N
is the sampling interval
in the frequency domain. It is also called the frequency resolution because
it tells us how close the frequency samples (or measurements) are.
□ EXAMPLE 5.3 Let x(n) ={0, 1, 2, 3}.
↑
a. Compute its discrete-time Fourier transform X(e jω ).
b. Sample X(e jω )atkω 1 = 2π k, k =0, 1, 2, 3 and show that it is equal to
4
˜X(k) inExample 5.1.
Solution
The sequence x(n) isnot periodic but is of finite duration.
a. The discrete-time Fourier transform is given by
∞∑
X(e jω )= x(n)e −jωn = e −jω +2e −j2ω +3e −j3ω
n=−∞
b. Sampling at kω 1 = 2π k, k =0, 1, 2, 3, we obtain
4
X(e j0 )=1+2+3=6= ˜X(0)
X(e j2π/4 )=e −j2π/4 +2e −j4π/4 +3e −j6π/4 = −2+2j = ˜X(1)
X(e j4π/4 )=e −j4π/4 +2e −j8π/4 +3e −j12π/4 =2= ˜X(2)
X(e j6π/4 )=e −j6π/4 +2e −j12π/4 +3e −j18π/4 = −2 − 2j = ˜X(3)
as expected.
□
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