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The Discrete Fourier Series 143
denote the complex exponential term, we express (5.3) and (5.2) as
˜X(k) = △ DFS[˜x(n)] = N−1 ∑
n=0
˜x(n) △ = IDFS[ ˜X(k)] = 1 N
˜x(n)W nk
N
N−1 ∑
k=0
˜X(k)W −nk
N
: Analysis or a
DFS equation
: Synthesis or an inverse
DFS equation
(5.5)
□ EXAMPLE 5.1 Find DFS representation of the periodic sequence
˜x(n) ={...,0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3,...}
↑
Solution The fundamental period of this sequence is N = 4. Hence W 4 = e −j 2π 4 =
−j. Now
Hence
˜X(k) =
3∑
n=0
˜x(n)W nk
4 , k =0, ±1, ±2,...
Similarly,
˜X(0) =
3∑
0
˜x(n)W 0·n
4 =
3∑
˜x(n) =˜x(0) + ˜x(1) + ˜x(2) + ˜x(3) =6
0
˜X(1) =
˜X(2) =
˜X(3) =
3∑
˜x(n)W4 n =
0
3∑
0
3∑
0
˜x(n)W 2n
4 =
˜x(n)W 3n
4 =
3∑
˜x(n)(−j) n =(−2+2j)
0
3∑
˜x(n)(−j) 2n =2
0
3∑
˜x(n)(−j) 3n =(−2 − 2j)
0
□
5.1.1 MATLAB IMPLEMENTATION
A careful look at (5.5) reveals that the DFS is a numerically computable
representation. It can be implemented in many ways. To compute each
sample ˜X(k), we can implement the summation as a for...end loop.
To compute all DFS coefficients would require another for...end loop.
This will result in a nested two for...end loop implementation. This is
clearly inefficient in MATLAB. An efficient implementation in MATLAB
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