Page 516 - Advanced engineering mathematics

496 CHAPTER 14 The Fourier Integral and Transforms

0.8

0.4

0
–4 –2 0 2 4 6 8 10
t
–0.4

FIGURE 14.7 f (t) = sin(t) for 0 ≤ t < 4, period 4.

To see the idea in action, consider the example of a function f having fundamental period
4, and f (t) = sin(t) for 0 ≤ t < 4. A graph of this function is shown in Figure 14.7. With p = 4
the Fourier coefficients are
1 4 1 4
d k = sin(ξ)e −2πikξ/4 dξ = sin(ξ)e −πikξ/2 dξ
4 0 4 0
cos(4) − 1 1 πk sin(4)
= + i
π k − 4 2 π k − 4
2 2
2 2
for k = 0,±1,±2,···.
Let N be a positive integer and subdivide [0,4] into N subintervals of equal lengths 4/N.
These subintervals are

4 j 4( j + 1)
, for j = 0,1,··· , N − 1.
N n
Select N numbers 4 j/N by choosing the left endpoint of each of these subintervals. These define
an N-point sequence u, where
4 j

u j = sin .
n
The N-point DFT of u is
N−1 N−1
4 j 4 j

U k = sin e −2πijk/4 = sin e −πijk/2 .
N N
j=0 j=0
Now
N−1
! 1 4 j
U k = sin e −πijk/2
N N N
j=0
is a Riemann sum for the integral defining d k .
We now ask: To what extent does (1/N)U k approximate d k ? In this example we have
an explicit expression for d k , so we can explore this question directly. We will evaluate
N−1
(1/N)U k using a = 4/N in the DFT of [sin( ja)] j=0 determined in Example 14.15. This
gives us

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October 14, 2010 16:43 THM/NEIL Page-496 27410_14_ch14_p465-504