Page 524 - Advanced engineering mathematics

504 CHAPTER 14 The Fourier Integral and Transforms

In the first integral on the right, set ζ = ξ + 2π L to obtain
2π L π L
ˆ
f (ω) ≈ f (ζ − 2π L)e −ik(ζ−2π L) dζ + f (ξ)e −ikξ/L dξ
π L 0
2π L π L

= f (ζ − 2π L)e −ikζ/L dζ + f (ξ)e −ikξ/L dξ.
π L 0
Write ξ for ζ as the variable of integration to write the last approximation as
2π L π L
ˆ
f (k/L) ≈ f (ξ − 2π L)e −ikξ/L dξ + f (ξ)e −ikξ/L dξ.
π L 0
Now define
⎧
⎪ f (t) for 0 ≤ t <π L
⎨
g(t) = ( f (π L) + f (−π L))/2for t = π L
⎪
f (t − 2π L) for π L < t < 2π L.
⎩
Then
2π L
ˆ
f (k/L) ≈ g(ξ)e −ikξ/L dξ
0
L
= g(2πt)e −2πikt/L (2π)dt (let ξ = 2πt)
0
L
= 2π g(2πt)e −2πikt/L dt.
0
Finally, approximate the last integral by a Riemann sum, subdividing [0, L], subdividing [0, L]
into L/N subintervals and choosing partition points t j = jL/N for j = 0,1,··· , N − 1. Then

2π L 2π jL
N−1
ˆ −2πijk/N
f (k/L) ≈ g e .
N N
j=0
As before, we assume that |k|≤ N/8. This approximates f (k/L) by a constant multiple of the
ˆ
N-point DFT of the sequence
N−1

2π jL
g .
N
j=0
SECTION 14.7 PROBLEMS

In each of Problems 1 through 4, make a DFT approxima- te −2t for t ≥ 0
3. f (t) =
tion to the Fourier transform of f at the given point, using 0 for t < 0,
N = 512 and the given L.
L = 3; ˆ f (2)

e −4t for t ≥ 0 e cos(t) for t ≥ 0
−t
1. f (t) = 4. f (t) =
0 for t < 0, 0 for t < 0,
L = 3; ˆ f (4) L = 4; ˆ f (4)

t cos(t) for 0 ≤ t ≤ 12π
2. f (t) =
0 for t < 0and t > 12π
L = 6; ˆ f (1)

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