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m14 - Evaluation of the DTFT by the DFT

Module by: C. Sidney Burrus

Summary: Samples of the DTFT can be calculated using the DFT

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Evaluation of the DTFT by the DFT

If the DTFT of a finite sequence is taken, the result is a continuous function of ω ω . If the DFT of the same sequence is taken, the results are N N evenly spaced samples of the DTFT. In other words, the DTFT of a finite signal can be evaluated at N N points with the DFT.

X ( ω ) = D T F T { x ( n ) } = ∑ n = − ∞ ∞ x ( n ) e − j ω n X ( ω ) = D T F T { x ( n ) } = ∑ n = − ∞ ∞ x ( n ) e − j ω n (1)
and because of the finite length
X ( ω ) = ∑ n = 0 N − 1 x ( n ) e − j ω n . X ( ω ) = ∑ n = 0 N − 1 x ( n ) e − j ω n . (2)
If we evaluate ω ω at N N equally space points, this becomes
X ( 2 π N k ) = ∑ n = 0 N − 1 x ( n ) e − j 2 π N k n X ( 2 π N k ) = ∑ n = 0 N − 1 x ( n ) e − j 2 π N k n (3)
which is the DFT of x ( n ) x ( n ) . By adding zeros to the end of x ( n ) x ( n ) and taking a longer DFT, any density of points can be evaluated. This is useful in interpolation and in plotting the spectrum of a finite length signal. This is discussed further in Chapter (Reference).

There is an interesting variation of the Parseval's theorem for the DTFT of a finite length- N N signal. If x ( n ) ≠ 0 x ( n ) ≠ 0 for 0 ≥ n ≥ N − 1 0 ≥ n ≥ N − 1 , and if L ≥ N L ≥ N , then

∑ n = 0 N − 1 | x ( n ) | 2 = 1 L ∑ k = 0 L − 1 | X ( 2 π k / L ) | 2 = 1 π ∫ 0 π | X ( ω ) | 2 ⅆ ω . ∑ n = 0 N − 1 | x ( n ) | 2 = 1 L ∑ k = 0 L − 1 | X ( 2 π k / L ) | 2 = 1 π ∫ 0 π | X ( ω ) | 2 ⅆ ω . (4)
The second term in (Equation 4) says the Riemann sum is equal to its limit in this case.

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