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# m14 - Evaluation of the DTFT by the DFT

Module by: C. Sidney Burrus. E-mail the author

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

## 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 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 ω .
(4)
The second term in (Equation 4) says the Riemann sum is equal to its limit in this case.

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