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Relations: DFT and DTFT

Module by: Richard Baraniuk

Summary: An overview of the relationship between the DFT and the DTFT.

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CASE 1: Periodic Signals

Figure 1
Figure 1 (fig1.png)

Considered as a finite-length signal xn x n has unnormalized DFT representation

xn=1N∑k=0N−1Xkⅇⅈ2πNkn x n 1 N k 0 N 1 X k 2 N k n (1)
Xk DFT =∑u=0N−1xnⅇ-ⅈ2πNkn X k DFT u 0 N 1 x n 2 N k n (2)
But xn x n and Equation 1 are also repetions for periodic signal, which is ∞ -length.

It's DTFT Xω DFT =∑n=-∞∞xnⅇ-ⅈωn X ω DFT n x n ω n Where xn=1N∑k=0N−1 Xk DFT ⅇⅈ2πNkn x n 1 N k 0 N 1 X k DFT 2 N k n from Equation 1 =1N∑n=-∞∞∑k=0N−1 Xk DFT ⅇⅈ 1 N n k 0 N 1 X k DFT Assuming we can reverse the order of the sums. =1N∑k=0N−1 Xk DFT ∑n=-∞∞ⅇⅈ2πNknⅇ-ⅈωn 1 N k 0 N 1 X k DFT n 2 N k n ω n Where ∑n=-∞∞ⅇⅈ2πNknⅇ-ⅈωn n 2 N k n ω n is the DTFT of sinusiod with frequency 2πNk 2 N k =2πδω−2πNk 2 δ ω 2 N k so... Xω DFT =2πN∑k=0N−1 Xk DFT δω−2πNk X ω DFT 2 N k 0 N 1 X k DFT δ ω 2 N k

Figure 2
Figure 2 (fig2.png)

CASE 2: Finite Length Signals

Figure 3
Figure 3 (fig3.png)

Assuming finite-length

Xk DFT =∑n=0N−1xnⅇ-ⅈ2πNkn X k DFT n 0 N 1 x n 2 N k n (3)
Assuming zero outside 0N−1 0 N 1
Xω DTFT =∑n=0N−1xnⅇ-ⅈωn X ω DTFT n 0 N 1 x n ω n (4)

Note:

Equation 3 and Equation 4 are related by Xk DFT = X2πk/N DTFT X k DFT X 2 π k / N DTFT ie: DFT =N N samples of DTFT in 02π 0 2 .
Figure 4
Figure 4 (fig4.png)

Extension

What if we compute zeropadded DFT?

Figure 5
Figure 5 (fig5.png)
That is, finer sampling

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