I'm trying to grasp some intuition about why zero-padding the time domain sequence $x[n]$ interpolates the frequency domain bins of the $DFT\{x[n]\} = X[k]$ and how does this relate to the $DTFT$ of $x[n]$.

Let's say we have $N$ point signal $x[n]$ and we pad it with 0's in the time domain as usual. So, as we append more and more zeros to $x[n]$ and perform the $DFT$, we know that $DFT$ makes the assumption that the signal is periodic, hence the period tends to infinity, hence we get frequency samples (more and more and padding increaes) of the $N$ point $DTFT$ applied to the original $x[n]$ ?

I'd love some insights and intution please. Thanks


1 Answer 1


Say a discrete signal $x(n)$ and its Z-transform $X(z)$, discrete time Fourier transform $X(e^{j\omega})$, and discrete Fourier transform $X(k)$. The relationships between DFT and ZT / DTFT are respectively:

$$ X(k) = X(z)|_{z=e^{j2\pi k/N}} \tag{1.1} $$

$$ X(k) = X(e^{j\omega})|_{\omega=2\pi k/N} \tag{1.2} $$

Since the original signal $x(n)$ can be perfectly reconstructed by IDFT, yes you can certainly obtain DTFT from DFT. However, you can't get DTFT by padding zeros as long as possible but by a technique called frequency domain interpolation. First is reconstruction of ZT from DFT:

$$ X(z) = \sum_{n=0}^{N-1}x(n)z^{-n}=\sum_{n=0}^{N-1}\Big[\frac{1}{N}\sum_{k=0}^{N-1}X(k)W_N^{-nk} \Big]z^{-n} \\ = \frac{1}{N}\sum_{k=0}^{N-1}X(k)\Big[ \sum_{n=0}^{N-1}W_N^{-nk}z^{-n} \Big] \\ = \frac{1}{N}\sum_{k=0}^{N-1}X(k)\frac{1-W_N^{-Nk}z^{-N}}{1-W_N^{-k}z^{-1}} = \frac{1-z^{-N}}{N} \sum_{k=0}^{N-1}\frac{X(k)}{1-W_N^{-k}z^{-1}} \tag{2} $$

where $W_N^{nk} = e^{j\frac{2\pi}{N}nk}$. Define the interpolation function

$$\varPhi_k(z) = \frac{1}{N}\frac{1-z^{-N}}{1-W_N^{-k}z^{-1}} = \frac{z^N-1}{Nz^{N-1}(z-W_N^{-k})} \tag{3}$$

and $X(z)$ can be written as

$$ X(z) = \sum_{k=0}^{N-1}X(k)\varPhi_k(z) \tag{4} $$

Now let $z = e^{j\omega}$ we can derive that

$$ X(e^{j\omega}) = \sum_{k=0}^{N-1}X(k)\varPhi_k(e^{j\omega}) \tag{5} $$

and the interpolation function

$$ \varPhi_k(e^{j\omega}) = \frac{1-e^{-j\omega N}}{N(1-W_N^{-k}e^{-j\omega})} = \frac{1-e^{-j\omega N}}{N(1-e^{-j(\omega-2\pi k/N)})}\\ = \frac{1}{N}\frac{\sin(N\omega/2)}{\sin\big[(\omega-2\pi k/N)/2\big]} e^{-j[\omega(N-1)/2 + k\pi/N]} \\ =\frac{1}{N}\frac{\sin(N\omega/2)}{\sin\big[(\omega-2\pi k/N)/2\big]} e^{jk\pi(N-1)/N} e^{-j(N-1)\omega/2} \tag{6} $$

Let's write it in a more convinient form as

$$ \varPhi_k(e^{j\omega}) = \varPhi(\omega - 2\pi k/N) \tag{7} $$

where $\varPhi(\omega)$ is the Fourier transform of a $N$-point rectangle window divided by $N$

$$ \varPhi(\omega) = \frac{1}{N} \frac{\sin(\omega N/2)}{\sin(\omega/2)} e^{-j(N-1)\omega/2} \tag{8} $$

Finally, we get the equation of reconstructing DTFT by interpolating DFT

$$ X(e^{j\omega}) = \sum_{k=0}^{N-1}X(k)\varPhi(\omega-2\pi k/N) \tag{9} $$

  • $\begingroup$ 1) So basicaly you are saying that for finite length sequence $x[n]$, there is an interpolation such that I can get the DTFT out of the DFT bins? 2) But how does this relate to zero padding? Isn't the zero padding sort of interpolation on its own? $\endgroup$
    – user3921
    Apr 9, 2021 at 21:10
  • 1
    $\begingroup$ 1) Yes, but recall the definition of DTFT, the summation extends from $-\infty$ to $\infty$. Mathematically, the interpolation can only reconstruct truncated DTFT. 2) And yes, the zero padding is a sort of interpolation approaching to DTFT, but the result it returns is still DFT (it's $X(k)$ not $X(e^{j\omega})$). $\endgroup$
    – ZR Han
    Apr 12, 2021 at 2:24

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