Getting the DTFT from the DFT samples
Myth: DTFT is Sinc-interpolated DFT.

Problem with the above statement: Sinc is not $2\pi$-Periodic function, but all DTFTs are. 

Correct Answer:


Theoretical, Continuous-$\omega$ $2\pi$-Periodic DTFT can be obtained by continuous Lagrangian-interpolation of the DFT Samples. So that the values at $\omega = 2\pi k/N$ will be the DFT Samples $X[k]$ for $k=0,1,...,N-1$ and the Interpolation-function's zero-crossings are at $2\pi k/N$.


In other words, DTFT will take same values at roots of unity as DFT Samples, but it will be a smooth interpolation of the DFT at other values of digital frequency $\omega$.

Mathematically, let $x[n]$ be N-Length sequence and $X[k]$ be it's N-point DFT. Now, DTFT is defined for infinite length sequences. So, lets derive DTFT of a finite length $x[n]$.

$$X(e^{j\omega}) = \sum^{N-1}_{n=0} x[n]e^{-j\omega n},$$ now write IDFT of $X[k]$ in place of $x[n]$.

$$X(e^{j\omega}) = \sum^{N-1}_{n=0} (1/N.\sum^{N-1}_{k=0}X[k]e^{j 2\pi k/N})e^{-j\omega n},$$ now bring summation w.r.t n inside,

$$X(e^{j\omega}) = \sum^{N-1}_{k=0} (X[k] (\sum^{N-1}_{n=0}1/N e^{j 2\pi k/N}e^{-j\omega n}))$$

$$ = \sum^{N-1}_{k=0} (X[k] (\sum^{N-1}_{n=0}1/N e^{-jn(\omega - 2\pi k/N)}))$$

So, basically, $$X(e^{j\omega}) = \sum^{N-1}_{k=0} X[k]. \Lambda(w - 2\pi k/N),$$ where $$\Lambda(w) = 1/N \sum^{N-1}_{n=0} e^{-jn\omega}.$$

What this means is each sample of $X[k]$ is multiplied a $2\pi k/N$ shifted copy of $\Lambda(\omega)$ and added together. Basically, $X[k]$ is interpolated by a continuous-$\omega$ and $2\pi$-Periodic function $\Lambda(\omega)$. And this function is not a Sinc-function but something else. Sure it looks like Sinc and it will approach to Sinc in the limit.

Further , $$\Lambda(\omega) = \frac{1}{N}*e^{-j\omega\frac{(N-1)}{2}} \frac{\sin(N\omega/2)}{\sin(\omega/2)}.$$ Plotting this function in $[-\pi,\pi]$ is below:

&gt;&gt; w = -pi:0.0001:pi;
&gt;&gt; y = 1/64 * sin(w*64/2)./sin(w/2);
&gt;&gt; plot(w,y)




I repeat, it is not a sinc interpolation. Sinc is not $2\pi$-Periodic function. There is no way we can get a DTFT by interpolating with sinc.


Practically, you can get DTFT by interpolating with the MATLAB snippet I have provided which approximates $\Lambda(\omega)$ function.


What you can check yourself is extending the above plot to $[-4\pi:4\pi]$ and see that it indeed is Periodic function.