Difference between Discrete Time Fourier Transform and Discrete Fourier Transform

Question

I have read many articles about DTFT and DFT but am not able to discern the difference between the two except for a few visible things like DTFT goes till infinity while DFT is only till $N-1$. Can anyone please explain the difference and when to use what? Wiki says

The DFT differs from the discrete-time Fourier transform (DTFT) in that its input and output sequences are both finite; it is therefore said to be the Fourier analysis of finite-domain (or periodic) discrete-time functions.

Is it the only difference?

Edit: This article nicely explains the difference

Answer 1

The discrete-time Fourier transform (DTFT) is the (conventional) Fourier transform of a discrete-time signal. Its output is continous in frequency and periodic. Example: to find the spectrum of the sampled version $x(kT)$ of a continous-time signal $x(t)$ the DTFT can be used.

The discrete Fourier transform (DFT) can be seen as the sampled version (in frequency-domain) of the DTFT output. It's used to calculate the frequency spectrum of a discrete-time signal with a computer, because computers can only handle a finite number of values. I would argue against the DFT output being finite. It is periodic as well and can therefore be continued infinitely.

To sum it up:

                DTFT                | DFT
       input    discrete, infinite  | discrete, finite *)
       output   contin., periodic   | discrete, finite *)

*) A mathematical property of the DFT is that both its input and output are periodic with the DFT length $N$. That is, although the input vector to the DFT is finite in practice, it's only correct to say that the DFT is the sampled spectrum if the DFT input is thought to be periodic.

Answer 2

Alright, I'm gonna answer this with an argument that "opponents" to my rigid nazi-like position regarding the DFT have.

First of all, my rigid, nazi-like position:

The Discrete Fourier Transform and Discrete Fourier Series are one-and-the-same.

The DFT maps one infinite and periodic sequence, $x[n]$ with period $N$ in the "time" domain to another infinite and periodic sequence, $X[k]$, again with period $N$, in the "frequency" domain. And the iDFT maps it back. and they're "bijective" or "invertible" or "one-to-one".

DFT: $$ X[k] = \sum\limits_{n=0}^{N-1} x[n] e^{-j 2 \pi nk/N} $$

iDFT: $$ x[n] = \frac{1}{N} \sum\limits_{k=0}^{N-1} X[k] e^{j 2 \pi nk/N} $$

That is most fundamentally what the DFT is. It is inherently a periodic or circular thing.

$$ x[n+N]=x[n] \qquad \forall n \in \mathbb{Z} $$ $$ X[k+N]=X[k] \qquad \forall k \in \mathbb{Z} $$

---

But the periodicity deniers like to say this about the DFT. It is true, it just doesn't change any of the above.

So, suppose you had a finite-length sequence $x[n]$ of length $N$ and, instead of periodically extending it (which is what the DFT inherently does), you append this finite-length sequence with zeros infinitely on both left and right. So

$$ \hat{x}[n] \triangleq \begin{cases} x[n] \qquad & \text{for } 0 \le n \le N-1 \\ \\ 0 & \text{otherwise} \end{cases} $$

Now, this non-repeating infinite sequence does have a DTFT:

DTFT: $$ \hat{X}\left(e^{j\omega}\right) = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] e^{-j \omega n} $$

$\hat{X}\left(e^{j\omega}\right)$ is the Z-transform of $\hat{x}[n]$ evaluated on the unit circle $z=e^{j\omega}$ for infinitely many real values of $\omega$. Now, if you were to sample that DTFT $\hat{X}\left(e^{j\omega}\right)$ at $N$ equally spaced points on the unit circle, with one point at $z=e^{j\omega}=1$, you would get

$$ \begin{align} \hat{X}\left(e^{j\omega}\right)\Bigg|_{\omega = 2 \pi\frac{k}{N}} & = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] e^{-j \omega n} \Bigg|_{\omega = 2 \pi\frac{k}{N}} \\ & = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] e^{-j 2 \pi k n/N} \\ & = \sum\limits_{n=0}^{N-1} \hat{x}[n] e^{-j 2 \pi k n/N} \\ & = \sum\limits_{n=0}^{N-1} x[n] e^{-j 2 \pi k n/N} \\ & = X[k] \\ \end{align} $$

That is precisely how the DFT and DTFT are related. Sampling the DTFT at uniform intervals in the "frequency" domain causes, in the "time" domain, the original sequence $\hat{x}[n]$ to be repeated and shifted by all multiples of $N$ and overlap-added. That's what uniform sampling in one domain causes in the other domain. But, since $\hat{x}[n]$ is hypothesized to be $0$ outside of the interval $0 \le n \le N-1$, that overlap-adding does nothing. It just periodically extends the non-zero part of $\hat{x}[n]$, our original finite-length sequence, $x[n]$.

Answer 3

If we were to compute a continuous DTFT, sample one cycle of it uniformly, and perform an inverse DFT, we would obtain one cycle of a periodic summation of the original infinite, aperiodic time sequence. Conversely, if we were to compute one cycle of a periodic summation of the original infinite, aperiodic time sequence, and perform a DFT, we would obtain samples of one cycle of the continuous DTFT.

Answer 4

If I am correct, even if DFT input is periodic, although the number of samples is finite, the mathematics behind it treats it as an infinite sequence which periodically commences the N samples after its termination. Please correct me if I am wrong.

Answer 5

Since DTFT output is continuous, it can not be processed with computers. So we have to convert this continuous signal into discrete form. It is nothing but DFT as a further advancement on FFT to reduce calculations.