0
$\begingroup$

i'm studying the practical utility of Fourier transforms and i have some questions. I hope to receive answers in layman terms.

1) Does the DTFT take only infinite input sequences?

2) If i apply the FT to a step function i get an infinite band spectrum. If i sample the step function and apply a DTFT does i get replicas of that infinite spectrum (obtained from FT) in the DTFT spectrum ?

$\endgroup$

1 Answer 1

2
$\begingroup$

I doubt that Hilmar will agree (maybe he will) but I will offer this answer as a counter to the one that has already been (prematurely) accepted by the questioner.

Hilmar's answer is wrong. It is factually wrong and analytically wrong.

1) Does the DTFT take only infinite input sequences?

No. There is nothing that prevents you from applying the formula to a finite sequence.

That's sorta true but it must be said that the "formula" is explicitly of an infinite sequence. However, there is nothing in that formula that says any particular sample must take on any particular value (like non-zero) other than a finite value. There is nothing that prevents all but a finite subset of samples having zero value.

The DTFT is:

$$ X(e^{j\omega}) = \sum\limits_{n=-\infty}^{\infty} x[n] \, e^{-j\omega n} $$

That's what the formula is and all of the $x[n]$ have a defined value, even if it's zero.

However, a finite time sequence has an infinite spectrum, so you can't sample it without some amount of aliasing

No one is sampling it (again). It has already been sampled. The question is: What happens if the length of the sequence (which is, in the DTFT, infinite in length) is shortened to a finite length?

And the answer is: It depends on the manner in which the sequence is shortened to a finite length. Or, if the sequence was never infinite in length and later applied to the DTFT, how are the infinite number of samples outside of the original finite length defined.

But if you do define all of those samples outside of your original finite length to be zero, that is saying:

$$ x[n] = 0 \qquad \text{for } x<0 \text{ and } x\ge N < \infty $$

Then the DFT is equivalent to:

$$ X(e^{j\omega}) = \sum\limits_{n=0}^{N-1} x[n] \, e^{-j\omega n} $$

And then it starts to look like the DFT. If $X(e^{j\omega})$ is sampled at $N$ equally-spaced values on the unit circle including $X(e^{j0})$, then it is equivalent to the DFT

$$ X[k] = X(e^{j\omega}) \Big|_{\omega = \frac{2 \pi k}{N}} = \sum\limits_{n=0}^{N-1} x[n] \, e^{-j2 \pi k n/N} $$

But something misleading in Hilmar's answer must be corrected:

Applying a finite-length sequence to the DTFT has nothing to do with aliasing. Nothing at all.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.