Are there general techniques to derive DTFTs? Given a bandlimited function $x(t)$, how do I find

$$X(\omega)=\sum_{n=-\infty}^\infty x[n]e^{-i\omega n}$$

Generally, it is easier to derive the continuous transform (never mind the constants):

$$X(f)=\int_{-\infty}^{\infty}x(t)e^{-i \omega t}\, dt$$

because we have a wealth of integration theory to fall back on. Ok, one can also calculate $X(\omega)$ from $X(f)$ by adding shifted copies, but we're again back to summing an infinite sequence. Any ideas on how to approach DTFT derivations in a smarter way? Or is sweating out the tiring summation the only way? I don't have any particular function in mind, but if one has to be chosen as an example, I'd pick $x(t) = \text{sech}(a t)$. The continuous FT of this is known and can be found in the table on Wikipedia (#208).

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    $\begingroup$ The "wealth of integration theory" works primarily for "nice" $x(t)$ for which the integrals can be computed analytically. There is similar theory for summation of series that works for "nice" sequences (cf. Deniz's answer) but is not as well-known to most people since the topic gets a lot less attention than integration techniques do. $\endgroup$ Sep 18, 2012 at 10:53

1 Answer 1


As an example, the rules of series can work. A typical example is the $x[n] = a^n u[n]$ where $|a| < 1$. For instance, say we want to find the DTFT of the signal $x[n] = 0.9^n u[n]$. Then,

$$X(e^{j\omega}) = \sum_{n = -\infty}^{\infty} 0.9^n u[n] e^{-j\omega n} \\ = \sum_{n = 0}^{\infty} (0.9 e^{-j\omega})^n$$

This is where the series work (geometric power series):

$$\sum_{n = 0}^{\infty} a^n = \frac{1}{1-a}$$

Then our DTFT is:

$$X(e^{j\omega}) = \frac{1}{1 - 0.9 e^{-j\omega}}$$


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