Consider the signal $ x \left[ n \right] $ and its DTFT transform $ X \left( {e}^{j \omega} \right) $.
Assume $ X \left( {e}^{j \omega} \right) $ is differentiable.

What is the Inverse DTFT of:

$$ j \frac{d}{d \omega} X \left( {e}^{j \omega} \right) $$

Express the result in terms of $ x \left[ n \right] $.


This is pretty straight forward using the definition of the Discrete Time Fourier Transform (DTFT).

The definition of the DTFT:

$$ X \left( {e}^{j \omega} \right) = \sum_{m = -\infty}^{\infty} x \left[ m \right] {e}^{-j \omega m} $$

Differentiating with respect to $\omega$:

$$\begin{align*} \frac{d}{d \omega} X \left( {e}^{j \omega} \right) & = \sum_{m = -\infty}^{\infty} \frac{d}{d \omega} x \left[ m \right] {e}^{-j \omega m} \\ & = \sum_{m = -\infty}^{\infty} \left( -j m \right) x \left[ m \right] {e}^{-j \omega m} \\ & = \frac{1}{j} \sum_{m = -\infty}^{\infty} m x \left[ m \right] {e}^{-j \omega m} \\ \end{align*}$$

Now, just plug it into the Inverse DTFT:

$$\begin{align*} \frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{1}{j} \sum_{m = -\infty}^{\infty} m x \left[ m \right] {e}^{-j \omega m} {e}^{j \omega n} d \omega & = \frac{1}{j} \sum_{m = -\infty}^{\infty} m x \left[ m \right] \frac{1}{2 \pi} \int_{0}^{2 \pi} {e}^{-j \omega m} {e}^{j \omega n} d \omega \\ & = \frac{1}{j} \sum_{m = -\infty}^{\infty} m x \left[ m \right] \delta \left[ n - m \right] \\ & = \frac{n}{j} x \left[ n \right] \end{align*}$$

This implies:

$$ j \frac{d}{d \omega} X \left( {e}^{j \omega} \right) = j \frac{n}{j} x \left[ n \right] = n x \left[ n \right] $$

  • $\begingroup$ With $d\omega$. $\endgroup$
    – Gilles
    Jul 5 '18 at 7:23
  • $\begingroup$ @Gilles, Indeed a typo :-) Fixed it. Thank You. $\endgroup$
    – Royi
    Jul 5 '18 at 8:13
  • $\begingroup$ Did you forget to multiply by j in the end? Also, I think one other way to calculate this would be just by using integration by parts (i.e. you can move derivative of X under d of the integral and switch). $\endgroup$
    – Dan M.
    Dec 30 '18 at 23:33
  • $\begingroup$ @DanM., where do you think the $ j $ is missing? $\endgroup$
    – Royi
    Dec 31 '18 at 8:59
  • $\begingroup$ @Royi the question asks for DTFT of $j$ multiplied by derivative, however you've dropped that. It's trivial to add, but someone might not notice it. $\endgroup$
    – Dan M.
    Dec 31 '18 at 12:55

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