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] $.


1 Answer 1


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
    Commented Jul 5, 2018 at 7:23
  • $\begingroup$ @Gilles, Indeed a typo :-) Fixed it. Thank You. $\endgroup$
    – Royi
    Commented Jul 5, 2018 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.
    Commented Dec 30, 2018 at 23:33
  • $\begingroup$ @DanM., where do you think the $ j $ is missing? $\endgroup$
    – Royi
    Commented Dec 31, 2018 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.
    Commented Dec 31, 2018 at 12:55

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.