1
$\begingroup$

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] $.

$\endgroup$
2
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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