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