# Calculate the Inverse DTFT of the DTFT Derivative in Terms of $x \left[ n \right]$

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:

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

• With $d\omega$. Jul 5 '18 at 7:23
• @Gilles, Indeed a typo :-) Fixed it. Thank You.
– Royi
Jul 5 '18 at 8:13
• 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). Dec 30 '18 at 23:33
• @DanM., where do you think the $j$ is missing?
– Royi
Dec 31 '18 at 8:59
• @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. Dec 31 '18 at 12:55