Using the given identities, find the inverse DTFT

Question

Using the given identities,

$ a^nu[n]$ <===> $\frac{1}{(1-ae^{-jw})}$

$\delta[n-k]$ <===> $e^{-jwk}$

Find the inverse DTFT of,

$ H(e^{jw}) = B \frac{e^{-jw}}{(1-ae^{-jw})}$

my attempt:

$ h[n] = B\delta[n-1]a^nu[n] $

It seems straight forward enough, just plug in the inverse dtft. This is not correct though, there is no delta to be found in the correct solution.

The correct answer is:

$h[n] = Ba^{n-1}u[n-1]$

It's like the delta disappeared and the delta convolution property was used.

($x[n]*\delta[n-1] = x[n-1]$)

But using the identities above, I do not want to use convolution. I am confused why the delta disappeared and the signals it was multiplied with became shifted, any help in understanding how to get the correct answer using the identities would be appreciated!

Answer 1

As pointed out in the comments, it is important for you to know and understand that multiplication in one domain (time or frequency) corresponds to convolution in the other domain. And, as a consequence, multiplication with $e^{-j\omega}$ in the Fourier domain corresponds to a delay of one sample in the time domain (i.e., convolution with $\delta[n-1]$). So you have

$$\begin{align}\mathcal{F}^{-1}\left\{B\frac{e^{-j\omega}}{1-ae^{-j\omega}}\right\}[n]&=B\cdot\mathcal{F}^{-1}\left\{\frac{e^{-j\omega}}{1-ae^{-j\omega}}\right\}[n]\quad (\textrm{due to linearity of }\mathcal{F}^{-1})\\&=B\cdot\mathcal{F}^{-1}\left\{\frac{1}{1-ae^{-j\omega}}\right\}[n-1]\\&=B\cdot a^{n-1}u[n-1]\end{align}$$

where I used $\mathcal{F}^{-1}$ to denote the inverse discrete-time Fourier transform.