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!