Using the given identities,

$$ a^nu[n] \Longleftrightarrow \frac{1}{(1-ae^{-jw})}$$ and $$\delta[n-k]\Longleftrightarrow e^{-jwk}$$

Find the inverse DTFT of,

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

my attempt:

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

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!

  • $\begingroup$ multiplication in frequency domain is....... in the time domain ? so why don’t you want to use convolution $\endgroup$
    – user28715
    Sep 2, 2018 at 2:44
  • $\begingroup$ Your attempt has a small mistake, which @StanleyPawlukiewicz pointed out. $\endgroup$
    – learner
    Sep 2, 2018 at 6:46
  • $\begingroup$ @StanleyPawlukiewicz. Thanks but what justifies the convolution: h(n) * x(n) and then switching back to u. In other words, could you explain more in detail what's happening with the math part. $\endgroup$
    – mark leeds
    Sep 2, 2018 at 7:33
  • $\begingroup$ You’re given a product in the frequency domain of two functions of w. THAT is the justification for the convolution in the time domain. Your text book explains why $\endgroup$
    – user28715
    Sep 2, 2018 at 7:48

1 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.


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