Every resource that I can find uses this identity when deriving impulse response:

$h[n] = IDTFT \Big\{H(e^{j\omega}) \Big\}$

Suggesting that the input signal was $e^{j\omega}$. But by definition Impulse Response is the result of applying the filter to delta signal. Why then do we use $e^{j\omega}$ instead?


1 Answer 1


I think you misunderstand the notation. Writing $H(e^{j\omega})$ does not mean that $e^{j\omega}$ is the input signal. It just means that the frequency response is a function of the complex variable $e^{j\omega}$, because, as you might know, it is defined by


where $h[n]$ is the impulse response. Eq. $(1)$ is the discrete-time Fourier transform (DTFT) of the sequence $h[n]$. Generalizing to a complex argument $z$ you get the definition of the $\mathcal{Z}$-transform:


Comparing $(1)$ and $(2)$ shows that the DTFT is equal to the $\mathcal{Z}$-transform evaluated on the unit circle, i.e., for $z=e^{j\omega}$. Note that evaluating $(2)$ on the unit circle only makes sense if the unit circle is inside the region of convergence of $(2)$, otherwise $H(z)$ doesn't converge for $|z|=1$ and in that case the DTFT either doesn't exist or it takes a form which is different from $H(z)$ with $|z|=1$.

Note that some people just use $\omega$ as the argument of the function that expresses the DTFT, so you might as well come across the notation $H(\omega)$. After all, it's just a matter of convention.

Concerning the notation $H(e^{j\omega})$ you can also refer to this question and its answer.

  • $\begingroup$ I see.. So It's not a (valid) mathematical notation but a DSP short-hand, right? $\endgroup$ Aug 26, 2018 at 8:37
  • $\begingroup$ If you look at the link, you could see the origin in the $z$-transform. And the Fourier transform is its evaluation on the unit disk ($|z|=1$). So $H(\omega)$ is also a reminder of how the rotation/angle disk is parametrized $\endgroup$ Aug 26, 2018 at 8:50
  • $\begingroup$ @StanislavBashkyrtsev: The notation is perfectly correct. It shows that $H$ is a function of a complex variable $z$ evaluated on the unit circle $|z|=1$, i.e., $z=e^{j\omega}$. $\endgroup$
    – Matt L.
    Aug 26, 2018 at 16:09
  • $\begingroup$ sorry Matt, i bumped you down. i will prepare an alternative answer. $\endgroup$ Aug 26, 2018 at 16:39
  • $\begingroup$ @MattL., well, when a function is defined, it names its argument which is then used in the body of the function. E.g. here is a function with argument $x$: $f(x) = x^2$. But in the formula above the parameter is not defined. It's replaced with $e^{j\omega}$ which as far as I understood describes the form $H(\omega)$ takes. $\endgroup$ Aug 26, 2018 at 16:51

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