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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?

Stanislav Bashkyrtsev
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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

$$H(e^{j\omega})=\sum_{n=-\infty}^{\infty}h[n]e^{-jn\omega}=\sum_{n=-\infty}^{\infty}h[n]\big(e^{j\omega}\big)^{-n}\tag{1}$$

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:

$$H(z)=\sum_{n=-\infty}^{\infty}h[n]z^{-n}\tag{2}$$

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.

Matt L.
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  • I see.. So It's not a (valid) mathematical notation but a DSP short-hand, right? – Stanislav Bashkyrtsev Aug 26 '18 at 08:37
  • 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 – Laurent Duval Aug 26 '18 at 08:50
  • @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., $z=e^{j\omega}$. – Matt L. Aug 26 '18 at 16:09
  • sorry Matt, i bumped you down. i will prepare an alternative answer. – robert bristow-johnson Aug 26 '18 at 16:39
  • @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. – Stanislav Bashkyrtsev Aug 26 '18 at 16:51
  • @robertbristow-johnson: That's perfectly alright, as long as you tell me what's wrong with this answer. – Matt L. Aug 26 '18 at 16:52
  • @StanislavBashkyrtsev: No, that's OK. The argument is a complex variable $z$, which takes the form $z=e^{j\omega}$. – Matt L. Aug 26 '18 at 16:53
  • $x[n]=e^{j\omega n}$ is the time-domain input. the (normalized) frequency-domain input is $X(?)=e^{j \omega}$ what goes into the argument is a matter of convention, but if we want to minimize the number of "things" (or superfluous definitions), i would recommend saying the frequency domain input is $X(e^{j \omega})=e^{j \omega}$. – robert bristow-johnson Aug 26 '18 at 17:09
  • @robertbristow-johnson: I don't follow your argument. The notation $H(e^{j\omega})$ is just about defining a function and its dependence on the independent variable. It doesn't say anything about any input, we're just talking about a function, which describes a system, but not any possible input to the system. – Matt L. Aug 26 '18 at 18:32
  • when the input $x[n]=e^{j\omega n}$ (which is a normalized eigenfunction), then the output is: $$\begin{align} y[n]&=H(e^{j \omega}) e^{j \omega n} \ &= H(e^{j \omega}) x[n] \end{align}$$

    or, we get to say,

    $$ Y(e^{j\omega}) = H(e^{j\omega}) \cdot X(e^{j\omega}) $$

     – robert bristow-johnson Aug 26 '18 at 18:38
  • @robertbristow-johnson: Yes sure, we know that. But still, I wouldn't say that the notation $H(e^{j\omega})$ says (or should say) anything about the input to the system which is described by that function. What if the input is $\delta[n]$? – Matt L. Aug 26 '18 at 18:40