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I know about the the shifting property of the impulse function in the time domain as can be seen in equation $(1)$.

$$ \int_{-\infty}^{\infty} f(x)\delta(x - a)dx = f(a)\tag{1} $$

But what is the effect of multiplication of a function by the impulse function in the Frequency domain? I.e

$$ X(\omega) = \delta(\omega - \omega_0)\cdot H(\omega) $$

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I think there is a slight typo in robert bristow-johnson's answer. Should be

\begin{align} x(t) &= \frac{1}{2 \pi} \int\limits_{-\infty}^{\infty} X(\omega) e^{j \omega t} \, \mathrm{d}\omega\\ \\ &= \frac{1}{2 \pi} \int\limits_{-\infty}^{\infty} H(\omega_0) \delta(\omega - \omega_0) e^{j \omega t} \, \mathrm{d}\omega\\ \\ &= \tfrac{1}{2 \pi} H(\omega_0) e^{j \omega_0 t} \end{align}

What this actually means is that during the inverse transform, $\delta(\omega - \omega_0)$ selects point frequency $\omega_0$ from the frequency response $H(\omega_0)$, and the value of $H(\omega_0)$ at that point determines the magnitude and phase of the time domain sinusoid $e^{j \omega_0 t}$.

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  • $\begingroup$ thank you for spotting and fixing it. $\endgroup$ – robert bristow-johnson Jun 6 '20 at 15:26
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All it means is

$$\begin{align} X(\omega) &= H(\omega) \delta(\omega - \omega_0)\\ &= H(\omega_0) \delta(\omega - \omega_0) \\ \end{align}$$

which means, in the time domain

$$\begin{align} x(t) &= \frac{1}{2 \pi} \int\limits_{-\infty}^{\infty} X(\omega) e^{j \omega t} \, \mathrm{d}\omega\\ \\ &= \frac{1}{2 \pi} \int\limits_{-\infty}^{\infty} H(\omega_0) \delta(\omega - \omega_0) e^{j \omega t} \, \mathrm{d}\omega\\ \\ &= \tfrac{1}{2 \pi} H(\omega_0) e^{j \omega_0 t} \end{align}$$

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For any function $f(x)$ that is continuous at $x=x_0$ the following holds:

$$f(x)\delta(x-x_0)=f(x_0)\delta(x-x_0)\tag{1}$$

So the result is a Dirac impulse at $x=x_0$ scaled by $f(x_0)$.

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  • $\begingroup$ So is it the same in the frequency domain ie: 1) F(w)δ(w−w0) = f(w0)δ(w−w0), Rather than: 2) F(w)δ(w−w0) = F(w0). Therefore, is (1) the correct answer? $\endgroup$ – John Jun 5 '20 at 10:48
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    $\begingroup$ @John: Yes, the domain doesn't matter, Eq. (1) is valid regardless of the interpretation of the independent variable $x$. The Dirac impulse only "disappears" when you integrate (1), otherwise it doesn't. $\endgroup$ – Matt L. Jun 5 '20 at 11:03

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