# What is the Effect of Multiplying a Function by the Unit Impulse Function in the Frequency Domain?

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

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

## 3 Answers

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 $$e^{j \omega_0 t}$$.

• thank you for spotting and fixing it. – robert bristow-johnson Jun 6 '20 at 15:26

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}

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 $$f(x_0)$$.

• 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? – John Jun 5 '20 at 10:48
• @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. – Matt L. Jun 5 '20 at 11:03