0
$\begingroup$

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

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

$\endgroup$
0

3 Answers 3

5
$\begingroup$

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 $e^{j \omega_0 t}$.

$\endgroup$
1
  • $\begingroup$ thank you for spotting and fixing it. $\endgroup$ Jun 6, 2020 at 15:26
2
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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 $f(x_0)$.

$\endgroup$
2
  • $\begingroup$ 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, 2020 at 10:48
  • 1
    $\begingroup$ @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, 2020 at 11:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.