0
$\begingroup$

I am solving a question, where a signal g(t) is passed through a squaring device and then through an LPF such that bandwidth of the LPF tends to 0.

I understand that since for this filter the bandwidth, $ \Delta f \to 0$ , the transfer function is practically an impulse. I also understand that the width of the impulse should be $ 2 * 2 \pi \Delta f = 4 \pi \Delta f $. But I don't understand why the transfer function is expressed as $ 4 \pi \Delta f \delta(f) $, that is, why is it expressed as an impulse function of area $4 \pi \Delta f $

Here is the question and the official solution to the question:

enter image description here

Solution:

enter image description here

$\endgroup$
1
$\begingroup$

I don't think that using a Dirac impulse really helps to see what's going on in this problem. Using the notation of the given solution we have

$$\mathcal{F}\left\{g^2(t)\right\}=A(\omega)\;\Longrightarrow\;A(0)=E_g$$

and

$$Y(\omega)=A(\omega)H(\omega)$$

where $H(\omega)$ is the frequency response of the ideal low pass filter.

The output signal is given by

$$\begin{align}y(t)&=\frac{1}{2\pi}\int_{-\infty}^{\infty}Y(\omega)e^{j\omega t}d\omega\\&=\frac{1}{2\pi}\int_{-\infty}^{\infty}A(\omega)H(\omega)e^{j\omega t}d\omega\\&=\frac{1}{2\pi}\int_{-2\pi\Delta f}^{2\pi\Delta f}A(\omega)e^{j\omega t}d\omega\\&\approx\frac{1}{2\pi}A(0)\cdot 4\pi\Delta f\\&=2E_g\Delta f\end{align}$$

with the approximation being justified for $\Delta f\to 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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