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:

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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




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 $\Delta f\to 0$.


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