# 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 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: Solution: 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 $$\Delta f\to 0$$.