# Energy in frequency and time domain

I know from Parseval's Theorem that, given a signal $$x(t)$$, with $$t$$ a variable in the time domain,

$$\int_{-\infty}^{+\infty} |x(t)|^2 dt = \int_{-\infty}^{+\infty}|X(f)|^2 df,$$

where $$|X(f)|$$ is the Fourier transform of the signal $$x(t)$$ in the frequency domain.

Now, consider the following signal:

$$s(t) = A \mathrm{sinc}^2(2t)\mathrm{cos}(100\pi t).$$

This is a modulated signal thus I can apply the modulation theorem for Fourier transform:

$$S(f) = \frac{X(f-f_0)+X(f+f_0)}{2},$$

where $$X(f) = \mathcal{F} [A\mathrm{sinc}^2(2t)] = \frac{A}{2} \Delta(\frac{f}{2})$$, so

$$S(f) = \frac{A}{4} \left[\Delta\left(\frac{f-50}{2}\right) + \Delta\left(\frac{f+50}{2}\right) \right].$$

Now, if I try to compute the energy in the frequency domain, I have that

$$E_s = \int_{-\infty}^{+\infty} E_s(f) df = \int_{-\infty}^{+\infty} |S(f)|^2 df = \frac{A^2}{6}$$

If I try to do the same thing in the time domain, I get

$$\int_{-\infty}^\infty |A(\mathrm{sinc}(2t))^2\mathrm{cos}(100\pi t)|^2 dt = \frac{A^2 \pi}{6}$$

So my question is, due to Parseval's Theorem, shouldn't the two integrals give the same result?

• The two integrals must indeed be the same. I don't see how you arrived at the last result with $\pi$ in the numerator. Could you show how you obtained that result? Jun 20, 2022 at 18:08
• Actually I did it using Wolfram Alpha. It's a lot easier to do in frequency domain and that was the first thing I did; then I usually check the result in the time domain by computint the integral with wolfram. Jun 20, 2022 at 19:10
• That explains the difference. WA uses a different definition of the sinc function than you did in your frequency domain computation. See my answer below. Jun 20, 2022 at 19:53

$$\textrm{sinc}(t)=\frac{\sin(\pi t)}{\pi t}\tag{1}$$
$$\textrm{sinc}(t)=\frac{\sin(t)}{t}\tag{2}$$