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?

  • $\begingroup$ 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? $\endgroup$
    – Matt L.
    Jun 20 at 18:08
  • $\begingroup$ 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. $\endgroup$
    – aghin
    Jun 20 at 19:10
  • $\begingroup$ 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. $\endgroup$
    – Matt L.
    Jun 20 at 19:53

1 Answer 1


The problem is the ambiguity in the definition of the sinc function. For the frequency domain solution you (implicitly) used

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

whereas for the time domain solution you (actually, Wolfram Alpha) used


  • $\begingroup$ I knew about the two definitions but I dind't think about what Wolfram could have used. Thank you for the reply! It's all clear now. $\endgroup$
    – aghin
    Jun 20 at 20:07

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