1
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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

0
$\begingroup$

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

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

$\endgroup$
1
  • $\begingroup$ 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

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.