The time-domain representation of $G(f) H(f)$, where $H(f)$ is an ideal brickwall filter of bandwidth $1/(2T)$ is:

$$ \int g(\tau) \operatorname{sinc}\left(\frac{t-\tau}{T}\right) d\tau $$

I want to show (using algebra, in the time domain) that the result of the above equation equals $g(t)$ if the bandwidth of $g$ is less than $1/(2T)$. This is something trivial to see in the frequency domain.

But how in the time domain, using convolution integrals?

By definition, if $g$ is bandlimited to $1/(2 T_g)$, I can expand it using the WKS interpolation formula, having the samples as coefficients:

$$ g(t) = \sum_{n=-\infty}^{\infty} g(n T_g) \operatorname{sinc}\left(\frac{t-n T_g}{T_g}\right) = \sum_{n-\infty}^{\infty} g[n] \operatorname{sinc}\left(\frac{t-n T_g}{T_g}\right) $$

Plugging this into the original equation:

$$ \int \sum_{n=-\infty}^{\infty} g(n T_g) \operatorname{sinc}\left(\frac{\tau-n T_g}{T_g}\right) \operatorname{sinc}\left(\frac{t-\tau}{T}\right) d\tau \\ = \sum_{n=-\infty}^{\infty} g(n T_g) \left[ \int \operatorname{sinc}\left(\frac{\tau-n T_g}{T_g}\right) \operatorname{sinc}\left(\frac{t-\tau}{T}\right) d\tau \right] $$

From that, the term in bracket would need to be equivalent to $\operatorname{sinc}\left(\frac{t-n T_g}{T_g}\right)$ if $T \leq T_g$. But I don't see this working out

  • $\begingroup$ Isn't Whittaker only applicable when filtering a sampled signal? It seems from your question that $g(t)$ is continuous. $\endgroup$
    – MBaz
    Sep 25, 2020 at 1:38
  • $\begingroup$ By definition, you can expand a bandlimited signal with the WKS interpolation formula by using its samples as the coefficients (see en.wikipedia.org/wiki/…). That's what I have done. $\endgroup$
    – divB
    Sep 25, 2020 at 2:02
  • $\begingroup$ I edited the question to clarify that. $\endgroup$
    – divB
    Sep 25, 2020 at 2:08
  • $\begingroup$ Yeah -- it seems like you reduced the problem to showing that the sinc is unaffected by the filter. An idea: replace the sinc by its definition in terms of a sine, and see if that makes the integral tractable. $\endgroup$
    – MBaz
    Sep 25, 2020 at 14:15
  • $\begingroup$ Maybe identity (31) can help here? mathworld.wolfram.com/SincFunction.html $\endgroup$
    – MBaz
    Sep 25, 2020 at 14:19


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