# Showing that filtering a signal with bandwidth B with a brickwall filter of bandwidth W>B has no effect in time domain

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

• Isn't Whittaker only applicable when filtering a sampled signal? It seems from your question that $g(t)$ is continuous. – MBaz Sep 25 '20 at 1:38
• 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. – divB Sep 25 '20 at 2:02
• I edited the question to clarify that. – divB Sep 25 '20 at 2:08
• 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. – MBaz Sep 25 '20 at 14:15
• Maybe identity (31) can help here? mathworld.wolfram.com/SincFunction.html – MBaz Sep 25 '20 at 14:19