The impulse response of an ideal low-pass filter can be determined by setting $H(\omega)=1$ in the Fourier-representation $$h(n) = \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} H(\omega)e^{j\omega n}d\omega$$

The solution will be a function of form $\sin(n)/n$. Now, the motivation behind the substitution above is the desire to amplify each frequency component before $\omega_c$ by an equal amount. Hence the term ideal response. Also, the fact that for some $\omega_0$ there is a corresponding amplitude $\vert H(\omega_0)\vert$ is not ( as far as I know ) interesting while performing spectral analysis. What matters are the amplitude relations between some $\omega_0$ and $\omega_1$, the frequency components of a signal perhaps.

Therefore, the substitution $H(\omega) = C$, for any constant $C$ should be equally valid right?


1 Answer 1


An additional gain (or attenuation) doesn't change the characteristic of a filter. So yes, any constant is fine. After all, it's matter of definition; there might be people who say that an ideal lowpass filter has unity gain. But that's quite a moot point in my opinion.

I do remember a case where a paper was submitted to a journal in which the author called a filter with a constant gain $\neq 1$ an "allpass filter". One of the reviewers, who is a very famous professor, known to everybody who has ever heard the term "filter bank", wrote in his review that an allpass filter only deserves its name when it has a gain of unity, and the author was strongly advised to make appropriate changes. So, some people take gain constants quite seriously.


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