Suppose there is a bounded continuous signal $f(t)$ that can be fourier-transformed into $F(\omega)$ of frequency contents. $|F(\omega)|$ is non-zero for only finite number of frequencies: $f_1,f_2,..f_n$. Using a filter, we want to retain zeroes of $|F(\omega)|$ at frequencies other than $f_1,..f_n$, but we want to make $|F(\omega)| = \alpha$ for some constant $\alpha$ at all frequencies $f_1,..f_n$.

Is there such a filter? It is fine to assume that phase of $F(\omega)$ is zero or constant for all frequencies.

  • $\begingroup$ Your question is not clear; please consder rephrasing it. Are you thinking of an all-pass filter? (en.wikipedia.org/wiki/All-pass_filter) $\endgroup$ – MBaz Mar 14 '15 at 17:17
  • $\begingroup$ No. I edited my question - hope this is clearer. $\endgroup$ – Signo Mar 14 '15 at 17:24

What you want is not possible, but the problem could be the problem formulation itself. If you have a signal which is non-zero only at a finite number of frequencies, then your signal is a sum of sinusoids (or complex exponentials). What you can do is make their amplitudes equal, but this does not mean that $F(\omega_i)=\alpha$ (where $\omega_i$ are the frequencies of the sinusoids/complex exponentials), because the values $F(\omega_i)$ are not defined. This is because $F(\omega)$ is a sum of Dirac delta impulses, which are only defined by their integrals:


with some non-zero weights $w_i$. If your goal is to transform (1) into


with $\alpha$ being independent of the index $i$, then this is possible by filtering the signal $f(t)$ with a linear time-invariant filter whose frequency response satisfies


But note that this does not mean that $G(\omega_i)=\alpha$ because the values $G(\omega_i)$ are not defined.

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