1
$\begingroup$

Suppose I have two signals $f,g$ which are orthogonal in the sense of the usual inner product on $\mathbb R^d$. Then if I'm not mistaken these signals have disjoint spectral support, i.e their fourier transforms have disjoint support. Suppose further that I know the spectral supports aswell. How can I make use of this information in order to sample only one of the signals, i.e, sample so I don't "see" the other at all?

The trivial solution seems involve filtering out the irrelevant frequencies using an ideal filter, but that's not an option; I am restricted to "physical" solutions.

It's fine to assume $f,g$ are well-behaved.

|improve this question
$\endgroup$
  • $\begingroup$ If you want disjoint frequency support then the cross correlation between the two signals has to vanish, i.e the inner product for every possible relative delay between the signals is 0. That's a very rare condition. If you are thinking of this, the answer to your question will be different. $\endgroup$ – Jazzmaniac Jan 22 '15 at 16:22
2
$\begingroup$

I'm afraid that your premise is wrong. The Fourier transforms of two orthogonal signals do not in general have disjoint support. You know that two signals $u(t)$ and $v(t)$ are orthogonal if

$$\int_{-\infty}^{\infty}u(t)v^*(t)dt=\int_{-\infty}^{\infty}U(f)V^*(f)df=0\tag{1}$$

where $U(f)$ and $V(f)$ are the Fourier transforms of $u(t)$ and $v(t)$, respectively, and $*$ denotes complex conjugation. One possibility for (1) to be true is indeed that the product $U(f)V^*(f)=0$, but this is of course a sufficient condition, not a necessary one.

|improve this answer
$\endgroup$
  • $\begingroup$ Thanks, it was silly of me to use intuition and not formally verify it. What can be done if indeed the spectral supports are disjoint? $\endgroup$ – Exterior Jan 21 '15 at 10:01
  • $\begingroup$ @Exterior: I'd say that if the spectra are indeed non-overlapping, filtering is still the most practical solution. $\endgroup$ – Matt L. Jan 21 '15 at 10:09
  • $\begingroup$ Please change your notation a little bit. In your $(1)$, the $f$ on the left side of the equation is not the same as the $f$ on the right side of the equation. $\endgroup$ – Dilip Sarwate Jan 22 '15 at 12:33
  • $\begingroup$ @DilipSarwate: Good catch! I usually use $\omega$, so as soon as I use $f$ things start going wrong ... $\endgroup$ – Matt L. Jan 22 '15 at 12:37

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.