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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.

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  • $\begingroup$ 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
    Commented Jan 22, 2015 at 16:22

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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.

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  • $\begingroup$ 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
    Commented Jan 21, 2015 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.
    Commented Jan 21, 2015 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. $\endgroup$ Commented Jan 22, 2015 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.
    Commented Jan 22, 2015 at 12:37

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