1
$\begingroup$

Given two signals $s_1,s_2$, I'm reasoning on the resolvability of one with respect to the other. Typically, I use two intuitions to proceed:

  • In time domain, consider as resolvable two signals orthogonal with respect to the product $$\left<s_1,s_2\right>\simeq\int dt s_1(t)s_2(t)$$
  • In frequency domain, consider as resolvable two signals whose Fourier transform is ortoghonal with respect to some product of the type $$\left<s_1,s_2\right>\simeq\int df s_1(f)s_2(f)$$

the two being connected by some suitable Fourier transforms. As an example, sine waves are distinguishable in Fourier domain if they have different frequency. Finite length time series are distinguishable (sufficient condition, not necessary) in time domain if they do not share any part of thei time domain.

I am wondering if there is some kind of base independent notion of resolvability. Please excuse me for the lack of previous research on the topic work, but I'm not that expert in this context.

$\endgroup$
2
$\begingroup$

Discretize the time domain and regard the time-sequences as $N\times 1$ matrices; then you have $\mathbf s_1^T~\mathbf s_2=0.$ This product is preserved by any invertible linear transform $\mathbf s\mapsto \mathbf C~\mathbf s$ as long as $\mathbf C^T~\mathbf C = \mathbf I,$ the orthogonal group. Complexify this by converting $\bullet^T\mapsto\bullet^\dagger$and you get a nice result; the Fourier transform looks like, $$s[f] =\mathcal F_{t\to f}~s(t) = \int_{-\infty}^\infty dt~e^{-2\pi ift}~s(t),\\s(t) =\mathcal F^{-1}_{f\to t}~s[f] = \int_{-\infty}^\infty df~e^{2\pi ift}~s[f].$$

Note that under the transpose $f\leftrightarrow t$ you don't actually see this $\mathcal F$ matrix become its inverse (it remains identical; it's a symmetric matrix), but upon conjugation $i \leftrightarrow -i$ it definitely does. So your Fourier matrix is a unitary matrix $\mathbf F^\dagger~\mathbf F = 1.$ (Actually looking over the discrete Fourier transform formula, you may need to have a constant $\alpha$ between them: but it looks like it's just a real number and so you can divide through and normalize appropriately.)

You can therefore see that $(\mathbf F ~ \mathbf s_1)^\dagger~(\mathbf F ~ \mathbf s_2) = \mathbf s_1^\dagger~\mathbf s_2 = 0,$ if and only if the two things are time-orthogonal. The cost is that you need to instead think of frequency-orthogonality as being $$\langle s_1,~s_2\rangle_f = \int df~s_1^*[f]~s_2[f].$$If you're not willing to do that then the best I can give you is this: $\mathbf C^T \mathbf C$ is a symmetric matrix and therefore in any basis the notion of orthogonality in some other basis becomes some symmetric kernel $g(t, t')$ with that inner product coming into the time basis as$$\langle s_1, s_2\rangle_g = \iint dt~ dt'~g(t, t')~s_1(t)~s_2(t').$$Of course multiplication in the frequency domain is going to look like convolution in the time domain, so probably for the Fourier transform, $g$ is some sort of Dirac $\delta$-function.

$\endgroup$

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.