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I have two signals $x(t)$ and $y(t)$ which I can sample at arbitrary $\Delta t$ and $N$. I am interested to the signals product time average $\langle x(t)y(t)\rangle_t$. In particular I want to understand when the time average is zero in term of the discrete Fourier coefficients, that is when the signals can be considered "orthogonal". For example, if $X(k)$ and $Y(k)$ are the discrete Fourier coefficients, thus the time average is well approximated by: $$ \langle x(t)y(t)\rangle_t\approx \frac{1}{N^2} \sum_{k=0}^{N-1}X(k)Y^*(k) $$ Thus vanishing cross spectrum represents a sufficient condition in order to have $\langle x(t)y(t)\rangle_t= 0$. My question is:

  • when this also a necessary condition?

Thank you very much

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  • $\begingroup$ Is N number of samples? $\endgroup$ – MimSaad Apr 29 '17 at 18:00
  • $\begingroup$ The answer to your question is probably easy to see using Parseval's Theorem (en.wikipedia.org/wiki/Parseval%27s_theorem), but it is hard to tell with your notation and lack of definitions. For example, are you considering continuous-time functions $x(t)$ and $y(t)$ or are they discrete-time? If they are continuous-time, they must be periodic since you are using discrete spectra $X(k)$ and $Y(k)$, right? Or is everything discrete and these are the DFT coefficients of two discrete sequences? All in all, I'd request better definitions so we can answer your questions more clearly. $\endgroup$ – hops May 1 '17 at 19:38
  • $\begingroup$ @hops Parseval's theorem states, among the others, that if two signals have vanishing cross spectrum (same as overlapping frequencies) thus $\langle x(t)y(t) \rangle_t = 0$. My question was: is the inverse true? That is: is it always true that two signals that have $\langle x(t)y(t) \rangle_t = 0$ thus have vanishing cross spectrum? In order to answer (negative) to the question it would be sufficient to find two signals such that $\langle x(t)y(t) \rangle_t = 0$ and have non-vanishing cross spectrum. $\endgroup$ – Fritz May 3 '17 at 19:23
  • $\begingroup$ That would not contradict the Parseval's theorem. Infact the sum (see Wiki ref.) $$ \sum _{n=-\infty }^{\infty }a_{n}{\overline {b_{n}}}$$ can vanish for two reasons: 1) there is no $n$ (Fourier component) such that $a_n$ and $b_n$ are both different from zero (spectra do not overlap). 2) there are $n$'s such that $a_n$ and $b_n$ are both different from zero, but when the sum is performed, this vanishes (that's the point I would like to understand) $\endgroup$ – Fritz May 3 '17 at 19:24
  • $\begingroup$ @Fritz, Parseval's theorem is an equality, so it works both ways. Can you provide a solid mathematical definition for $<x(t)y(t)>_t$ and add it to your question? This will help me understand your intent better. Also, are you assuming $x(t)$ and $y(t)$ are periodic signals? $\endgroup$ – hops May 3 '17 at 19:43
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I will infer from the "discrete Fourier coefficients" in your question that we are considering periodic signals, otherwise we should be considering a continuous spectrum (the Continuous-Time Fourier Transform). I will also be using the definition $$\left< x(t), y(t) \right>_t = \frac{1}{T}\int_{\alpha}^{\alpha +T} x^*(t) y(t) dt$$ where the superscript $*$ denotes complex conjugation, $T$ is the shortest period that both $x(t)$ and $y(t)$ share, and $\alpha$ is an arbitrary real number. If these definitions are not correct, please alter your question and provide the correct ones.

We know that we can decompose $x(t)$ and $y(t)$ using a Fourier series expansion $$ x(t) = \sum_{n=-\infty}^{\infty} c_{x,n} \exp\left(\frac{2 \pi n t}{T}\right)$$ and $$ y(t) = \sum_{n=-\infty}^{\infty} c_{y,n} \exp\left(\frac{2 \pi n t}{T}\right).$$

We know from Parseval's Theorem that $$ \sum_{n=-\infty}^{\infty} c^*_{x, n} c_{y, n} = \int_{-\frac{T}{2}}^{\frac{T}{2}} x^*(t) y(t) dt.$$ Notice the equality. Also notice that the right hand side is merely the product of $T$ and $\left< x(t), y(t) \right>_t$. This means that whenever the left hand side is zero, so is $\left< x(t), y(t) \right>_t$ and whenever $\left< x(t), y(t) \right>_t$ is zero so is the left hand side. I believe this resolves your question. Let me know if you have objections, and I will try to address them.

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  • $\begingroup$ Given that the right end side of the equation is zero, is it possible to distinguish if the left end side of the equation is zero because 1) $c^*_{x,n} c_{y,n}$ are identically zero for each $n$ (signals do not overlap in frequency) or because 2) $c^*_{x,n} c_{y,n}$ are different from $0$ for some $n$ and only the sum (possibly being sum of negative and positive addends) is zero? $\endgroup$ – Fritz May 4 '17 at 12:38
  • $\begingroup$ The only way I can see to make that distinction right now is by calculating the coefficients $c_{x,n}$ and $c_{y,n}$. There may be an indirect way, but knowing that the time average vanishes simply tells you that the cross spectrum vanishes, nothing more. $\endgroup$ – hops May 4 '17 at 13:34
  • $\begingroup$ @Fritz I do think this answers your question more or less as posed though, since it shows that the vanishing cross spectrum is both necessary and sufficient for a vanishing time average. $\endgroup$ – hops May 4 '17 at 13:37

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