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Suppose we have 2 orthogonal signals $x_{1}(t)$ and $x_{2}(t)$ and we add them up.

Can we always say that the resulting signal will be of this form:

$$x_{3}(t)=x_{1}(t)+jx_{2}(t)$$ ?

If that is true then we can write down the complex signal in the Fourier domain as $|X_{3}(j\omega)|e^{\phi(j\omega)}$.

What kind of info does $\phi(j\omega)$ give to us?

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  • $\begingroup$ The answer is obviously "no" as $x_1(t) + x_2(t) \neq x_1(t) + jx_2(t)$. Why do you think this would be different ? Ddi you confuse orthogonal and analytic ? $\endgroup$
    – Hilmar
    Feb 8 at 3:00
  • $\begingroup$ If you are getting this idea from one or a few texts, then re-read. If that really is what they are saying then please edit your question with references to them -- we'll either let you know why they're wrong, or how you're misreading them. $\endgroup$
    – TimWescott
    Feb 8 at 6:52

2 Answers 2

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No, clearly not. Because if you "add them up", you get $x_1+x_2$, and not $x_1+jx_2$.

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If you have the signal $x(t) = x_1(t) \cos \omega t + x_2(t) \sin \omega t$ then -- assuming that $x_1$ and $x_2$ are real-valued, and they have bandwidth significantly lower than $\omega$ -- you can represent them as $x(t) = x_3(t) e^{j\omega t}$, where $x_3(t) = x_1(t) + jx_2(t)$.

But the statement $x_1(t) + x_2(t) = x_1(t) + j x_2(t)$ is only true if $x_2(t) = 0$.

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