# Orthogonal signals basics [closed]

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?

• 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 ? Feb 8 at 3:00
• 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. Feb 8 at 6:52

No, clearly not. Because if you "add them up", you get $$x_1+x_2$$, and not $$x_1+jx_2$$.
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 $$x_2(t) = 0$$.