The ratio of single and double sided spectrum is mentioned equal to 2 in books.
But according to me it must be 4 because if you convert DS-PSD (double sided PSD) to SS-PSD you fold the negative frequencies to the positive ones and they must be correlated and thus we must get double the signal and 4 times the power.
Yes it is mentioned so in the books but I thought this must be 4 and not 2 because :
1) The sinusoid of frequency "f" is a sum of phasors at "+f" and "-f "
2) They will be correlated for the case of noise(actually its just f in real world and it's represented as "f" snd "-f" phasors just for mathematical convinience.
3) So "f" and "-f" phasors must interfere constructively and thus 4 times the power must be there in Singel sided PSD as compared to double sided PSD.
The only thing that must be true otherwise may be that in stochastics they develop this spectral mathematics in a different way - and not much like phasors.
then signal x coherently added signal power will be four times;
noise added non coherently since from two sourses;
single double side band is just a representation.
I am talking about the frequqency representation of noise itself. A noise at frequency "f" can be written in terms of phasors at "+f" and "-f" and they will be correlated (because we started with a noise and decomposed it in representation). So, when we revert back from double sided PSD to single sided PSD, then we must consider noise phasors at "+f" and "-f" correlated and we must add them constructively.
As I said the flaw must be in that we might not be having the stochastic spectral theory developed in terms of phasors. I need to confirm this and if anybody knows please confirm it.
As far as I understood it upto now is that the PSD diagram represents the power distribution over frequency. Since the total power of the signal in a frequency from f1 to f2 will be:
P = ∫S df |(f1 to f2 integral) + ∫S df |(-f2 to -f1)
= 2 ∫S df |(f1 to f2 integral) -> because for real signals the PSD function is even. So if we would want to represent as a single side function we must double it...
This'll hold if the signal at "+f" and "-f" are uncorrelated. But since the signals at "-f" and "+f" are just a represntation of the actual signal at "f" they must be correlated and thus we must add the signals to get four times the power (and not the powers to get twice the power).
I think since we represent the PSD of just 1 signal, half of its power is distibuted in the negative half axis of f and half on the positive axis of f. The total power is actually the sum of the 2 powers in the 2 axis. So it has to be double. This is all we are talking about is 1 signal. This is based on that the PSD of a real signal is an even function of f.
When we talk about 2 signals then we have to consider the thing that whether we have to add the magnitude of the signals(correlated signals) or the power of the signals (uncorrelated signals) to get the resultant PSD. Isn't this right?