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Single-sided Double-sided PSD

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advaita

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two-sided power spectral density

Hi

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.

Where am I wrong?
 

two sided power spectral density

I learn in my undergraduate course in EE:

A White Noise source maybe modeled by a Single Side Band PSD with a ratio 2 times Double Side Band, then we have the same Power.

Other example:


A sinusoid is a sum of two complex exponentials.

A complex exponencial in frequency is like a spectral line.


PSD of sinusoid is the sum of PSD of complex exponentials.
 

two sided psd

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.
 

two-sided psd

It is not the case that you add two psd. Only signals could be added.

Single side band is just a way of representing.

yopur argument is valid for below case

y1=x+n1;

y2=x+n2;

y=y1+y2=2x+(n1+n2)



then signal x coherently added signal power will be four times;


noise added non coherently since from two sourses;

hence noise power will be two times.


single double side band is just a representation.
 

double sided power spectral density

janath said:
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.
Click to expand...

Hi Janath

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.

Thanks
 

twosided psd

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...
 

single-sided power spectral density

aryajur said:
P = ∫S df |(f1 to f2 integral) + ∫S df |(-f2 to -f1)
Click to expand...

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).
 

single sided power spectral density

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?
 

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