PSD interpretation....
in [DSP]
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From: fisico32 on 4 May 2010 13:27 Hello forum, a real-valued power signal x(t) has a Fourier transform X(w) that involves both negative and positive frequencies w. The PSD corresponds to the S(w)=|X(w)^2|= |X(w) X*(w)|=|X(w)X(-w)| because of Hermitian symmetry. I have read that the function S(w) represents the power due to two frequencies whose sum equal to zero: w1+w2=0 where w1=w and w2=-w.... What does that really mean? What power does the integral INT S(w)dw between a frequency w and -w represent? The power due to all frequencies between 0 and w in the real signal? thanks, fisico32
From: Steve Pope on 4 May 2010 15:08 fisico32 <marcoscipioni1(a)n_o_s_p_a_m.gmail.com> wrote: >I have read that the function S(w) represents the power due to two >frequencies whose sum equal to zero: w1+w2=0 where w1=w and w2=-w.... >What does that really mean? Not much. sin(-wt) = -sin(wt) so unless you're analysing the signal as a complex signal there is no difference between positive and negative frequency components. (The question I find more interesting is what does it mean when the PSD evaluates to a negative value at a positivie frequency....but this is unrelated.) Steve
From: HardySpicer on 4 May 2010 15:23 On May 5, 5:27 am, "fisico32" <marcoscipioni1(a)n_o_s_p_a_m.gmail.com> wrote: > Hello forum, > > a real-valued power signal x(t) has a Fourier transform X(w) that involves > both negative and positive frequencies w. > > The PSD corresponds to the S(w)=|X(w)^2|= |X(w) X*(w)|=|X(w)X(-w)| because > of Hermitian symmetry. > > I have read that the function S(w) represents the power due to two > frequencies whose sum equal to zero: w1+w2=0 where w1=w and w2=-w..... > What does that really mean? > > What power does the integral INT S(w)dw between a frequency w and -w > represent? The power due to all frequencies between 0 and w in the real > signal? > > thanks, > fisico32 The integral is the variance (assuming zero dc) or total average power in the signal.
From: Alfred Bovin on 5 May 2010 03:33 "Steve Pope" <spope33(a)speedymail.org> wrote in message news:hrprb1$j81$5(a)blue.rahul.net... > (The question I find more interesting is what does it mean > when the PSD evaluates to a negative value at a positivie > frequency....but this is unrelated.) How can it do that?
From: Randy Yates on 5 May 2010 07:46
"fisico32" <marcoscipioni1(a)n_o_s_p_a_m.gmail.com> writes: > Hello forum, > > a real-valued power signal x(t) has a Fourier transform X(w) that involves > both negative and positive frequencies w. > > The PSD corresponds to the S(w)=|X(w)^2| If by X(w) you mean the FT of x(t), then this is not formally correct. The PSD is defined to be S(w) = FT(R(tau)), where R(tau) is the autocorrelation function of x(t). However, we commonly _estimate_ S(w) by |X(2)^2|. > = |X(w) X*(w)|=|X(w)X(-w)| because > of Hermitian symmetry. > > I have read that the function S(w) represents the power due to two > frequencies whose sum equal to zero: w1+w2=0 where w1=w and w2=-w.... > What does that really mean? I think it depends on how the author defines the power spectrum (one-sided or two-sided); the power in some positive bandwidth in a one-sided power spectrum (valid for a real input signal) includes the positive and negative frequencies (and don't get us started...). > What power does the integral INT S(w)dw between a frequency w and -w > represent? The power due to all frequencies between 0 and w in the real > signal? The way I view the world is that ALL signals (real or otherwise) have a two-sided PSD, in which case you would be correct. -- Randy Yates % "My Shangri-la has gone away, fading like Digital Signal Labs % the Beatles on 'Hey Jude'" mailto://yates(a)ieee.org % http://www.digitalsignallabs.com % 'Shangri-La', *A New World Record*, ELO |