On May 28, 7:15�pm, Jerry Avins <j...@ieee.org> wrote:> koyel.a...@gmail.com wrote: > > May be the details are missing here but though my source is a > > broadband, I am sampling it and digitizing the signal and then > > processing it for each and every spectral channel. Since the source > > for the two channels are same, there should be same frequency > > components only delayed by different amounts. My calculation yields > > one number for one spectral channel so different frequency components > > do not come into picture at a time. > > You don't account for different frequency components having different > delays. You correlate the whole signal, not the components separately.convollution(/correlation) in time domain is multilication in frequency domain. x (t-td1)conv y(t-td2) --FT-->X(W)Y(W)exp(-jw(td1+td2)).....W is one frequency at a time. We will have N number of such operations for N point FFT. So we multiply components in frequency domain which is equivalent to correlating the whole signal in the time domain as you said.
phase difference between two channels
Started by ●May 28, 2008
Reply by ●May 28, 20082008-05-28
Reply by ●May 28, 20082008-05-28
koyel.aphy@gmail.com wrote:> On May 28, 7:15 pm, Jerry Avins <j...@ieee.org> wrote: >> koyel.a...@gmail.com wrote: >>> May be the details are missing here but though my source is a >>> broadband, I am sampling it and digitizing the signal and then >>> processing it for each and every spectral channel. Since the source >>> for the two channels are same, there should be same frequency >>> components only delayed by different amounts. My calculation yields >>> one number for one spectral channel so different frequency components >>> do not come into picture at a time. >> You don't account for different frequency components having different >> delays. You correlate the whole signal, not the components separately. > > convollution(/correlation) in time domain is multilication in > frequency domain. > x (t-td1)conv y(t-td2) --FT-->X(W)Y(W)exp(-jw(td1+td2)).....W is one > frequency at a time. We will have N number of such operations for N > point FFT. So we multiply components in frequency domain which is > equivalent to correlating the whole signal in the time domain as you > said.Exactly. And since you correlate all components simultaneously, you get confusion if those components have different delays. At the very beginning of this thread, you wrote that you wanted to determine phase. You understand that different frequencies undergo different phase shifts for the same delay. You would likely have written more precisely had you been thinking more clearly. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●May 28, 20082008-05-28
>I want to find the phase difference between two channels(x and y) due >to instrumental effects. For that I am passing a signal from a noise >generator through the two channels. Then I switch off the noise >generator and note the outputs from the two channels. Subtraction of >the noise-off signal from the noise-on signal should remove any other >spurious signals so I get only the signal from the noise generator at >the output of the two channels. cross correlating the two channels now >would mean (xon - xoff)* conj(yon-yoff) in the frequency domain which >will yield the phase difference. However since it is a real time >problem, I have xon and yon at a time and not the off states so I can >perform a correlation like xon*conj(yon) and xoff*conj(yoff) and >subtract them. In the previous case I get two additional >multiplication terms i,e xon*conj(yoff) and xoff*conj(yoff). Since >these are uncorrelated, I should get same result but my simulation >shows different results when I plot the phases for the two cases. What >is wrong in the approach? How do I get better results? >From what I gather, you're performing an FFT on data from two channels and then performing Ch1 * conj(Ch2) for each FFT bin. The result of the multiplication should give a phasor from which you can estimate the phase difference between the two channels. Assuming your time delay is constant across frequency, the phase difference should be the same for all bins excluding any noise. Averaging the phasor values across all the bins will help if noise is causing problems. Also, and maybe a dumb question, but are you looking out for phase wrapping problems?
Reply by ●May 28, 20082008-05-28
"Impoliticus" <swiston@uiuc.edu> wrote in news:zuydnYBMhue3QqDVnZ2dnUVZ_q_inZ2d@giganews.com:> Assuming your time delay is constant > across frequency, the phase difference should be the same for all bins > excluding any noiseA constant time delay is a phase that is linear with frequency, not constant. -- Scott Reverse name to reply
Reply by ●May 28, 20082008-05-28
>"Impoliticus" <swiston@uiuc.edu> wrote in >news:zuydnYBMhue3QqDVnZ2dnUVZ_q_inZ2d@giganews.com: > >> Assuming your time delay is constant >> across frequency, the phase difference should be the same for all bins >> excluding any noise > >A constant time delay is a phase that is linear with frequency, not >constant. > > > > >-- >Scott >Reverse name to reply >Yep, the phase is linear, but the phase difference between the two channels should be a constant, no?
Reply by ●May 28, 20082008-05-28
"Impoliticus" <swiston@uiuc.edu> wrote in news:Np- dnWTloMX5fKDVnZ2dnUVZ_tLinZ2d@giganews.com:>>"Impoliticus" <swiston@uiuc.edu> wrote in >>news:zuydnYBMhue3QqDVnZ2dnUVZ_q_inZ2d@giganews.com: >> >>> Assuming your time delay is constant >>> across frequency, the phase difference should be the same for all bins >>> excluding any noise >> >>A constant time delay is a phase that is linear with frequency, not >>constant. >> >> >> >> >>-- >>Scott >>Reverse name to reply >> > > Yep, the phase is linear, but the phase difference between the two > channels should be a constant, no?I'm not sure what "phase difference between two channels" means. Phase difference is a function of frequency, and the phase difference will vary linearly as a function of frequency. -- Scott Reverse name to reply
Reply by ●May 28, 20082008-05-28
>"Impoliticus" <swiston@uiuc.edu> wrote in news:Np- >dnWTloMX5fKDVnZ2dnUVZ_tLinZ2d@giganews.com: > >>>"Impoliticus" <swiston@uiuc.edu> wrote in >>>news:zuydnYBMhue3QqDVnZ2dnUVZ_q_inZ2d@giganews.com: >>> >>>> Assuming your time delay is constant >>>> across frequency, the phase difference should be the same for allbins>>>> excluding any noise >>> >>>A constant time delay is a phase that is linear with frequency, not >>>constant. >>> >>> >>> >>> >>>-- >>>Scott >>>Reverse name to reply >>> >> >> Yep, the phase is linear, but the phase difference between the two >> channels should be a constant, no? > >I'm not sure what "phase difference between two channels" means. Phase >difference is a function of frequency, and the phase difference will vary>linearly as a function of frequency. > >-- >Scott >Reverse name to reply >Well if one channel has a longer time delay than the other channel (and that is all that is different between the two channels) and you send the same sinusoid through the two channels, the sinusoid at the output of one channel will be phase shifted from the sinusoid at the output of the other channel. I'm referring to that as a phase difference. If the time delay changes the phase difference will certainly change. If the time delay is constant, the phase difference between channels should be constant.
Reply by ●May 28, 20082008-05-28
"Impoliticus" <swiston@uiuc.edu> wrote in news:R5udnfgaj8-ue6DVnZ2dnUVZ_qLinZ2d@giganews.com:> and you send the > same sinusoid through the two channels, the sinusoid at the output of > one channel will be phase shifted from the sinusoid at the output of > the other channel. I'm referring to that as a phase difference.That's the phase difference at one frequency. You didn't specify that before. -- Scott Reverse name to reply
Reply by ●May 28, 20082008-05-28
>"Impoliticus" <swiston@uiuc.edu> wrote in >news:R5udnfgaj8-ue6DVnZ2dnUVZ_qLinZ2d@giganews.com: > >> and you send the >> same sinusoid through the two channels, the sinusoid at the output of >> one channel will be phase shifted from the sinusoid at the output of >> the other channel. I'm referring to that as a phase difference. > >That's the phase difference at one frequency. You didn't specify that >before. > >-- >Scott >Reverse name to reply >Actually you're right. I was having a brain fart and was wrong. Sorry folks. Where's the delete key :)
Reply by ●May 28, 20082008-05-28
"Impoliticus" <swiston@uiuc.edu> wrote in news:l_idnfoILJJPdqDVnZ2dnUVZ_vGdnZ2d@giganews.com:>>"Impoliticus" <swiston@uiuc.edu> wrote in >>news:R5udnfgaj8-ue6DVnZ2dnUVZ_qLinZ2d@giganews.com: >> >>> and you send the >>> same sinusoid through the two channels, the sinusoid at the output of >>> one channel will be phase shifted from the sinusoid at the output of >>> the other channel. I'm referring to that as a phase difference. >> >>That's the phase difference at one frequency. You didn't specify that >>before. >> >>-- >>Scott >>Reverse name to reply >> > > Actually you're right. I was having a brain fart and was wrong. Sorry > folks. Where's the delete key :) >No problem. Do it often, myself. In my own work, I've often wondered if the fairest comparison I can do is to subract out the steepest phase line I can to account for simple delay, and then work with the remaining lags from there. -- Scott Reverse name to reply
