dbell wrote:
> On Apr 9, 4:23 pm, J Elms <jeffrey.e...@gatech.edu> wrote:
>> bronx wrote:
>>>> You should post the equations for the frequencies, length of signals,
>>>> size of FFT, ...
>>>> I suspect this method is going to run into problems where there are a
>>>> lot of FFT bins where there is virtually no signal, such that the
>>>> phase is really based more on numerical noise than any signal
>>>> present.  In that case the phase difference between these bins would
>>>> not be dependent on the delay and would not contribute to your
>>>> result.  I don't know if you have this problem or not, but it would
>>>> somethng to check for.
>>>> Dirk
>>> here is my code in detail
>>> freq1 = 0.5e3;     %Frequency of incoming source in Hz
>> .
>> .
>> .
>>
>>
>>
>>> let me know if you know why the peaks don't show the same time delay
>>> index.
>>> Thanks
>> bronx,
>>
>> I had seen this in a few years ago in a toy radar problem. So I thought
>> I'd stab at this one. Two main points: noise and choice of signal.
>>
>> =Noise=
>> At first I thought it might be due to some numerical issue as Dirk
>> suggested so I added some noise, but to no avail. I left the noise code
>> in but notice the magnitude is set to a very small level compare to the
>> signal. Some noise is necessary to avoid any dividing by too small of
>> numbers. Try it with zero and see the difference. Also experiment with
>> larger noise to see how the accuracy decreases.
>>
>> =Signal selection=
>> I believe the main issue with your example is the signal.  From my
>> understanding you need a somewhat richer signal than a perfect cosine.
>>
>> What's a simple signal with rich frequency content?
>> A pulse, which is what some simple systems use (I've seen it in cheap
>> ultrasound). Better than that would be a chirp signal. If you are
>> looking at audio source location, most audio should be rich enough.
>>
>> =Misc=
>> On a side note, for efficiency it would be ideal if you made the FFT
>> lengths a power of two.  Currently they are an odd length, although it
>> looks you were aiming at that with the definition of N.
>>
>> Cheers,
>> J Elms
>>
>> PS: Maybe I shouldn't just give you the answer, but I'm feeling generous.
>>
>> Modified code:
>>
>> freq1 = 0.5e3;     %Frequency of incoming source in Hz
>> time = 20e-3;   %Total time of simulation in s
>> tau1 = 0.375e-3;     %time difference in arrival of 2 signals
>>
>> N = pow2(11);
>> %t = (0:N-1)*time;
>> t = linspace(0,time,N);
>>
>> % Signals arriving at input of microphones...with different phase
>> NoiseMag  = 1e-5;
>> tPulseLen = 1e-3; % 1ms length pulse
>> nPulseLen = round(tPulseLen/t(2)); % convert to samples
>>
>> BaseSig1 = ( (t>0) & (t<0+tPulseLen) );
>> BaseSig2 = ( (t>tau1) & (t<tau1+tPulseLen) );
>>
>> % MIC A
>> x1 = BaseSig1 + NoiseMag*rand(1,N);
>>
>> % MIC B
>> x2 = BaseSig2 + NoiseMag*rand(1,N);  %delay
>>
>> % figure(1);
>> % plot(t, x1, "r");
>> % hold on;
>> % plot(t, x2, "g");
>>
>> x1pad = [(x1) zeros(1,N)];
>> x2pad = [(x2) zeros(1,N)];
>>
>> NPC = length(x1pad);
>> X1 = fft(x1pad);
>> X2 = fft(x2pad);
>> XR_CC_num = X1.*conj(X2);
>> XR_CC_den = abs(XR_CC_num);
>> XR = (XR_CC_num)./(XR_CC_den);
>> %Phase Correlation
>> PhaseCorr = real(fftshift((ifft(XR))));
>> %Cross Correlation
>> CrossCorr = real(fftshift(ifft(XR_CC_num)));
>> CorrTime = time*linspace(-1,1, NPC);
>>
>> % figure(2);
>> % %Plot Phase correlation in ms
>> % plot(1e3*CorrTime,real(PhaseCorr),'r');
>>
>> % figure(3);
>> % %Plot Cross Correlation in ms
>> % plot(1e3*CorrTime,real(CrossCorr),'g');
>>
>> %the location of the peak of each result is the time delay between signal
>> %%x1 and signal x2
>>
>> [tmp,indPhase] = max(PhaseCorr);
>> [tmp,indCross] = max(CrossCorr);
>> tPhase = CorrTime(indPhase)
>> tCross = CorrTime(indCross)
>>
>> figure;
>>
>> %received and padded signals
>> subplot(3,1,1)
>> plot(x1pad )
>> hold on
>> plot(x2pad )
>> legend('x1pad','x2pad')
>>
>> subplot(3,1,2)
>> plot(abs(X1),':')
>> hold on
>> plot(abs(X2),'--')
>> legend('|FFT(x1pad)|','|FFT(x2pad)|')
>>
>> subplot(3,1,3)
>> plot(CrossCorr/max(CrossCorr) )
>> hold on
>> plot(PhaseCorr/max(PhaseCorr) )
>> legend('Normalized CrossCorr','Normalized PhaseCorr')- Hide quoted text -
>>
>> - Show quoted text -
> 
> Not only is the signal richer in frequency content, but over the
> interval they are defined, your delayed signal is really the original
> signal delayed, which was not true for the OP's signal. That makes the
> FFTs the same except for an approximately linear phase term which is
> important for the method to work. It is not quite linear phase due to
> the essentially random phase near zeros in the spectrum, but most of
> the phase values should be correct.
> 
> Dirk

That's a good point, so I tried it with a real audio clip. For 
simulation purposes, I picked samples in an interlaced fashion so no 
samples were exactly overlapped between the signals (highly correlated, 
but not identical). I got accuracy within .25 ms and within 1 sample 
agreement between the two methods.

I did encounter some anomalies with it detecting a zero delay when 
longer delays were used. But this begins to break the original 
assumption of this algorithm, which is that for the windows selected the 
data is highly correlated.

However, even in these long delay cases there was also a significant 
peak at the correct location.  An algorithm for peak detection could be 
used to improve results if these cases are encountered.

Notes: With a real signal I did not have to add any noise. I also tried 
without padding the signals and with windowing the signal.  These only 
had minor effects on the results.

J Elms

PS: Sorry if I ramble.

On Apr 9, 4:23&#4294967295;pm, J Elms <jeffrey.e...@gatech.edu> wrote:
> bronx wrote:
> >> You should post the equations for the frequencies, length of signals,
> >> size of FFT, ...
>
> >> I suspect this method is going to run into problems where there are a
> >> lot of FFT bins where there is virtually no signal, such that the
> >> phase is really based more on numerical noise than any signal
> >> present. &#4294967295;In that case the phase difference between these bins would
> >> not be dependent on the delay and would not contribute to your
> >> result. &#4294967295;I don't know if you have this problem or not, but it would
> >> somethng to check for.
>
> >> Dirk
>
> > here is my code in detail
>
> > freq1 = 0.5e3; &#4294967295; &#4294967295; %Frequency of incoming source in Hz
>
> .
> .
> .
>
>
>
> > let me know if you know why the peaks don't show the same time delay
> > index.
>
> > Thanks
>
> bronx,
>
> I had seen this in a few years ago in a toy radar problem. So I thought
> I'd stab at this one. Two main points: noise and choice of signal.
>
> =Noise=
> At first I thought it might be due to some numerical issue as Dirk
> suggested so I added some noise, but to no avail. I left the noise code
> in but notice the magnitude is set to a very small level compare to the
> signal. Some noise is necessary to avoid any dividing by too small of
> numbers. Try it with zero and see the difference. Also experiment with
> larger noise to see how the accuracy decreases.
>
> =Signal selection=
> I believe the main issue with your example is the signal. &#4294967295;From my
> understanding you need a somewhat richer signal than a perfect cosine.
>
> What's a simple signal with rich frequency content?
> A pulse, which is what some simple systems use (I've seen it in cheap
> ultrasound). Better than that would be a chirp signal. If you are
> looking at audio source location, most audio should be rich enough.
>
> =Misc=
> On a side note, for efficiency it would be ideal if you made the FFT
> lengths a power of two. &#4294967295;Currently they are an odd length, although it
> looks you were aiming at that with the definition of N.
>
> Cheers,
> J Elms
>
> PS: Maybe I shouldn't just give you the answer, but I'm feeling generous.
>
> Modified code:
>
> freq1 = 0.5e3; &#4294967295; &#4294967295; %Frequency of incoming source in Hz
> time = 20e-3; &#4294967295; %Total time of simulation in s
> tau1 = 0.375e-3; &#4294967295; &#4294967295; %time difference in arrival of 2 signals
>
> N = pow2(11);
> %t = (0:N-1)*time;
> t = linspace(0,time,N);
>
> % Signals arriving at input of microphones...with different phase
> NoiseMag &#4294967295;= 1e-5;
> tPulseLen = 1e-3; % 1ms length pulse
> nPulseLen = round(tPulseLen/t(2)); % convert to samples
>
> BaseSig1 = ( (t>0) & (t<0+tPulseLen) );
> BaseSig2 = ( (t>tau1) & (t<tau1+tPulseLen) );
>
> % MIC A
> x1 = BaseSig1 + NoiseMag*rand(1,N);
>
> % MIC B
> x2 = BaseSig2 + NoiseMag*rand(1,N); &#4294967295;%delay
>
> % figure(1);
> % plot(t, x1, "r");
> % hold on;
> % plot(t, x2, "g");
>
> x1pad = [(x1) zeros(1,N)];
> x2pad = [(x2) zeros(1,N)];
>
> NPC = length(x1pad);
> X1 = fft(x1pad);
> X2 = fft(x2pad);
> XR_CC_num = X1.*conj(X2);
> XR_CC_den = abs(XR_CC_num);
> XR = (XR_CC_num)./(XR_CC_den);
> %Phase Correlation
> PhaseCorr = real(fftshift((ifft(XR))));
> %Cross Correlation
> CrossCorr = real(fftshift(ifft(XR_CC_num)));
> CorrTime = time*linspace(-1,1, NPC);
>
> % figure(2);
> % %Plot Phase correlation in ms
> % plot(1e3*CorrTime,real(PhaseCorr),'r');
>
> % figure(3);
> % %Plot Cross Correlation in ms
> % plot(1e3*CorrTime,real(CrossCorr),'g');
>
> %the location of the peak of each result is the time delay between signal
> %%x1 and signal x2
>
> [tmp,indPhase] = max(PhaseCorr);
> [tmp,indCross] = max(CrossCorr);
> tPhase = CorrTime(indPhase)
> tCross = CorrTime(indCross)
>
> figure;
>
> %received and padded signals
> subplot(3,1,1)
> plot(x1pad )
> hold on
> plot(x2pad )
> legend('x1pad','x2pad')
>
> subplot(3,1,2)
> plot(abs(X1),':')
> hold on
> plot(abs(X2),'--')
> legend('|FFT(x1pad)|','|FFT(x2pad)|')
>
> subplot(3,1,3)
> plot(CrossCorr/max(CrossCorr) )
> hold on
> plot(PhaseCorr/max(PhaseCorr) )
> legend('Normalized CrossCorr','Normalized PhaseCorr')- Hide quoted text -
>
> - Show quoted text -

Not only is the signal richer in frequency content, but over the
interval they are defined, your delayed signal is really the original
signal delayed, which was not true for the OP's signal. That makes the
FFTs the same except for an approximately linear phase term which is
important for the method to work. It is not quite linear phase due to
the essentially random phase near zeros in the spectrum, but most of
the phase values should be correct.

Dirk

bronx wrote:
>> You should post the equations for the frequencies, length of signals,
>> size of FFT, ...
>>
>> I suspect this method is going to run into problems where there are a
>> lot of FFT bins where there is virtually no signal, such that the
>> phase is really based more on numerical noise than any signal
>> present.  In that case the phase difference between these bins would
>> not be dependent on the delay and would not contribute to your
>> result.  I don't know if you have this problem or not, but it would
>> somethng to check for.
>>
>> Dirk
>>
> 
> here is my code in detail
> 
> freq1 = 0.5e3;     %Frequency of incoming source in Hz

.
.
.

> 
> let me know if you know why the peaks don't show the same time delay
> index.
> 
> Thanks

bronx,

I had seen this in a few years ago in a toy radar problem. So I thought 
I'd stab at this one. Two main points: noise and choice of signal.

=Noise=
At first I thought it might be due to some numerical issue as Dirk 
suggested so I added some noise, but to no avail. I left the noise code 
in but notice the magnitude is set to a very small level compare to the 
signal. Some noise is necessary to avoid any dividing by too small of 
numbers. Try it with zero and see the difference. Also experiment with 
larger noise to see how the accuracy decreases.

=Signal selection=
I believe the main issue with your example is the signal.  From my 
understanding you need a somewhat richer signal than a perfect cosine.

What's a simple signal with rich frequency content?
A pulse, which is what some simple systems use (I've seen it in cheap 
ultrasound). Better than that would be a chirp signal. If you are 
looking at audio source location, most audio should be rich enough.

=Misc=
On a side note, for efficiency it would be ideal if you made the FFT 
lengths a power of two.  Currently they are an odd length, although it 
looks you were aiming at that with the definition of N.

Cheers,
J Elms

PS: Maybe I shouldn't just give you the answer, but I'm feeling generous.

Modified code:

freq1 = 0.5e3;     %Frequency of incoming source in Hz
time = 20e-3;   %Total time of simulation in s
tau1 = 0.375e-3;     %time difference in arrival of 2 signals

N = pow2(11);
%t = (0:N-1)*time;
t = linspace(0,time,N);

% Signals arriving at input of microphones...with different phase
NoiseMag  = 1e-5;
tPulseLen = 1e-3; % 1ms length pulse
nPulseLen = round(tPulseLen/t(2)); % convert to samples

BaseSig1 = ( (t>0) & (t<0+tPulseLen) );
BaseSig2 = ( (t>tau1) & (t<tau1+tPulseLen) );

% MIC A
x1 = BaseSig1 + NoiseMag*rand(1,N);

% MIC B
x2 = BaseSig2 + NoiseMag*rand(1,N);  %delay

% figure(1);
% plot(t, x1, "r");
% hold on;
% plot(t, x2, "g");

x1pad = [(x1) zeros(1,N)];
x2pad = [(x2) zeros(1,N)];

NPC = length(x1pad);
X1 = fft(x1pad);
X2 = fft(x2pad);
XR_CC_num = X1.*conj(X2);
XR_CC_den = abs(XR_CC_num);
XR = (XR_CC_num)./(XR_CC_den);
%Phase Correlation
PhaseCorr = real(fftshift((ifft(XR))));
%Cross Correlation
CrossCorr = real(fftshift(ifft(XR_CC_num)));
CorrTime = time*linspace(-1,1, NPC);

% figure(2);
% %Plot Phase correlation in ms
% plot(1e3*CorrTime,real(PhaseCorr),'r');

% figure(3);
% %Plot Cross Correlation in ms
% plot(1e3*CorrTime,real(CrossCorr),'g');

%the location of the peak of each result is the time delay between signal
%%x1 and signal x2

[tmp,indPhase] = max(PhaseCorr);
[tmp,indCross] = max(CrossCorr);
tPhase = CorrTime(indPhase)
tCross = CorrTime(indCross)

figure;

%received and padded signals
subplot(3,1,1)
plot(x1pad )
hold on
plot(x2pad )
legend('x1pad','x2pad')

subplot(3,1,2)
plot(abs(X1),':')
hold on
plot(abs(X2),'--')
legend('|FFT(x1pad)|','|FFT(x2pad)|')

subplot(3,1,3)
plot(CrossCorr/max(CrossCorr) )
hold on
plot(PhaseCorr/max(PhaseCorr) )
legend('Normalized CrossCorr','Normalized PhaseCorr')

On Apr 8, 2:02&#4294967295;pm, maury <maury...@core.com> wrote:
> On Apr 7, 4:31&#4294967295;pm, "bronx" <branko_blagoje...@hotmail.com> wrote:
>
>
>
>
>
> > >You should post the equations for the frequencies, length of signals,
> > >size of FFT, ...
>
> > >I suspect this method is going to run into problems where there are a
> > >lot of FFT bins where there is virtually no signal, such that the
> > >phase is really based more on numerical noise than any signal
> > >present. &#4294967295;In that case the phase difference between these bins would
> > >not be dependent on the delay and would not contribute to your
> > >result. &#4294967295;I don't know if you have this problem or not, but it would
> > >somethng to check for.
>
> > >Dirk
>
> > here is my code in detail
>
> > freq1 = 0.5e3; &#4294967295; &#4294967295; %Frequency of incoming source in Hz
> > time = 20e-3; &#4294967295; %Total time of simulation in s
> > tau1 = 0.375e-3; &#4294967295; &#4294967295; %time difference in arrival of 2 signals
>
> > N = pow2(11);
> > %t = (0:N-1)*time;
> > t = linspace(0,time,N);
>
> > % Signals arriving at input of microphones...with different phase
>
> > % MIC A
> > x1 = sin(2*pi*freq1*t);
>
> > % MIC B
> > x2 = sin(2*pi*freq1*(t + tau1));
>
> > figure(1);
> > plot(t, x1, "r");
> > hold on;
> > plot(t, x2, "g");
>
> > x1pad = [(x1) zeros(1,N-1)];
> > x2pad = [(x2) zeros(1,N-1)];
>
> > NPC = length(x1pad);
> > X1 = fft(x1pad);
> > X2 = fft(x2pad);
> > XR_CC_num = X1.*conj(X2);
> > XR_CC_den = abs(XR_CC_num);
> > XR = (XR_CC_num)./(XR_CC_den);
> > %Phase Correlation
> > PhaseCorr = fftshift((ifft(XR)));
> > %Cross Correlation
> > CrossCorr = fftshift(ifft(XR_CC_num));
> > CorrTime = time*linspace(-1,1, NPC);
>
> > figure(2);
> > %Plot Phase correlation in ms
> > plot(1e3*CorrTime,real(PhaseCorr),'r');
>
> > figure(3);
> > %Plot Cross Correlation in ms
> > plot(1e3*CorrTime,real(CrossCorr),'g');
>
> > %the location of the peak of each result is the time delay between signal
> > %%x1 and signal x2
>
> > let me know if you know why the peaks don't show the same time delay
> > index.
>
> > Thanks
>
> x1 is a delayed version of x2 by tau1. &#4294967295;However, the question I think
> you should be asking yourself is why you think the FFT of x1 would be
> different than the FFT of x2.
>
> Maurice Givens- Hide quoted text -
>
> - Show quoted text -

Do you think they should be the same?

You will notice that the OP never takes the FFT of x1 and x2.

Dirk

On Apr 8, 1:17&#4294967295;pm, maury <maury...@core.com> wrote:
> On Apr 8, 9:09&#4294967295;am, dbell <bellda2...@cox.net> wrote:
>
>
>
>
>
> > On Apr 7, 5:31&#4294967295;pm, "bronx" <branko_blagoje...@hotmail.com> wrote:
>
> > > >You should post the equations for the frequencies, length of signals,
> > > >size of FFT, ...
>
> > > >I suspect this method is going to run into problems where there are a
> > > >lot of FFT bins where there is virtually no signal, such that the
> > > >phase is really based more on numerical noise than any signal
> > > >present. &#4294967295;In that case the phase difference between these bins would
> > > >not be dependent on the delay and would not contribute to your
> > > >result. &#4294967295;I don't know if you have this problem or not, but it would
> > > >somethng to check for.
>
> > > >Dirk
>
> > > here is my code in detail
>
> > > freq1 = 0.5e3; &#4294967295; &#4294967295; %Frequency of incoming source in Hz
> > > time = 20e-3; &#4294967295; %Total time of simulation in s
> > > tau1 = 0.375e-3; &#4294967295; &#4294967295; %time difference in arrival of 2 signals
>
> > > N = pow2(11);
> > > %t = (0:N-1)*time;
> > > t = linspace(0,time,N);
>
> > > % Signals arriving at input of microphones...with different phase
>
> > > % MIC A
> > > x1 = sin(2*pi*freq1*t);
>
> > > % MIC B
> > > x2 = sin(2*pi*freq1*(t + tau1));
>
> > > figure(1);
> > > plot(t, x1, "r");
> > > hold on;
> > > plot(t, x2, "g");
>
> > > x1pad = [(x1) zeros(1,N-1)];
> > > x2pad = [(x2) zeros(1,N-1)];
>
> > > NPC = length(x1pad);
> > > X1 = fft(x1pad);
> > > X2 = fft(x2pad);
> > > XR_CC_num = X1.*conj(X2);
> > > XR_CC_den = abs(XR_CC_num);
> > > XR = (XR_CC_num)./(XR_CC_den);
> > > %Phase Correlation
> > > PhaseCorr = fftshift((ifft(XR)));
> > > %Cross Correlation
> > > CrossCorr = fftshift(ifft(XR_CC_num));
> > > CorrTime = time*linspace(-1,1, NPC);
>
> > > figure(2);
> > > %Plot Phase correlation in ms
> > > plot(1e3*CorrTime,real(PhaseCorr),'r');
>
> > > figure(3);
> > > %Plot Cross Correlation in ms
> > > plot(1e3*CorrTime,real(CrossCorr),'g');
>
> > > %the location of the peak of each result is the time delay between signal
> > > %%x1 and signal x2
>
> > > let me know if you know why the peaks don't show the same time delay
> > > index.
>
> > > Thanks
>
> > Notice that x2pad and x1pad are in no way delayed/advanced versions of
> > the other.
>
> > Take a step back and find the difference in the phases for the two
> > FFTs. &#4294967295;Is the phase what you expect?
>
> > Dirk- Hide quoted text -
>
> > - Show quoted text -
>
> Sure they are! &#4294967295;Plot them and take a look.
>
> Maurice Givens- Hide quoted text -
>
> - Show quoted text -

Look again. x2pad and x1pad contain delayed/advanced versions of the
same waveform WINDOWED. So shifting one does not give the other.

Dirk

On Apr 7, 4:31&#4294967295;pm, "bronx" <branko_blagoje...@hotmail.com> wrote:
> >You should post the equations for the frequencies, length of signals,
> >size of FFT, ...
>
> >I suspect this method is going to run into problems where there are a
> >lot of FFT bins where there is virtually no signal, such that the
> >phase is really based more on numerical noise than any signal
> >present. &#4294967295;In that case the phase difference between these bins would
> >not be dependent on the delay and would not contribute to your
> >result. &#4294967295;I don't know if you have this problem or not, but it would
> >somethng to check for.
>
> >Dirk
>
> here is my code in detail
>
> freq1 = 0.5e3; &#4294967295; &#4294967295; %Frequency of incoming source in Hz
> time = 20e-3; &#4294967295; %Total time of simulation in s
> tau1 = 0.375e-3; &#4294967295; &#4294967295; %time difference in arrival of 2 signals
>
> N = pow2(11);
> %t = (0:N-1)*time;
> t = linspace(0,time,N);
>
> % Signals arriving at input of microphones...with different phase
>
> % MIC A
> x1 = sin(2*pi*freq1*t);
>
> % MIC B
> x2 = sin(2*pi*freq1*(t + tau1));
>
> figure(1);
> plot(t, x1, "r");
> hold on;
> plot(t, x2, "g");
>
> x1pad = [(x1) zeros(1,N-1)];
> x2pad = [(x2) zeros(1,N-1)];
>
> NPC = length(x1pad);
> X1 = fft(x1pad);
> X2 = fft(x2pad);
> XR_CC_num = X1.*conj(X2);
> XR_CC_den = abs(XR_CC_num);
> XR = (XR_CC_num)./(XR_CC_den);
> %Phase Correlation
> PhaseCorr = fftshift((ifft(XR)));
> %Cross Correlation
> CrossCorr = fftshift(ifft(XR_CC_num));
> CorrTime = time*linspace(-1,1, NPC);
>
> figure(2);
> %Plot Phase correlation in ms
> plot(1e3*CorrTime,real(PhaseCorr),'r');
>
> figure(3);
> %Plot Cross Correlation in ms
> plot(1e3*CorrTime,real(CrossCorr),'g');
>
> %the location of the peak of each result is the time delay between signal
> %%x1 and signal x2
>
> let me know if you know why the peaks don't show the same time delay
> index.
>
> Thanks

x1 is a delayed version of x2 by tau1.  However, the question I think
you should be asking yourself is why you think the FFT of x1 would be
different than the FFT of x2.


Maurice Givens

On Apr 8, 9:09&#4294967295;am, dbell <bellda2...@cox.net> wrote:
> On Apr 7, 5:31&#4294967295;pm, "bronx" <branko_blagoje...@hotmail.com> wrote:
>
>
>
>
>
> > >You should post the equations for the frequencies, length of signals,
> > >size of FFT, ...
>
> > >I suspect this method is going to run into problems where there are a
> > >lot of FFT bins where there is virtually no signal, such that the
> > >phase is really based more on numerical noise than any signal
> > >present. &#4294967295;In that case the phase difference between these bins would
> > >not be dependent on the delay and would not contribute to your
> > >result. &#4294967295;I don't know if you have this problem or not, but it would
> > >somethng to check for.
>
> > >Dirk
>
> > here is my code in detail
>
> > freq1 = 0.5e3; &#4294967295; &#4294967295; %Frequency of incoming source in Hz
> > time = 20e-3; &#4294967295; %Total time of simulation in s
> > tau1 = 0.375e-3; &#4294967295; &#4294967295; %time difference in arrival of 2 signals
>
> > N = pow2(11);
> > %t = (0:N-1)*time;
> > t = linspace(0,time,N);
>
> > % Signals arriving at input of microphones...with different phase
>
> > % MIC A
> > x1 = sin(2*pi*freq1*t);
>
> > % MIC B
> > x2 = sin(2*pi*freq1*(t + tau1));
>
> > figure(1);
> > plot(t, x1, "r");
> > hold on;
> > plot(t, x2, "g");
>
> > x1pad = [(x1) zeros(1,N-1)];
> > x2pad = [(x2) zeros(1,N-1)];
>
> > NPC = length(x1pad);
> > X1 = fft(x1pad);
> > X2 = fft(x2pad);
> > XR_CC_num = X1.*conj(X2);
> > XR_CC_den = abs(XR_CC_num);
> > XR = (XR_CC_num)./(XR_CC_den);
> > %Phase Correlation
> > PhaseCorr = fftshift((ifft(XR)));
> > %Cross Correlation
> > CrossCorr = fftshift(ifft(XR_CC_num));
> > CorrTime = time*linspace(-1,1, NPC);
>
> > figure(2);
> > %Plot Phase correlation in ms
> > plot(1e3*CorrTime,real(PhaseCorr),'r');
>
> > figure(3);
> > %Plot Cross Correlation in ms
> > plot(1e3*CorrTime,real(CrossCorr),'g');
>
> > %the location of the peak of each result is the time delay between signal
> > %%x1 and signal x2
>
> > let me know if you know why the peaks don't show the same time delay
> > index.
>
> > Thanks
>
> Notice that x2pad and x1pad are in no way delayed/advanced versions of
> the other.
>
> Take a step back and find the difference in the phases for the two
> FFTs. &#4294967295;Is the phase what you expect?
>
> Dirk- Hide quoted text -
>
> - Show quoted text -

Sure they are!  Plot them and take a look.

Maurice Givens

On Apr 7, 5:31&#4294967295;pm, "bronx" <branko_blagoje...@hotmail.com> wrote:
> >You should post the equations for the frequencies, length of signals,
> >size of FFT, ...
>
> >I suspect this method is going to run into problems where there are a
> >lot of FFT bins where there is virtually no signal, such that the
> >phase is really based more on numerical noise than any signal
> >present. &#4294967295;In that case the phase difference between these bins would
> >not be dependent on the delay and would not contribute to your
> >result. &#4294967295;I don't know if you have this problem or not, but it would
> >somethng to check for.
>
> >Dirk
>
> here is my code in detail
>
> freq1 = 0.5e3; &#4294967295; &#4294967295; %Frequency of incoming source in Hz
> time = 20e-3; &#4294967295; %Total time of simulation in s
> tau1 = 0.375e-3; &#4294967295; &#4294967295; %time difference in arrival of 2 signals
>
> N = pow2(11);
> %t = (0:N-1)*time;
> t = linspace(0,time,N);
>
> % Signals arriving at input of microphones...with different phase
>
> % MIC A
> x1 = sin(2*pi*freq1*t);
>
> % MIC B
> x2 = sin(2*pi*freq1*(t + tau1));
>
> figure(1);
> plot(t, x1, "r");
> hold on;
> plot(t, x2, "g");
>
> x1pad = [(x1) zeros(1,N-1)];
> x2pad = [(x2) zeros(1,N-1)];
>
> NPC = length(x1pad);
> X1 = fft(x1pad);
> X2 = fft(x2pad);
> XR_CC_num = X1.*conj(X2);
> XR_CC_den = abs(XR_CC_num);
> XR = (XR_CC_num)./(XR_CC_den);
> %Phase Correlation
> PhaseCorr = fftshift((ifft(XR)));
> %Cross Correlation
> CrossCorr = fftshift(ifft(XR_CC_num));
> CorrTime = time*linspace(-1,1, NPC);
>
> figure(2);
> %Plot Phase correlation in ms
> plot(1e3*CorrTime,real(PhaseCorr),'r');
>
> figure(3);
> %Plot Cross Correlation in ms
> plot(1e3*CorrTime,real(CrossCorr),'g');
>
> %the location of the peak of each result is the time delay between signal
> %%x1 and signal x2
>
> let me know if you know why the peaks don't show the same time delay
> index.
>
> Thanks

Notice that x2pad and x1pad are in no way delayed/advanced versions of
the other.

Take a step back and find the difference in the phases for the two
FFTs.  Is the phase what you expect?

Dirk