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From: Greg Heath on 10 Apr 2010 21:19
On Apr 9, 7:05 am, Rune Allnor <all...(a)tele.ntnu.no> wrote:
> On 9 apr, 12:24, "Michal Kotze" <mkotz...(a)yahoo.de> wrote:
>
> > I know this is to do with using the max function and the floating point representation of values.
>
> No, it isn't.
>
> The problem is that the cross correlation produces a discrete
> series, and a naive search for the maximum amplitude will be
> constrained to the discrete delays. If the actual delay is
> not an integer number of samples, there will be poblems.
>
> You might want to refine your delay estimates by investigating
> the phase of the cross spectrum.

Rune,

I have not been able to get anywhere with this. If Rxy = fft(rxy),
is it a simple matter of dividing angle(Rxy) at the peak of
abs(Rxy) by 2*pi*f?

Greg

From: Mark Shore on 11 Apr 2010 08:40
A more appropriate signal to examine cross-correlation the way I'm guessing Michal wants would be something along the lines of :

k=400; m=10; n=20; p=50;
% m sine wave cycles with n samples/cycle padded with k leading and trailing zeros
CH1=[zeros(1,k) repmat(sin(2*pi/n*(1:n)),1,m) zeros(1,k)];
% shift CH1 by p sample intervals
CH2=[CH1(p:end) CH1(1:p-1)]; % no 1D equivalent of circshift in MATLAB(?)
plot(CH1,'b'); hold on; plot(CH2,'r'); hold off % optional plot

He should also consider the limiting case of a vector with a single non-zero value, say [1 zeros(1,999)] and its cross-correlation with a shifted equivalent.
From: Michal Kotze on 12 Apr 2010 04:04
Thanks Greg and Mark for your replies and for looking into my problem. Greg and Mark I will answer you questions soon but i am a bit busy now at university. Check the post round about Wednesday, 14 April 2010. Regards Michal Kotze
From: Greg Heath on 12 Apr 2010 04:45
On Apr 10, 9:19 pm, Greg Heath <he...(a)alumni.brown.edu> wrote:
> On Apr 9, 7:05 am, Rune Allnor <all...(a)tele.ntnu.no> wrote:
>
> > On 9 apr, 12:24, "Michal Kotze" <mkotz...(a)yahoo.de> wrote:
>
> > > I know this is to do with using the max function and the floating point representation of values.
>
> > No, it isn't.
>
> > The problem is that thecrosscorrelationproduces a discrete
> > series, and a naive search for the maximum amplitude will be
> > constrained to the discrete delays. If the actual delay is
> > not an integer number of samples, there will be poblems.
>
> > You might want to refine your delay estimates by investigating
> > the phase of thecrossspectrum.
>
> Rune,
>
> I have not been able to get anywhere with this. If Rxy = fft(rxy),
> is it a simple matter of dividing angle(Rxy) at the peak of
> abs(Rxy) by 2*pi*f?

OK. I see it now: Ideally

g = ifft(exp(i*angle(Rxy))

should yield a pulse whos position represents the time delay.

M = length(Rxy)
t = dt*(0:2*M-1);
[ gmax jmax0] = max(g) % [ 1 1500]
delay = max(t)-t(jmax0) % 0.5 sec

Greg
From: Greg Heath on 12 Apr 2010 04:49
On Apr 12, 4:45 am, Greg Heath <he...(a)alumni.brown.edu> wrote:
> On Apr 10, 9:19 pm, Greg Heath <he...(a)alumni.brown.edu> wrote:
>
>
>
>
>
> > On Apr 9, 7:05 am, Rune Allnor <all...(a)tele.ntnu.no> wrote:
>
> > > On 9 apr, 12:24, "Michal Kotze" <mkotz...(a)yahoo.de> wrote:
>
> > > > I know this is to do with using the max function and the floating point representation of values.
>
> > > No, it isn't.
>
> > > The problem is that thecrosscorrelationproduces a discrete
> > > series, and a naive search for the maximum amplitude will be
> > > constrained to the discrete delays. If the actual delay is
> > > not an integer number of samples, there will be poblems.
>
> > > You might want to refine your delay estimates by investigating
> > > the phase of thecrossspectrum.
>
> > Rune,
>
> > I have not been able to get anywhere with this. If Rxy = fft(rxy),
> > is it a simple matter of dividing angle(Rxy) at the peak of
> > abs(Rxy) by 2*pi*f?
>
> OK. I see it now: Ideally
>
> g = ifft(exp(i*angle(Rxy))
>
> should yield a pulse whos position represents the time delay.
>
> M = length(Rxy)
> t = dt*(0:2*M-1);
> [ gmax jmax0] = max(g)               % [ 1 1500]
> delay = max(t)-t(jmax0)              % 0.5 sec

For the test case

f0 = 1
x0 = sin(2*pi*f0*t0) ;
x1 = [x0 zeros(1,N)];
x2 = [zeros(1,N/2) x0 zeros(1,N/2)];
t = dt*(0:2*N-1);

Greg

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