On 27/01/11 12:31 PM, Giuseppe wrote: > I have a big problem: > I have 2 arrays "a" and "b" of size [1x1909]. How do I calculate the phase shift between this two signals? there is a .m files ready? fft(a), drop the first point (the constant term), angle() the result fft(b), drop the first point, angle() the result phase shift is then abs() of the difference in angles. If you have a pure sinusoidal then then angle should be very close to 0 except at the frequency of the wave. You could probably clip out noise via something like fft_a(abs(fft_a) < threshhold) = 0; and likewise for the fft of b.

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1/27/2011 9:46:57 PM

fft (a) I have an array with the first term (real) = 46,1373794890328 fft (b)--> the first term (real) = 804,447713718616 and now how should I proceed? thanks regards

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1/28/2011 3:27:09 AM

On 28/01/11 2:27 AM, Giuseppe wrote: > fft (a) I have an array with the first term (real) = 46,1373794890328 > > fft (b)--> the first term (real) = 804,447713718616 I do believe that I said to drop the first term. Drop means "throw it away". > > and now how should I proceed? Well, what I wrote before was: >> fft(a), drop the first point (the constant term), angle() the result >> fft(b), drop the first point, angle() the result >> phase shift is then abs() of the difference in angles. >> If you have a pure sinusoidal then then angle should be very close >> to 0 except at the frequency of the wave. You could probably clip >> out noise via something like >> fft_a(abs(fft_a) < threshhold) = 0; >> and likewise for the fft of b. fft_a = fft(a); %fft(a) fft_a = fft_a(2:end); %drop the first point angle_a = angle(fft_a); %angle() the result fft_b = fft(b); %fft(b) fft_b = fft_b(2:end); %drop the first point angle_b = angle(fft_b); %angle() the result phase_shift = abs(angle_a - angle_b); or with noise reduction, fft_a = fft(a); %fft(a) fft_a = fft_a(2:end); %drop the first point fft_a(abs(fft_a) < 1e-10) = 0; %some threshold angle_a = angle(fft_a); %angle() the result fft_b = fft(b); %fft(b) fft_b = fft_b(2:end); %drop the first point fft_b(abs(fft_b) < 1e-10) = 0; %some threshold angle_b = angle(fft_b); %angle() the result phase_shift = abs(angle_a - angle_b);

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1/28/2011 11:56:15 AM

Giuseppe <caesar_539@hotmail.it> wrote in message <1875866377.4904.1296153140587.JavaMail.root@gallium.mathforum.org>... > Hi, > I have a big problem: > I have 2 arrays "a" and "b" of size [1x1909]. How do I calculate the phase shift between this two signals? there is a .m files ready? > thanks for help > Regards! You have 2 vectors of samples of the same sinus signal with a different Phase? The best way to measure the phase would be using inner product.

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1/28/2011 12:06:03 PM

On 28/01/11 6:06 AM, Royi Avital wrote: > Giuseppe <caesar_539@hotmail.it> wrote in message > <1875866377.4904.1296153140587.JavaMail.root@gallium.mathforum.org>... >> I have a big problem: >> I have 2 arrays "a" and "b" of size [1x1909]. How do I calculate the >> phase shift between this two signals? there is a .m files ready? >> thanks for help >> Regards! > > You have 2 vectors of samples of the same sinus signal with a different > Phase? > The best way to measure the phase would be using inner product. Or put in to clumsy code, acos(dot(a,b) ./ (dot(a,a)*dot(b,b))) This would, though, not be able to handle the situation where the arrays contained a mix of frequencies.

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1/28/2011 12:22:06 PM

Thanks.. But I should get a number and not the array phase_shift! I can make the mean? (mean phase_shift)

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1/28/2011 2:36:13 PM

I tried the code acos(dot (a, b). / (dot (a, A) * dot(b,b))) but I do not have the exact result I should have 138° but I have 1.5200, then 180 * 1:52 / pi = 87° .. How Could I do? if you want i can send you the 2 array! thanks

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1/28/2011 4:45:58 PM

On 11-01-28 01:36 PM, Giuseppe wrote: > Thanks.. But I should get a number and not the array phase_shift! I can make the mean? (mean phase_shift) No, do *not* take the mean of the array. The dot product solution proposed by someone else, and which I gave code for in response, returns back the single number you are hoping for. The dot product solution assumes that if there are multiple frequencies present, then the phase shift is the same for all of them. The fft solution I gave is able to isolate down different phase shifts for different frequencies.

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1/28/2011 9:01:07 PM

> acos(dot(a,b) ./ (dot(a,a)*dot(b,b))) > > This would, though, not be able to handle the situation where the arrays > contained a mix of frequencies. I think this code can not approximate if the shift is positive or negative, for example: w = 2; t = 0:0.01:10; Y = -pi/6; f1 = sin(w*t)+cos(w*t); u1 = sin(w*t+Y)+cos(w*t+Y); acos(dot(f1,u1)/(norm(f1)*norm(u1))) = pi/6 instead of -pi/6 Is there a way to obtain correct sign?

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1/31/2016 10:56:04 PM

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Hi to all, I am presently facing problem with finding the phase shift in two Digital signals in MATLAB. I have two signals one is standard sine wave and other is output of my mechanical system which is again sine wave with some phase shift. can any body tell me how can I find exact phase shift in MATLAB. Thank you in advance with regards suhas deshmukh IIT Bombay India suhas.deshmukh@gmail.com wrote: > Hi to all, > I am presently facing problem with finding the phase shift in two > Digital signals in MATLAB. I have two signals one is standard sine wave > and other is output of...

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