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#### How do I calculate the phase shift between two sinusoidal signals?

```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!
```
 0
1/27/2011 1:31:47 PM
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```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.
```
 0
Think
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
```
 -1
Giuseppe
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);
```
 3
Think
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.
```
 0
Royi
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.
```
 0
Think
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)
```
 0
Giuseppe
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
```
 0
Giuseppe
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.
```
 0
Think
1/28/2011 9:01:07 PM
```Giuseppe <caesar_539@hotmail.it> wrote in message <711823938.11516.1296251189039.JavaMail.root@gallium.mathforum.org>...
> 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

Good Morning Guiseppe I also tried the above equation and got similiar results when I google for phase difference this solution comes up time and time again.

I think you want to use
acos(dot(a,b)/(norm(a)*norm(b)))

--Marc

I just stated to work on this issue if I find out that I am incorrect I will eat humble pie and repost a better solution
```
 0
12/31/2012 11:13:20 AM
```> 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)))

Is there a way to obtain correct sign?
```
 0
Baha
1/31/2016 10:56:04 PM

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