Today's Deals (in Amazon.com)

 Question:

How do I get the phase angle from Cross Correlation?
  • Learner12

    Im trying to cross correlate two signals in matlab and get the phase difference between the signals.

    For cross correlation (the idea is to do it without xcorr) I used:

    Cxx=fftshift(ifft(fft(x,N).*conj(fft(y,N))))/(norm(x) * norm(y));
    

    I get the result and not sure about the reference point I have to take for phase calculation Now which point should I take as zero on time scale and how do I get the phase difference from this result.

    Thanks.

  •  Answers:

  • Dilip Sarwate

    There is no such thing as the phase angle between two signals unless they both consist of a single sinusoid at the same frequency, that is, $x(t) = A\cos(\omega t+\psi)$ and $y(t) = B\cos(\omega t + \phi)$. If you have $N$ samples of these signals $x(t)$ and $y(t)$, taken at times $0$, $T$, $2T, \ldots$, $(N-1)T$, so that $$x[n] = x(nT), ~~ y[n] = y(nT), 0 \leq n < N,$$ and either $N\omega T$ is an integer multiple of $2\pi$ or $N\omega T \gg 1$ then the phase angle between the two sinusoids is $$\theta = \arccos\left(\frac{\displaystyle \sum_{n=0}^{N-1} x[n]y[n]}{\sqrt{\displaystyle \sum_{n=0}^{N-1}(x[n])^2\displaystyle \sum_{n=0}^{N-1}(y[n])^2}}\right). \tag{1}$$ For complex-valued sinusoids $x(t) = Ae^{j(\omega t+\psi)}$ and $y(t) = BAe^{j(\omega t+\phi)}$, $(1)$ should be replaced by $$\theta = \arccos\left(\frac{\displaystyle \sum_{n=0}^{N-1} x[n](y[n])^*}{\sqrt{\displaystyle \sum_{n=0}^{N-1}|x[n]|^2\displaystyle \sum_{n=0}^{N-1}|y[n]|^2}}\right). \tag{2}$$

    Of course, these formula can be used for arbitrary signals, not just for pure sinusoids, but then, what you get is the _angle between the two vectors $\mathbf x = (x[0], x[1], \ldots, x[N-1])$ and $\mathbf y =(y[0], y[1], \ldots, y[N-1])$ in the $N$-dimensional spaces $\mathbb R^N$ or $\mathbb C^N$, and not a phase angle between two sinusoids at the same frequency. Note that the three points $\mathbf 0 = (0,0,\ldots,0)$, $\mathbf x$ and $\mathbf y$ lie in a (two-dimensional) plane in $N$-dimensional space and the $\theta$ that you get is the angle between the line segments with endpoints $\mathbf 0$ and $\mathbf x$ and endpoints $\mathbf 0$ and $\mathbf y$ which lie in this plane. Another way to think about this is that $$\langle\mathbf x, \mathbf y\rangle = ||\mathbf x||\cdot||\mathbf y||\cdot\cos(\theta)$$ and thus $(1)$ is obtained from $$\cos(\theta) = \frac{\langle\mathbf x, \mathbf y\rangle}{||\mathbf x||\cdot||\mathbf y||} = \frac{\displaystyle \sum_{n=0}^{N-1} x[n](y[n])^*}{\sqrt{\displaystyle \sum_{n=0}^{N-1}|x[n]|^2}\sqrt{\displaystyle \sum_{n=0}^{N-1}|y[n]|^2}}.$$

    If you must use FFTs because that's the way you have been told to do it, then you have $$\theta = \arccos\left(\frac{\displaystyle \sum_{n=0}^{N-1} X[n](Y[n])^*}{\sqrt{\displaystyle \sum_{n=0}^{N-1}|X[n]|^2\displaystyle \sum_{n=0}^{N-1}|Y[n]|^2}}\right)\tag{3}$$ so that you have the ineffable pleasure of not only needing to compute two FFTs first, but also of using complex multiplications in $(3)$ instead of the real multiplications in $(1)$ (for real-valued signals). This is overkill in my estimation, but as usual, YMMV, and what your boss insists on is always right, regardless of what people write on Internet forums.

  • hotpaw2

    If you can use the FFTs of x and y to get some sort of periodicity estimates from these two signals, and they are similar (or you have the periodicity a-priori), then one phase angle difference measure might be 2pi times the ratio between the cross-correlation lag and your periodicity estimate. Note that this works even if the signals are not sine waves or even have a missing fundamental component (by using the FFTs with a pitch estimation method such a cepstral or HPS).