Timeline for Practical cross-spectrum estimation using Blackman-Tukey approach
Current License: CC BY-SA 3.0
10 events
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| Oct 8, 2016 at 9:36 | comment | added | msm | @vvv I am glad it helped. Just updated the answer as you suggested. | |
| Oct 8, 2016 at 9:35 | history | edited | msm | CC BY-SA 3.0 |
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| Oct 8, 2016 at 9:21 | vote | accept | vvv | ||
| Oct 8, 2016 at 9:21 | comment | added | vvv | Okay got it now, many thanks for your great help! Just a thought, it might be helpful to add your previous comment to your original answer as well, at least for me it did the trick (was confused about how to shift vectors). | |
| Oct 8, 2016 at 3:03 | comment | added | msm |
@vvv The answer I provided will require you to use linear shift, not circular shift. If you wan to use circular shift, then do it like the following: x=w.*rhat; phi=fft(circshift(x',M+1)'); where M is even and w and rhat are just the symmetric vectors around M+1 (for instance w=[2 4 6 4 2] is symmetric around M=3).
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| Oct 7, 2016 at 18:18 | comment | added | vvv | Thanks again. I'm almost there; I circularly shifted both vectors and ran the FFT. What I am wondering though is the scaling term. Consider, for example, that we are dealing with an autocorrelation function, i.e. $\hat{r}_{yy}$. My understanding is that in this case the periodogram values should be real-valued. This indeed is the case after running the FFT (up to very small imaginary terms due to numerical imprecisions, I suppose). However, if I scale these results with $e^{i \omega (M+1)}$ the results are obviously complex valued figures. Any idea what's up with this? | |
| Oct 7, 2016 at 10:47 | comment | added | msm | You're welcome @vvv. We need $w'(k)=w(k-M-1)$ and $\hat{r}'_{yu}(k)=\hat{r}_{yu}(k-M-1)$. It means, both $w(k)$ and $\hat{r}_{yu}(k)$ should be shifted from $[-M, M]$ to $[1, 2M+1]$ such that the start of both are from $1$ rather than $-M$. When you did that, the inside of sum is consistent with the definition of FFT and you can calculate it. At the end, scale it with $e^{i\omega(M+1)}$ | |
| Oct 7, 2016 at 10:37 | comment | added | vvv | Thank's alot for your answer! Just to be clear, with "shifted vectors" do you mean a circular shift of the original vector so that the first M values are taken out and shifted to the end, so that the shifted vector starts with the value corresponding to M+1th value in the original vector? Or maybe something else? Now, practically speaking, if I were to calculate the part inside the sum notation with fft should I scale the result with a complex exponent $e^{i \omega (M+1)}$? | |
| Oct 6, 2016 at 22:55 | history | edited | msm | CC BY-SA 3.0 |
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| Oct 6, 2016 at 22:44 | history | answered | msm | CC BY-SA 3.0 |