Greetings, I am currently writing my diploma-thesis on new spectroscopic methods, and one of them (2D-Correlation-Spectroscopy) involves cross-correlation of fourier-transforms. I have encountered a strange issue, for which even my professor could not supply me with a satisfying answer: When the cross-correlation of two Fourier-transformed signals is to be calculated, why do I have to take the complex conjugate of the second signal? This would mean that the cross-correlation of one fourier-transformed signal with the complex conjugated fourier-transform of the second signal is performed, but I always thought this would mean mirroring the imaginary part of the signal - which would be senseless when the goal is to compare the two signals. It would be really great if somebody could enlighten me on that issue. Thanks in advance, Xuttuh

# Question about Cross-Correlation in Frequency-domain

Started by ●August 8, 2007

Reply by ●August 8, 20072007-08-08

On Aug 8, 1:33 pm, "Xuttuh" <peter_rehb...@gmx.de> wrote:> Greetings, > > I am currently writing my diploma-thesis on new spectroscopic methods, and > one of them (2D-Correlation-Spectroscopy) involves cross-correlation of > fourier-transforms. I have encountered a strange issue, for which even my > professor could not supply me with a satisfying answer: When the > cross-correlation of two Fourier-transformed signals is to be calculated, > why do I have to take the complex conjugate of the second signal? This > would mean that the cross-correlation of one fourier-transformed signal > with the complex conjugated fourier-transform of the second signal is > performed, but I always thought this would mean mirroring the imaginary > part of the signal - which would be senseless when the goal is to compare > the two signals. > > It would be really great if somebody could enlighten me on that issue. > Thanks in advance, > > XuttuhThe usual definition of cross-correlation contains a complex conjugate on one of the terms: http://en.wikipedia.org/wiki/Cross-correlation And, it makes sense to do this; remember that j^2 = -1, so by inverting the sign of the imaginary part on one of the signals, you get a positive contribution to the correlation value from terms that have the same sign on the imaginary part. Jason

Reply by ●August 8, 20072007-08-08

On Aug 8, 1:33 pm, "Xuttuh" <peter_rehb...@gmx.de> wrote:> Greetings, > > I am currently writing my diploma-thesis on new spectroscopic methods, and > one of them (2D-Correlation-Spectroscopy) involves cross-correlation of > fourier-transforms. I have encountered a strange issue, for which even my > professor could not supply me with a satisfying answer: When the > cross-correlation of two Fourier-transformed signals is to be calculated, > why do I have to take the complex conjugate of the second signal? This > would mean that the cross-correlation of one fourier-transformed signal > with the complex conjugated fourier-transform of the second signal is > performed, but I always thought this would mean mirroring the imaginary > part of the signal - which would be senseless when the goal is to compare > the two signals. > > It would be really great if somebody could enlighten me on that issue. > Thanks in advance, > > XuttuhI would normally call what you are doing auto-correlation. Perhaps I am mistaken. If you look into the concept of correlation and matched filtering, you should be able to find a good answer. Phil

Reply by ●August 8, 20072007-08-08

>On Aug 8, 1:33 pm, "Xuttuh" <peter_rehb...@gmx.de> wrote: >> Greetings, >> >> I am currently writing my diploma-thesis on new spectroscopic methods,and>> one of them (2D-Correlation-Spectroscopy) involves cross-correlationof>> fourier-transforms. I have encountered a strange issue, for which evenmy>> professor could not supply me with a satisfying answer: When the >> cross-correlation of two Fourier-transformed signals is to becalculated,>> why do I have to take the complex conjugate of the second signal? This >> would mean that the cross-correlation of one fourier-transformedsignal>> with the complex conjugated fourier-transform of the second signal is >> performed, but I always thought this would mean mirroring theimaginary>> part of the signal - which would be senseless when the goal is tocompare>> the two signals. >> >> It would be really great if somebody could enlighten me on that issue. >> Thanks in advance, >> >> Xuttuh > >The usual definition of cross-correlation contains a complex conjugate >on one of the terms: > >http://en.wikipedia.org/wiki/Cross-correlation > >And, it makes sense to do this; remember that j^2 = -1, so by >inverting the sign of the imaginary part on one of the signals, you >get a positive contribution to the correlation value from terms that >have the same sign on the imaginary part. > >Jason > >Thanks for all the quick answers! @ Jason: The hint that the imaginary correlation becomes positive if one of the signals is complex conjugated was a great help! So this all comes from the so-called "Convolution-Theorem" and the corresponding "Correlation-Theorem". Is there any source I can study on the mathematics which lead to those theorems or is there any way one could "illustrate" the process somehow, so that I can understand it more easily? Perhaps there is even a website which illustrates this? Please excuse my questions, I am just an bioengineering-student, and our math-courses did not concentrate very much on signal processing, so I am trying to get around it myself. @ Phil: No, it is not an auto-correlation, it is a correlation of two different signals, which are both Fourier-transformed, just one of them is "complex-conjugate-fourier-transformed", then they are multiplied and the product is integrated.

Reply by ●August 9, 20072007-08-09

On Aug 8, 10:33 am, "Xuttuh" <peter_rehb...@gmx.de> wrote:> Greetings, > > I am currently writing my diploma-thesis on new spectroscopic methods, and > one of them (2D-Correlation-Spectroscopy) involves cross-correlation of > fourier-transforms. I have encountered a strange issue, for which even my > professor could not supply me with a satisfying answer: When the > cross-correlation of two Fourier-transformed signals is to be calculated, > why do I have to take the complex conjugate of the second signal? This > would mean that the cross-correlation of one fourier-transformed signal > with the complex conjugated fourier-transform of the second signal is > performed, but I always thought this would mean mirroring the imaginary > part of the signal - which would be senseless when the goal is to compare > the two signals. > > It would be really great if somebody could enlighten me on that issue. > Thanks in advance, > > XuttuhFor a discussion of the practical aspects of the 1D cases look at: http://www.bksv.com/pdf/Bv0013.pdf and http://www.bksv.com/pdf/bv0014.pdf Dale B. Dalrymple http://dbdimages.com

Reply by ●August 9, 20072007-08-09

>On Aug 8, 10:33 am, "Xuttuh" <peter_rehb...@gmx.de> wrote: >> Greetings, >> >> I am currently writing my diploma-thesis on new spectroscopic methods,and>> one of them (2D-Correlation-Spectroscopy) involves cross-correlationof>> fourier-transforms. I have encountered a strange issue, for which evenmy>> professor could not supply me with a satisfying answer: When the >> cross-correlation of two Fourier-transformed signals is to becalculated,>> why do I have to take the complex conjugate of the second signal? This >> would mean that the cross-correlation of one fourier-transformedsignal>> with the complex conjugated fourier-transform of the second signal is >> performed, but I always thought this would mean mirroring theimaginary>> part of the signal - which would be senseless when the goal is tocompare>> the two signals. >> >> It would be really great if somebody could enlighten me on that issue. >> Thanks in advance, >> >> Xuttuh > >For a discussion of the practical aspects of the 1D cases look at: >http://www.bksv.com/pdf/Bv0013.pdf >and >http://www.bksv.com/pdf/bv0014.pdf > >Dale B. Dalrymple >http://dbdimages.com > >

Reply by ●August 9, 20072007-08-09

Hi,>> why do I have to take the complex conjugate of the second signal?Intuitively: to rotate back each frequency bin. For example, if the signals match perfectly, the components of the product end up all pointing into the same direction. Then we sum them up, and since they are aligned, the abs. value of the sum is maximized. -Markus