Question about Cross-Correlation in Frequency-domain

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

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,
>
> 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

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,
>
> Xuttuh

I 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

>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,
>>
>> 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. 

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,
>
> 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

>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,
>>
>> 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
>
>

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