Difference between revisions of "Dictionary:Convolution theorem"

From SEG Wiki
Jump to: navigation, search
(Convolution in the Frequency domain)
Line 46: Line 46:
  
 
Similarly, multiplication in the time domain may be interpreted as convolution in the frequency domain, to within a constant factor. Such
 
Similarly, multiplication in the time domain may be interpreted as convolution in the frequency domain, to within a constant factor. Such
a frequency domain convolution representation is useful if we were interested in finding the Fourier transform of the product of functions
+
a frequency domain convolution representation is useful, for example, if we were interested in finding the Fourier transform of the product of functions
 
of known Fourier transform.
 
of known Fourier transform.
  
Line 74: Line 74:
  
 
There is an extra factor of <math> 2 \pi. </math> Thus, if we were representing the Fourier transform of the product <math> f(t)g(t) </math>  
 
There is an extra factor of <math> 2 \pi. </math> Thus, if we were representing the Fourier transform of the product <math> f(t)g(t) </math>  
of the functions as the Frequency domain convolution of the respective Fourier transforms of <math> f(t) </math> and <math> g(t), </math> we
+
as the Frequency domain convolution of their respective Fourier transforms,  we
 
would need to include a factor of <math> 1/2 \pi </math> in the convolution
 
would need to include a factor of <math> 1/2 \pi </math> in the convolution
  
 
<center> <math> \frac{1}{2 \pi}(F \star G )(\omega) =  \int_{-\infty}^{\infty}f(t) g(t) e^{i \omega t } \; dt. </math> </center>
 
<center> <math> \frac{1}{2 \pi}(F \star G )(\omega) =  \int_{-\infty}^{\infty}f(t) g(t) e^{i \omega t } \; dt. </math> </center>
 +
 +
 +
To come full circle, if we were to perform the inverse Fourier transform of <math> F(\omega) </math> we would need to treat this as a contour integral, and choose
 +
the contour along the real axis to pass over any poles that would be on the real axis, to obtain the causal function <math> f(t). </math>

Revision as of 16:05, 6 December 2016

The Fourier transform of the convolution of two functions is equal to the product of their individual transforms (or multiplying their amplitude spectra and summing their phase spectra). See Figures F-20 and F-22.

Integral definition

The process of convolution of two functions and is defined in one dimension, as


Fourier domain equivalent

We may replace and by their Fourier domain representations

and

where and are the Fourier transforms of and respectively.

Substituting these representations into the original integral representation of convolution yields

We may rearrange the order of integrations

Recognizing the factor in as the frequency domain representation of the Dirac delta function,


permits us to write the equivalent expression

The integral may be performed, exploiting the sifting property of the delta function to convert the to yields the equivalence of multiplication in the frequency domain to convolution in the time domain


Convolution in the Frequency domain

Similarly, multiplication in the time domain may be interpreted as convolution in the frequency domain, to within a constant factor. Such a frequency domain convolution representation is useful, for example, if we were interested in finding the Fourier transform of the product of functions of known Fourier transform.

Paralleling the derivation above, we write the convolution in the frequency domain

As above, we substitute the Fourier representations of and

and

As in the derivation above, we substitute the Fourier representations of and and rearrange the terms to yield


We recognize the term in as the Fourier form of the Dirac delta function

As before, we apply the sifting property of the delta function, in this case to perform the integration to yield

There is an extra factor of Thus, if we were representing the Fourier transform of the product as the Frequency domain convolution of their respective Fourier transforms, we would need to include a factor of in the convolution


To come full circle, if we were to perform the inverse Fourier transform of we would need to treat this as a contour integral, and choose the contour along the real axis to pass over any poles that would be on the real axis, to obtain the causal function