Difference between revisions of "Dictionary:Convolution theorem"

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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 [[Dictionary:Fig_F-20|F-20]] and [[Dictionary:Fig_F-22|F-22]].
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{{#category_index:C|convolution theorem}}
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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 [[Special:MyLanguage/Dictionary:Fig_F-20|F-20]] and [[Special:MyLanguage/Dictionary:Fig_F-22|F-22]].
  
=== Integral definition ===
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=== Integral definition === <!--T:2-->
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<!--T:3-->
 
The process of convolution of two functions <math> f(t)</math> and <math> g(t) </math> is defined in one dimension, as
 
The process of convolution of two functions <math> f(t)</math> and <math> g(t) </math> is defined in one dimension, as
  
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<!--T:4-->
 
<center> <math> (f \star g )(t) =  \int_{-\infty}^{\infty}f(\tau) g(t - \tau) \; d \tau. </math> </center>
 
<center> <math> (f \star g )(t) =  \int_{-\infty}^{\infty}f(\tau) g(t - \tau) \; d \tau. </math> </center>
  
  
=== Fourier domain equivalent ===
 
  
We may replace <math> f(\tau) </math> and <math> g(t - \tau) </math> by their [[Dictionary:Fourier transform|Fourier domain]] representations
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=== Fourier domain equivalent === <!--T:5-->
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<!--T:6-->
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We may replace <math> f(\tau) </math> and <math> g(t - \tau) </math> by their [[Special:MyLanguage/Dictionary:Fourier transform|Fourier domain]] representations
  
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<!--T:7-->
 
<center><math> f(\tau) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{-i \omega \tau } \; d \omega </math> </center>
 
<center><math> f(\tau) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{-i \omega \tau } \; d \omega </math> </center>
  
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<!--T:8-->
 
and
 
and
  
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<!--T:9-->
 
<center><math> g(t-\tau) = \frac{1}{2 \pi}  \int_{-\infty}^{\infty} G(\Omega) e^{-i \Omega (t - \tau) } \; d \Omega </math> </center>
 
<center><math> g(t-\tau) = \frac{1}{2 \pi}  \int_{-\infty}^{\infty} G(\Omega) e^{-i \Omega (t - \tau) } \; d \Omega </math> </center>
  
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<!--T:10-->
 
where <math> F(\omega) </math> and <math> G(\omega) </math> are the Fourier transforms of <math> f(t) </math> and <math> g(t),  </math>  respectively. Here we have used the symbol <math> \Omega </math> to represent frequency in the second integral as a dummy
 
where <math> F(\omega) </math> and <math> G(\omega) </math> are the Fourier transforms of <math> f(t) </math> and <math> g(t),  </math>  respectively. Here we have used the symbol <math> \Omega </math> to represent frequency in the second integral as a dummy
 
variable in the integration, to avoid ambiguity when combining the integral representations in the next step.
 
variable in the integration, to avoid ambiguity when combining the integral representations in the next step.
  
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<!--T:11-->
 
Substituting these representations into the original integral representation of convolution yields  
 
Substituting these representations into the original integral representation of convolution yields  
  
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<!--T:12-->
 
<center> <math> (f \star g)(t) = \int_{-\infty}^{\infty} \left( \frac{1}{2 \pi}  \int_{-\infty}^{\infty} F(\omega) e^{-i \omega \tau } \; d \omega \right) \left(\frac{1}{2 \pi}  \int_{-\infty}^{\infty} G(\Omega) e^{-i \Omega (t - \tau) } \; d \Omega  \right)  \; d \tau. </math> </center>
 
<center> <math> (f \star g)(t) = \int_{-\infty}^{\infty} \left( \frac{1}{2 \pi}  \int_{-\infty}^{\infty} F(\omega) e^{-i \omega \tau } \; d \omega \right) \left(\frac{1}{2 \pi}  \int_{-\infty}^{\infty} G(\Omega) e^{-i \Omega (t - \tau) } \; d \Omega  \right)  \; d \tau. </math> </center>
  
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We may rearrange the order of integrations
 
We may rearrange the order of integrations
  
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<!--T:14-->
 
<center> <math> (f \star g) (t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \; d \omega  \int_{-\infty}^{\infty} \; d \Omega \; F(\omega) G(\Omega) e^{-i \Omega t} \left[\frac{1}{2 \pi}  \int_{-\infty}^{\infty}  e^{-i (\omega -\Omega) \tau } \; d \tau \right]. </math> </center>
 
<center> <math> (f \star g) (t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \; d \omega  \int_{-\infty}^{\infty} \; d \Omega \; F(\omega) G(\Omega) e^{-i \Omega t} \left[\frac{1}{2 \pi}  \int_{-\infty}^{\infty}  e^{-i (\omega -\Omega) \tau } \; d \tau \right]. </math> </center>
  
Recognizing the factor in <math> [ ... ] </math> as the frequency domain representation of the [[Dictionary:Dirac function|Dirac delta function]],
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<!--T:15-->
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Recognizing the factor in <math> [ ... ] </math> as the frequency domain representation of the [[Special:MyLanguage/Dictionary:Dirac function|Dirac delta function]],
  
  
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<!--T:16-->
 
<center> <math> \delta ( \omega - \Omega )  = \delta( \Omega - \omega ) = \frac{1}{2 \pi}  \int_{-\infty}^{\infty}  e^{-i (\omega -\Omega) \tau } \; d \tau . </math> </center>
 
<center> <math> \delta ( \omega - \Omega )  = \delta( \Omega - \omega ) = \frac{1}{2 \pi}  \int_{-\infty}^{\infty}  e^{-i (\omega -\Omega) \tau } \; d \tau . </math> </center>
  
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<!--T:17-->
 
permits us to write the equivalent expression
 
permits us to write the equivalent expression
  
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<!--T:18-->
 
<center> <math> (f \star g) (t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \; d \omega  \int_{-\infty}^{\infty} \; d \Omega \; F(\omega) G(\Omega) e^{-i \Omega t} \delta(\Omega - \omega ). </math> </center>
 
<center> <math> (f \star g) (t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \; d \omega  \int_{-\infty}^{\infty} \; d \Omega \; F(\omega) G(\Omega) e^{-i \Omega t} \delta(\Omega - \omega ). </math> </center>
  
The <math> \Omega </math> integral may be performed, exploiting the [[Dictionary:Dirac function|sifting property]] of the delta function to convert the <math> \Omega </math> to <math> \omega </math>  yields the equivalence of
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<!--T:19-->
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The <math> \Omega </math> integral may be performed, exploiting the [[Special:MyLanguage/Dictionary:Dirac function|sifting property]] of the delta function to convert the <math> \Omega </math> to <math> \omega </math>  yields the equivalence of
 
multiplication in the frequency domain to convolution in the time domain
 
multiplication in the frequency domain to convolution in the time domain
  
  
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<!--T:20-->
 
<center> <math> (f \star g) (t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty}    F(\omega) G(\omega) e^{-i \Omega t} \; d \omega . </math> </center>
 
<center> <math> (f \star g) (t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty}    F(\omega) G(\omega) e^{-i \Omega t} \; d \omega . </math> </center>
  
=== Convolution in the Frequency domain ===
 
  
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=== Convolution in the Frequency domain === <!--T:21-->
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<!--T:22-->
 
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, for example, 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.
  
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<!--T:23-->
 
Paralleling the derivation above, we write the convolution in the frequency domain
 
Paralleling the derivation above, we write the convolution in the frequency domain
  
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<!--T:24-->
 
<center> <math> (F \star G )(\omega) =  \int_{-\infty}^{\infty}F(\Omega) G(\omega - \Omega) \; d \Omega </math> </center>
 
<center> <math> (F \star G )(\omega) =  \int_{-\infty}^{\infty}F(\Omega) G(\omega - \Omega) \; d \Omega </math> </center>
  
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<!--T:25-->
 
As above, we substitute the Fourier representations of <math> F(\omega) </math> and <math> G(\omega) </math>  
 
As above, we substitute the Fourier representations of <math> F(\omega) </math> and <math> G(\omega) </math>  
  
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<!--T:26-->
 
<center><math> F(\Omega) = \int_{-\infty}^{\infty} f(t) e^{i \Omega t} \; d t </math> </center>
 
<center><math> F(\Omega) = \int_{-\infty}^{\infty} f(t) e^{i \Omega t} \; d t </math> </center>
  
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and
 
and
  
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<center><math> G(\omega - \Omega) =  \int_{-\infty}^{\infty} g(\tau)  e^{i  (\omega - \Omega) \tau } \; d \tau.</math> </center>
 
<center><math> G(\omega - \Omega) =  \int_{-\infty}^{\infty} g(\tau)  e^{i  (\omega - \Omega) \tau } \; d \tau.</math> </center>
  
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<!--T:29-->
 
As in the derivation above, we substitute the Fourier representations of <math> F(\omega) </math> and <math> G(\Omega) </math> and rearrange the terms to yield
 
As in the derivation above, we substitute the Fourier representations of <math> F(\omega) </math> and <math> G(\Omega) </math> and rearrange the terms to yield
  
  
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<!--T:30-->
 
<center> <math> (F \star G) (\omega) = K \int_{-\infty}^{\infty} \; d t \int_{-\infty}^{\infty} \; d \tau \; f(t) g(\tau) \; \left[ \int_{-\infty}^{\infty} e^{i \Omega (t-\tau)}  \; d \Omega \right]. </math> </center>
 
<center> <math> (F \star G) (\omega) = K \int_{-\infty}^{\infty} \; d t \int_{-\infty}^{\infty} \; d \tau \; f(t) g(\tau) \; \left[ \int_{-\infty}^{\infty} e^{i \Omega (t-\tau)}  \; d \Omega \right]. </math> </center>
  
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<!--T:31-->
 
We recognize the term in <math> [ ... ] </math> as the Fourier form of the Dirac delta function
 
We recognize the term in <math> [ ... ] </math> as the Fourier form of the Dirac delta function
 
<center> <math> 2 \pi \delta ( t- \tau )  = 2 \pi \delta( \tau - t ) =  \int_{-\infty}^{\infty}  e^{i \Omega (t -\tau ) } \; d \Omega . </math> </center>
 
<center> <math> 2 \pi \delta ( t- \tau )  = 2 \pi \delta( \tau - t ) =  \int_{-\infty}^{\infty}  e^{i \Omega (t -\tau ) } \; d \Omega . </math> </center>
  
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<!--T:32-->
 
As before, we apply the sifting property of the delta function, in this case to perform the <math> \tau </math> integration to yield
 
As before, we apply the sifting property of the delta function, in this case to perform the <math> \tau </math> integration to yield
  
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<!--T:33-->
 
<center> <math> (F \star G) (\omega) = (2 \pi) \int_{-\infty}^{\infty} f(t) g(t) e^{i \omega t } \; dt </math> </center>
 
<center> <math> (F \star G) (\omega) = (2 \pi) \int_{-\infty}^{\infty} f(t) g(t) e^{i \omega t } \; dt </math> </center>
  
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<!--T:34-->
 
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>  
 
as the Frequency domain convolution of their respective Fourier transforms,  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
  
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<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>
  
(The extra factor of <math> 1/2 \pi </math> echoes what is seen in other results involving the Fourier transform, such as [[Parseval's relation]].
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<!--T:36-->
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(The extra factor of <math> 1/2 \pi </math> echoes what is seen in other results involving the Fourier transform, such as [[Special:MyLanguage/Parseval's relation|Parseval's relation]].
  
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<!--T:37-->
 
To come full circle, if we were to perform the inverse Fourier transform of <math> F(\omega), </math> considered as a purely causal function,
 
To come full circle, if we were to perform the inverse Fourier transform of <math> F(\omega), </math> considered as a purely causal function,
 
we would need to treat this as a contour integral, and choose
 
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>
 
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>
 
given the definition of the Fourier transform we are using here.
 
given the definition of the Fourier transform we are using here.
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Latest revision as of 08:27, 20 June 2017

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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. Here we have used the symbol to represent frequency in the second integral as a dummy variable in the integration, to avoid ambiguity when combining the integral representations in the next step.

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

(The extra factor of echoes what is seen in other results involving the Fourier transform, such as Parseval's relation.

To come full circle, if we were to perform the inverse Fourier transform of considered as a purely causal function, 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 given the definition of the Fourier transform we are using here.