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Given $x(t) = f(t) \cdot g(t)$, what is the Fourier transform of $x(t)$? If possible, please explain your answer.

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Is this homework or why do you ask? –  AD. Nov 29 '10 at 17:24
    
The motivation behind the question is homework, but this is a basic principle in the class that I never quite grasped properly. My current homework builds upon the principle of the question. So answering this question by no means will be doing my homework for me. This is why I am asking the general case. –  Cory Klein Nov 29 '10 at 17:35
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To start with, you need some conditions on $f$ and $g$, for example their product $f\cdot g$ should be integrable - can you find such conditions? Also, what integral do you use in the course? –  AD. Nov 29 '10 at 17:44
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Sure, I'm happy to, but may I ask why the reasoning behind the question is relevant to whether somebody decides to answer? –  Cory Klein Nov 29 '10 at 17:46
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f and g don't represent any specific functions. If there are restraints determining whether the transform can be performed, I would appreciate mention of those in the answer. I'm not sure what you mean by "what integral", I was unaware that there are more than one kind of "integral". –  Cory Klein Nov 29 '10 at 17:50
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2 Answers

Transforms such as Fourier transform or Laplace transform, takes a product of two functions to the convolution of the integral transforms, and vice versa.

This is called the Convolution Theorem, and is available with proof at wikipedia.

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Fourier transform of a product is the convolution of the corresponding transforms. For details on conditions on the functions refer links below

http://en.wikipedia.org/wiki/Fourier_transform

http://en.wikipedia.org/wiki/Fourier_transform#Convolution_theorem

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Maybe I am reading your answer or the wikipedia article incorrectly. You seem to be stating that the Fourier transform of x is the convolution of Fourier(f) and Fourier(g). But your second link appears to state that Fourier(x) = Fourier(f) x Fourier(g), where the transforms of f and g are multiplied, not convolved. Perhaps I am missing something. –  Cory Klein Nov 29 '10 at 18:02
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@nomoreink: The link states that if $h$ is the convolution of $f$ and $g$ (not $*$ denotes convolution not multiplication), then the Fourier transform of $h$ is the product of the fourier transforms of $f$ and $g$. Conversely, (this is the part relevant to your question), if $h$ can be decomposed as a product of two integrable functions, then the Fourier transform of $h$ is the convolution of the transforms of $f$ and $g$. –  Timothy Wagner Nov 29 '10 at 18:05
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