On Factors of 2-pi in the Fourier Transform

Posted on November 4, 2007
Filed Under Signal Processing |

Every physicist is exposed during his or her undergraduate degree to the Fourier Transform, defined as:

\hat{f}(\omega) = \int_{-\infty}^\infty f(t) e^{-\imath \omega t} dt

f(t) = \frac{1}{2\pi} \int_{-\infty}^\infty f(\omega) e^{\imath \omega t} d\omega

However, a source of great confusion is that of factors. Different books define the Fourier Transform using varying factors in front of the integral:

\hat{f}(\omega) = \frac{1}{2\pi} \int_{-\infty}^\infty f(t) e^{-\imath \omega t} dt

\hat{f}(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(t) e^{-\imath \omega t} dt

Are those definitions valid, and if so, how can we settle the conflict between them?
The answer to the first question is yes, they are all, indeed, valid. To resolve the conflict, we need to dig in and ask ourselves where does the Fourier theorem come from at all? Upon opening our math textbooks, we learn that the Fourier theorem is derived by showing that, given any function sufficiently smooth and finite, we may write the function as:

f(t) = \frac{1}{2\pi} \int_{-\infty}^\infty \int_{-\infty}^\infty f(t') e^{-\imath \omega (t-t')} d\omega dt'

Here comes the important part: this seemingly innocent and useless looking identity can be turned, as if by magic, into the Fourier theorem stated at the beginning of this chapter, provided we define:

\hat{f}(\omega) = \int_{-\infty}^\infty f(t') e^{-\imath \omega t'} dt'

Using this definition, the theorem becomes

f(t) = \frac{1}{2\pi} \int_{-\infty}^\infty f(\omega) e^{\imath \omega t} d\omega

.
Now, it is important to grasp that \hat{f}(\omega), as we’ve defined it, is just a definition. We could have equally defined

\frac{1}{2\pi} \hat{f}(\omega) = \int_{-\infty}^\infty f(t) e^{-\imath \omega t} dt

which would’ve yielded

f(t) = \int_{-\infty}^\infty f(\omega) e^{\imath \omega t} d\omega

.
As you can see, it does not matter how you define \hat{f}(\omega), as long as multiplying the factors of f(t) and \hat{f}(\omega) yield

\frac{1}{2\pi}

which is the factor that appears in the theorem stated above (which we have not proven, in case you were left scratching your head - for that you should open a math textbook. Hey, I’m a physicist, give me a break). So the important thing is to check the factors in front of both f(t) and its Fourier transform and check that their product gives the correct result. Looking at just the factor in front of the Fourier transform is meaningless!

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