Fourier transform using convolution product

convolutionfourier analysisfourier transformharmonic-analysis

The question asks to find the Fourier transform of the function:

$$ I(x) = \int^{1/2}_{0} e^{-(x-t)^2} dt$$
using the theorem about convolution products. I know that the theorem states that $\mathcal{F} \{f *g\} = \mathcal{F} \{f\} \mathcal{F}\{ g \}$.

My textbook does not do a great job explaining Fourier transforms so if anyone could give a detailed explanation of how to solve a problem such as this one it would be greatly appreciated.

Best Answer

Hint: Write $I(x)=f*g(x)$, where $f(x)=e^{-x^2}$ and $g(x)=\chi_{[0, 1/2]}$. Then using the convolution rule you get:

$$\mathcal{F}\{I\}(t)=\frac{1}{\sqrt{2}} e^{-t^2/4} \times \frac{i (1-e^{i t/2})}{\sqrt{2 \pi} t}.$$

Related Solutions

[Math] fft of point wise product

You need to do circular convolution.

 1. fft(w.*a) 
 2. cconv(fft(w), fft(a), 8)/8

Try above two lines in MATLAB, and you will find they generate the same results. In MATLAB, the circular convolution can be performed by cconv(). Find the explanation yourself, it is quite simple.

Basically, taking the fft of the two individually, then circularly convolve them together, restrict the length (assuming your two sequence are of same length, otherwise you zero-pad the shorter one) to be the same as the sequences, and finally inversely scale the result by the sequence length. Then you will have the same result as the fft of the point-wise product.

[Math] Convolution and the square root of the Fourier Transform

To take it one step further than DisintegratingByParts's answer:

The Fourier transform $\mathcal F$ is unitary on $L^2({\mathbb R})$ with spectrum $\{1,i,-1,-i\}$. This means that $\mathcal F$ effectively behaves like a finite-dimensional diagonal operator. In particular, (basically) any square root of $\mathcal F$ can be written as a polynomial of degree $3$ in $\mathcal F$ itself. These polynomials can be deduced by Lagrange interpolation and equal

$$p(x) = \pm \frac{1}{4} (x+1)(x^2+1) \pm \frac{1+i}{4 i \sqrt{2}} (x^2-1)(x+i) \pm \frac{i}{4} (x-1)(x^2+1) \pm \frac{-1+i}{4 i \sqrt{2}} (x^2-1)(x-i).$$

So the answers are:

On $\mathbb R$, the basis functions are Hermite functions like DisintegratingByParts mentioned. In higher dimensions, $L^2(\mathbb R^d)$ can be decomposed into invariant subspaces whose angular parts are given by spherical harmonics. The book of Stein and Weiss has a very nice exposition.
Generally convolution, pointwise multiplication, differentiaion, and multiplication by monomials will not transform into anything nice simply because the different terms in $p({\mathcal F})$ will transform differently. (It's easy to check that in none of the cases do coefficients magically vanish to produce unexpected symmetry).
Because it's a polynomial in $\mathcal F$, it rather directly extends to distributions.

Best Answer

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[Math] fft of point wise product

[Math] Convolution and the square root of the Fourier Transform

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