This is a only a summary of knowledge.
Fourier transform
There are many conventions as in where to put the normalisation \(2\pi\) factor.
- Forward Fourier transform (Fourier analysis)
\[ \tilde{f}(k)=\int_{-\infty}^\infty f(x)e^{-ikx}\,dx \]
- Inverse Fourier transform (Fourier synthesis)
\[ f(x)=\frac{1}{2\pi}\int_{-\infty}^\infty \tilde{f}(k) e^{ikx}\,dx \]
Another common convention is to put \(\frac{1}{\sqrt{2\pi}}\) before both integrals (balanced convention).
Properties
Linearity
\[\begin{align} h(x)=\alpha f(x)+\beta g(x) && \iff && \tilde{h}(k)=\alpha \tilde{f}(k)+\beta \tilde{g}(k) \end{align}\]
Rescaling (real \(\alpha\))
\[\begin{align} g(x)=f(\alpha x) && \iff && \tilde{g}(k)=\frac{1}{|\alpha|} \tilde{f}(\frac{k}{\alpha}) \end{align}\]
Shift / exponential (real \(\alpha\))
\[\begin{align} g(x)= f(x-\alpha) && \iff && \tilde{g}(k)=e^{-ik\alpha} \tilde{f}(k)\\ g(x)= e^{ik\alpha}f(x) && \iff && \tilde{g}(k)= \tilde{f}(k-\alpha) \end{align}\]
Differentiation / multiplication
\[\begin{align} g(x)= f'(x) && \iff && \tilde{g}(k)=ik \tilde{f}(k)\\ g(x)= xf(x) && \iff && \tilde{g}(k)= i\tilde{f}'(k) \end{align}\]
Duality
\[\begin{align} g(x)= \tilde{f}(x) && \iff && \tilde{g}(k)=2\pi f(-k)\\ \end{align}\]
Complex conjugation and parity inversion
\[\begin{align} g(x)= [f(x)]^* && \iff && \tilde{g}(k)= [\tilde{f}(-k)]^*\\ \end{align}\]
Symmetry - even & odd
\[\begin{align} f(-x)= \pm f(x) && \iff && \tilde{f}(-k)=\pm \tilde{f}(k)\\ \end{align}\]
FT of special functions
Top-hat function
FT is \(\mathrm{sinc}\).
e.g. \(f(x)\) is a top hat of strength 1, \((-1,1)\), then \(\tilde{f}(k)=\frac{2\sin k}{k}\).
Gaussian
FT is another Gaussian, with inversely proportional width, \(\sigma_x \sigma_k=1\).
Done by completing the square of the exponent. Also need Jordan’s lemma.
Delta function
Gives a definite position, hence should give no information about the momentum - a constant. Then we have the identity \[ \int_{-\infty}^\infty e^{\pm ikx} \, dx = 2\pi \delta(k) \]
Convolution theorem & such
What is convolution
Definition
The convolution \(h\) of two function \(f\) and \(g\) \[\begin{align} h=f*g && \iff && h(x)=\int_{-\infty}^\infty f(y)g(x-y) \, dx \end{align}\] This is a symmetric operation, i.e. \(f*g=g*f\).
Concept
Fix one function in space, reverse the other function in space. Then convolution is a function of the position of the reversed function, with the value of the integral of the product of the two functions.
Function / plot intuition
Copy and (continuously) pasting \(f\) at \(x\) with the weight \(g(x)\).
The simplist illustration of this idea is if \(g\) is a delta function, or a sum of several delta functions. This is the same as “lattice & motif” idea.
Probability
If \(f(x)\) and \(g(y)\) are two (independent) probability distribution functions, then \(h(z)\) is the PDF of \(z=x+y\).
Convolution theorem
\[\begin{align} h=f*g && \iff && \tilde{h}(k)=\tilde{g}(k)\tilde{f}(k) \end{align}\]
So, it is easiest to manipulate a convolution in the frequency space, as a product.
- Deconvolution is possible as a division in the frequency space.
Correlation
The correlation of two functions is \[\begin{align} h=f\otimes g && \iff && h(x)=\int_{-\infty}^\infty [f(y)]^*g(x+y) \, dx \end{align}\]
- If two signals are in phase, the correlation is positive;
- Vice versa
- If two signals are completely unrelated (e.g. noises), then correlation is 0.
Parseval’s theorem
\[ \int_{-\infty}^\infty [f(x)]^*g(x) \, dx=\frac{1}{2\pi}\int_{-\infty}^\infty [\tilde{f}(k)]^*\tilde{g}(k) \, dk \]
or more commonly, when \(f=g\), \[ \int_{-\infty}^\infty |f(x)|^2 \, dx=\frac{1}{2\pi}\int_{-\infty}^\infty |\tilde{f}(k)|^2 \, dk \]
Derive from: either the Fourier transform of a correlation; or change the order of integration and use the delta function identity.
Fourier transform preserves the inner product of functions, hence it is a unitary transformation.