Friday, March 2, 2012

Fourier Transforms

F=1/T and X=1/K (K is wave number)

Fourier series motivation
A Fourier transform is the continus limit of a Fourier series.

Think of a function f(x) which satisfies some boundary conditions f(0)=0 and f(pi)=0.  I have drawn some example functions in blue.  The function sin(nx) satisfies the boundary conditions for n=integer.  Any sum of sin(nx) will also satisfy the boundary conditions.

The inverse is true, that any continuous function satisfying the boundary conditions can be written as a sum of sin(nx) with coefficients a_{n}.

There are 2 ways to describe the function f(x):

  • Tell me the value of f(x) for every x
  • Tell me the Fourier coefficients a_{1}, a_{2}, a_{3}....

Waves in a box
One particular wavelength or distance X, corresponds to a single value in K space.  Unless we have an infinite decomposition, our K spectrum is discretised.

 To fill in K space values for a continuous spectrum you have to count waves that go into the box a non-integer number of times.  eg what wavelength x fits in the box 1.1 times, 1.2, 1.3, 1.4 etc and all the waves in between.

The size of the box in X space limits the largest wavelength mode you can see.  The size of the pixels in X space limits the smallest wavelength mode you can see, which corresponds to the highest frequency mode you can resolve in K space.
(Text reads:  Lowest frequency mode you can resolve
                    Highest frequency mode you can resolve
                          Can't resolve this very high frequency mode

What happens to the K space box if you cut out a piece of the X space box?

Size of box in X space not equal to size of box in K space, but you do get same number of pieces of information.  Highest mode in K space is inverse to smallest spacing in X.  Longest wavelength mode in X is inverse to smallest spacing in K.  Cutting the box decreases resolution in K space and information about the longest wavelength modes is lost.

(Thanks Jeremy Mardon, Adam Brown)

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