Looking At Fourier Transforms

 The Fourier transform is one of the most useful mathematical tools available to the physicist and we explore some basic aspects of it in this post.   Probably the most logical place to start is that these transforms can be used to represent a function.   Thus, a function f(x) may be represented by a power series of the form:

   å ^¥ _{n = 0}    a _n  (x)  ⁿ   

provided it satisfies the necessary conditions for this expansion.   If the set of coefficients ( a _n ) is given then the function f(x) can be constructed - at least within the radius of convergence.   In other words a function can be defined in terms of the coefficients  a _n  .   

Similarly if a function f(x)  is representable by a Fourier series:

f(x)  =    å^¥ _{n = - 0o}    c _n  exp [i(n p x/L) ]     (-L   <   x  <   L)

We can say that the set of coefficients  { c _n  }  also defines  f(x).    In effect, the set of numbers -  a _n   or  c  _n  - can be regarded as a function of the variable n written as c(n).   The particular function here is defined for a discrete set of values for the independent variable (x) but this is not essential.  

Indeed, in previous posts, i.e. for the Laplace transform of a function f(t),  e.g.

• Revisiting Basic Laplace Transforms In Solving Dif...

We saw cases such as:  F(s) =  £ {F(t) }

where the representation is over the continuous variable, t.  Thus, one can say that F(x) represents f(t) since f(t) can be defined uniquely if it is continuous by F(x), e.g.

f(t)  =   1/2 pi  ò _a ^b   F(s)  exp(is) ds 

where:  a = g   -   ia

    and   b =  g   +   ia
 

Hence, we don't regard F(x) as a different function from f(t) but we say rather than F(x) is f(t) in the Laplace transform representation.   Going back to the Fourier series case the function c(n) is often called the Fourier spectrum of f(x).  The simple plotting of this is left as an exercise for the energetic reader, i.e. using c(n) as the ordinate and n ('wave number') as the abscissa.    

A plot can also be made of c vs.  k =  n p / L.  In this case if L is large then the frequencies will be closely spaced given that:

 Δk  =  (p / L) Δn   =   p / L

Note then that if L is large,  Δk is  small  and the function more closely approaches a continuous spectrum, i.e.

Note we can multiply each term of the Fourier series by:            (L/p)Δk    and write:

f(x)  =    å ^¥ _{n = - 0o}   ( L/ p )  c _n  exp [i(n p x/L) ]  Δk

Where:

( L/ p )  c _n  = 1/2 pi  ò _-L  ^+L   f(x) exp [i(n p x/L) ] dx

We can switch here completely to k-notation and write:

( L/ p )  c _n    =   c _L  (k)

Then obtain:

 c _L  (k)  =  ò _-L  ^+L   f(x) exp (-ik x )  dx

 and:

f(x)  =  å^¥ _{Lk / p  = - 0o}     c _L  (k)   (k) exp (-ik x )   Δk

Now, let k  -> oo

Then the sum goes to an integral and we obtain:

c(k)  =    c _L  (k)  =  lim _{L ® oo}      1/2 pi  ò ^¥_-¥   f(x) exp (-ik x )  dx  

Whence:  f(x)  =  ò ^¥_-¥   c(k) exp (-ik x )  dx  

This set of formulas is known as the Fourier transformation.  For a more 'streamlined' update we define a function:

F(k)  =  Ö (2 p)   c(-k)   

Then the formulas now appear:

1) F(k)  =  1/ Ö 2 p     ò ^¥_-¥   f(x) exp (-ik x )  dx  

2) f(x) =  1/ Ö 2 p     ò ^¥_-¥   F(k)  exp (-ik x )  dx  

The function F(k)  is known as the Fourtier transform of the function f(x).   

Problem for the Math aficionado:

Find the Fourier transform of:  

f(x) =  exp(-  lx ²) cos  b x