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Secrets FFTs revealed!!
- Subject: Secrets FFTs revealed!!
- From: Peter Brooker <ra5589(at)email.sps.mot.com>
- Date: Thu, 04 May 2000 16:07:29 -0700
- Newsgroups: comp.lang.idl-pvwave
- Organization: Motorola Semiconductor Products Sector
- Xref: news.doit.wisc.edu comp.lang.idl-pvwave:19486
I have just been through a learning curve on FFTs. Much thanks to Alan
Barnett for putting me on the right track. I think I have them figured
out and now want to write a reference that captures my present level of
understanding. Realize that I have learned only as much of the FFT
theory as needed. My motivation is that I am going to be applying the
FFT functions to modeling the effect of a finite lens size on the image.
(The finite lens size will chop off the higher order frequencies).
Perhaps somebody else will want to expand/improve this reference. These
are only my best guesses to how everything works. Perhaps this is
something for the IDL FAQ.
This reference is organized as follows:
PART#1:Relate complex expansion to real Fourier series
PART#2:IDL form of complex expansion
PART#1: Relate complex expansion to real Fourier series.
Assume you have a function f(t) that is periodic in t with a period T.
Then there exists coefficients a_n & b_n such that
f(t) = a_o + sum_n(a_n*cos(2*pi*n*t/T)+b_n*sin(2*pi*n*t/T)) , n=1,2,...
This is just the Fourier series of function with period X. Nothing new
here. See eq #4, section 10.3, Advanced Engineering Mathematics,
Kreyszig. This expansion though assumes -T/2 < t < T/2
Now consider an alternate form of the Fourier series expansion.
In order for me to be comfortable with this expansion I need to see how
this expansion relates to the expansion above. In particular, how do the
complex An relate to the real a_n and b_n?
Consider the following:
II= sum_n(A_n*exp(j*2*pi*n*t/T)), n=0,1,2,... j*j = -1
= A_o + sum_n( aa_n*cos() - bb_n*sin() + j*(bb_n*cos()+aa_n*sin())
Real(II) = Real(A_o) + sum_n( aa_n*cos(2*pi*n*t/T)-bb_n*sin(2*pi*n*t/T))
Comparing to the first expansion we see that
Real(A_o)=a_o, aa_n=a_n, -bb_n=b_n
To me, this proves existence of the complex expansion. Knowing one, you
can figure out the other. Part #1 is complete.
Part #2: IDL form of complex expansion
Let f(t) be a periodic function with period T defined on an interval
Then there exist complex A_n such that
f(t)= sum_n(A_n*exp(j*2*pi*n*t/T)), n=0,1,2,... j*j = -1
Divide the interval into N sections. t~t_i = i*T/N
f( t_i ) = sum_n(A_n*exp(j*2*pi*n*t_i/T))
= sum_n( A_n*exp(j*2*pi*n*i/N) ) , n=0,1,...
This is exactly what is found in the IDL manual under the section for
FFT. The only difference is that t has been replaced by u and A_n has
been replaced by F(u). Note that the period T has dropped out. Also note
that t has been replaced by t_i = i*T/N. In order for this to happen,
the interval over which t is defined must be from [0,T]. This is
different from the definition of t being defined over the interval
[-T/2,T/2]. Perhaps this is why b_n = -bb_n.
********UNFORTUNATELY IT IS WRONG****************
What is wrong is the values of n in the sum. IDL does not use the values
of n=0,1,2,... IDL actually uses n= -N/2+1, -N/2+2, ...-1,0,1,...,N/2
The reason for doing this must have to do with FFT theory. Note also
that the number of values of n is N.
It gets more complicated. From the manual we have
F(u) = 1/N*sum_x(f(x)*exp(-j*2pi*ux/N)) , x=0,1,...N-1
First thing to realize is that F(u) is really F_n. Where n is an
integer. This comes from the fact that f(x) is periodic in x.
The manual also mentions that the "frequencies" are
After trial and error I have determined that the value of the ns range
for -N/2 to N/2. Futhermore, the F_n are stored in the order associated
with the following values of n
0,1,2,...,N/2,-(N/2-1),-(N/2-1),...,-1 <== this is bizarre!!
Let N=8. Then N/2=4
The F_n would be stored in an array. The array of n values associated
with this array would be:
Part #3: Specific Example
Consider the interval t = [0,1]. This choice of interval implies T=1.
Let f(t) = sin ( 4*pi*t)
f(t_i)=sum_n(A_n*exp(-j*2pi*n*i/N)) , n=-N/2,...-1,0,1,...N/2
= A_nN/2... + A_n2*(cos(2pi*(-2)*i/N)+j*sin(2pi*(-2)*i/N))+
=... + A_n2*cos(2pi*2*i/N)+A_2*cos(2pi*2*i/N) +
+A_n2*j*(-1)*sin(2pi*2*i/N)+A_2*j*sin(2pi*2*i/N)) + ....
= ... + (A_n2+A_2)*cos(2pi*2*i/N)+j*( -A_n2 + A_2)*sin(2pi*2*i/N)
where A_n2 stands for A_n where n= -2
Equating the series to sin(2pi*2*i/n) we conclude
A_n = 0 for all n except n = -2 or n = 2.
j*(-A_n2 + A_2) = 1
Let A_n2=(a_n2+j*b_n2) and A_2=(a_2 + j*b_2)
The above equations imply
(a_n2 + a_2) + j*( b_n2+b_2) = 0 &
j*[( -a_n2 + a_2) + j*( -b_n2 + b_2)] = 1
==> a_n2 + a_2=0, b_n2+b_2 =0 ==> a_n2= - a_n2, b_n2 = -b_n2
==> 2*a_n2=0 ==> a_n2=a_2 = 0
==> j*j*(2*b_2)=1 ==> 2*b_2 = -1/2, b_2 = 1/2
A_n2 = 0 + j*(1/2)
A_2 = 0 + j*(-1/2)
We now have calculated the solutions.
The following code calculates this and displays the correct answers. It
shows how to plot A_n vs n correctly.
plot,t,f_t, title='f(t) vs t'
A_n=fft(f_t,-1) ; complex fourier coefficients
c=[a,b] ; c=[-N/2+1,-N/2+2, ...,-1,0,1,...,N/2]
plot,c(sub),float(A_n(sub)),yrange=[-.5,.5],title='float(A_n) vs n'
title='imaginary(A_n) vs n'
title='imaginary(An) vs n' ; finer x scale