IDL 1D FFT tests Friday 03 Oct 03
PRO fft2dteste
; why are frequency peaks shifted from analytic w?
IDL> n=100
IDL> j=findgen(n)
IDL> dt=0.1
IDL> t=j*dt
IDL> yt=sin(t)+ 2*sin(3*t)
IDL> plot,t,yt, title='yt = sin(t) + 2 sin(3t) vs t'
IDL> dx=0.2
IDL> x=j*dx
IDL> yx=sin(2*x)
IDL> plot,x,yx, title='yx=sin(2x) vs x'
IDL> fw=FFT(yt,-1)
IDL> plot, ABS(fw), title='ABS(fw)'
IDL> fk=FFT(yx,-1)
IDL> plot, ABS(fk),title='ABS(fk)'
Expect one peak at 1, a bigger peak at 3
Fine, but frequencies are at about 1*pi and 3*pi.
1=dt*sqrt(n)=0.1*10
Expect one peak at 2
Fine, but frequency is at about 2*pi/2
2=dx*sqrt(n)=0.2*10
IDL> PRINT, 'Maximum value in array fw is:', MAX(fw, I)
Maximum value in array fw is: 0.686987
IDL> PRINT, 'The location of the maximum value is', I
The location of the maximum value is
10
IDL> semimax=max(fw[0:5],i)
IDL> print,semimax,i
0.506744
3
Guess general relationship for 1D FFT: frequency spectrum peaks at (w*pi/m) where m= dx*sqrt(n)
Check: first, find actual values of frequency peaks here, then plot some cases with n.ne.100
Increase resolution of the same case to see if the peaks at 1,3 are actually at pi, 3*pi.
IDL> n=1000
IDL> j=findgen(n)
IDL> dt=0.01
IDL> t=j*dt
IDL> yt=sin(t)+ 2*sin(3*t)
IDL> plot,t,yt, title='yt = sin(t) + 2 sin(3t) vs t'
IDL> fw=FFT(yt,-1)
IDL> plot, ABS(fw), title='ABS(fw)'
Analytic peaks at w=1 and w=3
Guess FFT peaks will be at w’=(w*pi/m) where m= dt*sqrt(n)
M=dt*sqrt(n)=0.01*31.6227766=0.316, so pi/m=0.9935 <10
Then peaks should be at about 10*pi =31.2 and 10*3*pi=93.6
WRONG: here is the IDL result – just the same as with lower resolution.
INPUT
OUTPUT
; Find location of highest peak
Maximum value in array fw is: 0.683548
PRINT, 'Maximum value in array fw is:', MAX(fw, i)
The location of the highest peak is
10
PRINT, 'The location of the highest peak is', i
;
; Find location of second-highest peak
Value of peak between fw[0:5] is: 0.508961
; (determined to lie between elements 0-5, in testd).
The location of this lower peak is
3
semimax=max(fw[0:5],i)
print,'Value of peak between fw[0:5] is:', semimax,i
PRINT, 'The location of this lower peak is', i
3
NEW GUESS: Resolution and number of points is irrevelant. Frequencies are simply multiplied by
Pi.
TESTF: Try a different function and see how the frequencies shift.
n=100
j=findgen(n)
; Construct a function varying in time, with higher resolution
dt=0.2
t=j*dt
yt=sin(2* t)+ 2*sin(4*t)
Maximum value in array fw is: 0.883451
The location of the highest peak is
13
Value of peak between fw[0:10] is: 0.436424
The location of this lower peak is
6
6
2*pi = 6.28 ~ 6, and 4*pi = 12.6 ~ 13
New guess looks correct. Try one more case... TEST G
PRO FFTtestG
; Try different frequencies and resolutions to find how
; spectral peaks compare to analytic frequencies,
; in general.
n=200
j=findgen(n)
dt=0.2
t=j*dt
yt=sin(t) + 2*sin(2* t)+ 3*sin(3*t)
Maximum value in array fw is:
1.48652
The location of the highest peak is
19
Value of peak between fw[0:10] is:
The location of this lower peak is
0.437406
6
6
Analytic frequencies: 1, 2, 3
FFT peaks: 6, 13, 19 = 2 pi, 4 pi, 6 pi.
So number of points matters after all. n=200 -> w’ = 2Pi*w.
Two more tests:
Check if w’=3Pi*w for n=300 and
Redo n=200 for dt = 0.1
TestG1 = same as testG above, except dt = 0.1 instead of 0.2:
PRO FFTtestG1
; same as testG except dt = 0.1 instead of 0.2
n=200
j=findgen(n)
dt=0.1
t=j*dt
yt=sin(t) + 2*sin(2* t)+ 3*sin(3*t)
Value and location of highest peak is:
1.08077
10
The location of the highest peak is
10
Value and location of peak between fw[0:8] is: 0.929409
Value and location of peak between fw[0:4] is: 0.622931
Analytic f=1, 2, 3
FFT peaks = 3, 6, 10 = 1*pi, 2*pi, 3*pi
6
3
TEST G3:
Check if w’=3Pi*w for n=300
PRO FFTtestG3
; same as testG2 except dt = 0.1 and n=300 instead of 200
n=300
j=findgen(n)
dt=0.1
t=j*dt
yt=sin(t) + 2*sin(2* t)+ 3*sin(3*t)
Value and location of highest peak is:
1.22145
14
Value and location of peak between fw[0:10] is: 0.649463
Value and location of peak between fw[0:5] is: 0.391206
9
5
PRO FFTtestG3b
; same as testG3 except dt = 0.2 instead of 0.1
n=300
j=findgen(n)
dt=0.2
t=j*dt
yt=sin(t) + 2*sin(2* t)+ 3*sin(3*t)
INPUT
; Find location of highest peak
PRINT, 'Value and location of highest peak is:',
MAX(fw, i), i
;
; Find location of second-highest peak
n2=3*i/4
semimax=max(fw[0:n2],i2)
print,'Value and location of peak between
fw[0:',n2,'] is:', semimax,i2
;
; Find location of third-highest peak
n3=n2/2
semimax=max(fw[0:n3],i3)
print,'Value and location of peak between
fw[0:',n3,'] is:', semimax,i3
;
OUTPUT
Value and location of highest peak is:
1.20775
29
Value and location of peak between fw[0:
0.982564
19
21] is:
Value and location of peak between fw[0:
0.329043
9
10] is: