FFT tests Monday 6 Oct.03 (FFTtest3)
For a timeseries y(t) constructed of analytic functions of w, call its FFT Y(w), which has peaks at
values f. Caution: meanings of f and w depend on context. For example, f is a label for the FFT
peaks, whatever they turned out to be, ordinarily NOT w/2.
Last week, I ran many test cases to investigate why the FFT frequency peaks (f) were shifted
from the analytic frequencies (w) of source timeseries. I did learn to write increasingly nice IDL
programs, but I did not find a general relation between f and w. The shift depends on
the number of points (n) and the timestep(dt).
Today I looked up articles on the web about DFT (Discrete Fourier Transforms).
 DFT/FFT by Paul Bourke,1993 (http://astronomy.swin.edu.au/~pbourke/analysis/dft)
addresses the location of frequency peaks – my first priority
 Inferring signal amplitudes from discrete (FFT) power values, by Thomas C. Ferree,
PhD, 2000 (www.egi.Technotes/SignalAmplitudes.pdf) – of secondary interest
 Calculating the Power Spectral Density (PSD) in IDL, by Thomas C. Ferree, PhD, 1999
(www.csi.uoregon.edu/members/ferree/tutorials/PowerDensitySpectrum.pdf) – read later
Some of these points summarized below are demonstrated in last week’s work (cf. FFTtest2e),
and some are for this week’s investigation
 Y(w) has the same number of frequencies as f(t) has timepoints: N
 A periodic y(t) has transform Y(w) which is symmetric about N/2 = Nyquist frequency
 FFT is designed for functions with N=2p, p an integer
 DC (steady) y(t) has one peak in Y(w), at f=0
 The first harmonic of y(t) – with one period in N timepoints - has its Y(w) peak at f=1
 The second harmonic of y(t) – with two periods in N – has its Y(w) peak at f=2
New tests:
 choose N=2p, p an integer
 choose integral numbers of periods for each y(t) function
 Check last two points above
 See if locations of Y peaks depends on dt.
Result of today’s test – I get the right frequency peaks (at c), if the argument of the sine is
(c*wt) where w=2 pi /T and period T=N*dt.
Bonus: The amplitudes are right, too.
Test: y=sin(wt) should yield Y with peak at 1.
PRO FFTest3
; See FFTtest3.doc and green notebook p.27, Mon.6.Oct.03
; Should get FFT peak at 1 for sine argument below,
; regardless of N (that is, p) or dt.
p=6
n=2^p
j=findgen(n)
dt=0.01
t=j*dt
;tt = period of fundamental
tt=n*dt
w=2*!PI/tt
print,'Number of points=',n,' period=',tt,' w=',w
;Frequency factors c take the place of w input in earlier codes
c1=1
yt=sin(c1*w*t)
Number of points=
64, period=
Amp and freq of highest FT peak is:
0.640000, w=
0.500000
fw=ABS(FFT(yt,-1))
9.81748
1
Good, this is the way it should be. Now try it with two frequencies, then vary p and dt.
Test: y=sin(wt) should yield Y with peak at 1, even if I change p to 5 or dt to 0.1
IDL> fftest3
p=
5, dt= 0.0100000
Number of points=
32, period= 0.320000, w=
Amp and freq of highest FT peak is: 0.500000
19.6350
1
IDL> fftest3
p=
5, dt= 0.200000
Number of points=
32, period=
6.40000, w=
Amp and freq of highest FT peak is: 0.500000
0.981748
1
Test: y=sin(wt) + 3 sin (3wt) should yield FT with peaks at 1 and 3.
PRO FFTest3
; See FFTtest3.doc and green notebook p.27, Mon.6.Oct.03
; Should get FFT peak at 1 for sine argument below,
; regardless of N (that is, p) or dt.
p=5
n=2^p
j=findgen(n)
dt=0.2
t=j*dt
;tt = period of fundamental
tt=n*dt
w=2*!PI/tt
print, 'p=',p,', dt=',dt
print,'Number of points=',n,', period=',tt,', w=',w
;Frequency factors c take the place of w input in earlier codes
c1=1
c2=3
A1=1
A2=2
yt= A1*sin(c1*w*t) + A2*sin(c2*w*t)
IDL> fftest3
p=
5, dt= 0.200000
Number of points=
32, period=
6.40000, w= 0.981748
c1=
1 ,c2=
3
A1=
1 ,A2=
2
Amp and freq of highest FT peak is:
1.00000
3
Amp and freq of peak between fw[0:
2] is: 0.500000
fw=ABS(FFT(yt,-1))
1
Good, this resolves last week’s question.
Now go back to last week’s data and practice extracting actual frequencies from location of
peaks, using T and n.
Another question – are amplitudes as input, now?
PRO FFTest3b
; See FFTtest3.doc and green notebook p.27, Mon.6.Oct.03
; Should get FFT peak at 1 for sine argument below,
; regardless of N (that is, p) or dt.
print,'Enter a power of 2 and a time interval, separated by commas:'
read,p,dt
n=2^p
j=findgen(n)
t=j*dt
;tt = period of fundamental
tt=n*dt
w=2*!PI/tt
print,'Number of points=',n,', period=',tt,', w=',w
print,'Enter three frequencies:'
read,c1, c2, c3
print,'Enter three amplitudes, from lowest to highest:'
read,A1, A2, A3
yt= A1*sin(c1*w*t) + A2*sin(c2*w*t) + A3*sin(c3*w*t)
; Find the frequency spectrum
fw=ABS(FFT(yt,-1))
;
set_plot, 'metafile'
device, filename='plotyt.emf'
plot,t,yt, title='yt = A1 sin(c1*wt) + A2 sin(c2*wt)'
device, /close
;
device, filename='plotfw.emf'
plot,fw, title='fw=|FFT(yt)|'
device, /close
;
; Find location of highest peak
PRINT, 'Amp and freq of highest FT peak is:', MAX(fw, i), i
;
; Find location of middle peak
semimax=max(fw[0:i-1],i2)
print,'Amp and freq of peak between fw[0:',i-1,'] is:', semimax,i2
; Find location of lowest peak
semimax=max(fw[0:i2-1],i3)
print,'Amp and freq of peak between fw[0:',i2-1,'] is:', semimax,i3
END
Where is the c=1 peak?
IDL> fftest3b
Enter a power of 2 and a time interval, separated by commas:
: 5, 0.2
Number of points=
32.0000, period=
6.40000, w= 0.981748
Enter three frequencies:
: 1, 2, 4
Enter three amplitudes, from lowest to highest:
: 1, 2, 3
Amp and freq of highest FT peak is:
1.50000
4
Amp and freq of peak between fw[0:
3] is:
1.00000
2
Amp and freq of peak between fw[0:
1] is: 0.500000
1
The peaks are indistinguishable because the resolution is too
low.
IDL> fftest3b
Enter a power of 2 and a time interval, separated by commas:
: 5, 0.1
Number of points=
32.0000, period=
3.20000, w=
1.96350
Enter three frequencies:
: 2, 3, 4
Enter three amplitudes, from lowest to highest:
: 1, 1.1, 1.2
Amp and freq of highest FT peak is: 0.600000
4
Amp and freq of peak between fw[0:
3] is: 0.550000
3
Amp and freq of peak between fw[0:
2] is: 0.500000
2
The peaks are still not distinguishable, though the resolution is
higher. How come?
IDL> fftest3b
Enter a power of 2 and a time interval, separated by commas:
: 5, 0.01
Number of points=
32.0000, period= 0.320000, w=
19.6350
Enter three frequencies:
: 2, 3, 4
Enter three amplitudes, from lowest to highest:
: 1, 1.1, 1.2
Amp and freq of highest FT peak is: 0.600000
4
Amp and freq of peak between fw[0:
3] is: 0.550000
3
Amp and freq of peak between fw[0:
2] is: 0.500000
2