N13
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Fourier Transforms
Discrete Fourier Transform (DFT)
find frequency content inherent within a signal
guess specific frequencies fk and assess how well each matches the given signal
8 Hz 5 V and 29 Hz 1 V - fs = 100 Hz, n=64
Voltage
5
0
-5
Phase [deg]
Amplitude [V]
0
0.1
0.2
0.3
Time (sec)
0.4
0.5
0.6
5
0
0
10
20
30
40
50
60
Frequency [Hz]
70
80
90
0
10
20
30
40
50
60
Frequency [Hz]
70
80
90
100
0
-100
N13
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digitize xi for i = 1 to n at fixed sampling frequency fs = 1/h
choose frequency fk
create guess gi for i = 1 to n using frequency fk and compare to xi
g  sin 2 f k t
t  i h  i / fs
g i  sin 2
fk
i
fs
xi
h = 10 msec
+3V
i
-3V
gi
i
correlation coefficient
C xg 
2 n
 xigi
n i 1
correlation coefficient for sine at frequency fk
f
2 n 
C k   x i  sin 2 k
n i 1 
fs

i 

Ck is approximate amplitude at frequency fk inherent within digitized signal xi
N13
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Ck = -amplitude
xi
i
gi
i
Ck = 0
xi
i
gi
i
N13
page 4 of 12
Ck = 0
xi
i
gi
i
Ck ≠ 0
xi
i
gi
i
N13
correlation coefficient
2 n
 xigi
n i 1
C xg 
x  Vo sin 2 f o t
unknown signal at frequency fo
g  sin 2 f k t
perfect guess
 sin
2

fk = fo
au du 
2Vo
C
T
2 T
x g  dt
T 0
C
convolution integral over time sample T
guess
page 5 of 12
C
2 T
Vo sin 2 f o t sin 2 f k t   dt
T 0
C
u 1
 sin au
2 4a
2Vo
T
ut
 sin
T
2
0

2 f o t dt
a  2f o
T
t


2Vo  T
1
1
1
sin 2 f o t  
sin 2 f o T  Vo  Vo
sin 2 f o T
 
 
T  2 8 f o
4 f o T
 2 8 f o
o

time T for m = 1, 2, 3, … complete cycles at frequency fo
C  Vo  Vo
1
sin 2 m
4 m
T
m
fo
C  Vo
for m cycles plus (or minus) some small portion 0    1 of another cycle
C  Vo  Vo
1
1
sin 2 m     Vo  Vo
sin 2 
4 m   
4 m   
T fo  m  
N13
page 6 of 12
Ck = 0
xi
i
gi
i
DFT at guess fk requires correlation for both sine and cosine
Ak 
f
2 n 
x i  cos 2 k

n i 1 
fs

i 

Bk 
f
2 n 
x i  sin 2 k

n i 1 
fs

i 

amp k  A 2k  B 2k
DC fk = 0
tan  k 
Ak = 2 mean(xi)
Bk
Ak
Bk = 0
N13
page 7 of 12
1) single DFT at one given fk
2) multiple DFT at several discrete frequencies over band
3) complete DFT at n discrete frequencies
spectral resolution
f 
f k  k f
f1 < fk < f2
for k = 0 to n
fs
n
small number of samples n causes coarse resolution in frequency domain
why n discrete frequencies ?
Order of Computational Complexity
FLOP = floating point operation = one multiply and one add
assume sine and cosine functions have been pre-computed for given number of samples n
O[ ] = number of FLOPS for computation
O[ Ak ] = n
O[ Bk ] = n
O[ single DFT ] = 2n
O[ complete DFT ] = 2n(n+1) ≈ 2n2
larger sample size n provides finer spectral resolution
larger sample size n increases computation time by n2
f 
fs
n
N13
Fast Fourier Transform (FFT)
Cooley and Tukey (1965)
radix-2 algorithm breaks n point DFT into two smaller (n/2) point DFTs
compute DFT of even index points ( x2 x4 x6 … )
compute DFT of odd index points ( x1 x3 x5 … )
combine results using symmetric “butterfly” operations
recursively break each (n/2) DFTs into smaller (n/4) DFTs
must use n = 2P
for P = integer
n = …, 1024, 2048, 4096, …
provides significantly faster performance over DFT
O[ complete DFT ] = 2 n2
O[ FFT ] = n ln(n)
n = 1024 requires 2,097,152 FLOPS
n = 1024 requires 7,098 FLOPS
What about 3000 data points?
1) discard samples at beginning or end and use n = 2048
2) pad extra values onto the end and use n = 4096
a) pad with zeros
b) pad with last value
3) use non-radix-2 FFT (slower)
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N13
page 9 of 12
Practical considerations of using FFT
1) not all FFT libraries normalize by 2/n
2) most FFT libraries use complex values
3) DFT and FFT are fully reversible
real = Ak
imaginary = Bk
N13
% try_fft.m - example use of FFT
% HJSIII - 13.11.18
clear
% create synthetic data
% 2048 samples at 1000 Hz
n = 2048;
fs = 1000;
% units [Hz]
% 30 Hz sine, +/- 5 V
f = 30;
V = 5;
% units [Hz]
% units [volts]
% optional
% n=2048 and fs=1000 creates incomplete final period
%
spectral resolution misses 30 Hz narrow peak, FFT peaks ~ 3.5 V
%
cleaner phase transitions
% n=2048 and fs=1024 creates integer number of periods
%
spectral resolution hits 30 Hz exactly, FFT peak = 5 V
%
noisier phase transitions
%fs = 1024;
% units [Hz]
% time
h = 1 / fs;
t = [ 0:(n-1) ]' * h;
% units [sec]
% units [sec]
% synthetic sine wave
DC = 0;
x = DC + V * sin( 2 * pi * f * t );
% units [volts]
% optional
% white noise with LP filter
%x = V * randn(n,1);
%x = ( x([ 2:n 1 ]) + 2*x + x([ n 1:(n-1) ]) ) / 4;
% optional
% square wave
%x = DC + V * sign(x);
% units [volts]
% optional
% 100 mV RMS noise
noise = 0.100;
%x = x + noise * randn( size(x) );
% units [volts]
% units [volts]
% optional
% exponential decay
tau = 2.0;
envelope = exp( -3 * t / tau );
%x = x .* envelope;
% FFT
% MATLAB FFT scaled by 2/n - DC component scaled by 1/n
a = fft(x) * 2 / n;
% complex number - units [volts]
a(1) = a(1) / 2;
% units [volts]
amp = abs( a );
% units [volts]
phase = angle( a ) * 180 / pi;
% units [deg]
df = fs / n;
% units [Hz]
freq = [ 0:(n-1) ]' * df;
% units [Hz]
% show results
disp( ' ' )
disp( 'FFT' )
disp( ' freq [Hz] amplitude' )
[ y,i1 ] = max(amp);
i1 = i1 - 5;
i2 = i1 + 10;
disp( [ freq(i1:i2) amp(i1:i2) ] )
% plot time domain
figure( 1 )
% units [Hz] [volts]
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N13
subplot( 3,1,1 )
plot( t,x )
% units [sec] [volts]
axis( [ 0 t(n) -2*V 2*V ] )
xlabel( 'Time (sec)' )
ylabel( 'Voltage' )
title( [ num2str(f) ' Hz, +/- ' num2str(V) 'V, ' ...
num2str(n) ' samples at ' num2str(fs) ' Hz' ] )
% plot amplitude
subplot( 3,1,2 )
plot( freq,amp )
axis( [ 0 freq(n) 0 2*V ] )
xlabel( 'Frequency [Hz]' )
ylabel( 'Amplitude [V]' )
% plot phase
subplot( 3,1,3 )
plot( freq,phase )
axis( [ 0 freq(n) -180 180 ] )
xlabel( 'Frequency [Hz]' )
ylabel( 'Phase [deg]' )
% units [Hz] [volts]
% units [Hz] [deg]
% power spectrum AND power spectral density
ipwr = 0;
if ipwr==1,
figure( 2 )
% power spectrum = amplitude squared / 2
ps = amp .* amp / 2;
% units [volts^2]
% this method is the same
% ps = 4 * fft(x) .* conj(fft(x)) / n / n / 2;
subplot( 3,1,1 )
plot( freq, ps )
% units [Hz] [volts^2]
xlabel( 'Frequency (Hz)' )
ylabel( 'FFT^2 / 2' )
disp( ' ' )
disp( 'FFT^2 / 2' )
disp( ' freq [Hz]
ps' )
[ y,i1 ] = max(ps);
i1 = i1 - 5;
i2 = i1 + 10;
disp( [ freq(i1:i2) ps(i1:i2) ] )
% units [Hz] [volts^2]
% power spectral density = power spectrum / spectral bandwidth
%
for equally spaced spectral lines from FFT, spectral
%
bandwidth = df = fs/n = constant
psd_fft = ps / df;
% units [volts^2.sec]
subplot( 3,1,2 )
plot( freq, psd_fft )
% units [Hz] [volts^2.sec]
xlabel( 'Frequency [Hz]' )
ylabel( 'FFT^2 / 2 / df' )
disp( ' ' )
disp( 'FFT^2 / 2 / df' )
disp( ' freq [Hz]
psd_fft' )
[ y,i1 ] = max(psd_fft);
i1 = i1 - 5;
i2 = i1 + 10;
disp( [ freq(i1:i2) psd_fft(i1:i2) ] )
% MATLAB PSD function
% [Pxx,F] = PSD(X,NFFT,Fs,WINDOW,NOVERLAP)
nfft = 256;
[ px,fx ] = psd( x, nfft, fs, nfft, 0 );
subplot( 3,1,3 )
plot( fx, px )
xlabel( 'Frequency [Hz]' )
% units [Hz] [volts^2.sec]
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N13
ylabel( 'MATLAB PSD output' )
disp( ' ' )
disp( 'MATLAB PSD output' )
disp( ' freq [Hz]
psd' )
[ y,i1 ] = max(px);
i1 = i1 - 5;
i2 = i1 + 10;
disp( [ fx(i1:i2) px(i1:i2) ] )
% bottom of iwpr segment
end
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