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Fourier AnalysesTime series
,t)1N( t....., ,t2 t,t t,t
,s ....., ,s ,s ,s
0000
1-N210
t
t N
Sampling interval
Total period
Question: How perturbations with different frequencies contribute to the turbulent kinetic energy?
Decomposition illustration
Fourier Transform
a. What is a Fourier Series?
Decompose a single using a series of sine and cos waves
Time-Amplitude domain
Frequency-Amplitude domain
3Hz 10Hz 50Hz
.....)22sin(t)2sin(tsin(t)F(t) 2T:Period
21
T1f:Frequency
1f2:frequencyAngular
)]sin[2(t)2t2sin(sin(2t)F(t) T:Period
1
T1f:Frequency
2f2:frequencyAngular
t)sin(F(t) t)cos(F(t)
How to find frequencies of a signal?
xesin(x)cos(x) ii
...t)sin(b...t)2sin(b sin(t)b
t)cos(a...cos(2t)acos(t)aaF(t)
21
210
00t)sin(bt)cos(aF(t)
0f
ft2f
0
t ececF(t)
ii
(Euler’s formula)
sin(2t)cos(t)F(t)
Discrete Fourier Transform
Observations: N
Sampling interval: t
t NPeriod
tN1
T1
tkt
First harmonic frequency:
... , ..., , ,tN
ntN
2tN
1All frequency:
nth harmonic frequency:tN
nf
1N
0n
tk2tN
nc(n)et)F(kF(t)
i
1N
0n
2Nnk
c(n)eF(k)i
time at kth observation:
1N
0n
2Nnk
c(n)eF(k)i
Time-Amplitude domain F(k)
Frequency-Amplitude domain c(n)
If time series F(k) is known, then, the coefficient c(n) can be found as:
Forward Transform:
1N
0k Nkn2
N1N
0k Nkn2
N1
1N
0k
2N
F(k)
)F(k)sin( )F(k)cos(
ec(n) Nnk
i
i
Example:
Index (k): 0 1 2 3 4 5 6 7 Time (UTC): 1200 1215 1230 1245 1300 1315 1330 1345Q(g/kg): 8 9 9 6 10 3 5 6
1N
0kN1 7A(k)c(0)
i
i
i
03.1 0.28
......)]cos(6)cos(5)cos(3)cos(10)cos(6)cos(9)cos(98[
)A(k)sin( )A(k)cos(c(1)
814
812
810
88
86
84
82
81
1N
0kN
k2N
1N
0kN
k2N1
n 0 1 2 3 4 5 6 7
c(n) 7.0 0.28-1.03i 0.5 -0.78-0.03i 1.0 -0.78+0.03i 0.5 0.28+1.03i
For frequencies greater than 4, the Fourier transform is just the complex Conjugate of the frequencies less than 4.
1N
0k Nkn2
N1N
0k Nkn2
N1
1N
0k
2N
F(k)
)F(k)sin( )F(k)cos(
ec(n) Nnk
i
i
1N
0n
2Nnk
c(n)eF(k)i
c(0) =7.0 c(1)=0.28-1.03i c(2)=0.5 c(3)=-0.78-0.03i c(4)=1.0 c(5)=-0.78+0.03i c(6)=0.5 c(7)=0.28+1.03i
)sin(03.1
)cos(28.0
1
4
4k
k
n
Aliasing
We have ten observations (10 samples) in a second and two different sinusoids that could have produced the samples.
Red sinusoid has 9 cycles spanning 10 samples, so the frequency Blue sinusoid has 1 cycle spanning 10 samples, so the frequency
Hz 9fred
Hz 1fblue Which one is right?
Two data points are required per period to determine a wave.2 observations: 1 wave 4 observations: 2 waves
Two-point rule
If sampling rate is , the highest wave frequency can be resolved is , which is called
sf
2fs Nyquist frequency
Folding occurs at Nyquist frequency.
What problem does folding cause?
Folding
What will cause aliasing or folding?
• The sensor can respond to frequencies higher than the rate that the sensor is sampled.• The true signal has frequencies higher than the sampling rate.
Leakage
Fast Fourier Transform (FFT)
FFT is nothing more than a discrete Fourier transform that has been restructured to take advantage of the binary computation processes of digital computer. As a result, everything is the same but faster!
1N
0n
2 Nnk
c(n)eF(k) i
1N
0k
2N
A(k) Nnk
ec(n)i
1N
(inverse) 0n(forward) 0k
nk ZYX
1N
0kNkn2
N
1N
0kNkn2
N1
1N
0k
Nnk2
NA(k) )A(k)sin( )A(k)cos(ec(n) ii
Relationship between decimal and binary numbers
0 1 2 3 4 5 6 7 8 9 100 1 10 11 100 101 110 111 1000 1001 1010
The decimal numbers n and k can be represented by 1,0n ,n2n j0j
jj
If N=8, then, j=0, 1, 2, 3 0123 nn2n4n8n 7: binary 1 1 1; 5: binary 101; 3: binary 11
1
0k
1
0k
1
0k
)kk2)(4knn2(4n012012
2 1 0
012012 Z)k,k,Y(k )n,n,X(n
Energy Spectrum2
imag2
real2 (n)c(n)c|c(n)|
1-N
1n
21N
0k
2kN
12A |c(n)|)A(A
Note that n starts from 1, because the mean (n=0) does not contribute any information about the variation of the signal.
For frequencies higher than Nyquist frequency, values are identically equal to those at the lower frequencies. They are folded back and added to the lower frequencies.
2|c(n)|
Discrete spectral intensity (or energy)
f2
f2
f2
A
nnfor ,|c(n)|
evenN if ,1n1,....,nfor ,|c(n)|2
oddN if ,n1,....,nfor ,|c(n)|2
n)(E
Example Index (k): 0 1 2 3 4 5 6 7 Time (UTC): 1200 1215 1230 1245 1300 1315 1330 1345Q(g/kg): 8 9 9 6 10 3 5 6
1.0 1.22 0.5 2.28 S(n)
1.0 1.22 0.5 2.28 E(n)
1.14 0.25 0.61 1.0 0.61 0.25 1.14 |c(n)|
1.030.28 0.5 0.030.78 1.0 0.030.78 0.5 03.10.28 7 c(n)
7 6 5 4 3 2 1 0 n
2
iiii