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