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ComputingFourierSeriesandPower

SpectrumwithMATLAB

ByBrianD.Storey

1.Introduction

frequencyranges.

2.TheMathPart

timedomaintothefrequencydomain. f(t)=a0+1X n=1(ansin(2¼nt)+bncos(2¼nt))(2.1) T

EXPaper

2FourierSeries

end(i.ef(t=0)=f(t=1)). Z 1 0 Z 1 0 Z 1 0 Z 1 0 Z 1 0 oftheequationtheequalitywillhold. Z 1 0Ã f(t)=a0+1X n=0(ansin(2¼nt)+bncos(2¼nt))! sin(2¼mt)dt:(2.7)

2.7willreducetoZ1

0 f(t)sin(2¼nt)dt=an

2:(2.8)

toobtaintheanalogousresult: Z 1 0 f(t)cos(2¼nt)dt=bn

2:(2.9)

interval0FourierSeries3

3.Someexamples

result. 2Z 1 0 sin(2¼t)sin(2¼nt)dt=an:(3.1) 2 Z 1=2 0 sin(2¼nt)dt=an:(3.2)

¡2cos(2¼nt)

2¼n¯

¯1=2

0 =an(3.3) andevaluatingatt=0andt=1=2yields

1¡cos(¼nt)

¼n=an:(3.4)

Thereforewhennisodd,

a n=2

¼n(3.5)

andwhenniseven a n=0(3.6)

2sin(2¼nt)

2¼n¯

¯1=2

0 =bn(3.7) andevaluatingatt=0andt=1=2yields sin(¼nt)

¼n=bn;(3.8)

which,foralln,issimply b n=0(3.9)

4FourierSeries

00.10.20.30.40.50.60.70.80.91-0.2

0 0.2 0.4 0.6 0.8 1 1.2 time f(t) n=20 n=2,000 untilweincludemanypointsintotheseries. thetimeinterval a 0=Z 1=2 0 dt=1=2:(3.10) toconstructthesignalaccurately.

N=2000;

x=[0:100]/100; f=ones(1,101)*1/2; fori=1:2:N a=2/pi/i; f=f+a*sin(2*pi*i*x); end plot(x,f)

4.FourierTransformofDiscreteData

FourierSeries5

00.10.20.30.40.50.60.70.80.91-1

-0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 time f(t)t1t2t3t4t5t6t7t 8A 1A2A3 A 4 A 5A6A 7 up. seeFigure2 A j=f(tj)+f(tj+1)

2(tj+1¡tj)(4.1)

¢t.Thereforethetotalareais

A=¢tµf(t1)+f(t2)

(4.2) whichgeneralizesto Z f(t)dt=¢t0 @f(t1)=2+f(tn)=2+N¡1X j=2f(tj)1

A(4.3)

a j=2sin(2¼ntj)f(tj)](4.4)

6FourierSeries

theintegralRx(N) datapoints. thesquarewave,equation3.5.

Fouriercoe±cientsvs.frequency.

5.TheFFT

FourierSeries7

whatthealgorithmhasgivenyouback. (a)DecipheringtheFFTData numbers(i=p bitaboutcomplexnumbers.

¡1.Acomplex

number,c,istypicallywritten c=a+bi(5.1) cc=(a+bi)(a+bi)=a2¡b2+2abi:(5.2)

¹c=(a¡bi):(5.3)

8FourierSeries

-1-0.8-0.6-0.4-0.200.20.40.60.81-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

0.3 + 0.5 i

-0.1 - 0.8 i real imaginary distanceofthenumberfromtheorigin.

Theabsolutevalueofcisgivenas

jcj=p c¹c=pa2+b2(5.4) numberfromtheorigin.

0.0000

-0.0000-4.0000i

0.0000-0.0000i

0.0000

0.0000+0.0000i

-0.0000+4.0000i

FourierSeries9

a n=¡1

2Nimag(cn);0 b n=1
2Nreal(cn);0 -0.0000 -0.0000+0.0000i
4.0000-0.0000i

0.0000+0.0000i

0.0000

0.0000-0.0000i

4.0000+0.0000i
-0.0000-0.0000i
Nowtrythefollowingcommands
>>t=[0:7]'/8; >>f=sin(2pit); >>f=fft(f); >>f(1) small,butneverexactly0. >>t=[0:7]'/7; >>f=sin(2pit);
10FourierSeries
>>f=fft(f)
Theresultshouldappearas
-0.00000000000000
1.36929808581104-3.30577800969654i
-0.62698016883135+0.62698016883135i -0.50153060757593+0.20774077960317i -0.48157461880753 -0.50153060757593-0.20774077960317i -0.62698016883135-0.62698016883135i
1.36929808581104+3.30577800969654i
properresultback.
6.PowerSpectrum
commands. >>N=8;%%numberofpoints >>t=[0:N-1]'/N;%%definetime >>f=sin(2pit);%%definefunction tainedatafrequencyof1.
FourierSeries11

02468101214161820
10-15 10-10 10-5 100
frequency (Hz) power
N=10000;%%numberofpoints

T=3.4;%%definetimeofinterval,3.4seconds
t=[0:N-1]/N;%%definetime t=tT;%%definetimeinseconds semilogy(freq,p);%%plotonsemilogscale axis([02001]);%%zoomin
TheresultofthesecommandsisshowninFigure4
microphoneandspeaker.
Fs=44100;
y=wavrecord(5Fs,Fs); wavplay(y,Fs);
12FourierSeries

0510152025300
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 frequency (Hz) power timedomain. asinglefrequency.
7.Fouriertransformasa¯lter
beforetakingthe®taddtheline f=f+4*randn(1,N); 5. extractsignalsburiedinanoisybackground.
FourierSeries13
wouldbetheprimarysignalwithnonoise. andplayit. runningthesignalthroughafewresistors.
AppendixA.
startwithZ1 0 sin(2¼nt)sin(2¼mt)dt(A1) and
Thereforewecanwrite

Usingthisexpressionwewritetheintegralas
1 2Z 1 0 whichreducesto 1 2Z 1 0
14FourierSeries

Performingtheintegralyields
1 2µ ¯1
0:(A7)

Evaluatingatzeroandonegivesyou
sin(2¼(n¡m)) weobtain, 1 2Z 1 0 (1¡cos(2¼nt+2¼nt))dt(A9)
Performingtheintegralis
1
2¡sin(4¼tn)4¼n¯
¯1 0=1
2:(A10)

Toobtaintheresultfortheintegral
Z 1 0 cos(2¼nt)cos(2¼mt)dt(A11)
2cos(a)cos(b)=cos(a¡b)+cos(a+b):(A12)

Finallywewillcomputetheintegral
Z 1 0 sin(2¼nt)cos(2¼mt)dt:(A13)
Forthisexpressionweusethetrigidentity

Usingthisexpressionwewritetheintegralas
1 2Z 1 0
Performingtheintegralis
1 2µ ¯1
0:(A16)

Evaluatingatzeroandonegivesyou
cos(2¼(n¡m))¡1
FourierSeries15
weobtain, 1 2Z 1 0 sin(4¼nt)dt=0:(A18) tionoftheFourierseries.quotesdbs_dbs14.pdfusesText_20

[PDF] Computing Fourier Series and Power Spectrum with MATLAB

ComputingFourierSeriesandPower

SpectrumwithMATLAB

ByBrianD.Storey

1.Introduction

2.TheMathPart

EXPaper

2FourierSeries

2.7willreducetoZ1

2:(2.8)

2:(2.9)

3.Someexamples

¡2cos(2¼nt)

2¼n¯

¯1=2

1¡cos(¼nt)

¼n=an:(3.4)

Thereforewhennisodd,

¼n(3.5)

2sin(2¼nt)

2¼n¯

¯1=2

¼n=bn;(3.8)

4FourierSeries

00.10.20.30.40.50.60.70.80.91-0.2

N=2000;

4.FourierTransformofDiscreteData

FourierSeries5

00.10.20.30.40.50.60.70.80.91-1

2(tj+1¡tj)(4.1)

¢t.Thereforethetotalareais

A=¢tµf(t1)+f(t2)

A(4.3)

6FourierSeries

Fouriercoe±cientsvs.frequency.

5.TheFFT

FourierSeries7

¡1.Acomplex

¹c=(a¡bi):(5.3)

8FourierSeries

0.3 + 0.5 i

Theabsolutevalueofcisgivenas

0.0000

0.0000-0.0000i

0.0000-0.0000i

0.0000

0.0000+0.0000i

0.0000+0.0000i

FourierSeries9

4.0000-0.0000i

0.0000+0.0000i

0.0000

0.0000-0.0000i

4.0000+0.0000i

Nowtrythefollowingcommands

10FourierSeries

Theresultshouldappearas

1.36929808581104-3.30577800969654i

1.36929808581104+3.30577800969654i

6.PowerSpectrum

FourierSeries11

02468101214161820

N=10000;%%numberofpoints

T=3.4;%%definetimeofinterval,3.4seconds

TheresultofthesecommandsisshowninFigure4

Fs=44100;

12FourierSeries

0510152025300

7.Fouriertransformasa¯lter

FourierSeries13

AppendixA.

Thereforewecanwrite

Usingthisexpressionwewritetheintegralas

14FourierSeries

Performingtheintegralyields

0:(A7)

Evaluatingatzeroandonegivesyou

Performingtheintegralis

2¡sin(4¼tn)4¼n¯

2:(A10)