# MATH348: Advanced Engineering Mathematics

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MATH348: Advanced Engineering MathematicsNori Nakata

Sep. 7, 20121Fourier Series (sec: 11.1)1.1General concept of Fourier Series (10 mins)Show some figures by using a projector. Fourier analysis is a method to decompose afunction into sine and cosine functions. Explain a little bit about Gibbs phenomenon.1.2Who cares? frequency domain (spectral analysis, noise separation, convolution) ODE and PDE1.3Periodic functionWhen the period is p,f (x p) f (x).(1)The period of trigonometric series 2π.1.4Fourier Series (10 mins)f (x) a0 !(an cos nx bn sin nx)n 1where Fourier coeﬃcients a0 , an , and bn are" π1a0 f (x)dx2π π"1 πan f (x) cos nxdxπ π"1 πbn f (x) sin nxdx.π πThe meaning of Fourier coeﬃcients:a0 average of the function in the intervalan amplitude for each cosine wave (frequency)bn amplitude for each sine wave (frequency)I will do the derivation of these coeﬃcients on Sep. 17.—1—(2)

1.5Examples (25 mins)In all examples, π x π.Even functionf (x) #1 if π/2 x π/20 if x π/2, π/2 x(3)ANSWER" π211a0 dx 2π π22 %π" π'π (' π ()1 21 sin nx 21 &2nπan cos nxdx sin n sin n sinπ ππnnπ22nπ2 π22" π'π (' π ()1 2 1 * cos nx π2 1 &bn sin nxdx cosn cosn 0π ππnnπ22 π22f (x) a0 1 21 2 !(an cos nx bn sin nx)n 1 ,!n 1 ,!n 12nπsincos nxnπ2( 1)n 1-2cos(2n 1)x(2n 1)πOdd function 0 x π/2 1 iff (x) 1 if π/2 x 0 0 if x π/2, π/2 x—2—(4)

ANSWER# "" π 2021a0 dx dx 02π π20# "2# % %π 2" π0211sin nx 0sin nx 2an cos nxdx cos nxdx ππnn π20 π20&'()1ππ sin( n ) sin(n ) 0nπ#22 2# % %π 2" 0" π211 cos nx 0 cos nx 2bn sin nxdx sin nxdx ππnn π20 π20&'('()1nπnπ cos 0 cos( ) cos( ) cos 0 0nπ22()1 &'nπ ( 'nπ2 'nπ ( 1 cos( ) cos( ) 1 1 cosnπ22nπ2 !f (x) a0 (an cos nx bn sin nx)n 1 ,!2 'nπ ( 1 cossin nxnπ2n 1Compare these two examples and mention odd and even functions.—3—

MATH348A AEMWorksheet: Fourier SeriesSep. 7, 2012Find the Fourier series of the function f (x) of period p 2π.1)f (x) #1 if 0 x π/20 if x 0, π/2 x2) 1 1f (x) 1 1if π x π/2if π/2 x 0if0 x π/2ifπ/2 x π

MATH348A AEMWorksheet: Fourier Series (Answer)Sep. 7, 20121)" π211dx 2π 04" π1 21nπan cos nxdx sinπ 0nπ2" π'1 21nπ (bn sin nxdx 1 cosπ 0nπ2 &')%!11nπnπ (f (x) sincos nx 1 cossin nx4nπ22a0 n 12)a0 0an 01bn πf (x) 3" π ,!n 1 π2sin nxdx "0 π2sin nxdx "π202 *nπ 1 cos nπ 2 cossin nxnπ2sin nxdx "ππ24sin nxdx 2*nπ 1 cos nπ 2 cosn2

Sep. 10, 20122Functions of any period p 2L (sec: 11.2)2.1Worksheet on Sep. 7 (15 mins)Maybe ask students to solve.2.2Even & Odd Functions (10 mins)Even function (f ( x) f (x)):f (x) a0 '!an cosn 1Lnπ (xL(5)"1a0 f (x)dxL 0"2 Lnπan f (x) cosxdxL 0Le.x., cos x, x2 , 2x 2 x , · · ·Odd function (f ( x) f (x)):f (x) '!bn sinn 12bn L"Lnπ (xLf (x) sin0nπxdxLe.x., sin x, x, 2x 2 x , · · ·(Even) · (Even) (Even)(Even) · (Odd) (Odd)(Odd) · (Odd) (Even)Therefore, we need to compute only the half range.—6—(6)

2.3p 2L (15 mins)Exercise (2) can solve p π rather than p 2π. Introduce x f (x) a0 πLx '!nπnπ (an cosx bn sinxLLof equation 2:(7)n 1" L1f (x)dx2L L"1 Lnπan f (x) cosxdxL LL"1 Lnπbn f (x) sinxdxL LLa0 Explain why we want to solve of period p 2L.2.4Who cares?The period of most of periodic functions is not 2π. If L .2.5Exercises (20 mins)Solve Exercise (2) with p π and get the same answer. Solve Exercise (1) with 5 x 5 (take 10 mins for students to solve it) and get the diﬀerent answer.ANSWERa0 0, an 0" π4 22bn sin 2nxdx (1 cos nπ)π 0nπ !2f (x) (1 cos nπ) sin 2nxnπn 1Calculations with p 2π and p π are the same. ,!2 *nπ f (x) 1 cos nπ 2 cossin nxnπ2n 144sin 2x 0 0 0 sin 6x · · ·π3π !2f (x) (1 cos mπ) sin 2mxmπ 0 m 1 44sin 2x 0 sin 6x · · ·π3π—7—

Exercise (1) with 5 x 5 (p 2L 10):" 111a0 dx 10 010"1 1nπ1nπan cosxdx sin5 05nπ5" 1'1nπ1nπ (bn sinxdx 1 cos5 05nπ5% &'!11nπnπnπ (nπ )f (x) sincosx 1 cossinx10nπ5555n 1—8—

MATH348A AEMWorksheet: Fourier series of period p 2LSep. 10, 2012Find the Fourier series of function f (x) of period p 2L.1)f (x) x, p 62)f (x) x2 , p 6

MATH348A AEMWorksheet: Fourier series of period p 2L (ANSWER)Sep. 10, 20121)a0 0, an 0"2 3nπbn x sinxdx3 035 6%" 323xnπ 33nπ cosx cosxdx3nπ330 nπ056, -2 *293nπ 3 cos nπ sinx3nπnπ306cos nπnπ !!6nπ6nπf (x) cos nπ sinx ( 1)n 1sinxnπ3nπ3 n 1n 1

2)"1 3 2bn 0, a0 x dx 33 0"2 3 2nπan x cosxdx3 035 6%23 2nπ 3nπ3 6 x sinx int0 x sinxdx3nπ3nπ305 6%" 343xnπ 33nπ cosx cosxdxnπnπ330 nπ05 % 64933nπ 3 cos nπ sinxnπnπnπ nπ3036( 1)n(nπ)2 !36nπ36π92πf (x) 3 ( 1)n cosx 3 2 cos x 2 cos x2(nπ)3π3π3 n 1

Sep. 12, 20123Even & Odd Functions. Half-range Expansions3.1In-class exercise (20 mins)3.2Review Even & Odd Functions (10 mins)Even function (f ( x) f (x)) (Fourier cosine series):f (x) a0 '!an cosn 1Lnπ (xL(8)"1a0 f (x)dxL 0"2 Lnπan f (x) cosxdxL 0Le.x., cos x, x2 , 2x 2 x , · · ·Odd function (f ( x) f (x)) (Fourier sine series):f (x) '!bn sinn 12bn L"Lnπ (xLf (x) sin0(9)nπxdxLe.x., sin x, x, 2x 2 x , · · ·Think a little bit aboutf (x) 2x 3x23.3(10)Half-range Expansions (10 mins)We can pretend a function (0 x L) is odd or even functions and find Fourier series.For example, !f (x) Kn sinn 1nπx.L(11)If we assume f (x) is a periodic function ( L x L: we extend a string to L x 0area), we know2Kn L"Lf (x) sin0— 12 —nπxdx.L(12)

Name:MATH348A AEMIn-class Exercise: Even & Odd Functions.Sep. 12, 2012Find the Fourier series of periodic function f (x).1)The period of f (x) is p 2π.f (x) #3 if π x 0 3 if 0 x π

2)The period of f (x) is p 20.#f (x) 3 if 5 x 50 if 10 x 5, 5 x 10

MATH348A AEMIn-class Exercise: Even & Odd Functions. (ANSWER)Sep. 12, 20121)"2 π6bn 3 sin nxdx (cos nπ 1)π 0nπ !6f (x) (cos nπ 1) sin nxnπn 12)" 513a0 3dx 10 02" 51nπ6nπan 3 cosdx sin5 010n2 !36nπnπxf (x) sincos2nπ210n 1

MATH348A AEMWorksheet: Even & Odd Functions. Half-range ExpansionsSep. 12, 20121)Find the Fourier sine and cosine series of function f (x).1.1f (x) 1(0 x 2)1.2f (x) 2 x(0 x 2)2)Find the Fourier series of fuction f (x) of period p 2π.#x π2 if π x 0f (x) x π2 if 0 x π

MATH348A AEMWorksheet: Even & Odd Functions. Half-range Expansions(ANSWER)Sep. 12, 20121.1sine seriesbn "20sinnπ2xdx (1 cos nπ)2nπ !2nπf (x) (1 cos nπ) sinxnπ2n 1cosine series"1 2a0 dx 12 0" 2nπan cosxdx 020f (x) 11.2sine series"2nπxdx20 %" 2 2nπ 2 2nπ (2 x) cosx cosxdxnπ220 nπ04 nπ !4nπf (x) sinxnπ2bn (2 x) sinn 1

cosine series"1 2a0 (2 x)dx 12 0" 2nπan (2 x) cosxdx20, -22 (1 cos nπ)nπ ,!2 2nπf (x) 1 (1 cos nπ) cosxnπ2n 12)"1 π'π( x dx 0π 02"2 π'π(an x cos nxdxπ 02, '%π " π21π(1 x sin nx sin nxdxπn20 n02 2 (1 cos nπ)n π !2f (x) (1 cos nπ) cos nxn2 πn 1,411 cos x 2 cos 3x 2 cos 5x · · ·π35a0

Sep. 14, 20124Fourier-series exercise4.1Review half-range expansion (10 mins)Explain Fourier sine and cosine functions. Connect to physics (string with length L).4.2Review in-class exercise (10 mins) Students forget formula of Fourier series when they stop using. For exam or quiz, they need to remember correctly, but otherwise, they can checka book or Internet, or ask someone. They need to remember when we can do by using Fourier analysis. Solving PDE is for next month. Spectral analysis.Good Students have some sense for Fourier series.Need to improve Practice for integration ”Find the Fourier series” means ”f (x) · · · ”, not a0 , an , and bn . If p 2L, don’t forget cos(nπx)/L (not cos nx).Emphasize what one does by finding Fourier series (draw figures for n 1, 2, 3, · · · ).4.3Fourier-series exercise using a handout (25 mins)— 19 —

MATH348A AEMWorksheet: Fourier SeriesSep. 14, 20121) Find the Fourier series. p 6.f (x) x 1, ( 3 x 3)2) Find the Fourier series. p 2π#f (x) sin x if 0 x π0if π x 03) Find the Fourier series. p 6f (x) 1 x2 , ( 3 x 3)4) Find the Fourier series. p 4f (x) ex , ( 2 x 2)5) π x π. Show !sin nxn 1n #(π x)/2 if 0 x π (π x)/2 if π x 06) Showπ1 1 1 1 ···43 5 7Hint: Find the Fourier series of f (x) 1 in π/2 x π/2 with p 2π.

MATH348A AEMWorksheet: Fourier Series (ANSWER)Sep. 14, 20121)f (x) #x 1 if 0 x 3 x 1 if 3 x 0f (x) is the even function.1a0 32an 3 f (x) 30"350 %31 x21(x 1)dx x 3 220(x 1) cosnπxdx36%" 33nπx 33nπx(x 1) sin sindxnπ3 030 nπ56, -2 *263nπx 3sin nπ cos3 nπnπ3 0, 2 3 2(cos nπ 1)3 nπ6(cos nπ 1)(nπ)2 1 ! 6nπ (cos nπ 1) cosx22(nπ)32 3 "n 1

2)" π111 cos π1a0 sin xdx [ cos x]π0 2π 02π2ππ %π" π" π1111a1 sin x cos xdx sin 2xdx cos 2x 0π 02π 02π20n ̸ 1an b1 n ̸ 1bn f (x) ""1 π1 π sin(n 1)x sin(n 1)xsin cos nxdx dxπ 0π 02 %π1 1 1cos(n 1)x cos(n 1)x2π n 1n 1,- 0 1 cos(n 1)π 1 cos(n 1)π 1 2πn 1n 1, 1 cos nπ 1 cos nπ 1 2πn 1n 11 cos nππ(1 n2 ) %π""1 π 21 π 1 cos 2x111sin xdx dx x sin 2x π 0π 022π220"1 πsin x sin nxdxπ 0" π1(cos(n 1)x cos(n 1)x) dx2π 0 %π111sin(n 1)x sin(n 1)x 02π n 1n 10 !1 11 cos nπ sin x cos nxπ 2π(1 n2 )n 2

3)f (x) is the even function.bn 0"1 3(1 x2 )dx 23 0"2 3nπan (1 x2 ) cosxdx3 033 4%" 323nπ 33nπx2 (1 x ) sinx ( 2x) sindx3 nπ330 nπ0 %" 32 6nπx x sindx3 nπ 033 4%" 34 3nπx 33nπx x cos cosdxnπnπ3 030 nπ ,%4 99 *nπx 3 cos nπ sinnπnπ(nπ)23 0 36 cos nπ(nπ)2 !36nπf (x) 2 cos nπ cosx2(nπ)3a0 n 1

4)1a0 4"817 2e e 24 2 %" 2" 21nπx1 2 * xnπx 22nπxxxan e cosdx e sin e sindx2 222 nπ2 2 nπ 22" 21nπx ex sindxnπ 223 4%" 21 2 xnπx 2 2 xnπx e cos e cosdxnπnπ2 22 2 nπ %" 212nπx2 2 2 cos nπx ( e e ) e cosdxnπnπnπ 22 %" 212 cos nπ2nπx ( e2 e 2 ) ex cosdxnπnπnπ 222 22(e e ) cos nπ (nπ)2 4" 21nπxbn ex sindx2 22 (e2 e 2 )nπ cos nπ (nπ)2 4% e2 e 2 ! 2(e2 e 2 ) cos nπnπx (e2 e 2 )nπ cos nπnπxf (x) cos sin4(nπ)2 42(nπ)2 42n 134 1 ! cos nπ 'nπxnπx ( (e2 e 2 ) 2cos nπsin4(nπ)2 4222ex dx n 15)f (x) #(π x)/2 if 0 x π (π x)/2 if π x 0f (x) is the odd function."2 π11bn (π x) sin nxdx π 0 2n !sin nxf (x) nn 1

6)f (x) 1 ( π/2 xπ/2, p 2π)f (x) is the even function.a0 an 1π2π"π/20"π/20dx 12cos nxdx 2nπsinnπ2 1 ! 2nπ sincos nπ2nπ2n 1,1 211 cos x cos 3x cos 5x · · ·2 π35f (x) If x 0, f (x) 1 and,1 21 11 1 ···2 π3 5π1 1 1 1 ···43 5 7You can use diﬀerent functions for deriving this expression.

Name:MATH348A AEMHomework: Fourier SeriesDue date: Sep. 17, 20121)Find the Fourier series. p 6f (x) x3 , ( 3 x 3)Draw the first four Fourier series (n 1 4).

2)Showπ1 1 1 1 ···43 5 7using the Fourier series of the function#f (x) where p 2π.x if 0 x π,0 if π x 0

MATH348A AEMHomework: Fourier Series (ANSWER)Due date: Sep. 17, 20121)a0 0, an 0"2 3 3nπxbn x sindx3 033 4%" 32 3 3nπx 33nπx2 x cos 3x cosdx3nπ3 0 nπ 0333 44%3" 32 8193x2nπx3nπx cos nπ sin 2x sindx3 nπnπnπ3 0 nπ 03 %" 546 6 3nπx cos nπ x sindxnπnπ nπ 03# 2%" 3 5436 3xnπx 33nπx cos nπ cos cosdxnπ(nπ)2nπ3 030 nπ" 3 5436 9108nπx cos nπ cos nπ cosdx23nπ(nπ) nπ(nπ) 03 % 543241083nπx 3 cos nπ cos nπ sinnπ(nπ)3(nπ)3 nπ3 0324 54(nπ)2cos nπ(nπ)3 !324 54(nπ)2nπxf (x) cos nπ sin3(nπ)3 n 1

2)" π1πa0 xdx 2π 04 %π%" π"1111 πan x cos nxdx x sin nx sin nxdxπ 0π nn 001 2 (cos nπ 1)n "π %π%"1 π1 11 πbn x sin nxdx cos nπ cos nxdxπ 0πnn 00 %1 π1 cos nπ 2 [sin nx]π0π nn1 cos nπn% π ! 11f (x) (cos nπ 1) cos nx cos nπ sin nx4n2 πnn 1π11111f (x) ( 2) cos x ( 1) sin x (0) cos 2x sin 2x ( 2) cos 3x ( 1) sin 3x4 π4π29π31111 (0) cos 4x sin 4x ( 2) cos 5x ( 1) sin 5x16π425π5πx 2ππ1 1 1 ···243 5π1 1 1 ···43 5

MATH348A AEMFeedback to Nori NakataYour comments are so important for me (Nori). Thank you for your participation.What are good points of my teaching? As a teacher, what are my strengths?What have I done as a teacher that has helped you learn in this class?Do you have any suggestion for me to improve my teaching skills? Whatcould I do in this class to better help you understand the material or improveyour skills?

Sep. 17, 20125Derivation of Euler Formulas. Orthogonal Functions5.1Review homework and worksheet (10 mins)5.2Derivation of Fourier seriesI gave students the concept of Fourier series. Here, I teach how to derive the formula.5.3Fourier series '!nπnπ (f (x) a0 an cosx bn sinxLLn 1" L1a0 f (x)dx2L L"1 Lnπan f (x) cosxdxL LL"1 Lnπbn f (x) sinxdxL LLWe need to show a0 , an , and bn can be shown by these expressions.5.4For a0We integrate from L to L on both sides of expression 14,"""nπnπ (x bn sinx dxLL L L L n 1" L" L" L" L ,!nπnπf (x)dx a0dx ancosxdx bnsinxdxLL L L L LLf (x)dx La0 dx 'L !an cosn 1"πcos nxdx 0, π"πsin nxdx 0 πTherefore,"Lf (x)dx 2La0 La0 12L"— 31 —L Lf (x)dx(13)

5.5For an and bnWe first need to show orthogonal functions.5.6Orthogonal functionsShow"Lcosnπmπx sinxdx 0,LL(14)cosnπmπx cosxdx 0, (n ̸ m)LL(15)nπmπx sinxdx 0, (n ̸ m)LL L" L" Lnπ2 nπcosxdx sin2xdx LLL L L(16) L" L L" Lsinwhere n, m 1, 2, 3 · · · .Kronecker delta" L" Lnπmπnπmπcosx cosxdx sinx sinxdx LδmnLLLL L LANSWEREquation 14 Solution 1sin(α β) sin(α β)sin α cos β 2%" L" nπxmπx1 L(m n)πx(m n)πxcossindx sin sindx 0LL2 LLL LSolution 2" LnπxmπxI cossindxLL L %" LLnπxmπx LL Lnπxmπx sinsin sincosdxnπLL LLL L nπ mπ# 2%" LL2 Lnπxmπx LL Lnπxmπx 0 coscos cossindxmnπ 2nπLx LLL L nπ mπ, 2 -2L Imnπ 2I 0— 32 —(17)

Solution 3eix cos x i sin x, e ix cos x i sin xeix e ixeix e ixsin x , cos x 22" L" L'('(nπxnπxmπxmπxnπxmπx1cossindx ei L e i Lei L e i L dxLL4 L L"((n m)πx( n m)πx (n m)πx1 L ' i (n m)πxLLL e ei L ei eidx4 L %L" L(n m)πx(n m)πxLiiLLedx ei(n m)π L L ei(n m)π e (n m)π 2 sin(n m)π 0therefore,"Lcos Lnπxmπxsindx 0LLEquation 15cos(α β) cos(α β)cos α cos β 2" L" ,nπmπ1 L(n m)πx(n m)πxcosx cosxdx cos cosdx 0LL2 LLL LEquation 16 cos(α β) cos(α β)sin α sin β 2" L" ,nπmπ1 L(n m)πx(n m)πxsinx sinxdx cos cosdx 0LL2 LLL LEquation 171 cos 2α1 cos 2αcos2 α , sin2 α 22" L" L2nπx1 cos Lnπcos2xdx dx LL2 L L" L" L1 cos 2nπx2 nπLsinxdx dx LL2 L Lf (x) #sin x if 0 x π0if π x 0— 33 —(18)

5.7For an and bnWe multiply cos(mπx)(L) and integrate from L to L on both sides of expression 14,4" L" L3 '!mπxnπxnπx (mπxf (x) cosdx a0 an cos bn sincosdxLLLL L Ln 1" L" L" L" L ,!mπxmπxnπxmπxnπxmπxf (x) cosdx a0cosdx ancoscosdx bnsincosdxLLLLLL L L L Ln 1" L" L ,!mπxnπxmπxf (x) cosdx ancoscosdxLLL L Ln 1"n m" Lnπxnπxf (x) cosdx ancos2dxLL L L" Lnπxf (x) cosdx LanL L"1 Lnπxan f (x) cosdxL LLLWe multiply sin(mπx)(L) and integrate from L to L on both sides of expression 14,4" L" L3 '!mπxnπxnπx (mπxf (x) sindx a0 an cos bn sinsindxLLLL L Ln 1" L" L" L" L ,!mπxmπxnπxmπxnπxmπxf (x) sindx a0sindx ancossindx bnsinsindxLLLLLL L L L Ln 1" L" L ,!mπxnπxmπxf (x) sindx bnsinsindxLLL L Ln 1"n m" Lnπxnπxf (x) sindx bnsin2dxLL L L" Lnπxf (x) sindx LbnL L"1 Lnπxbn f (x) sindxL LLL— 34 —

Name:MATH348A AEMHomework: Fourier seriesDue date: Sep. 19, 2012Find the Fourier series of function f (x). p 2π and π x π.Hint:9π0f (x) sin x sin x cos nxdx ̸ 0.

MATH348A AEMHomework: Fourier series (ANSWER)Due date: Sep. 19, 2012f (x) is an even function."1 πa0 sin xdxπ 0 %π" π21 1a1 sin x cos xdx cos 2x 0π 0π 20n ̸ 1"2 πan sin x cos nxdxπ 0 " π%1 {sin(n 1)x sin(n 1)x} dxπ 0 2(cos nπ 1) π(n2 1) 2 ! 2(cos nπ 1)f (x) cos nxππ(n2 1)n 2

Sep. 19, 20126Fourier integral transformp 2π(Real) p 2L(Real) p 2L(complex) p (complex)6.1Complex Fourier seriesReal Fourier series '!nπnπ (f (x) a0 an cosx bn sinxLLn 1" L1a0 f (x)dx2L L"1 Lnπan f (x) cosxdxL LL"1 Lnπbn f (x) sinxdxL LLEuler formulaeiθ cos θ i sin θeiθ e iθ2eiθ e iθsin θ 2icos θ f (x) a0 a0 !5n 1 ,!n 1 ,!aneinπxL e 2inπxL bneinπxLinπxL e 2i-6an ibn inπx an ibn inπxe L e L22an ibn inπx a n ib n i( n)πx a0 e L e L22n 1 , 1 ,!!a n ib n inπxan ibn inπxLL a0 e e22n n 1— 37 —

We define cn asc0 a0an ibncn , n 02a n ib n cn , n 02Then f (x) isf (x) c0 !cn einπxL cn einπxLn n 1 ! 1!cn einπxLn What is cn ?c0 an n 012L"Lf (x)dx L5"6inπxinπxinπxinπx" LLan ibn1e L e Le L e Lcn f (x)dx if (x)dx22L22i L L" Linπx1 f (x)e L dx2L Ln 05"6i n πxi n πxi n πxi n πx" LLa n ib n 1e L e Le L e Lcn f (x)dx if (x)dx22L22i L L" Li n πx1 f (x)e L dx2L L" Linπx1 f (x)e L dx2L LTherefore, complex Fourier series isf (x) !cn einπxLn 1cn 2L"Lf (x)e L— 38 —inπxLdx

6.2Fourier transformnπL nππ k LLk We multiply this to the complex Fourier series, thenf (x) L!cn eikx kπ L (partly: just for summation to integral)"L f (x) cn eikx dkπ " 1cn f (x)e ikx dx2L We define F (k) as:2Lcnπ:1 πcn F (k)L 2F (k) : "L1 π f (x) F (k)eikx dkπ L 2 " 1 F (k)eikx dk2π :" 2 1F (k) Lf (x)e ikx dxπ 2L " 1 f (x)e ikx dx2π Fourier transform1F (k) 2π"1f (x) 2π"Inverse Fourier transform f (x)e ikx dx F (k)eikx dk If x is ”time”, k is ”frequency”.— 39 —

6.3time (s) frequency (1/s), space (m) wavenumber (1/m)k nπLTherefore, the unit of k is the unit of 1/x. Fourier analysis is important because wecan understand which frequency waves are including. Then we can analyze predominantfrequency of a structure.6.4Several interesting functions Ricker wavelet— 40 —

Sep. 21, 20127Fourier Integral: spectrumF (ω) ReF (ω) iImF (ω)7.1Amplitude spectrum F (ω) 7.2;F (ω)F (ω)Power spectrum F (ω) 2 F (ω)F (ω)Total energy" 7.3 F (ω) 2 dωPhase spectrumFind the similarity. Sometimes amplitude is not reliable.argF (ω) arctan— 41 —ReF (ω)ImF (ω)

Name:MATH348A AEMIn-class Exercise: Fourier seriesSep. 21, 2012Find the Fourier series of function f (x) in Figure 1. The period of function f (x) is 9 x 9 (Hint: Can you find a shorter period?).30 9 6 30x3Figure 1. Function f (x).69

MATH348A AEMIn-class Exercise: Fourier series (ANSWER)Sep. 21, 2012Find the Fourier series of function f (x) in Figure 1. The period of function f (x) is 9 x 9 (Hint: Can you find a shorter period?).30 9 6 30x36ANSWERPeriod is 6 (p 6) and f (x) is an even function."1 33a0 xdx 3 02" 32nπ6an x cosxdx (cos nπ 1)3 03(nπ)2 3 !6nπf (x) (cos nπ 1) cosx2(nπ)23n 19

Name:MATH348A AEMHomework: Fourier transformDue date: Sep. 24, 20121)Find the Fourier seires of function f (x) ex if π x π and f (x 2π) f (x).2)Find the Fourier transform (F (k)) of function f (x).1#1 if x 1f (x) 0 if x 12f (x) #1 if 0 x 10 if x 0, 1 x

MATH348A AEMWorksheet: Fourier transform (ANSWER)Due date: Sep. 24, 20121)" π1cn ex e inx dx2π π" π1 e(1 in)xdx2π π'(1 e(1 in)π e(1 in)( π)2π(1 in)81 in 7 π inπ π inπ ee ee2π(1 n2 )81 in 7 π π e ecos nπ2π(1 n2 ) !81 in 7 πf (x) e e π cos nπeinx22π(1 n ) 56 !!eπ e π1 in1 inn inx n inx 1 ( 1) e ( 1) e2π1 n21 n2n 1n 156 !?eπ e π( 1)n inx inx 1 (1 in)e (1 in)e2π1 n2n 156 π π!?e e( 1)n inx 1 e e inx ineinx ine inx2π1 n2n 156 π π!e e( 1)n 1 {2 cos nx 2n sin nx}2π1 n2n 12)11F (k) 2π"1 1 ikxe: ikx %11ee ik eik 2i sin k2 sin k dx π k2π ik 1 ik 2π ik 2π

21F (k) 2π"01e ikx dx 1 e ik ik 2π

Sep. 24, 20128ExerciseDelta functionGauss function— 47 —

MATH348A AEMWorksheet: Fourier transformSep. 24, 2012F: Fourier transform, F 1 : Inverse Fourier transform.1)Find the Fourier transform (F (k)) of function f (x).2) 0 x 1 1 iff (x) 1 if 1 x 0 0 if x 1, 1

MATH348: Advanced Engineering Mathematics Nori Nakata. Sep. 7, 2012 1 Fourier Series (sec: 11.1) 1.1 General concept of Fourier Series (10 mins) Show some ﬁgures by using a projector. Fourier analysis is a method to decompose a function into sine and cosine functions. Explain a little bit about Gibbs phenomenon. 1.2 Who cares? frequency domain (spectral analysis, noise separation .

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