Andy Buffler Department of Physics University of Cape Town

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1 University of Cape Town Department of Physics PHY014F Vibrations and Waves Part Coupled oscillators Normal modes of continuous systems The wave equation Fourier analysis covering (more or less) French Chapters, 5 & 6 Andy Buffler Department of Physics University of Cape Town 1

2 Problem-solving and homework Each week you will be given a take-home problem set to complete and hand in for marks... In addition to this, you need to work through the following problems in French, in you own time, at home. You will not be asked to hand these in for marks. Get help from you friends, the course tutor, lecturer,... Do not take shortcuts. Mastering these problems is a fundamental aspect of this course. The problems associated with Part are: -, -3, -4, -5, -6, 5-, 5-8, 5-9, 6-1, 6-, 6-6, 6-7, 6-10, 6-11, 6-14 You might find these tougher: 5-4, 5-5, 5-6, 5-7

3 French page 0 The superposition of periodic motions Two superimposed vibrations of equal frequency x = A cos( ω t+ φ ) x = A cos( ω t+ φ ) 0 combination can be written as x= Acos( ω t+ φ) 0 Using complex numbers: ω t A β A 1 + φ 0 1 A φ φ j( 0t 1) z1 = Ae ω + φ 1 j 0t z = Ae ω + φ ( ) z = z1+ z { } 1 z= e A+ Ae 1 j( ω t+ φ ) j( φ φ ) Phase difference Then and φ = φ φ1 A = A + A + AA cos( φ φ ) ( ) Asin β = A sin φ φ 1 3

4 Superposed vibrations of slightly different frequency: Beats If we add two sinusoids of slightly different frequency we observe beats ω1 ω ω1+ ω cosω1t+ cosωt = cos t cos t cosω1t cosωt x 1 x ω1 and ω t French page x 1 +x t cosωt + cosω t 1 T beat = π ω ω 1 ω1 ω cos t 4

5 Combination of two vibrations at right angles x= A cos( ωt+ φ ) y = A cos( ω t+ φ )??? French page 9 Consider case where frequencies are equal and let initial phase difference be φ Write x= A cos( ω t) 1 0 and y = A cos( ω t+ φ) 0 Case 1 : φ = 0 x= A1cos( ω0t) y = A cos( ω t) 0 y A A = x Rectilinear motion 1 Case : φ = π x= A1cos( ω0t) y = A cos( ω t+ π ) = A sin( ω t) 0 0 x y 1 A + A = Elliptical path in clockwise direction 1 5

6 Combination of two vibrations at right angles Case 3 : φ x A cos( ω t) y = A cos( ω t+ π) = A cos( ω t) = π = y = A A 1 x Case 4 : φ 3π x= A cos( ω t) = 1 0 y = A cos( ω t+ 3π ) =+ A sin( ω t) 0 0 x y 1 A + A = Elliptical path in anticlockwise direction 1 Case 5 : φ = π 4 x= A1cos( ω0t) y = A cos( ω t+ π 4) 0 Harder to see use a graphical approach 6

7 Superposition of simple harmonic vibrations at right angles with an initial phase difference of π 4 7

8 Superposition of two perpendicular simple harmonic motions of the same frequency for various initial phase differences. 8

9 Abbreviated construction for the superposition of vibrations at right angles see French page 34. 9

10 Perpendicular motions with different frequencies: Lissajous figures See French page 35. Lissajous figures for ω = ω1 with various initial phase differences. φ = 0 π 4 π 3π 4 π 10

11 ω : ω1 1:1 Lissajous figures 1: 1:3 :3 3:4 3:5 4:5 5:6 φ = 0 π 4 π 3π 4 π 11

12 French page 11 Coupled oscillators When we observe two weakly coupled identical oscillators A and B, we see: x A t x B t these functions arise mathematically from the addition of two SHMs of similar frequencies so what are these two SHMs? These two modes are known as normal modes which are states of the system in which all parts of the system oscillate with SHM 1 either in phase or in antiphase.

13 Coupled oscillators A B x A x B t t 13

14 The double mass-spring oscillator k m x A m x B k Individually we know that For both oscillators: ω 0 = Now add a weak coupling force: mx A = kxa and mx B = kxb k m k m x A k c m x B k For mass A: mx A = kxa + kc( xb xa) k k or xa = ω0 xa +Ω ( xb xa) where ω 0 = c, Ω = m m 14

15 For mass A: For mass B: The double mass-spring oscillator x x x x A = ω0 A +Ω ( B A) x x x x B = ω0 B Ω ( B A) two coupled differential equations how to solve? Adding them: d ( x ) ( ) A + xb = ω0 xa + xb dt Subtract B from A: d ( x ) ( ) ( ) A xb = ω0 xa xb Ω xa xb dt Define two new variables: q1 = xa + xb q = xa xb called normal coordinates Then dq1 dt = ω q 0 1 and dq = ( ω + Ω ) q dt 0 15

16 The double mass-spring oscillator 3 The two equations are now decoupled dt Write dq dt dq1 = ω s f q = ω q 1 ω ω = ω s 0 f = ω0 + Ω s = slow f = fast which have the solutions: q = Ccos( ω t+ φ ) 1 s 1 q = Dcos( ω t+ φ ) f Since q1 = xa + xb and q = xa xb We can write A 1 1 = ( + ) x = ( q q ) x q q 1 and B 1 16

17 Then The double mass-spring oscillator 4 C D xa = cos( ωst+ φ1) + cos( ωft+ φ) C D xb = cos( ωst+ φ1) cos( ωft+ φ) So x A and x B have been expressed as the sum and difference of two SHMs as expected from observation. C, D, φ 1 and φ may be determined from the initial conditions. when x A = x B,then q = 0 there is no contribution from the fast mode and the two masses move in phase the coupling spring does not change length and has no effect on the motion ωs = ω0 when x A = x B,then q 1 = 0 there is no contribution from the slow mode the coupling spring gives an extra force each mass experiences a force ( k+ k c ) x giving k+ k ω c f = m 17 = ω 0 + Ω

18 The double mass-spring oscillator 5 symmetric mode antisymmetric mode mixed mode 18

19 The double mass-spring oscillator 6 We now have a system with two natural frequencies, and experimentally find two resonances. Amplitude Frequency 19

20 Pitch and bounce oscillator French page 17 d k k m L Pitching x A = x B x A Centre of mass stationary τ = I θ 1 1 kd( θd) = ml θ 1 θ = Two normal modes (by inspection): Bouncing x = x A B Restoring force = d x m = kx dt 6kd ml θ x B θ x A kx ω = bounce pitch ω = 6k d m L x B k m 0

21 N = 1π ω1 = ω0sin = ω0 1 ( + ) π ω = ω0sin = 3ω0 1 ( + ) 1

22 N = 3 N = 4

23 French page 136 fixed l N-coupled oscillators Tension T fixed y 1 3 p 1 p p+1 N Each bead has mass m consider transverse displacements that are small. α α p 1 p 1 3 p 1 p p+1 N Transverse force on p th particle: Fp = Tsinαp 1 + Tsinαp y y y y = T + T l l for small α p p 1 p+ 1 p 3

24 p N-coupled oscillators d yp yp yp 1 yp+ 1 yp dt l l F = m = T + T d yp ω ( ) 0yp ω 0 yp+ 1 yp = dt T where ω 0 =, p = 1, N ml a set of N coupled differential equations. Normal mode solutions: y = A sinωt Substitute to obtain N simultaneous equations or ( ) ω ω ( ) 0 Ap ω 0 Ap+ 1 Ap 1 p + + = 0 A + A ω + ω p+ 1 p 1 0 = Ap ω0 p 4

25 N-coupled oscillators 3 From observation of physical systems we expect sinusoidal shape functions of the form A = Csin pθ p Substitute into A + A ω + ω p+ 1 p 1 0 = Ap ω0 And apply boundary conditions A and 0 = 0 A N + 1 = 0 nπ find that θ = n = 1,, 3, N (modes) N + 1 There are N modes: and sin sin pn π ypn = Apn ωnt = Cn sinωnt N + 1 ω n = ω nπ 0 sin 1 ( N + ) 5

26 N-coupled oscillators 4 For small N: ω n = ω nπ 0 sin 1 ( N + ) ω n ω N+1 n 6

27 N-coupled oscillators 5 In many systems of interest N is very large and we are only interested in the lowest frequency modes. ω n ω 0 linear region For n << N : then ω n i.e. ωn n 0 n << N N+1 n nπ nπ sin = 1 1 ( N + ) ( N + ) nπ πω n + 1 N = ω0 = ( N ) for n << N 7

28 N-coupled oscillators 6 N coupled oscillators have N normal modes and hence N resonances response ω 8

29 Continuous systems 9

30 Continuous systems Consider a string stretched between two rigid supports x = 0 tension T String has mass m and mass per unit length x = L µ = ml Suppose that the string is disturbed in some way: x y The displacement y is a function of x and t : yxt (,) 30

31 Normal modes of a stretched string Consider the forces on a small length of string T θ + θ French page 16 θ T y x x + x Restrict to small amplitude disturbances then θ is small and y cosθ = 1 sinθ = tanθ = θ = x The tension T is uniform throughout the string. Net horizontal force is zero: Tcos( θ + θ) Tcosθ = 0 Vertical force: F = Tsin( θ + θ) Tsinθ Then y y F = T x+ x T x x x x 31

32 y y F = T x+ x T x x x Use dg gx ( + x) gx ( ) = dx x Then y F = T x x µ x y y = T x t x or ( ) giving Write Normal modes of a stretched string y µ y = x T t y 1 y = x v t One dimensional wave equation µ: mass per unit length Check: µ T has the dimensions 1 v Then v= T µ is the speed at which a wave propagates along the string see later 3

33 Normal modes of a stretched string 3 Look the standing wave (normal mode) solutions Normal mode: all parts of the system move in SHM at the same frequency Write: yxt (, ) = f( x)cosωt f( x) is the shape function substitute into wave equation y xt d f x (,) () = cos ωt x dx y (,) xt t ( ω ωt) = f( x) cos d f( x) 1 cos t = f( x)cos t dx ω ω ω v which must be true for all t then d f ω = f( x) dx v 33

34 Normal modes of a stretched string 4 d f ω = f( x) dx v which has the same form as the eq. of SHM: d x has general solution: x= Asin( ω0t+ φ) ω Thus we must have: f( x) = Asin x+φ v Apply boundary conditions: y = 0 at x = 0 and x = L dt = ω x 0 f (0) = 0 and f( L ) = 0 ω x = 0, f =0 : 0 = Asin 0 +Φ i.e. Φ=0 v ω ω x = L, f =0 : 0 = Asin L i.e. L= nπ v v n = 1,,3, 34

35 Normal modes of a stretched string 5 nπ v Write ω n = n = 1,,3, L xnπv nπx Therefore f( x) = Ansin = Ansin v L L shape function, or eigenfunction 1 ( π ) f( x) = Asin xl ( π ) f( x) = Asin xl 3 ( π ) f( x) = Asin 3 xl 4 ( π ) f( x) = Asin 4 xl 5 ( π ) f( x) = Asin 5 xl x = 0 x = L n = 1 n = n = 3 n = 4 n = 5 ω1 = πvl ω = πvl ω3 = 3πvL ω4 = 4πvL ω5 = 5πvL 35

36 Normal modes of a stretched string n = 1 n = n = 3 n = 4 36

37 37

38 Normal modes of a stretched string 6 Full solution for our standing waves: nπ x yxt (, ) = Ansin cosωnt L ω = The mode frequencies are evenly spaced: ωn = nω1 n nπ v L ω n ω 3 ω ω 1 (recall the beaded string) n This continuum approach breaks down as the wavelength approaches atomic dimensions also if there is any stiffness in the spring which adds an additional restoring force which is more pronounced in the high frequency modes. 38

39 Normal modes of a stretched string 7 All motions of the system can be made up from the superposition of normal modes nπ x yxt (, ) = Ansin cos( ωnt+ φn) n= 1 L with ω = n nπ v L Note that the phase angle is back since the modes may not be in phase with each other. 39

40 Whispering galleries best example is the inside dome of St. Paul s cathedral. If you whisper just inside the dome, then an observer close to you can hear the whisper coming from the opposite direction it has travelled right round the inside of the dome. 40

41 Longitudinal vibrations of a rod x x+ x section of massive rod French page 170 F 1 x + ξ x F + ξ + x+ ξ section is displaced and stretched by an unbalanced force ξ Average strain = ξ x Average stress = Y x Y : Young s modulus (stress) stress at x+ x = (stress at x) + x x 41

42 Longitudinal vibrations of a rod If the cross sectional area of the rod is α ξ F1 = αy ξ ξ and F x = αy + αy x x x ξ F1 F = αy x = ma x ξ αy x= ρα x x ξ t ξ ρ ξ or = x Y t ξ 1 ξ = x v t v = Y ρ 4

43 Longitudinal vibrations of a rod 3 Look for solutions of the type: ξ( xt, ) = f( x)cosωt ω where f( x) = Asin x+φ v Apply boundary conditions: one end fixed and the other free x = 0 : ξ (0, t) = 0 i.e. Φ= 0 ξ x = L : F = αy = 0 x ωl then cos = 0 v ω 1 or L= ( n ) π n = 1,,3, v 1 1 ( n ) πv ( n ) π Y The natural angular frequencies ωn = = L L ρ 43

44 x = 0 x = L n = 1 ω = 1 π L Y ρ n = ω = 3π L Y ρ n = 3 ω = 3 5π L Y ρ n = 4 n = 5 44

45 Normal modes for different boundary conditions Simply supported Clamped one end Free both ends Clamped both ends n = 1 n = n = 3 45

46 l The elasticity of a gas French page 176 A ρ, p Bulk modulus: K dp = V dv m 1 Kinetic theory of gases: Pressure p = 3 ρvrms = vrms 3Al E 1 If Ek = mvrms then p = k 3A l Now move piston so as to compress the gas work done on gas: W = pa l = Ek Ek l l 5 l p = E ( ) k = pa l p = p 3A l 3Al 3A l 3 l Then ( ) giving p 5 Kadiabatic = V = p V 3 and K v = = ρ p ρ 46

47 French page 174 Sound waves in pipes A sound wave consists of a series of compressions and rarefactions of the supporting medium (gas, liquid, solid) In this wave individual molecules move longitudinally with SHM. Thus a pressure maximum represents regions in which the molecules have approached from both sides, receding from the pressure minima. wave propagation 47

48 Longitudinal wave on a spring 48

49 Standing sound waves in pipes t = 0: Pressure p p 0 Flow velocity u 0 x x t = T : p x u x 49

50 Standing sound waves in pipes Consider a sound wave in a pipe. At the closed end the flow velocity is zero (velocity node, pressure antinode). At the open end the gas is in contact with the atmosphere, i.e. p = p 0 (pressure node and velocity antinode). Open end pressure node p Closed end pressure antinode p 0 u 0 velocity antinode velocity node 50

51 Standing sound waves in pipes 3 Pipe closed at both ends Pipe open at both ends Pipe open at one end L nλ nv = = f nv f = L nπ v ω n = L L ( 1) λ ( 1) n n v = = 4 4 f ( n 1) v f = 4L ( n 1) πv ωn = 51 L

52 Sound Audible sound is usually a longitudinal compression wave in air to which the eardrum responds. Velocity of sound (at NTP) ~ 330 m s -1 By considering the transport of energy by a compression wave, can show that P = π f ρavs m where A is cross sectional area of air column and s m is maximum displacement of air particle in longitudinal wave P Then intensity = = π f ρvsm unit: W m - A 5

53 Sound The human ear detects sound from ~10-1 W m - to ~1 W m - use a logarithmic scale for I : I Intensity level or loudness : β = 10log10 decibels I0 where I 0 = reference intensity = 10-1 W m - 53

54 Musical sounds Waveforms from real musical instruments are complex, and may contain multiple harmonics, different phases, vibrato,... Pitch is the characteristic of a sound which allows sounds to be ordered on a scale from high to low (!?). For a pure tone, pitch is determined mainly by the frequency, although sound level may also change the pitch. Pitch is a subjective sensation and is a subject in psychoacoustics. The basic unit in most musical scales is the octave. Notes judged an octave apart have frequencies nearly (not exactly) in the ratio :1. Western music normally divides the octave into 1 intervals called semitones... which are given note names (A through G with sharps and flats) and designated on musical scales. 54

55 Musical sounds... Timbre is used to denote tone quality or tone colour of a sound and may be understood as that attribute of auditory sensation whereby a listener can judge that two sounds are dissimilar using any criteria other than pitch, loudness or duration. Timbre depends primarily on the spectrum of the stimulus, but also on the waveform, sound pressure and temporal characteristics of the stimulus. One subjective rating scale for timbre (von Bismarck, 1974) dull sharp compact scattered full empty colourful colourless 55

56 Two dimensional systems y French page 181 Consider an elastic membrane clamped at its edges Δx Δy x the membrane has mass per unit area σ, and a surface tension S which gives a force SΔl perpendicular to a length Δl in the surface The forces on the shaded portion are SΔy SΔx SΔy SΔx 56

57 Two dimensional systems If the membrane is displaced from the z = 0 plane then a cross section through the shaded area shows: θ SΔy z SΔy θ + θ x x + x x looks exactly like the case of the stretched string. z The transverse force on the element will be S y x x And if we looked at a cross section perpendicular to this the transverse force will be z S x y y 57

58 Two dimensional systems 3 The mass of the element is σ x y. Thus z z z S y x+ S x y = σ x y x y t or z z σ z + = x y S t a two dimensional wave equation with the wave velocity v = S σ 58

59 Two dimensional systems 4 Look for normal mode solutions of the form: zxyt (,, ) = f( xg ) ( y)cosωt z d f = g y t ( )cosω x dx z d g = ( )cos f x ω t y dy z = f( xgy ) ( ) cos t ( ω ωt) d f dx d g g( y)cos ωt+ f( x)cosωt = dy ω f( xgy ) ( )cosωt v 1 d f 1 d g ω i.e. + = f dx g dy v In a similar fashion to the 1D case, find n1π x nπ y f( x) = An sin 1 and g( y) = Bn sin L x L y 59

60 Two dimensional systems 5 n1πx nπy then zxyt (,, ) = Cnn sin sin cos 1 ωn 1, nt L x L y where the normal mode frequencies are ω n, n 1 1 = + L x L y nπv n πv e.g. for a membrane having sides 1.05L and 0.95L then ωn 1, n πv n1 n = + L

61 Normal modes of a rectangular membrane 1,1 up down,1, 3,1 3, 61

62 Normal modes of a circular membrane 1,0,0 3,0 1,1,1, 6

63 Modes of vibration of a 38 cm cymbal. The first 6 modes resemble those of a flat plate... but after that the resonances tend to be combinations of two or more modes. 63

64 Normal modes of a circular drum 64

65 Chladni plates 65

66 Soap films 66

67 Holographic interferograms of the top and bottom plates of a violin at several resonances. 67

68 Holographic interferograms of a classical guitar top plate at several resonances. 68

69 Holographic interferograms showing the vibrations of a 0.3 mm thick trombone driven acoustically at 40 and 630 Hz. 69

70 Time-average hologram interferograms of inextensional modes in a C 5 handbell 70

71 n v L 3 v L v L 1 v L Normal modes of a square membrane f 4,3 area per point = f n, n 1 v L one point per normal mode 1 nv nv = + L L 0 0 v 1 L v L 3 v L 4 v L 5 v 0 L Normal modes having the same frequency are said to be degenerate n 1 71

72 Normal modes of a square membrane for large n 1 and n n df area = 1 ( π f ) df 4 f area per mode = v L Number of modes with frequencies between f and (f + df) = = 1 4 n 1 ( π f ) df π L f df v L v 7

73 French page 188 Three dimensional systems Consider some quantity Ψ which depends on x, y, z and t, e.g. the density of air in a room. In three dimensions: Ψ Ψ Ψ 1 Ψ + + = x y z v t which can be written: Ψ= 1 v Ψ t The solutions for a rectangular enclosure: ω n, n, n 1 3 nπv n πv n3π v L x L y Lz 1 = + + πv and for a cube: ω n1 n n = n + n + n 3 L,,

74 How many modes are there with frequencies in the range f and (f + df)? Set up an imaginary cubic lattice with spacing v L n f df Three dimensional systems n 1 and consider positive frequencies only. Volume of shell = Volume per mode = 1 (4 π f ) df 8 3 v L n 3 Number of modes with frequencies between f and (f + df) = = 1 8 (4 f ) π 4 π L f v 3 3 df df L v 3 74

75 Three dimensional systems 3 Number of modes with 4 πv f df frequencies between f and (f + df) = 3 v holds for any volume V provided its dimensions are much greater than the wavelengths involved. need to multiply by a factor of when dealing with electromagnetic radiation ( polarization states) Ultraviolet catastrophe for blackbody radiation Equipartition theorem: in thermal equilibrium each mode has an average energy kt in each of its two energy stores 1 B Hence, energy density of radiation in a cavity: 4 πv f df 1 µ df = 3 ( kt ) µ c 8 π f or µ = kt 3 c experiment!? f 75

76 Planck was able to show, effectively by assuming that energy was emitted an absorbed in quanta of energy hf, that the average energy of a cavity mode was not kt but hf hf kt e 1 Then where Planck s constant h = J K -1 µ df = 8 π c f df 3 e hf hf kt Planck s law 1 energy density no. of modes in range f to f +df average energy per mode which agrees extremely well with experiment. 76

77 77

78 French page 189 Introduction to Fourier methods We return to our claim that any physically observed shape function of a stretched string can be made up from normal mode shape functions. f( x) x nπ x i.e. f( x) = Bn sin n= 1 L a surprising claim? first find Bn 1 multiply both sides by sin n π x L and integrate over the range x = 0 to x = L L L n1πx n1πx nπ x f ( x)sin dx = sin Bn sin dx L L n= 1 L

79 Fourier methods L L 1 1 ( )sin sin n sin 0 0 n= 1 nπx nπx nπ x f x dx = B dx L L L If the functions are well behaved, then we can re-order things: L L 1 1 ( )sin n sin sin 0 n= 1 0 nπx nπx nπ x f x dx = B dx L L L [n 1 is a particular integer and n can have any value between 1 and.] L Integral on rhs: ( ) π ( + ) π π n n x n n πx sin sin = cos cos L L L L L n1 x n x 1 dx 1 1 dx

80 Both (n 1 + n) and (n 1 n) are integers, so the functions ( n1 ± n) π x cos L on the interval x = 0 to L must look like from which it is evident that ( ± ) L 1 cos 0 0 n n π x dx = L Fourier methods 3 Except for the special case when n 1 and n are equal then ( ) n n π x = L 1 cos 1 and L 0 cos ( ) n1 n π x dx L L = 80

81 L Fourier methods 4 Thus all the terms in the summation are zero, except for the single case when n 1 = n i.e. L n1π x 1 n1 n πx n1+ n πx f ( x)sin dx = B n cos cos dx 1 L L L L = B 0 0 n 1 ( ) ( ) i.e. L n1π x Bn = f ( x)sin dx 1 L L 0 We have found the value of the coefficient for some particular value of n 1 the same recipe must work for any value, so we can write: L nπ x Bn = f ( x)sin dx L L 0 81

82 Fourier methods 5 The important property we have used is that the functions 1 sin n π x and sin n π x L L are orthogonal over the interval x = 0 to x = L. i.e. L 0 n1π x nπ x dx sin sin L L = 0 if n1 n L if n1 = n Read French pages

83 Fourier methods 6 The most general case (where there can be nodal or antinodal boundary conditions at x = 0 and x = L) is A0 nπx nπx f( x) = + Ancos + Bnsin n= 1 L L where L nπ x An = f ( x)cos dx L L 0 L nπ x Bn = f ( x)sin dx L L 0 83

84 Fourier methods 7 One of the most commonly encountered uses of Fourier methods is the representation of periodic functions of time in terms of sine and cosine functions π Put Ω= T This is the lowest frequency in f() t clearly there are higher frequencies by the same method as before, write A0 πnt πnt f( t) = + Ancos + Bnsin n= 1 T T A 0 = + n= 1 ncos Ω+ nsin Ω T A n t B n t Bn = f t nωt dt T where A f ( t)cos( n t) dt and ( )sin( ) n 0 f() t = Ω T T T 0 84 t

85 Waveforms of... a flute a clarinet an oboe a saxophone 85

86 Fast Fourier transform experiments, 10 March

87 Fast Fourier transform experiments, 10 March

88 Fast Fourier transform experiments, 10 March

89 Fast Fourier transform experiments, 10 March

90 Fast Fourier transform experiments, 10 March

91 Odd functions An odd periodic function f( t) = f() t where T < t < T f() t f() t f() t t t t can be expressed as a sum of sine functions f( t) = Bn sin nωt n= 0 T Bn = f ( t)sin( nωt ) dt T 0 Ω= π T 91

92 Even functions An even periodic function f( t) =+ f() t where T < t < T f() t f() t f() t t t t can be expressed as a sum of cosine functions A f t = + A nωt ( ) 0 n= 1 n cos T An = f ( t)cos( nωt ) dt T 0 Ω= π T 9

93 Fourier methods Example Find Fourier coefficients for the case: f() t 1-1 This is an odd function: f( t) = f() t T Bn = f ( t)sin( nωt ) dt T 0 T T t f( t) = Bn sin nωt = (1) sin ( n t) dt ( 1) sin ( n t) dt T Ω + T Ω 0 T 1 T 1 = (1) cos ( nω t) + ( 1) cos ( nωt) T nω T nω n= 1 0 T 93 T

94 Ω= π T Fourier methods Example cont. 1 1 B = n n n n nπ + nπ [ cos π cos 0] [ cos π cos π ] For even n: cos nπ = cos nπ = 1 Beven = 0 For odd n: cos nπ = 1 and cos nπ = B = odd n ( 1 1) (1 1) nπ + nπ + = nπ n f( t) = sin Ω+ t sin3ω+ t sin5 Ω+ t... π

95 Fourier sums Example 3 terms T T 0 terms 4 terms 50 terms 8 terms 00 terms 95

96 Fourier sums Example 1 terms 0 T 8 terms 3 terms 0 terms 4 terms 50 terms 96

97 Fourier sums Example terms 0 T 0 terms 4 terms 50 terms 8 terms 00 terms 97

98 Time domain Frequency spectrum Fourier transforms t t f 1 7 f 1 f f t f t f t t f1 3 f1 5 f1 7f1 f1 3 f1 5 f1 7f1 f f 98

99 99

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