Chapter 3 Mathematical Methods Slides to accompany lectures in Vibro-Acoustic Design in Mechanical Systems 0 by D. W. Herrin Department of Mechanical Engineering Lexington, KY 40506-0503 Tel: 859-8-0609 dherrin@engr.uky.edu
Euler s Formula The Power of Complex Numbers θ e j = cosθ + j sinθ Taylor s Series Proof 3 4 x x x f +! 3! 4! 3 4 n ( x) = f ( a) + xf ( a) + f ( 0) + f ( 0) + f ( 0) + f ( 0) θ cosθ =! 3 θ sinθ = θ 3! 4 θ + 4! 5 θ + 5! 6 θ + 6! 7 θ + 7! 3 4 5 e j θ θ θ θ θ = + θj j + +! j +! 3! 4! 5 n x n!
Complex Numbers - Example Addition and Subtraction z = 3 j + z = 5 j z + z = 7 + j z z = 3 4 + j 3
Complex Numbers - Example Multiplication and Division (Rationalization) z = 3 j + z = 5 j z z = 0+ 5j j + 3 = 3 3j + ( + 3 j)( 5 + j) 7 = ( 5 j)( 5 + j) 6 * z zz + 7 j = = * z z z 4
Close and Frederick, 00 Complex Numbers - Review 5
Close and Frederick, 00 Complex Numbers - Review 6
Close and Frederick, 00 Polar Form - Review x = y = z z cosθ sinθ z = x + θ = arctan z = z e jθ y y x 7
Close and Frederick, 00 Complex Number Notations - Review Rectangular Form z = x + jy ( cosθ jsinθ ) z = z + Polar Form z = z θ z = z e jθ 8
Polar Form - Review j j e z j = + = = sin cos π π π j j e z j = + = = 3 sin 3 cos 3 π π π ( ) ( ) ( ) sin cos = + = = π π π j e z j ( ) ( ) ( ) sin cos = + = = π π π j e z j 9
Polar Form - Examples z = 3 j z = 5 j + z = 3e 0.983 j z = e 6 0.97 j π j z z 338e 4 = = 3 + 3 j z z * 0.983 j zz 3e. 8 j 7 + 7 j = = = e = * 0.97 j zz 6e 6 0
Polar Form - Examples z = 3 j z = 5 j + z = 3 56. 3 = 6.3 = 6 348. 7 z z z ( 56.3.3 ) = 338 = 338 45 z z 3 56.3 = = 6 6.3 ( 56.3.3 ) = 67.
Polar Form Complex Conjugate z = 3 j + * z = 3 j z = 3e z = e 0.983 j * 0.983 j 3 z z * 0.983 j + 0.97 j zz 3e 6e. 8 j 7 + 7 j = = = e = * 0.97 j + 0.97 j zz 6e 6e 6
MATLAB Commands r = abs(z) theta = angle(z) x = real(z) y = imag(z) 3
Example Write the harmonically varying force as a complex force. ( ) = ˆF cos( ωt) ( ) = ˆFe jωt = ˆF cos( ωt) + jsin( ωt) F t F t F t ( ) = Re ˆFe jωt ( ) ( ) 4
Example Write the harmonically varying force as a complex force. F ( t) = ˆF cos( ωt) ˆF sin( ωt) F( t) = ( ˆF + j ˆF )e jωt = ˆF + j ˆF ( ) cos ωt ( ) + jsin( ωt) ( ) F ( t) = ˆF cos( ωt) ˆF sin( ωt) + j ˆF sin( ωt) ˆF cos( ωt) ( ) 5
Example 3 Write the harmonically varying force as a complex force. F ( t) = ˆF cos( ωt) ˆF sin( ωt) = ˆF cos( ωt +ϕ) ˆF = ˆF + ˆF ϕ = arctan ( ˆF ) ˆF F( t) = ˆFe jωt+ jϕ = F t ( ) = Re ˆFe jωt ( ) ˆFe jωt 6
Displacement, Velocity and Acceleration v( t) = ˆve j ( ωt+ϕ v) a( t) = d dt v ( t ) = jω ˆve j ( ωt+ϕ v ) = ω ˆve j ( ωt+ϕ v+π /) x t dt = jω ˆve j ωt+ϕ v ( ) = v( t) ( ) = ω ˆve j ( ωt+ϕ v π /) 7
Complex Conjugate F = FF * j( ωt+ϕ ) j( ωt+ϕ = ˆFe ˆFe ) = ˆF ˆF = ˆF Re( F) = ( F + F* ) = ˆF cos ωt ( ) + jsin( ωt) ( ) jsin( ωt) ( ( ) + ˆF ( cos ωt )) = ˆF cos( ωt) 8
Derivation Mechanical Power ( ( )) F ( t) = ˆF cos( ωt +ϕ F ) = Re ˆFe j ωt+ϕ F ( ( )) v( t) = ˆvcos( ωt +ϕ v ) = Re ˆve j ωt+ϕ v W ( t) = F ( t)v( t) = Re F( t) W t W t W t W t W t ( ) = F ( t ) + F( t) * ( ) = 4 F t ( )Re v t ( ( )) ( ) ( v ( t ) + v( t) * ) ( ( )v( t) + F( t) * v( t) * + F( t) * v( t) + F( t)v( t) * ) ( ( ( )v( t) ) + Re( F( t) * v( t) )) ( ) = Re F t ( ( )) + Re ˆFˆve j ( ϕ F ϕ v ) ( ) = Re ˆFˆve j ωt+ϕ F+ϕ v ( ) ( ) ( ) = ˆFˆv cos( ωt +ϕ F +ϕ v ) + cos( ϕ F ϕ v ) 9
Derivation Mechanical Power W t ( ) = ˆFˆv cos( ωt +ϕ F +ϕ v ) + cos( ϕ F ϕ v ) W = T W = T ( ) W ( t )dt = 0 Re( ) Fv* = Re ( F* v) ˆFˆvcos( ϕ F ϕ v ) 0
The Transfer Function H i = a i = output F input Measure using impact testing or shaker. Simulate by using unit force and determining acceleration. a i F
Transfer Functions Measure using impact testing or shaker. Simulate by using unit force and determining acceleration. Unit Force
Assumption Superposition + = F F a i I F a i II a i I + a i II F a i I + a i II = H i F + H i F 3
Assumption Linearity Doubling the force will double the response. a i F a i F 4
Linear Superposition + In Operation F F = + + F 3 F 4 5
Single DOF Systems Equation of Motion: M + x Bx + Kx = f ( t) = F cos( ωt) f ( t) x( t) M K B Total response = Transient Response + Steady State Response Transient response depends on (a) the system parameters (M, B, and K) and (b) the I.C s Steady state response depends on (a) the system parameters and (b) the amplitude and frequency of the forcing function 6
Transient and Steady State Response Transient Total Response Steady State Forcing Function f(t) and x(t) 0 4 6 8 0 x Time (s) ζωnt ( t) = Ae cos( ω t + θ ) + X cos( ωt φ) D 7
Complex Exponential Form M x + B x + Kx = f t x( t) = ˆXe jωt f ( t) = ( ) = ˆF cos( ωt) ˆFe jωt ( Mω + jωb + K) ˆXe jωt = ˆFe jωt f ( t) x( t) M K B ˆX = x t ˆF ( K Mω ) + jωb = ˆF K Mω ( ) = Re jωt ( ˆXe )= " ωb ϕ = arctan$ # K Mω ( ) + ωb ˆF ( ) e jϕ ( ) cos ωt ϕ ( K Mω ) + ( ωb) % ' & 8
The Amplification Factor ω 0 = δ = B M g t K M ( ) = F ( t ) M φ = ˆX ω ˆX = ĝ ( ω 0 ω ) + ( δω) " δω % ϕ = arctan$ ω '+ nπ, n = 0,,,... # ω 0 & The Amplification Factor ( ) ˆX ( ω = 0) = " $ # ( ω ω0 ) % ' & + 4( ) δ ω0 ( ωω0 ) 9
Sinusoidal Steady State Response f ( t) = F cos( ωt) Vibrating System with transfer function H(s) x( t) = X cos( ωt ϕ) Frequency Response Function H ( s) s= jω H ( jω ) ϕ X ϕ = F H ( jω ) ϕ Input, Output.5 0.5 0-0.5 - -.5 f(t) x(t) 0 0.005 0.0 0.05 0.0 Time (s) 30
Frequency Response Functions ( ) = Y ( ω ) X ( ω) H ω Dynamic Flexibility or Receptance Mobility or Mechanical Admittance Accelerance Dynamic Stiffness Mechanical Impedance Acoustic Impedance Specific Impedance H ( ω) = x( ω) F ( ω) H ( ω) = v( ω) F ( ω) A( ω) = a( ω) F ( ω) κ ( ω) = F ( ω) x( ω) Z ( ω) = F ( ω) v( ω) Z ( ω) = p( ω) Q( ω) Z ( ω) = p( ω) u( ω) 3
Frequency Response Functions ( ) = X ( ω ) F ( ω) = H ω ω M + jωmδ + K = ω ω 0 K ( ) + j ωδ ω 0 ( ) δ = ω 0 δ > ω 0 δ < ω 0 δ = 0 Critically Damped Overdamped Underdamped Undamped Often a damping ratio is defined ξ = δ ω 0 3
Dynamic Flexibility Magnitude 33
Dynamic Flexibility Phase 34
Nyquist Diagram 35
Derivation of Loss Factor E kin = M ω x cos ( ωt +ϕ) E pot = K x sin ( ωt +ϕ) E dis = dx F d dx = B dt dx = T = Bx ω cos ( ωt +ϕ)dt 0 η = E dis π max( E pot ) = ωb K = ωδ ω 0 B " $ # dx dt % ' & dt = Bx ω T = Bx ω π ω = Bx ωπ 36
Material Damping Creep Test σ ε σ 0 ε 0 σ 0 J(t) ε t t Relaxation Test ε σ σ 0 ε 0 ε 0 E(t) σ t t 37
Material Damping The Kelvin-Voigt -Parameter Model E R Creep Compliance ε J ( t) = J( t) σ 0 E t ( ) = R t e E Stress Relaxation σ E ( t) = ε 0 E( t) ( t) = E 38
The 3-Parameter Model E 0 R Creep Compliance ( ) ( ) ( ) = = t q q e q p q q t J t J t 0 0 0 0 σ ε E Constants 0 0 0 E E E R q E q E R p + = = = Stress Relaxation ( ) ( ) ( ) = = t p e p q q q t E t E t 0 0 ε 0 σ Material Damping 39
Material Damping For Low Frequency Harmonic Processes the Kelvin-Voigt Model is Applicable E ( ω) E 0 + jωr = E 0 ( + jη) The Material Loss Factor (Steel ~.000) Energy Dissipated per Cycle η ( ω) = W D ωr πu E 0 Maximum Potential Energy 40
Component and Structural Damping Component loss factor differs because stress and distortion conditions are position-dependent. Unless particular measures are taken to increase component damping, damping at the contact surfaces of joints (bolted, riveted, clamped) is the predominant damping mechanism. Damping is due to relative motion of the mating surfaces. This can happen in the form of macro-slip (relative motion across the entire face of the joint), micro-slip (interfacial slip of small areas), and gas pumping losses. η ~ 0.0 4
Cook et al., 00 Structural Damping Damping in structures is normally not viscous. It is due to mechanisms such as hysteresis in the material and slip in connections. In reality, due to computation easiness, damping in structures are approximated by viscous damping. Phenomenological damping methods physical dissipative mechanisms are modeled. Spectral damping methods viscous damping is introduced by means of specified fractions of critical damping. 4
Models for Damped Structures C x M f a (t) K Mx + Cx + Kx = x + ξω x + ω x n n f ( t) = M f ( t) x f jωt ( t) = Xe j t ( t) = Fe ω 43
Models for Damped Structures Viscous Damping Representation x + ξω x + ω x = ω X n + ξω n n M F M ( jωx ) + ω X = F n At Resonance η = ξ Loss Factor Representation Mx + x + K M K ω X ( + jη) x = f ( t) ( + jη) x = f ( t) + jηω X n M + ω X n = M F Used in SEA Codes Used in FEM Codes 44
VDI 3830 Models for Damped Structures [ M ] + x [ C] x + [ K] x = { f ( t) } [ C] =α [ K] + β[ M] a Assuming special damping matrices (which simplify calculation) with viscous damping or with structural damping, the equations of motion can be decoupled by modal transformation. This yields equations of motion for simple damped systems which can be solved in the familiar manner. If this assumption is not justified by the physical damping behavior of a structure, then the coupling effects should be incorporated in the model. 45
Models for Damped Structures Proportional Damping Representation Mx + x + ( αk + βm ) x + Kx = f ( t) ( ) αω + β x + ω x = f ( t) n x + ξω x + ω x = n n n M f M ( t) Relation to Viscous Damping Ratio ξω ξ = n = ( αω + β ) n αωn + β ωn 46
Cook et al., 00 Rayleigh or Proportional Damping Design Spectrum Estimating α and β α = ( ξ ω ξ ω ) ω ω ξ ξ ξ = β αω + ω β = ωω ω ( ξ ω ξ ω ) ω ω ω 47
Structural Vibration at a Single DOF At a single degree of freedom u = ϕ cos ωt ( ) u =displacement of a nodal DOF ϕ =amplitude 48
Structural Vibration for the Entire Structure { u} = { ϕ}cos( ωt) { u} =Vector of nodal displacements { ϕ} =vector of amplitudes for each DOF 49
Mode Shapes The vector of amplitudes is a mode shape φ φ φ 3 φ 4 5 φ 6 φ 7 φ 8 Use subscript i to differentiate mode shapes and natural frequencies { u} = { ϕ} i cos( ω i t) 50
Appropriate Initial Conditions If the I.C.s are a scalar (a) multiple of a specific mode shape then the structure will vibrate in the corresponding mode and natural frequency ICs = a ϕ { } i 5
Determining Natural Frequencies Consider a multi DOF system [ M ]{ u }+ [ K] { u} = { 0} Note that the system is ü Undamped ü Not excited by any external forces 5
Modal Analysis If the system vibrates according to a particular mode shape and frequency { u} = { ϕ} i cos( ω i t) { ϕ} i =mode shape i ω i =natural frequency i 53
Modal Analysis g First derivative (Velocity) u { } = ω i ϕ { } i sin ω i t ( ) g Second derivative (Acceleration) { u } = ω i { φ} cos( ω i i t) 54
Modal Analysis Plug velocity and acceleration into the equation of motion ( " # M $ % ω + " K$ i # %) φ { } = { 0} i Two possible solutions { φ} = { 0} i ( ) = 0 det " # M $ % ω + " K$ i # % 55
The Eigen Problem ( ) = 0 det " # M $ % ω + " K$ i # % The Eigen Problem!A " # $ x { } = λ x { } Eigenvalue Eigenvector λ { x} 56
Modal Analysis An Eigen Problem ( " # M $ % ω + " K$ i # %) φ!k " # $ φ! " { } = ω i M " # $ % { } = 0 # $ φ { }!M " #! " $ K # $ { φ } = ω i { φ} Natural frequencies are eigenvalues The mode shapes are eigenvectors ω i { φ} 57
Modal Analysis An Eigen Problem!M " #! " $ K # $ { φ } = ω i { φ}!a " # $ { x } = λ{ x} Solving the eigen problem!a " # $ { x } λ{ x} = 0 (!A " # $ λ! " I# $ ) x { } = 0 det(! " A # $ λ! " I# $ ) = 0 58
Example Eigen Problem! # " 3 3 '( det )* () $! # &" % $# 3 3 %! # x % # & = λ # i " & '# $# x '# +! - λ 0 $ * i # &, " 0, % = 0 + x x 59
Find Eigenvalues )# det + % * $ λ 3 3 λ &, (. ' = 0 - ( λ) ( λ) 9 = λ 4λ 5 Eigen values λ = λ = 5 60
For Eigen Vector λ = )# + % * $ )# + % * $ 3 3 + 3 3 + & # ( ( ) 0 &,/ % ( ' $ 0. 0 '- &,/ x 3 ( /. 0 4 = 0 '- x 5 x x 0 0 3 4 = 5 3 4 5 / 0 0 0 3 4 5 6
For Eigen Vector '! )# (" 3 3 3 3 $ *- / &,. % + 0/ x x / = 3/ -/. 0/ 0 0 / 3/ x can be any value! # " $# x x % # & = '#!# " $# %# & '# =!# ".707 $#.707 %# & '# 6
For Eigen Vector λ = 5 )" + $ *# )" + $ *# 3 3 % ' 5 & ( ) 5 3 3 5 " $ # 0 0 %,/ '. 0 &- x x %,/ '. 0 &- 3 / 4 = 0 5 x x 3 4 = 5 0 0 3 4 5 / 0 0 0 3 4 5 63
For Eigen Vector (" * $ )# 3 3 3 3 % +. 0 ' -/ &, 0 x x 0 3 = 40. 0 / 0 0 0 0 3 40 x = x x can be any value! # " $# x x % # & = '#!# " $# %# & '# =!# ".707 $#.707 %# & '# 64
Example M = 5 kg M =0 kg K = 800 N/m K = 400 N/m x x K K M M Equations of motions M x + K x K x = 0 M x + K x + K x K x = 0 65
Matrix Form! # # "! # " M 0 0 M 5 0 0 0 $ ' ) &( %*) $ ' & ) ( & %*) x x x +! ) K, + K # x # -) " K K + K + )!, + 800 800 $ ' ) # &( -) " 800 00 %*) $ ' & ) ( & %*) x x x x + ), = -) + ), = -) ') ( *) 0 0 ') ( *) + ), -) 0 0 + ), -) 66
Set Up Eigen Problem!M "!M " #! " $ K # +! det -(,"! " $ K # $ { φ } = ω { i φ} # $! = 60 60 ( " 80 0 60 60 80 0 # ) $ #! ) λ( 0 $ " 0 #. ) 0 $ = 0 / 67
Solve Eigen Problem )# det + % * $ 60 λ 60 80 0 λ &, (. ' = 0 - ω = f = 5.0 π λ = 5.0 rad/s =.7975 Hz " $ # %$ φ φ & $ ' = ($ " $ # %$ 0.7645 0.6446 & $ ' ($ ω = f = 5.965 π λ =5.965 rad/s =.5409 Hz " $ # %$ φ φ & $ ' = ($ " $ # %$ 0.860 0.50 & $ ' ($ 68
Mode 0.7645 0.6446 K K M M 69
Mode -0.860 0.50 K K M M 70
Weighted Orthogonality! # " $# ϕ ϕ T % # & '# M [ ]! # " $# ϕ ϕ % # & = 0 '#! # " $# ϕ ϕ T % # & '# K [ ]! # " $# ϕ ϕ % # & = 0 '# 7
Modal Vector Scaling g Unity Modal Mass! # " $# ϕ ϕ T % # & '# M [ ]! # " $# ϕ ϕ % # & = '# 7
Modal Vector Scaling g Unity Modal Mass! " #.7645X.6446X T $ % & ' ) ( 5 0 0 0 *!," + #.7645X.6446X $ % & = X =.376! # " $# ϕ ϕ % # & = '#! " $.87.4 % & ' 73
Example Vibrating Machine Isolation m =00 kg m = 500 kg κ = 5E6 N/m κ =E6 N/m d v =00 kg/s d v = 00kg/s 74
Eigenvalues and Eigenvectors ω = 40.7 rad/s f = 6.48 Hz ψ =! " # 5.8 $ % & ω = 45.6 rad/s f = 39. Hz ψ = " # $ 0.034 % & ' 75
Forces With / Without Excitation F F x p x p ( t) = ˆF e jωt ( t) = ˆF e jωt ( t) = ˆx p e jωt ( t) = ˆx p e jωt ˆF without = ˆF exc ˆF with = κ ˆx e jωt + jωd vˆx e jωt 76
Simultaneous Differential Equations m ω ˆx p + jω ( d v + d v ) ˆx p + ( κ +κ ) ˆx p jωd v ˆx p κ ˆx p = ˆF exc jωd v ˆx p κ ˆx p m ω ˆx p + jωd v ˆx p +κ ˆx p = 0 77
Simultaneous Differential Equations ( )(( κ +κ ) + jω ( d v + d v ) ω m ) κ + jωd v γ ( ω) = ˆF without = κ + jωd v ω m ˆF with ( κ + jωd v ) κ + jωd v ( ) ( ) 78
The Frequency Spectrum Dynamic signals can be represented in the frequency domain by a frequency spectrum time domain 0 frequency domain Fundamental (first harmonic) Second harmonic amplitude vs. time f (Hz) amplitude vs. frequency 79
Fourier Analysis Used to determine the frequency spectrum of dynamic signals spectrum analyzer hardware y( t) = A o A 0 = T A n = T ( ( ) + B n sin( nωt )) + A n cos nωt n= ( ( )) = A o + C cos nωt ϕ n n T 0 T 0 n= y( t)dt y( t)cos( nωt )dt B n = T y( t)sin( nωt )dt C n = A n + B n ϕ n = tan B n A n T = π/ω Is the period of the signal, ω is the fundamental frequency (first harmonic), ω is the second harmonic, etc. T 0 ( ) 80
A T Example y(t) t A n = T i.e., T 0 B n = T T 0 y t -A ( ) cos( nωt )dt = T Acos( nωt )dt T 0 ( ) y t sin( nωt )dt = # % $ % T = A T sin( nωt ) nω 0 0 Acos( nωt )dt + 4A nπ, n odd 0, n even T T = A # nω sin ) % + n π $ * T C n = A n + B n = B n ϕ n = tan ( B n A n ) = π / 0 & ( A)cos( nωt )dt( '( = 0 T, &. 0( = 0 - ' Phase between A n & B n 8
Example Fourier Series Expansion y 4A π 3 5 ( t) = sin( ωt) + sin( 3ωt ) + sin( 5ωt ) + + sin( nωt) + n 4A π 4A 3π 8
Example Fourier Series Expansion 83
y Example Fourier Line Spectrum 4A π 3 5 ( t) = sin( ωt) + sin( 3ωt ) + sin( 5ωt ) + + sin( nωt) + n time domain frequency domain A T 4A π y(t) -A t ω 4A 3π 4A 5π 3ω 5ω etc. amplitude vs. time amplitude vs.frequency 84
Typical Situation How much is the vibration? Peak = 0.5 mm Peak-to-peak =.04 mm Average = - 8.x0-4 mm rms = 0.5 mm 85
Typical Situation Spectrum reveals several harmonics under 00 Hz, probably related to engine rpm and firing frequency. 86
Digitization of Analog Signals Time signals are sampled using an analog-to-digital converter (a digital logic hardware device) to yield a digitized version of the waveform for further processing. Time (s) Voltage (V) 0.0 0.0 ADC 0.00 0.5 0.00 0.8 0.003 0.8 Analog voltage signal from transducer etc. Voltage values stored in computer memory 87
Sampling Parameters y(t) y(rδt) y(t) Δt f s = sample interval = sample rate = / Δt (samples/sec or Hz) N = total number of samples T = total sample period = r = sampleindex number (,, 3 N) t = rδt NΔt = time of any given sample 88
Discrete Fourier Transform Continuous (analog) form: Discrete (digital) form: The discrete form will yield frequency values up to N/ only. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = = + + = = T n T n n n n o dt t n t y T B dt t n t y T A t n B t n A A t y 0 0 sin cos sin cos ω ω ω ω ( ) ( ) ( ) Δ = = Δ = Δ Δ Δ Δ = = = = N rn t r y N B N n N rn t r y N t t r n t N t y t N A N r n N r N r r n π π π sin,, cos cos 89
Frequencies in Discretely Sampled Signals. Like its analog counterpart, the discrete Fourier series contains frequency components spaced /T apart. This is called the frequency resolution Δf : Δf = /T (Hz) Fourier Coefficients Δf = /T f 90
Frequencies in Discretely Sampled Signals. Signals must be sampled at a minimum rate f s twice the highest frequency of interest (the Nyquist frequency f Nyq ). The highest frequency resolved in a DFT is: f = Nyq fs If this is not done, aliasing error occurs. 9
The Aliasing Phenomenon (a) f s = f (b) f > f s > f The signal f will appear as f/3 in the spectrum (c) f s = f (minimum sample rate) 9
The Aliasing Phenomenon f Nyq f Nyq True spectrum f Measured spectrum f Aliasing results in harmonics above the Nyquist frequency being folded back onto their neighbors. Commercial spectrum analyzers set the sampling frequency at least.5 times the highest frequency of interest. Signal is low-pass filtered before sampling. 93
Non Periodic (transient, random) Signals The Fourier Integral (Transform): F(ω) = f (t) = π f (t)e jωt dt F(ω)e jωt dω F(ω) is continuous, complex lim t f ( t) = 0 f(t) t Time Domain F(ω) ω Frequency Domain 94
The Fourier Integral Example: f(t) A T -A t F(ω) = T / Ae jωt 0 T dt + ( A)e jωt dt = ja ω e jωt / e jωt T / ( ) F(ω) = F(ω)F *(ω) = A ω 6 8cos ( π ω ω o) + cos( π ω ω o ) where ω o = π / T 95
The Fourier Integral Example: 5 F(ω) A /ω o 4 3 Compare with Fourier series results 0 0 3 4 5 6 7 8 9 0 ω/ω o 96
Frequency Response Functions (FRF s) An FRF is the ratio of two Fourier transforms: H (ω) = F (ω) F (ω) What is the FRF between the sound pressures at two points in a duct carrying a +x wave? A x x 97
Frequency Response Functions (FRF s) A x x H (ω) = P (ω) P (ω) = Ae jkx Ae jkx = e jk(x x ) Note: k(x x ) = ωδt where Δt is the propagation time 98