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Vibration of single degree of freedom systems y Assoc. Prof. Dr. Pelin Gundes Bakir [email protected] Course Schedule • A short h review i on the h d dynamic i b behaviour h i off the h single i l d degree off ffreedom d systems • A short review on the dynamic behaviour of multi‐degree of freedom structures • Objectives for vibration monitoring • Fourier Series Expansion, Fourier Transforms, Discrete Fourier Transform • Digital signal processing, problems associated with analog‐to‐digital conversion, sampling, aliasing, l leakage, l k windowing, d filters fl • Steps in instrumenting a structure, selection and installation of instruments,maintenance, vibration ib ti iinstrumentation, t t ti exciters, it ttransducers, d performance f specification, ifi ti d data t acquisition i iti systems, strong‐motion data processing Course Schedule • Random R d variables, i bl stochastic h i processes, statistical i i l analysis, l i correlation l i and d convolution, l i coherence, time and frequency domain representation of random dynamic loads • Dynamic response of single and multi degree of freedom systems to random loads • Modal analysis • Applications in bridges, buildings, mechanical engineering and aerospace structures • MATLAB exercises • Term Projects References Modal M d l analysis: l i • Heylen W., Lammens S. And Sas P., ‘Modal Analysis Theory and Testing’, Katholieke Universiteit Leuven, 1997. • Ewins D D.J., J ‘Modal Testing Testing, Theory Theory, Practice Practice, and Application’ (Mechanical Engineering Research Studies Engineering Design Series), Research Studies Pre; 2 edition (August 2001) ISBN‐13: 978‐0863802188 • Maia, N. M. M. and Silva, J. M. M.Theoretical and Experimental Modal Analysis, Research Studies Press Ltd,, Hertfordshire, 1997, 488 pp.,ISBN 0863802087 Signal processing: • Blackburn,, James A,, Modern instrumentation ffor scientists and engineers, g , New York : Springer, 2001 • Stearns S. D. and David, R. A., Signal Processing Algorithms in Matlab, Prentice‐Hall Inc, 1996 • Mitra S.K., ‘Digital Signal Processing’, A Computer based approach, Mc‐Graw Hill, 3rd Edition, 2006. • Heylen W., Lammens S. And Sas P., ‘Modal Analysis Theory and Testing’, Katholieke Universiteit Leuven, 1997. • Keith Worden ‘Signal Processing and Instrumentation’, Lecture Notes, http://www.dynamics.group.shef.ac.uk/people/keith/mec409.htm References Signal processing: • Lynn, P. A. Introductory Digital Signal processing With Computer Applications. John Wiley & Sons, 1994. • Stearns D. D. and David, R. A., Signal Processing Algorithms in Matlab, Prentice‐Hall Inc, 1996 • Ifeachor E.C. and Jervis B.W. Digital Signal Processing: A Practical Approach, Addison‐Wesley, 1997 General vibration theory • Rao S.S., ‘Mechanical vibrations’, Pearson, Prentice Hall, 2004. • Inman D.J., ‘Engineering Vibration’, Prentice Hall, 1994. • Meirovitch L., ‘Fundamentals of vibrations’, McGrawHill, 2001. References R d Random vibrations: ib i • Bendat J.S. and Piersol A.G., ‘Random data analysis and measurement procedures’, Wiley Series in Probabilityy and Statistics,, 3rd Edition,, 2004. • Lutes L.D. and Sarkani S., ‘Random Vibrations: Analysis of structural and mechanical systems’, Elsevier, 631 pp, 2004. • Newland D.E., ’An introduction to random vibrations, spectral and wavelet analysis’, Longman, 1975/1984/1993. • Soong T.T. and Grigoriu, ‘Random vibration of mechanical and structural systems’, Prentice Hall, 1993. • Wirsching P.H. and Paez T.L. and Ortiz K. ‘Random vibrations: Theory and Practice’, John Wiley and sons, 1995. 199 • Bendat J.S. and Piersol A.G. ‘Engineering applications of correlation and spectral analysis’, John Wiley and Sons,, 2nd Edition,, 1993. References Vib i Instrumentation: Vibration I i • Vibration, monitoring, testing and instrumentation handbook, CRC Press, Taylor and Francis, Edited by by: Clarence W W. De Silva Silva. • Aszkler C., ‘Acceleration, shock and vibration sensors’, Sensors handbook, Chapter 5, pages 137 159 137‐159. • McConnell K.G., ‘Vibration testing, theory and practice’, John Wiley and Sons, 1995. • Prerequisites: Basic knowledge on structural analysis. Vibration based health monitoring B i Information: Basic I f ti • Instructor: Assoc. Prof. Dr. Pelin Gundes Bakir (http://atlas.cc.itu.edu.tr/~gundes • Email: [email protected] • Office hours TBD by email appointment • Website: W b it http://atlas.cc.itu.edu.tr/~gundes/lectures • Lecture time: Wednesday 14.00 14 00‐17 17.00 00 • Lecture venue: NH 404 Vibration based health monitoring ‘Vibration based str structural ct ral health monitoring’ is i a multidisciplinary ltidi i li research topic. The course is suitable both for undergraduate and graduate students as well as the following departments: • Civil engineering • Earthquake engineering • Mechanical engineering • Aerospace engineering • Electrical and electronic engineering Vibration based health monitoring Basic Information: • 70 % attendance is required. Grading: • Quiz+homeworks: 35% • Mid‐term project:25% • Final project:40% Introduction •Concepts Concepts from vibrations •Degrees of freedom •Classification of vibration Concepts from vibrations NEWTON’S ’ LAWS First law: If there are no forces acting upon a particle, then the particle will move in a straight line with constant velocity. Second law: A particle acted upon by a force moves so that the force vector is equal to the time rate of change of the linear momentum vector. Third law: When two particles exert forces upon one another, the forces lie along the line joining the particles and the corresponding force vectors are the negative of each other. Definition • The h minimum number b off independent d d coordinates d required d to d determine completely the positions of all parts of a system at any instant of time defines the degree of freedom of the system. A single degree of freedom system requires only one coordinate to describe its position at any instant of time. Single degree of freedom system • FFor the th simple i l pendulum d l in i th the fi figure, the th motion ti can be b stated t t d either ith in i terms t off θ or x and d y. If th the coordinates x and y are used to describe the motion, it must be recognized that these coordinates are not independent. They are related to each other through the relation x2 + y2 = l 2 where l is the constant length of the pendulum. Thus any one coordinate can describe the motion of the pendulum. In this example, we find that the choice of θ as the independent coordinate will be more convenient than the choice of x and y. Two degree of freedom system • SSome examples l off ttwo d degree off freedom f d systems t are shown h iin th the fi figure. Th The fi firstt fi figure shows h a ttwo mass – two spring system that is described by two linear coordinates x1 and x2. The second figure denotes a two rotor system whose motion can be specified in terms of θ1 and θ2. The motion of the system in the third figure can be described completely either by X and θ or by x,y and X. Discrete and continuous systems • A large l number b off practical ti l systems t can be b d described ib d using i a finite fi it number b off d degrees off ffreedom, d such h as the simple system shown in the previous slides. • Some systems, especially those involving continuous elastic members, have an infinite number of degrees of freedom as shown in the figure. Since the beam in the figure has an infinite number of mass points, we need an infinite number of coordinates to specify its deflected configuration. The infinite number of coordinates defines its elastic deflection curve. Thus, the cantilever beam has infinite number of degrees of freedom. Discrete and continuous systems • SSystems t with ith a fi finite it number b off d degrees off ffreedom d are called ll d discrete di t or lumped parameter systems, and those with an infinite number of degrees of freedom are called continuous or distributed systems. • Most of the time, continuous systems are approximated as discrete systems, and solutions are obtained in a simple manner. Although treatment of a system as continuous gives exact results, the analytical methods available for dealing with continuous systems are limited to a narrow selection of problems, such as uniform beams, slender rods and thin hi plates. l • Hence,, most of the p practical systems y are studied byy treatingg them as finite lumped masses, springs and dampers. In general, more accurate results are obtained by increasing the number of masses, springs and dampers‐ that is by increasing the number of degrees of freedom. Classification of vibration • Free vibration: ibration If a system, t after ft an iinitial iti l disturbance di t b iis lleft ft tto vibrate on its own, the ensuing vibration is known as free vibration. No external force acts on the system. The oscillation of a simple pendulum is an example of free vibration. vibration • Forced vibration: If a system is subjected to an external force (often a repeating ti type t off force), f ) the th resulting lti vibration ib ti is i known k as forced f d vibration. – If the frequency of the external force coincides with one of the natural frequencies of the system, a condition known as resonance occurs, and the system undergoes dangerously large oscillations. Failures of such structures as buildings buildings, bridges bridges, turbines turbines, and airplane wings have been assoicated with then occurrence of resonance. Classification of vibration • • • Undamped d d vibration: ib i Iff no energy is lost l or d dissipated d in ffriction or other h resistance during oscillation, the vibration is known as undamped vibration. If any energy is lost in this way however, it is called damped vibration. While the spring p g forms a physical p y model for storingg kinetic energy gy and hence causing vibration, the dashpot, or damper, forms the physical model for dissipating energy and damping the response of a mechanical system. A dashpot p consists of a piston p fit into a cylinder y filled with oil. This piston p is perforated with holes so that motion of the piston in the oil is possible. The laminar flow of the oil through the perforations as the piston moves causes a damping p g force on the piston. p Classification of vibration • Linear vibration: Li ib ti If allll th the b basic i components t of a vibratory system‐the spring, the mass, and the damper, behave linearly, the resulting vibration is known as linear vibration. The differential equations that govern the behaviour of vibratory linear y are linear. Therefore,, the principle p p of systems superposition holds. • Nonlinear vibration: If however, any of the basic components behave nonlinearly, the vibration is called ‘nonlinear vibration’. The differential equations that govern the behaviour of vibratory non‐linear systems are non‐linear. Therefore, the principle of superposition does not hold. Classification of vibration Linear and Li d nonlinear li vibrations ib ti contd: td • The nature of the spring force can be deduced by performing a simple static experiment With no mass attached experiment. attached, the spring stretches to a position labeled as xo=0 in the figure. • As successively more mass is attached to the spring, the force of gravity causes the spring to stretch further. If the value of the mass is recorded, along with the value of the displacement of the end of the spring each time more mass is added, the plot of the force (mass denoted by m, times the acceleration due to gravity, denoted by g), versus this displacement denoted by x, yields a curve similar to that shown in the figure. Classification of vibration Linear and d nonlinear l vibrations b contd: d • Note that in the region of values for x between 0 and about 20 mm,, the curve is a straight line. This indicates that for deflections less than 20 mm and forces less than 1000 N, N the force that is applied by the spring to the mass is proportional to the stretch of the spring. • The constant of proportionality is the slope of the straight g line. Classification of vibration • Deterministic i i i vibration: ib i Iff the h value l or magnitude i d off the h excitation i i (f (force or motion) acting on a vibratory system is known at any given time, the excitation is called ‘deterministic’. The resulting vibration is known as ‘deterministic vibration’. • Nondeterministic vibration: In some cases, the excitation is non‐deterministic or random; the value of excitation at a given time cannot be predicted. In these cases, a large collection of records of the excitation may exhibit some statistical regularity. It is possible to estimate averages such as the mean and mean square values of the excitation. Classification of vibration • • Examples l off random d excitations are wind d velocity, l road d roughness, h and d ground motion during earthquakes. If the excitation is random,, the resultingg vibration is called random vibration. In the case of random vibration, the vibratory response of the system is also random: it can be described only in terms of statistical quantities. quantities Mathematical background •Homogeneous Homogeneous linear ODEs with constant coefficients •Nonhomogeneous N h ODEs ODE Introduction • Th The dynamic d i behaviour b h i off mechanical h i l systems t iis d described ib d b by what h t we call second order Ordinary Differential Equations. • The input to the mechanical structure appears on the right hand side of the equation and is the Force and the solution of the equation gives the output which is usually the displacement. • In order to be able solve these equations, it is imperative to have a solid background g on the solution of homogeneous g and nonhomogeneous Ordinary Differential Equations. • Homogeneous Ordinary Differential Equations represent the ‘Free Free Vibrations’ and the non‐homogeneous Ordinary Differential Equations represent ‘Forced Vibrations’. Homogeneous linear ODEs with constant coefficients • We shall h ll now consider d second‐order d d h homogeneous llinear ODEs whose h coefficients a and b are constant. y′′ + ay′ + byy = 0 • The solution of a first order linear ODE: y′ + ky = 0 • • By separating variables and integrating, we obtain: dy = −kdx y ln y = − ∫ kdx + c * Taking exponents on both sides: y ( x) = ce ∫ − kdx • = ce − kx Let’s tryy the above solution in the first equation. q Usingg a constant coefficient k: λx y=e Substituti Subs u ngg itss de derivative v ves : y′ = λe λx (λ2 + aλ + b)e λx = 0 andd y′′ = λ2 e λx Homogeneous linear ODEs with constant coefficients • Hence iff λ is a solution l off the h important characteristic h equation ((or auxiliary equation) λ2 + aλ + b = 0 • Then the exponential solution y = e λx is a solution of the y′′ + ay′ + byy = 0 • Now from elementary algebra we recall that the roots of this quadratic equation are: 1 λ = (−a + a 2 − 4b) 1 2 1 λ2 = (−a − a 2 − 4b) 2 • The functions below are solutions to y′′ + ay′ + by = 0 y1 = e λ1x and y 2 = e λ2 x Homogeneous linear ODEs with constant coefficients • From algebra l b we know k that h the h quadratic d equation below b l may have h three h kinds of roots: λ2 + aλ + b = 0 • • • Case I: Two real roots if a 2 − 4b > 0 Case II: A real double root if a 2 − 4b = 0 Case III: Complex conjugate roots if a 2 − 4b < 0 • CASE I: In this case case, a basis of solutions of y′′ + ay′ + by = 0 λx λ x in any interval is: y1 = e and y2 = e because y1 and y2 are defined and real for all x and their q quotient is not constant. The corresponding general solution is: 1 2 y = c1e λ1x + c2 e λ2 x Homogeneous linear ODEs with constant coefficients CASE II: Reall double d bl root λ=‐a/2 λ / • If the discriminant a 2 − 4b is zero, we see from 1 2 1 λ2 = (−a − a 2 − 4b) 2 λ1 = (−a + a 2 − 4b) that we get only one root: λ = λ1 = λ2 = − a / 2, hence only one solution : y1 = e −( a / 2) x TTo obtain b i a second d independent i d d solution l i y2 needed d d for f a basis, b i we use the h method of order of reduction. Setting y2 = uy1 , Substituting this and its derivatives y 2 ' = u ' y1 + uy1 ' and y2 ' ' into y′′ + ay′ + by = 0 Homogeneous linear ODEs with constant coefficients CASE II: Reall double d bl root λ=‐a/2 λ / • We have : (u′′y1 + 2u′y1′ + uy1′′) + a(u′y1 + uy1′ ) + buy1 = 0 • Collecting terms u′′y + u′(2 y′ + ay ) + u( y′′ + ay′ + by ) = 0 • This expression in the last paranthesis is zero, since y1 is a solution of 1 1 1 1 1 1 y′′ + ay′ + by = 0 • • 2 y′ = − ae The expression in the second paranthesis is zero too since We are thus left with 1 u ′′y1 = 0 Hence u ′′ = 0 By two integrations u = c1 x + c2 − ax / 2 = −ay1 Homogeneous linear ODEs with constant coefficients CASE II: Reall double d bl root λ=‐a/2 λ / To get a second independent solution y2=uy1, we can simply choose c1=1 and c2=0 and take u=x. Then y2=xyy1. Since these solutions are not proportional, they form a basis. Hence in the case of a double root of λ2 + aλ + b = 0 a basis b i off solutions l i off on any interval is: • y′′ + ay′ + by b =0 e − ax / 2 , xe − ax / 2 The corresponding general solution is: y = (c1 + c2 x)e − ax / 2 Homogeneous linear ODEs with constant coefficients CASE III: Complex roots –a/2+iω a/2+iω and –a/2‐iω a/2 iω • This case occurs if the discriminant of the characteristic equation λ2 + aλ + b = 0 is negative. In this case, the roots of the above equation and thus the solutions of the ODE come at first out complex. However, we show that from them we can obtain a basis of real solutions: y′′ + ay′ + by = 0 y1 = e − ax / 2 cos ω x , where y 2 = e − ax / 2 sin i ωx 1 4 ω 2 = b − a2 • This is proved in the next slides. It can be verified by substitution that these are solutions in the present case. They form a basis on any interval since their quotient cotωx is not constant. Hence, a real general solution l ti iin C Case III iis: y = e − ax / 2 ( A cos ω x + B sin ω x ) ( A , B arbitrary ) Homogeneous linear ODEs with constant coefficients: Proof • Complex l number b representation off h harmonic motion: Since r r X = OP this vector can be represented as a complex number: r X = a + ib r where i = − 1 and a and b denote x and y components of X . Components a and b are also called the real and the imaginary parts of the vector X. If A denotes the modulus or the absolute value of the vector X, and θ denotes r the argument g of the angle g between the vector and the x‐axis,, then X can also be expressed as: Homogeneous linear ODEs with constant coefficients: P f Proof Complex number representation of harmonic motion r r X = OP r X = a + ib r X Homogeneous linear ODEs with constant coefficients: Proof • A apparentt we have As h two t complex l roots. t These Th are: 1 2 1 2 λ1 = a + iω and λ1 = a − iω • We know from basic mathematics that a complex exponential function can be expressed as: r + it r it r e • = e e = e (cos t + i sin t ) Thus the roots of the second order Ordinary Differential Equation can be e λ x = e − ( a / 2) x +iωx = e − ( a / 2) x (cos ωx + i sin ωx) expressed as: 1 e λ2 x = e −( a / 2 ) x −iωx = e −( a / 2) x (cos ωx − i sin ωx) • • We now add these two lines and multiply the result by ½. This gives: y1 = e − ax / 2 cos ω x Then we subtract the second line from the first and multiply the result by 1/2i. This gives: y 2 = e − ax / 2 sin ω x Homogeneous linear ODEs with constant coefficients Case Roots I Distinct real Basis General solution e λ1x , e λ2 x y = c1e λ1x + c2 e λ2 x λ1, λ2 II Real double root λ=‐a/2 III e − ax / 2 , xe − ax / 2 y = (c1 + c2 x)e − ax / 2 Complex conjugate 1 2 1 λ2 = − a − i ω 2 λ1 = − a + iω y1 = e − ax / 2 cos ωx y2 = e − ax / 2 sin ωx y = e − ax / 2 ( A cos ω x + B sin ω x ) Nonhomogeneous ODEs • In this h section, we proceed d from f homogeneous h to nonhomogeneous h ODEs. ′′ ′ y + p ( x) y + q ( x) y = r ( x) • The general solution consists of two parts: y ( x) = yh ( x) + y p ( x) where yh = c1 y1 + c2 y2 is a general solution of the homogeneous ODE. ODE Term in r(x) Choice for yp(x) keγx Ceγx kx n (n = 0,1,2,...) K n x n + K n −1 x n −1 + .... + K1 x + K 0 k cos ωx k sin ωx keαx cos ωx keαx sin ωx K cos ω x + M sin ω x eαx ( K cos ωx + M sin ωx) Nonhomogeneous ODEs Choice rules for the method of undetermined coefficients a) Basic rule: If r(x) is one of the functions in the first column in the Table, choose yp in the same line and determine its undetermined coefficients by substituting yp and its derivatives into y′′ + p ( x) y′ + q ( x) y = r ( x) b) Modification rule: If a term in your choice for yp happens to be a solution of the homogeneous ODE corresponding to the above equation, multiply yyour choice of yp byy x ((or x^2 if this solution corresponds p to a double root of the characteristic equation of the homogeneous ODE) c) Sum rule: If r(x) is a sum of functions in the first column of the table, choose for yp the sum of the functions in the corresponding lines of the second column. Free Vibration of Single Degree of Freedom Systems y •Harmonic Motion •Free vibration of undamped SDOF systems •Free vibration of damped p SDOF systems y Harmonic motion • O Oscillatory ill t motion ti may repeatt it itself lf regularly, l l as in the case of a simple pendulum, or it may display considerable irregularity irregularity, as in the case of ground motion during an earthquake. • If the motion is repeated after equal intervals of time, it is called periodic motion. • The simplest type of periodic motion is harmonic motion. Harmonic motion • Shown in the figure is a vector OP that rotates counterclockwise with g velocityy ω. constant angular • At any time t, the angle that OP makes with the horizontal is θ=ωt. • Let y be the projection of OP on the vertical axis. Then y=A sin ωt. Here y, a function of time is plotted p versus ωt. Harmonic motion • A particle that experiences this motion is said to have harmonic motion. • The maximum displacement of a vibrating body from its equilibrium position is called the amplitude of vibration Amplitude A is shown in vibration. the figure. • Range 2A is the peak to peak displacement. • Now consider the units of θ. Let C be the circumference of the circle shown in the figure. Harmonic motion • Thus C=2πA C=2πA. Or we can write C=Aθ, C=Aθ where θ=2π for one revolution. Thus defined, θ is said to be in radians and is equivalent to 360°. Therefore, one radian is approximately equal to 58.3 58.3°.. • In general, for any arc length, s=A θ , where θ is in radians. It follows that ω in the figure would be in radians per second. • As seen in the figure, the vectorial method of representing harmonic motion requires the description of both the horizontal and vertical components. • The ti Th time ttaken k to t complete l t one cycle l of motion is known as the period of oscillation or time period and is denoted by τ. The period is the time for the motion to repeat (the value of τ in the figure). Harmonic motion • Note that ω τ=2 π where ω denotes the angular velocity of the cyclic motion. The angular velocity ω is also called the circular i l frequency. f • The movement of a vibrating body from its undisturbed or equilibrium position to its extreme position in one direction, then to the equilibrium position, then to its extreme position in the other direction, and back to equilibrium position i i is i called ll d a cycle l off vibration. ib i • One revolution (i.e., angular displacement p of 2π radians)) of the p pin P in the figure or one revolution of the vector OP in the figure constitutes a cycle. Cycle is the motion in one period, as shown in the figure. Harmonic motion • Frequency is the h number b off cycles l per unit time. • The most common unit of time used in vibration analysis is seconds. seconds Cycles per second is called Hertz. • The time the cycle takes to repat itself is the period T. In terms of the period, the frequency is: 1 f = τ ω • The frequency f is related to ω: f = 2π ω = 2πf Harmonic motion • Phase h angle: l Consider d two vibratory b motions denoted d d by: b x1 = A1 sin ωt x2 = A2 sin(ωt + φ ) • These two harmonic motions are called synchronous because they have the same frequency or angular velocity ω. Two synchronous oscillations need not have the same amplitude, and they need not attain their g maximum values at the same time as shown in the figure. Harmonic motion • In this h figure, f the h second d vector OP2 leads l d the h first f one OP1 by b an angle l φ known as the phase angle. This means that the maximum of the second vector would occur φ radians earlier than that of the first vector. These two vectors are said to have a phase difference of φ. Harmonic motion • • From introductory d physics h and d dynamics, d the h fundamental f d l kinematical k l quantities used to describe the motion of a particle are displacement, velocity and acceleration vectors. The acceleration of a particle is given by: dv d 2 x a= = 2 = &x& dt dt • Thus, displacement Thus displacement, velocity velocity, and acceleration have the following relationships in harmonic motion: x = A sin ωt v = x& = Aω cos ωt a = &x& = − Aω 2 sin ωt Operations on harmonic functions • r Using complex l number b representation, the h rotating vector X can b be written as: r where ω denotes the circular frequency (rad/sec) of rotation of the vector X in counterclockwise direction. The differentiation of the harmonics given b the by h above b equation i gives: i • Thus the displacement, velocity and acceleration can be expressed as: Operations on harmonic functions • It can be b seen that h the h acceleration vector leads the velocity vector by 90 degrees and the velocity vector leads the displacement vector by 90 degrees. g Harmonic motion • • Naturall frequency: f Iff a system, after f an initiall disturbance, d b is left l f to vibrate on its own, the frequency with which it oscillates without external forces is known as its natural frequency. As will be seen, a vibratory system having n degrees of freedom will have, in general, n distinct natural frequencies of vibration. Beats: When two harmonic motions, motions with frequencies close to one another, are added, the resulting motion exhibits a phenomenon known as beats. For example if: x1 (t ) = X cos ωt x2 (t ) = X cos(ω + δ )t where δ is a small quantity. The addition of these two motions yield: x(t ) = x1 (t ) + x2 (t ) = X [cos ωt + cos(ω + δ )t ] Harmonic motion Beats: x(t ) = x1 (t ) + x2 (t ) = X [cos ωt + cos(ω + δ )t ] Using the relation ⎛ A+ B ⎞ ⎛ A− B ⎞ cos A + cos B = 2 cos⎜ ⎟ cos⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠ The first equation can be written as: x(t ) = 2 X cos δt δ cos(ω + )t 2 2 Harmonic motion Beats: • It can be seen that the resulting motion x(t) represents a cosine wave δ ω + with frequency q y which is approximately pp y equal q to ω and with a varying y g 2 δt amplitude 2 X cos 2 . Whenever, the amplitude reaches a maximum it is called a beat. • In machines and in structures,the structures the beating phenomenon occurs when the forcing frequency is close to the natural frequency of the system. We will later return to this topic. Harmonic motion • O Octave: t Wh the When th maximum i value l off a range off frequency is twice its minimum value, it is known as an octave band. band • For example example, each of the ranges 75‐150 75 150 Hz, Hz 150‐ 150 300 Hz, and 300‐600Hz can be called an octave band. • In each case case, the maximum and minimum values of frequency, which have a ratio of 2:1, are said to differ byy an octave. Free vibration of undamped SDOF systems • Consider d the h single‐degree‐of‐freedom l d ff d ((SDOF)) system shown h in the h ffigure. The spring is originally in the unstretched position as shown. It is assumed that the spring obeys Hooke’s law. The force in the spring is proportional to displacement with the proportionality constant (spring constant) equal to k. Free vibration of undamped SDOF systems • The h stiffness ff in a spring can b be related l d more directly d l to materiall and d geometric properties of the spring. A spring like behaviour results from a variety of configurations, including longitudinal motion (vibration in the direction of the length), transverse motion (vibration perpendicular to the length), and torsional motion(vibration rotating around the length). Free vibration of undamped SDOF systems • A spring is generally ll made d off an elastic l material.l For a slender l d elastic l material of length l, cross‐sectional area A and elastic modulus E (or Young’s modulus), the stiffness of the bar for vibration along its length is given by: EA k= • l The modulus E has the units of Pascal (denoted Pa) which are N/m2 N/m2. Free vibration of undamped SDOF systems • When h the h mass m (weight ( h W)) is applied, l d the h spring willll deflect d fl to a static equilibrium position δst. • At this position, we find that: W = mg = kδ st • If the mass is perturbed and allowed to move dynamically, the displacement x, measured from the equilibrium position, will be a function of time. Here,, x(t) ( ) is the absolute motion of the mass and the force in the spring can be expresssed as: − k ( x + δ st ) • TTo d determine i the h position i i as a ffunction i off time, i the h equations i off motion i are employed; the free body diagrams are drawn as shown in the figure. Note that x is measured positive downward. Free vibration of undamped SDOF systems • Applying l Newton’s ’ second d law, l W − k ( x + δ st ) = m&x& • But from the static condition condition, note that W=kδst. st Thus, Thus the equation of motion becomes: m&x& + kx = 0 • With the standard form of: • Since in the above equation: k x=0 m • This is Case III that has complex roots where the general solution was computed as: x = e − ax / 2 ( A cos ω t + B sin ω t ) a=0 &x& + 1 4 ω 2 = b − a2 Free vibration of undamped SDOF systems • It is shown h that h x(t ) = A cos ωnt + B sin ωnt where A and B are constants of integration and k (rad / sec) m Here ωn defines the natural frequency of the mass. This is the frequency at which the mass will move regardless of the amplitude of the motion as long as the spring in the system continues to obey Hooke Hooke’ss law law. The natural frequency in Hertz is: ωn = fn = 1 2π k m ( Hz ) Free vibration of undamped SDOF systems • The h initiall conditions d x = xo at t=0 0 andd x& = x& o at t = 0 are usedd to evaluate l the constants of integration A and B. When substituted into the equation x(t ) = A cos ωnt + B sin ωnt we get: A = xo and B = • x&o ωn and consequently x(t ) = xo cos ω n t + The sum in the equation x& o ωn sin ω n t x(t ) = A cos ωnt + B sin ωnt C = A2 + B 2 can also be combined to a p phase shifted cosine with amplitude p and phase angle φ=arctan(B/A). For this purpose let: A = C cos φ and B = C sin φ Free vibration of undamped SDOF systems • Introducing d the h new values l off A and d B into x(t ) = A cos ωnt + B sin ωnt we get: x(t ) = C cos φ cos ωnt + C sin φ sin ωn t • Since • x(t) ( ) can b be expressed d as: • Where φ φ=arctan(B/A) arctan(B/A) and consequently: cos( x − y ) = cos x cos y + sin x sin y x(t ) = C cos(ωnt − φ ) ⎛ x&o ⎞ ⎛ x&o ⎞ 2 2 2 ⎜⎜ ⎟⎟ ⎜ ⎟ = + = + C A B x φ = arctan⎜ and o ⎟ ⎝ ωn xo ⎠ ⎝ ωn ⎠ 2 Free vibration of undamped SDOF systems • The h equation x(t ) = xo cos ω n t + x& o ωn sin i ω n t is a hharmonic ffunction off time. Thus time Thus, the spring‐mass spring mass system is called a harmonic oscillator. oscillator The nature of harmonic oscillation is shown in the figure. If C denotes a vector of magnitude C, which makes an angle ωnt‐φ with respect to the vertical x axis then the solution x(t ) = C cos(ωnt − φ ) can be seen to be the axis, projection of vector C on the x axis. Damping • • • • Undamped U d d and dd damped d vibration: ib ti Th response off a spring‐mass The i model d l predicts di t that the system will oscillate indefinitely. However, everyday observation indicates that most freely oscillating systems eventually die out and reduce to zero motion. Th choice The h i off representative i model d l ffor the h observed b dd decay iin an oscillating ill i system is based partially on physical observation and partially on mathematical convenience. The theory of differential equations suggests that adding a term to equation m&x&(t ) + kx(t ) = 0 of the form cx& , where c is a constant, constant will result in a solution x(t) that dies out. Physical observation agrees fair well with this model and it is used very successfully f ll to t model d l th the d damping i or d decay iin a variety i t off mechanical h i l systems. t This type of damping is called the viscous damping. Damping The h llaminar fl flow off the h oill through h h the h perforations f as the h piston moves causes a damping force on the piston. • The force is p proportional p to the velocityy of the p piston,, in a direction opposite that of the piston motion. This damping force has the form: f c = cx& (t ) where h c iis a constant off proportionality i li related l d to the h oilil viscosity. i i The h constant c, called the damping coefficient, has units of Ns/m, or kg/s. • Damped free vibration of SDOF system • Consider d the h spring‐mass system with h an energy dissipating d mechanism h described by the damping force as shown in the figure. It is assumed that the damping force FD is proportional to the velocity of the mass, as shown; the damping coefficient is c. When Newton’s second law is applied, this model for the damping force leads to a linear differential equation, m&x& + cx& + kx k =0 Damped free vibration of SDOF system • The h ODE is homogeneous h llinear and d has h constant coefficients. ff The h characteristic equation is found by dividing the below equation by m: s2 + • c k s+ =0 m m By the roots of a quadratic equation, we obtain: s1 = −α + β , s2 = −α − β , where c α= 2m • and β = 1 c 2 − 4mk 2m It is now most interesting that depending on the amount of damping (much, medium or little) there will be three types of motion corresponding to the three cases I, II and III. Damped free vibration of SDOF system C Case I Roots c > 4mk 2 Definition fi i i Distinct real roots Overdamping s1, s2 II c 2 = 4mk Real double root Critical damping p g III c 2 < 4mk Complex conjugate roots Underdamping Damped free vibration of SDOF system • Define f the h criticall d damping coefficient ff cc as that h value l off c that h makes k the h radical equal to zero, cc = 2m • Define the damping factor as: ξ= • k = 2mωn m c c = cc 2mωn Introducing the above equation into 2 s1, 2 • • k c ⎛ c ⎞ =− ± ⎜ ⎟ − 2m ⎝ 2m ⎠ m ( ) We find: s1, 2 = − ξ ± ξ 2 − 1 ωn Then the solution can be written as: x(t ) = Ae ⎛⎜ −ζ + ζ 2 −1 ⎞⎟ω t ⎝ ⎠ n + Be ⎛⎜ −ζ − ζ 2 −1 ⎞⎟ω t ⎝ ⎠ n Three cases of damping • Heavy damping when c > cc • Critical damping c = cc • Light damping 0 < c < cc Heavy damping (c > cc or ζ>1) • The h roots are b both h real.l The h solution l to the h differential d ff l equation is: x(t ) = Ae s1t + Be s2t where A and B are the constants of integration. integration Both s1 and s2 will be 2 2 2 negative because α > 0, β > 0, and β = α − k / m < α . Since s1 = −α + β , s2 = −α − β , where α = c 2m and β = 1 c 2 − 4mk 2m Thus, given any initial displacement, the mass will decay to the equilibrium position without vibratory motion. An overdamped system does not oscillate but rather returns to its rest p position exponentially. p y 2 2 x(t ) = Ae ( −ζ + ζ −1 )ωnt + Be ( −ζ − ζ −1 )ωnt Critical damping (c = cc, or ζ=1) • • 1 c 2 − 4mk is zero in this Since β = h case, s1=s2=‐α=‐cc/2m=‐ω / n. 2m −ω t Both roots are equal and the general solution is: x(t ) = ( A + Bt )e . Substitutingg the initial conditions,, x = xo at t=0 and x& = x& o at t = 0 A = xo and B = x&o + ωn xo n and the solution becomes : x(t ) = [xo + ( x&o + ωn xo )t ]e −ωnt • The motion is again not vibratory and decays to the equilibrium position. Light damping (0 < c < cc or ζ<1) • This h case occurs iff the h damping d constant c is so smallll that h c 2 < 4mk imaginary • Then β is no longer real but pure imaginary. 1 k c2 2 4mk − c = β = iω * where ω* = − 2m m 4m 2 • The roots of the characteristic equation are now complex conjugate: s1 = −α + iω*, s2 = −α − iω * with α= • c 2m Hence the corresponding general solution is: x = e − α t ( A cos ω * t + B sin ω * t ) = Ce − α t cos( ω * t − φ o ) where C 2 = A 2 + B 2 and tan φ o = B/A Light damping (0 < c < cc ) • The h solution l can also l be b expressed d as: ( x(t ) = e −ζω nt A cos 1 − ζ 2 ωnt + B sin 1 − ζ 2 ωnt • ) The roots are complex. It is easily shown, using Euler’s formula that the −ξω t general solution is: x(t ) = [C cos(ωd t − φo )]e where C and φ are the constants of integration. The damped natural 2 frequency ωd is given by ωd = ωn 1 − ξ n Light damping (0 < c < cc ) • For the h initiall conditions d x(t = 0) = xo x& (t = 0) = x&o • The equation ( x(t ) = e −ζω nt A cos 1 − ζ 2 ωnt + B sin 1 − ζ 2 ωnt can be expressed as: x(t ) = e where −ζω n t ) ⎛ ⎞ x&o + ζω ζ n xo 2 2 ⎜⎜ xo cos 1 − ζ ωnt + sin 1 − ζ ωnt ⎟⎟ ωd ⎝ ⎠ ζ = c 2mωn Nature of the roots in the complex plane • FFor ζ=0, ζ 0 we obtain bt i th the iimaginary i roots t iωn and ‐iωn and a solution of x(t ) = A cos ωnt + B sin ωnt • For 0<ζ<1, the roots are complex conjugate and are located symmetrically about the real axis. • As the value of ζ approaches 1, both roots approach the point ‐ωn on the real axis. axis • If ζ is greater than 1, both roots lie on the real axis, one increasing and the other decreasing. Logarithmic decrement • The h logarithmic l h d decrement represents the h rate at which h h the h amplitude l d off a free damped vibration decreases. It is defined as the natural logarithm of the ratio of any two successive amplitudes. • Let t1 and t2 denote the times corresponding to two consecutive amplitudes (displacements) measured one cycle apart for an underdamped system as shown in the figure. Logarithmic decrement • ξ t Using x(t ) = [A cos(ωd t − φ )]e −ξω , we can form f the h ratio: n x1 X o e −ζω nt1 cos(ωd t1 − φo ) = x2 X o e −ζω nt2 cos(ωd t 2 − φo ) • But t2=t1+τd , hence Logarithmic decrement • The h llogarithmic h d decrement can b be ffound d ffrom: • For small damping, the above equation can be approximated as: Logarithmic decrement • Figure shows h the h variation off the h logarithmic decrement δ with ζ as shown in the equations: • It can be noticed that for values up to ζ = 0.3, the two curves are difficult to distinguish. Logarithmic decrement • The logarithmic decrement is dimensionless and is actually another form of the dimensionless damping ratio ζ. Once δ is known, ζ can be found by solving: g • If we use instead of we have: Logarithmic decrement • Iff the h damping d in the h given system is not known, k we can determine d it experimentally by measuring any two consecutive displacements x1 and x2. By taking the natural logarithm of the ratio x1 and x2, we obtain δ. By using we can compute the damping ratio ζ. • In fact the damping p g ratio ζ can also be found byy measuringg two displacements separated by any number of complete cycles. If x1 and xm+1 denote the amplitudes corresponding to times t1 and tm+1=t1+mτd where m is an integer integer, we obtain: Logarithmic decrement • Since any two successive displacements d l separated d by b one cycle l satisfy f the h equation: the equation becomes: • The above equation yields which can be substituted into the either of the equations q to obtain the viscous damping ratio ζ: Forced Vibration •Harmonic Harmonic excitation •Base excitation Harmonically excited vibration • • • • A mechanical h l or structurall system is said d to undergo d f forced d vibration b whenever external energy is supplied to the system during vibration. External energy can be supplied to the system through either an applied force or an imposed displacement excitation. The applied force or displacement excitation may be harmonic, nonharmonic but periodic, periodic nonperiodic or random in nature. nature The response of a system to harmonic excitation is called harmonic response. The nonperiodic excitations may have a long or short duration. The response off a d dynamic i system to suddenly dd l applied li d nonperiodic i di excitations i i is called transient response. In this p part of the course,, we shall consider the dynamic y response p of a single degree of freedom system under harmonic excitations of the form F (t ) = Fo ei (ωt +φ ) or F(t) = Fo cos (ωt + φ) or F(t) = Fo sin (ωt + φ) h where Fo is the amplitude, ω is the frequency, and φ is the phase angle of the harmonic excitation. Forced‐Excited System: Harmonic Excitation • Some single l degree d off freedom f d (SDOF) ( ) systems with h an externall force f are shown in the figure. Force can be applied both as an external force F(t), or as a base motion y(t), as shown. The coordinate x(t) is the absolute motion of the mass. The forces W and kδst are ignored in the free‐body diagrams as we know they will add to zero in the equation of motion. Forced‐Excited System: Harmonic Excitation • • • Consider d the h force‐excited f d system off the h figure, f where h the h applied l d force f is F (t ) = Fo sin ωt harmonic, Applying pp y g Newton’s second law,, the equation q of motion becomes: m&x& + cx& + kx = Fo sin ωt The general solution for this second‐order nonhomogeneous linear diff differential i l equation i is i x(t ) = xh (t ) + x p (t ) where xh is the complementary solution or solution to the homogeneous equation. q But this solution dies out soon . Our interest focuses on xp, the particular solution. In vibration theory, the particular solution is also called the steady‐state solution. Forced‐Excited System: Harmonic Excitation • • • The h variations off h homogeneous, particular, l and d generall solutions l with h time for a typical case are shown in the figure. ( ) becomes xp((t)) after some time ((τ It can be seen that xh((t)) dies out and x(t) in the figure). The part of the motion that dies out due to damping (the free vibration part) is called transient. transient The particular solution represents the steady state vibration and is present as long as the forcing function is present. Forced‐Excited System: Harmonic Excitation • The h verticall motions off a mass‐spring system subjected b d to an externall force r(t) can be expressed as: mx′′ + cx′ + kx = r ((t ) • • • Mechanically, this means that at each time instant t, the resultant of the internal forces is in equilibrium with r(t). The resulting motion is called a forced motion with forcing function r(t) r(t), which is also known as the input force or the driving force, and the solution x(t) to be obtained is called the output or the response of the system to the driving force. Of special interest are periodic external forces, and we shall consider a driving force of the form: r (t ) = Fo cos ωt Then we have the nonhomogeneous ODE: m x ′′ + c x ′ + kx = Fo cos ω t Forced‐Excited System: Harmonic Excitation Solving the nonhomogeneous ODE To find yp, we use the method of undetermined coefficients: x p (t ) = a cos ω t + b sin i ωt x ′p (t ) = −ω a sin ω t + ω b cos ω t x ′p′ (t ) = −ω 2 a cos ω t − ω 2 b sin ω t Substituting the above equations into m x ′′ + c x ′ + kx = Fo cos ω t And collecting the cosine and the sine terms, we get [( k − m ω 2 ) a + ω cbb ] cos ω t +[ −ω ca + ( k − m ω 2 )b ] sin ω t = Fo cos ω t The cosine terms on both sides must be equal, and the coefficient of the sine term on the left must be zero since there is no sine term on the right. Forced‐Excited System: Harmonic Excitation • This h gives the h two equations: (k − mω 2 )a + ωcb = Fo − ωca + (k − mω 2 )b = 0 for determining the unknown coefficients a and b. b This is a linear system. system We can solve it by elimination to find: k = ωo m , we obtain: k − mω 2 a = Fo (k − mω 2 ) 2 + ω 2 c 2 ωc b = Fo (k − mω 2 ) 2 + ω 2 c 2 m(ωo2 − ω 2 ) a = Fo 2 2 m (ωo − ω 2 ) 2 + ω 2 c 2 ωc b = Fo 2 2 m (ωo − ω 2 ) 2 + ω 2 c 2 • If we set • We thus obtain the general solution of the nonhomogeneous ODE in the form: x(t ) = xh (t ) + x p (t ) Forced‐Excited System: Harmonic Excitation Case I: Undamped d d forced f d oscillations: ll • If the damping of the physical system is so small that its effect can be neglected g over the time interval considered,, we can set c=0. Then a = Fo m(ωo2 − ω 2 ) m 2 (ωo2 − ω 2 ) 2 + ω 2 c 2 b = Fo ωc m (ω − ω 2 ) 2 + ω 2 c 2 reduces to 2 a= 2 o Fo m(ωo2 − ω 2 ) b=0 Hence becomes x p (t ) = a cos ω t + b sin ω t x p (t ) = Fo Fo = cos cos ωt ω t ω 2 m(ωo2 − ω 2 ) k[1 − ( ) ] ωo Forced‐Excited System: Harmonic Excitation • We thus h have h the h generall solution l off the h undamped d d system as: x(t ) = C cos(ωo t − δ ) + • Fo cos ωt m(ωo2 − ω 2 ) We see that this output is a superposition of two harmonic oscillations of the natural frequency ωo/2π [cycles/sec] of the system, which is the f frequency off the h undamped d d motion i and d the h ffrequency ω/2 / π [cycles/sec] [ l / ] of the driving force. Harmonic motion Beats: • As mentioned before, if the frequency of the forcing function and the frequency of the system are very close to each other, then again beating effect should be expected. • If we for example take the particular soluton: x(t ) = Fo (cos ωt − cos ωot ) 2 2 m(ωo − ω ) (ω ≠ ωo ) which can be rewritten as : x(t ) = • 2 Fo ⎛ ωo + ω ⎞ ⎛ ωo − ω ⎞ t ⎟ sin ⎜ t⎟ sin ⎜ m(ωo2 − ω 2 ) ⎝ 2 2 ⎠ ⎝ ⎠ Since ω is close to ωo, the difference ωo‐ ω is small. Hence the period of the last sine function is large. This is because the greater the quantity under the sine, the smaller the period is. Harmonic motion Beats: Tb = 2TTo 4π = ωo − ω T − To 4π ω + ωo x(t ) = Fo ⎛ ωo + ω ⎞ ⎛ ωo − ω ⎞ sin t ⎟ sin ⎜ t⎟ ⎜ 2 2 m(ωo − ω ) ⎝ 2 ⎠ ⎝ 2 ⎠ Harmonic motion Beats: This phenomenon is also frequently observed in lightly damped sytems with close coupling of the torsional and translational frequencies. A typical example is the thirteen storey steel framed Santa Clara County Office Building as reported in : Celebi and Liu, ‘Before and after retrofit‐ response of a building during ambient and strong motions’, Journal of Wind Engineering and Industrial Aerodynamics, 77&78 (1998) 259‐268. The proximity of the torsional frequency at 0.57 Hz to the translational frequency at 0.45 Hz causes the observed coupling and beating effect in this structure. Harmonic motion Comparison of the frequencies of the building Thirteen storey steel framed Santa Clara County Office Building as reported in : Celebi and Liu, ‘Before and after retrofit‐ response of a building during ambient and strong motions’, Journal of Wind Engineering and Industrial Aerodynamics Aerodynamics, 77&78 (1998) 259 259‐268. 268 Resonance • If the th damping d i off th the system t is i so smallll that th t its it effect ff t can be b neglected l t d over the time interval considered, we can set c=0. Then the particular response can be expressed by: xp = • Fo Fo cos ωt = cos ωt 2 m(ω o − ω ) ⎡ ⎛ ω ⎞2 ⎤ k ⎢1 − ⎜⎜ ⎟⎟ ⎥ ⎢⎣ ⎝ ωo ⎠ ⎥⎦ 2 Puttingg cosωt=1,, we see that the maximum amplitude p of the p particular solution is: F ao = o k ρ where ρ = If • 1 2 ⎛ω ⎞ 1 - ⎜⎜ ⎟⎟ ⎝ ωo ⎠ ω → ωo , then ρ and ao tend to infinity. This excitation of large oscillations by matching input and natural frequencies is called resonance. Resonance F In the h case off resonance and d no d damping, the h ODE &x& + ωo2 x = o cos ωot m becomes: m&x& + cx& + kx = Fo cos ωt • Then from the modification rule, rule the particular solution becomes: • x p (t ) = t (a cos ωot + b sin ωot ) • By substituting this into the second equation, we find: x p (t ) = • Fo t sin ωot 2mωo We see that because of the factor t, the amplitude becomes larger and larger Practically speaking, larger. speaking systems with very little damping may undergo large vibrations that can destroy the system. Forced‐Excited System: Harmonic Excitation Case II: Damped d forced f d oscillations: ll Iff the h damping d off the h mass‐spring system is not negligibly small, we have c>0 and a damping term cx’ in m&x& + cx& + kx = r ((t ) m&x& + cx& + kx = 0 Then the general solution yh of the homogeneous ODE approaches zero as t goes to infinity. Practically, it is zero after a sufficiently long time. Hence the transient solution given by y (t ) = yh (t ) + y p (t ) approaches pp the steadyy state solution yp. This p proves the following: g Steady state solution: After a sufficiently long time, the output of a d damped d vibrating ib i system under d a purely l sinusoidal i id l driving d i i force f will ill practically be a harmonic oscillation whose frequency is that of the input. Response of a damped system under harmonic force • Iff the h forcing f function f is given by b becomes: the h equation off motion • The particular solution is also expected to be harmonic; we assume it in the following form: where X and φ are constants to be determined that denote the amplitude phase angle g of the response, p , respectively. p y Byy substitutingg the and the p second equation into the first: • U i the Using h trigonometric i i relations l i b below l iin the h above b equation i Response of a damped system under harmonic force • We obtain: b • If we solve the above equation equation, we find: • If we insert the above into the particular solution. Using: , we find the r= ω ωn Response of a damped system under harmonic force • We obtain: b Forced‐Excited System: Harmonic Excitation • The h quantity M=X/δ /δst is known k as the h magnification f factor, f amplification lf factor or the amplitude ratio. The amplitude of the forced vibration becomes smaller with increasing values of the forcing frequency (that is, M 0 as r ∞) Xo 1 = Fo / k (1 − r 2 ) 2 + (2ξr ) 2 Forced‐Excited System: Harmonic Excitation • A fact f that h creates d difficulty ff l ffor d designers is that h the h response can b become large when r is close to 1 or when ω is close to ωn. This condition is called resonance. The reduction in M in the presence of damping is very significant at or near resonance. Forced‐Excited System: Harmonic Excitation • For an undamped d d system (ζ (ζ=0), ) the h phase h angle l is zero for f 0<r<1 and d 180 degrees for r>1. This implies that the excitation and response are in phase for 0<r<1 and out of phase for r>1 when ζ=0. ⎛ 2ζr ⎞ 2 ⎟ ⎝1− r ⎠ φ = arctan⎜ Forced‐Excited System: Harmonic Excitation • For ζ>0 ζ and d 0<r<1, the h phase h angle l is given by b 0<φ<90, φ implying l that h the h response lags the excitation. ⎛ 2ζr ⎞ 2 ⎟ ⎝1− r ⎠ φ = arctan⎜ Forced‐Excited System: Harmonic Excitation • For ζ>0 ζ and d r>1, the h phase h angle l is given by b 90<φ<180, φ implying l that h the h response leads the excitation. ⎛ 2ζr ⎞ 2 ⎟ ⎝1− r ⎠ φ = arctan⎜ Response p of a damped p system y under F (t ) = Fo e iωt • Using complex l algebra, l b let l the h harmonic h force f be: b F (t ) = Fo eiωt where Fo is a real constant and i is the imaginary unit. unit Assume that the response has the same frequency as the force, but is, in general, out of phase with the force ~ x(t ) = X o e i (ωt +φ ) = Xe iωt ~ where Xo is the amplitude of the displacement and X is the complex displacement, p , ~ iφ X = X oe Substituting this into the differential equation of motion ~ (−mω 2 + icω + k ) Xeiωt = Fo eiωt Forced‐Excited System: Harmonic Excitation • Define f the h transfer f function f ((or ffrequency response function) f ) H(ω) ( ) as the h complex displacement due to a force of unit magnitude (Fo=1). Thus, H (ω ) = • 1 (k − mω 2 ) + icω Rationalizing, the transfer function becomes: (k − mω 2 ) − icω H (ω ) = (k − mω 2 ) 2 + (cω ) 2 • This is also the ratio between the complex p displacement p response p and the complex input forcing function. Forced‐Excited System: Harmonic Excitation • Define f the h gain function f as the h modulus d l off the h transfer f function f H (ω ) = H (ω ) H * (ω ) = (Re H ) 2 + (Im H ) 2 where H* is the complex conjugate. For the force excited system under consideration, H (ω ) = • 1 (k − mω 2 ) 2 + (cω ) 2 The gain function is the amplitude of the displacement for Fo=1. Thus, Xo = H (ω ) Fo Forced‐Excited System: Harmonic Excitation • It is convenient to develop d l a nondimensional d l form f off the h gain function. f First define the frequency ratio ω r= ωn Multiplying the equation H (ω ) = 1 (k − mω 2 ) 2 + (cω ) 2 by k and employing the definitions for ζ, cc and ωn, it is easily shown that Xo 1 = Fo / k (1 − r 2 ) 2 + (2ξr ) 2 and the p phase angle g is: ⎛ 2ζr ⎞ 2 ⎟ ⎝1− r ⎠ φ = arctan⎜ Forced‐Excited System: Harmonic Excitation • TTransfer f ffunctions ti expressing i the th velocity l it and d acceleration l ti responses can be written based on equation x(t ) = X o e by multiplying H(ω) in equation H (ω ) = i (ωt +φ ) ~ iω t = Xe 1 (k − mω 2 ) 2 + (cω ) 2 by iω and (iω)^2=‐ω^2 respectively. The resulting gain function for the velocity output would be derived by multiplying both sides of the above equation and the equation Xo 1 = Fo / k (1 − r 2 ) 2 + (2ξr ) 2 by ω and noting that ωXo is the amplitude of the velocity. Similarly, the gain function for the acceleration can be obtained by multiplying both sides by ω^2. Quality factor and bandwidth • For smallll values l off d damping we can take: k ⎛ X ⎜⎜ ⎝ δ st ⎞ ⎛ X ⎟⎟ ≅ ⎜⎜ ⎠ max ⎝ δ st ⎞ 1 ⎟⎟ = =Q 2 ζ ⎠ ω =ω n where X denotes the amplitude of the response and F δ st = o = deflection under the static force Fo k • The value Th l off the h amplitude li d ratio i at resonance is i called ll d the h Q factor f or the h quality factor of the system. The points R1 and R2, where the amplification factor falls to Q/√2, are called half power points because the power absorbed (ΔW) by the damper (or by the resistor in an electrical circuit), responding harmonically at a given frequency, is proportional to the q of the amplitude. p square Quality factor and bandwidth • The h d differences ff b between the h frequencies f associated with the half power points R1 and R2 is called the bandwidth of the system. system • To find the values of R1 and R2, we set so that or • From which, we can get: Quality factor and bandwidth • For smallll values l off ζ, ζ the h roots can be simplified as: Then Quality factor and bandwidth • Using the h relation l in the h equation we find that the bandwidth Δω is given by: Combining the above equation and the equation ⎛ X ⎜⎜ ⎝ δ st ⎞ ⎛ X ⎟⎟ ≅ ⎜⎜ ⎠ max ⎝ δ st ⎞ 1 ⎟⎟ = =Q 2 ζ ⎠ ω =ω n W obtain: We b i • It can be seen that the quality factor Q can be used for estimating the equivalent viscous damping in a mechanical system. Electrical systems • An electronic circuit is a closed path formed by the interconnection of electronic components through which an electric current can flow. • We have just seen that linear ODEs have important applications in mechanics. Similarly, they are models of electric circuits as they occur as portions of large networks in computers and elsewhere. • The circuits we shall consider here are basic building blocks of such networks. • They contain three kinds of components, namely, resistors, inductors and capacitors. • Kirchhoff’s Voltage Law (KVL): The voltage (the electromotive force) impressed on a closed loop is equal to the sum of the voltage drops across p the other elements of the loop. Electrical systems • Figure shows Fi h such h a RLC circuit. i i IIn iit a resistor i off resistance i R Ω (ohms), an inductor of inductance L H (Henrys) and a capacitor of capacitance C F (farads) are wired in series as shown,, and connected to an electromotive force E(t) () V(volts) (a generator for instance), sinusoidal as shown in the figure or some other kind. • R, L, C, and E are given and we want to find the current I(t) A(Amperes) in the circuit. • An ODE for the current I(t) in the RLC circuit in the figure is obtained from the Kirchhoff’s Voltage Law. • In the figure, the circuit is a closed loop and the impressed voltage E(t) equals the sum of the voltage drops across the three elements R,L,C, of the loop. Electrical systems • Voltage l drops: d Experiments show h that h the h current I fl flowing through h ha resistor, inductor and capacitor causes a voltage drop (voltage difference, measured in volts) at the two ends. These drops are: • • • RI (Ohm’s law) Voltage drop for a resistor of resistance R ohms Ω dI LI ′ = L l d drop ffor an iinductor d off iinductance d Lh henrys ((H)) dt Voltage Q Voltage drop for a capacitor of capacitance C farads (F) • Here Q coulombs is the charge on the capacitor, related to the current by C I (t ) = dQ equivalently Q(t) = ∫ I(t)dt dt Electrical systems • Table bl Elements l in an RLC circuit Name • Symbol Notation Unit Voltage Drop Ohm’s resistor Ohm’s resistance, R Ohms (Ω) RI Inductor Inductance, L henrys (H) LdI/dt Capacitor p Capacitance, p C farads (F) Q/C According to Kirchhoff’ voltage law we thus have an RLC circuit with electromotive force E(t)=E ( ) o sinωt ((Eo constant)) as a model for the ‘integro g differential equation’. 1 LI ′ + RI + C∫ Idt = E (t ) = Eo sin ωt Electrical systems • T gett rid To id off th the iintegral t l iin LI ′ + RI + • 1 Idt = E (t ) = Eo sin ωt C∫ We differentiate the above equation with repect to t, obtaining: LI ′′ + RI ′ + • • 1 I = E ′(t ) = Eoω cos ωt C This shows that the current in an RLC circuit is obtained as the solution of this nonhomogeneous second‐order ODE with constant coefficients. 1 Using LI ′ + RI + ∫ Idt = E (t ) = Eo sin ωt and noting that I=Q’ and I’=Q’’, we C have directly: 1 LQ′′ + RQ′ + • C Q = Eo sin ωt But in most practical problems, the current I(t) is more important than the charge Q(t) and for this reason, we shall concentrate on the below q rather than the above. equation LI ′′ + RI ′ + 1 I = E ′(t ) = Eoω cos ωt C Solving ODE for the current • A general solution of LI ′′ + RI ′ + 1 I = E ′(t ) = Eoω cos ωt C is the sum II=IIh+Ip , where Ih is a general solution of the homogeneous ODE corresponding to the above equation and Ip is a particular solution. We first determine Ip by the method of undetermined coefficients. We substitute: I p = a cos ωt + b sin ωt I ′p = ω (−a sin ωt + b cos ωt ) I ′p′ = ω 2 (−a cos ωt − b sin ωt ) into the first equation. Then we collect the cosine terms and equate them to Eoωcosωt on the right, and we equate the sine terms into zero because there i no sine is i term on the h right. i h a = Eoω C b Lω 2 (−b) + Rω (− a) + = 0 C Lω 2 (− a) + Rωb + (Cosine terms) (Sine Si terms t ) Solving ODE for the current • TTo solve l this thi system t for f a and db b, we fi firstt iintroduce t d a combination bi ti off L and dC C, called the reactance: 1 S = ωL − • ωC Dividing the previous two equations by ω, ordering them and substituting S gives: − Sa + Rb = E o − Ra − Sb = 0 • We now eliminate b by multiplying the first equation by S and the second by R, and adding. Then we eliminate a by multiplying the first equation by R and second by –S, and adding. This gives: a= • − Eo S R2 + S 2 b= Eo R R2 + S 2 I p = a cos ωt + b sin ωt I ′p = ω (−a sin ωt + b cos ωt ) I ′p′ = ω 2 (−a cos ωt − b sin ωt ) Equation for Ip with coefficients a and b as given above is the desired particular solution of the nonhomogeneous ODE governing the current I in an RLC circuit with sinusoidal electromotive force. Solving ODE for the current • Using a = − Eo S R2 + S 2 b= Eo R R2 + S 2 we can write Ip in terms off physically h ll visible quantities, quantities namely, namely amplitude Io and phase lag θ of the current behind the electromotive force, that is, I p (t ) = I o sin(ωt − θ ) where Io = a2 + b2 = tan θ = − Eo R2 + S2 a S = b R The q quantity y R 2 + S 2 is called the impedance. p Our formula shows that the impedance p equals q the ratio This is somewhat analogous to E/I = R (Ohm' s law) Eo . Io Solving ODE for the current • A generall solution l off the h h homogeneous equation corresponding d to LI ′′ + RI ′ + is: 1 I = E ′(t ) = Eoω cos ωt C I h = c1e λ1t + c2 e λ2t where λ1 and λ2 are the roots of the characteristic equation: λ2 + • R 1 λ+ =0 L LC We can write the roots in the form λ1=‐α+β α+β and λ2=‐α‐β, α β, where R R2 1 1 4L 2 α= , β= − = − R 2L 4 L2 LC 2 L C • Now in an actual circuit, R is never zero (hence R>0). From this, it follows that pp zero,, theoreticallyy as t→∞,, but practically p y after a short time. Ih approaches Solving ODE for the current • Hence the h transient current I=Ih+Ip tends d to the h steady d state current Ip and d after some time the output will practically be a harmonic oscillation, which is given by: I p (t ) = I o sin(ωt − θ ) and whose frequency is that of the input (of the electromotive force) Analogy of electrical and mechanical quantities • EEntirely ti l diff differentt physical h i l or other th systems t may have h the th same mathematical model. For instance, the ODE of a mechanical system and the ODE of an electric RLC circuit can be expressed by: LQ′′ + RQ′ + • • • 1 Q = Eo sin ωt C m y ′′ + c y ′ + ky = Fo cos ω t The inductance L corresponds p to the mass,, and indeed an inductor opposes a change in current, having an inertia effect similar to that of a mass. The resistance R corresponds to the damping constant c and a resistor causes loss of energy, just as a damping dashpot does. This analogy is strictly quantitative in the sense that to a given mechanical system we can construct an electrical circuit whose current will give the exact values of the displacement in the mechanical system when suitable scale factors are used. Analogy of electrical and mechanical quantities • • The h practical i l importance i off this hi analogy l is i almost l obvious. b i The h analogy l may be b used for constructing an ‘electrical model’ of a given mechanical model, resulting in substantial savings of time and money because electric circuits are easy to assmeble, bl and d electric l t i quantities titi can be b measured d much h more quickly i kl and d accurately than mechanical ones. 1 m y ′′ + c y ′ + ky = Fo cos ω t LQ′′ + RQ′ + Q = Eo sin ωt C Table: Analogy of electrical and mechanical quantities Electrical system Mechanical system Inductance, L Mass m Resistance, R Damping c Reciprocal of capacitance capacitance, 1/C Spring modulus k Electromotive force Eo sinωt Driving force Focosωt Current,, I(t)=dq/dt ( ) q/ Velocityy , v(t)=dy/dt ( ) y/ Charge, Q(t) Displacement, y(t) Base excited systems: absolute motion • Consider d the h base‐excited b d system of the figure. The goal of the analysis will be to determine the absolute response x(t) (typically acceleration or displacement of the mass) given the base motion y(t). • FFrom the h free‐body f b d diagram, di application li i of Newton’s second law leads directly to the differential equation: m&x& + cx& + kx = ky + cy& Base excited systems: absolute motion • Assume that h the h base b motion is harmonic, h y (t ) = Yo e iωt • And assume that the response will be harmonic, harmonic ~ x(t ) = Xe iωt ~ where X is the complex response. The transfer function and the gain function are derived in the same manner as for the force excited system. The transfer function is: k + icω H (ω ) = (k − mω 2 ) + icω • The gain function is Xo k 2 + (cω ) 2 1 + (2ζr ) 2 = H (ω ) = = 2 2 2 2 Yo (k − mω ) + (cω ) (1 − r 2 ) 2 + (2ζr ) Base excited systems: absolute motion • • Xo 1 + (2ξr ) 2 In nondimensional d l form f = Yo (1 − r 2 ) 2 + (2ξr ) 2 The gain function for the absolute displacement for the base‐excited base excited system is shown in the figure. Base excited systems: absolute motion • • • • The h value l off Td is i unity i at r=0 0 and d close l to unity i for f smallll values l off r. For an undamped system ζ=0, Td ∞ at resonance (r=1). The value of Td is less than unity (Td <1) for values of r >√2 (for any amount of damping ζ) The value of Td is equal to unity (Td=1) for all values of ζ at r=√2 Xo 1 + (2ξr ) 2 = Yo (1 − r 2 ) 2 + (2ξr ) 2 Base excited systems: absolute motion • The h equations Xo 1 + (2ξr ) 2 = Yo (1 − r 2 ) 2 + (2ξr ) 2 Xo k 2 + (cω ) 2 = H (ω ) = Yo (k − mω 2 ) 2 + (cω ) 2 can be interpreted as the gain functions for acceleration output given acceleration input input. And again note that the transfer function for velocity and acceleration responses can be derived by multiplying the equation k + icω H (ω ) = (k − mω 2 ) + icω by iω and ‐ω^2, respectively. Base excited system: Relative motion Base‐excited • Consider C id th the free f body b d diagram di in i the th figure. Now the response variable under consideration will be the relative displacement, displacement z (t ) = x(t ) − y (t ) • In the model, the spring represents a structural element. The stress in that element will be proportional to z. Thus, this problem would be relevant to designers of structures subjected to base motions, for example, earthquakes. • Letting z=x-y in the equation of motion eads d directly ec y to o m&z& + z& + kz = − m&y& leads Base excited system: Relative motion Base‐excited • Assuming that h the h base b motion is harmonic, h y (t ) = Yo e iωt • And assuming the response is also harmonic, ~ z (t ) = Z eiωt • Following the procedure as described above, the transfer function is: mω 2 H (ω ) = (k − mω 2 ) + icω • And the gain function is Zo mω 2 = H (ω ) = Yo (k − mω 2 ) 2 + (cω ) 2 Base excited system: Relative motion Base‐excited • • 2 Z r o In nondimensionless d l fform, = Yo (1 − r 2 ) 2 + (2ξr ) 2 The gain function for the relative motion for the base base‐excited excited system is shown in the figure: Base excited system: Relative motion Base‐excited • Again note that h the h transfer f function f for f relative l velocity l and d acceleration l responses can be derived by multiplying mω 2 H (ω ) = (k − mω 2 ) + icω by iω and ‐ω^2, respectively. • The gain functions for velocity and acceleration responses can be obtained by multiplying both sides of the equations Zo mω 2 Zo r2 = H (ω ) = = 2 2 2 Yo Yo (k − mω ) + (cω ) (1 − r 2 ) 2 + (2ξr ) 2 by ω and ω^2, respectively. Gain functions for force excited systems Gain functions for base‐excited systems Example • • A fi fixed db bottom tt offshore ff h structure t t is i subjected bj t d to t oscillatory ill t storm t waves. In a first approximation, it is estimated that the waves produce a harmonic force F(t) having amplitude F=122 kN. The period of these waves is τ=8 sec The structure is modeled as having a lumped mass of 110 tons sec. concentrated in the deck. The weight of the structure itself is assumed to be negligible. The natural period of the structure was measured as being τn= 4 4.0 0 sec sec. It is assumed that the damping factor is ζ=5%. ζ=5% It is required to determine the steady state amplitude of the response of the structure. Solution: As modeled, this will be a force‐excited system, and the response can be computed from the gain function of Xo 1 = Fo / k (1 − r 2 ) 2 + (2ξr ) 2 The problem reduces to one of finding the frequency ratio r and the stiffness k. Example • Because r is the h ratio off the h forcing f frequency f to the h naturall frequency, f it follows that r will also be the ratio of the natural period to the forcing period. Thus, τ 4 r= • 1 τn = 0.25 Hz. Then noting that the expression for the natural frequency is fn = • = 0 .5 τ 8 To compute k, first note that the natural frequency is fn = • = n We compute k=27667 N/m 1 2π k W /g Example • Finally ll substituting b into Xo 1 = Fo / k (1 − r 2 ) 2 + (2ξr ) 2 • Xo=5.87 m. Resonance • For resonance please download the Millenium g and Tacoma Bridge g videos in the bridge website of the course. Background for response of SDOF system y to random forces •Response of SDOF system to impulsive forces •Response of single degree of freedom system to arbitrary loading •Relationship between the impulse response and the transfer function •Relationship p between the Fourier transform of displacement p and force Response under a nonperiodic force • We will see that periodic forces of any general wave form can be represented by Fourier series as a superposition of harmonic components of various frequencies. • The response of a linear sytem is then found by superposing the harmonic response to each of the exciting forces forces. • When the exciting force F(t) is nonperiodic, such as that due to blast from an explosion, a different method of calculating the response is required. • Various methods can be used to find the response of the system to an arbitrary excitation as follows: – Representing the excitation by a Fourier integral – Using the method of convolution integral – Using the method of Laplace transforms – First approximating F(t) by a suitable interpolation model and then using a numerical procedure – Numerically integrating the equations of motion. Response of single degree of freedom system to impulsive forces • A nonperiodic exciting force usually has a magnitude that varies with time time; it acts for a specified period of time and then stops. • The simplest form is the impulsive force‐ force a force that has a large magnitude F and acts for a very short period of time Δt. • From dynamics dynamics, we know that impulse can be measured by finding the change in momentum of the system. • The unit impulse F acting at t=0 is also denoted by the Dirac delta function, δ(t). The Dirac delta function at time t=τ, denoted as δ (t − τ ) has the properties ∞ ∞ 0 0 ∫ δ (t − τ )dt = 1 ∫ δ (t − τ ) F (t )dt = F (τ ) where 0<τ<∞. Thus an impulsive force acting at t=τ can be denoted as: F (t ) = Fδ (t − τ ) Response of single degree of freedom system to impulsive forces • Consider d a SDOF system subjected b d to impulsive l lloading d as shown h in ffigure. The external force is: F (t ) = Foδ (t ) where δ(t) is the Dirac delta function. Response of single degree of freedom system to impulsive forces • The h equation off motion off the h mass willll b be similar l to m&x& + cx& + kx = Fo sin ωt with the impulsive force of F (t ) = Foδ (t ) on the right hand side. The unit impulse is defined as Fo=1. The response x(t) to the unit impulse is denoted as h(t): mh&& + ch& + kh = (1)δ (t ) • Physically speaking speaking, for t≈0, t≈0 a radical change in the system motion takes place when the short duration high amplitude force excites an initial motion in the system. But for t>0, the response will be free vibration. U i elementary Using l mechanics, h i F(Δt)=m(Δv) F(Δ ) (Δ ) iit can b be shown h that h the h velocity of the system just after the impulse is: 1 h&(0 + ) = m Response of single degree of freedom system to impulsive forces • U i th Using the iinitial iti l conditions: diti 1 h&(0 + ) = m h(0) = 0 • ⎞ x&o + ζω n xo −ζω n t ⎛ 2 2 ⎜ ⎟ = − + − x ( t ) e x cos 1 ζ ω t sin 1 ζ ω t In the equation: n n ⎟ ⎜ o ωd ⎝ ⎠ • The free vibration response is: h(t ) = • 1 −ξωnt e sin ωd t mωd t >0 Here h(t) is known as the force‐excited absolute displacement response, impulse response function of the single‐degree‐of‐freedom system. Note that h(t) characterizes a system just like the transfer function H(ω) does. does The velocity and acceleration impulse response functions can also be obtained as derivatives of h(t). Response of single degree of freedom system to impulsive forces • If the magnitude of the impulse is F instead of unity, the initial velocity x&o is F/m and the response of the system becomes: x(t ) = • F −ξωnt e sin ωd t = Fg (t ) mωd If the impulse F is applied at an arbitrary time t=τ by an amount F/m as shown in the figure, it will change the velocity at t=τ by an amount F/m. Assuming that x=0 until the impulse is applied, the displacement h at any subsequent time t, caused by a change in the velocity at time τ is given by the above equation with t replaced by the time elapsed after the application of the impulse, that is, t‐ τ. As shown in Fig.b, we obtain x(t ) = Fg (t − τ ) Response of Single‐Degree‐of‐ Freedom System to Arbitrary Loading • For a linear l system, the h impulse l response ffunction can be b used d to derive d the response of a system under an arbitrary loading history. Consider the force shown in the figure Response of Single‐Degree‐of‐ Freedom System to Arbitrary Loading • The h iimpulse l d during i Δτ Δ is i F(τ) ( ) Δτ. Δ The h response to this impulse at any time t>τ is approximately [F(τ) Δτ]h(t‐ τ). Th the Then th response att t is i the th sum off the responses due to a sequence of impulses. In the limit as Δτ→0 t x(t ) = ∫ F (τ )h(t − τ )dτ −∞ where the input F(t) is accounted for as t→‐∞, for example,F(t) could be defined as zero for t<0. The expression for x(t) is called the convolution integral. Response of Single‐Degree‐of‐ Freedom System to Arbitrary Loading • Note that h(t‐τ) =0 when τ>t. Thus, we can expand the limits to the interval (‐∞,∞): ∞ x(t ) = ∫ F (τ )h(t − τ )dτ −∞ • Another useful form is obtained by letting θ=t‐τ: t x(t ) = ∫ F (t − θ )h(θ )dθ −∞ Response of Single‐Degree‐of‐ Freedom System to Arbitrary Loading • B substituting By b tit ti th the equation ti h(t ) = into 1 −ξωnt e sin ωd t mωd t >0 ∞ x(t ) = ∫ F (τ )h(t − τ )dτ −∞ we obtain: 1 x(t ) = mωd t ∫ F (τ )e −ζω n ( t −τ ) sin ωd (t − τ )dτ 0 which represents the response of an underdamped single degree of freedom system to the arbitrary excitation F(t). • Note that the above equation does not consider the effect of initial conditions of the system. • The integral in either of the two above equations is called the convolution or Duhamel integral. Relationship between h(t) and H(ω) • An important result l ffrom Fourier transform f theory h is that h h(t) h( ) and d H(ω) ( ) form a Fourier transform pair. This relationship is useful when deriving responses of dynamic systems to random vibration inputs. Let, F (t ) = eiωt • x(t ) = H (ω )e iωt Then, h ∞ = ∫ h(t − τ )e iωτ dτ −∞ ∞ = ∫ h(θ )e iω (t −θ ) dθ −∞ =e iωt ∞ ∫ h(θ )e −iωθ dθ −∞ which implies that h(t) and H(ω) form a Fourier transform pair. ∞ H (ω ) = ∫ h(θ )e −∞ −iωθ dθ → 1 h(t ) = 2π ∞ ∫ H (ω )e −∞ iωt dω Relationship between h(t) and H(ω) • Based B d on th the FFourier i transform t f off a convolution, l ti the th expression i for f the th response to an arbitrary input in equation ∞ x(t ) = and representation in ∫ F (τ )h(t − τ )dτ −∞ ∞ H (ω ) = ∫ h(θ )e −iωθ dθ → −∞ 1 h(t ) = 2π ∞ ∫ H (ω )e i ωt dω −∞ it is clear that we can also express the response to an arbitrary input as: 1 x(t ) = 2π ∞ ∫ F (ω ) H (ω )e iωt dω −∞ where F(ω) is the Fourier transform of F(t). This expression is useful for the analysis or numerical computation of system response or as the basis for random vibration computations. Relationship between X(ω) and F (ω) • The h relationship l h b between the h Fourier transforms f off x(t) ( ) and d F(t) ( ) is used d to derive responses of dynamic systems to random vibration input in random vibration theory. Take the Fourier transform of both sides of t x(t ) = ∫ F (t − θ )h(θ )dθ −∞ to find fi d ⎡∞ ⎤ − i ωt F t θ h θ d θ ( − ) ( ) ⎥ e dt ∫−∞⎢⎣−∫∞ ⎦ Let τ = t − θ , dt = dτ 1 X (ω ) = 2π ∞ ⎡∞ ⎤ −iω (τ +θ ) F ( τ ) h ( θ ) d θ dτ ⎥e ∫−∞⎣⎢−∫∞ ⎦ Rearranging : 1 X (ω ) = 2π ∞ ∞ X(ω ) = ∫ h(θ )e -∞ -iωθ 1 dθ 2π ∞ ∫ F (τ )e −∞ −iωτ dτ Relationship between X(ω) and F (ω) • From ∞ H (ω ) = ∫ h(θ )e −iωθ dθ −∞ and the basic relationship of the Fourier transform: 1 u (τ ) = 2π +∞ −iωt ( ) g t e dt ∫ −∞ it follows that X(ω ) = H(ω )F(ω )