Lecture 16 Tuesday March 17 2015 An application of convolution For forces expressed analytically eg harmonic forces or power series or exponentials such as we have considered above the use of the convolution is usually unhelpful The methods described in previous lectures are more convenient Nevertheless the convolution would work it would give the same answer obtained using the other methods including whatever complementary part would be needed so that the total expression satisfied quiescent initial conditions More commonly we apply convolutions to cases in which the forcing is not an analytic function perhaps it is given by a table or taken from some laboratory data Here is such a case Let F Fo for T1 t T2 and zero otherwise a square pulse We write form 3a t x t F G t d 0 corresponding to quiescent initial conditions at the time of the lower limit 0 in this case Our mass spring damper systems have impulse response function G t H t exp nt sin dt m d In the following example we will simplify it by taking c 0 ie G t H t sin nt m n Evaluation Recognize that F is piece wise analytic and that therefore we can expect x to take on a different analytic form in the three regions I II and III t 1 In region I for t T1 the convolution looks like F 0 G t d 0 0 In region II for t between t T1 and t T2 we decompose the integration into an integration over in regions I and II in which F is has a different form t T 2 t T1 F G t d F 0 G t d 0 Fo m n t 0 Fo sin t d m n T1 T1 t F Fo G t d 0 Fo G t d T1 t 2 n cos n t T 1 Fo 1 cos n t T1 m n2 We see that x t is in region II the sum of a constant and a sinusoid ie xparticular and xcomplementary 146 In region III where t T2 the integration over is decomposed into regions I II and III in each of which F has a different form t F G t d 0 T2 T1 t T2 F 0 G t d F F G t d F 0 G t d 0 F G t d 0 o 0 o T1 Fo m n T2 T2 Fo sin t d m n T1 T2 2 n cos n t T 1 T1 Fo cos n t T2 cos n t T1 m 2n We see that for t in region III x t is the sum of two sinusoids As should have been anticipated it is a free vibration at the natural frequency Our ODE Lx F t has so far been solved in a number of ways Finding an xp t for the case of a power series F t and supplementing with xh t to match initial conditions Finding an xp t for the case of a exponential F t exp st and supplementing with xh t to match initial conditions exp i t If the excitation is harmonic F t Re Fo exp i t we write xss Re Fo G and supplement if desired with xh t to match initial conditions extendable to Fourier Series where F t is a sum of harmonic terms If F t is impulsive F t I t to we constructed x t I G t to corresponding to quiescent initial conditions If F t is arbitrary we can write F t as a superposition of impulses and construct x t as a convolution of F and G t that corresponds to quiescent initial conditions wherever you put the lower limit of the integration t F G t d lower lim it Another technique is to do an integral transform such as the Laplace or the Fourier Transform Laplace Transform is one of many types of linear integral transformations that associate a function of one variable for example time with a different function of a different variable e g s 147 f s I f t exp st dt 0 The new function f s will exist if f t doesn t grow too fast at large times The new function is defined uniquely and thus in principle you can get f t from knowing f s over a class of functions that includes the condition that f t vanishes at negative times Note that the Laplace transform of f t is a different function f of a different variable s N B s has units of inverse time f has units of f ness times time The Laplace transform is a linear operation such that the LT of the sum of two functions is the sum of the LTs af bg a f s bg s But the Laplace Transform s most noteworthy property is that it transforms the operation of differentiation with respect to time into multiplication by s 0 0 df dt f df dt exp st dt f t exp st 0 s f t exp st dt f 0 s f s which is simply s f s if f t 0 0 Similarly f sf 0 f 0 s 2 f s which is simply s 2 f s if f t 0 f t 0 0 Differentiation has been transformed into algebra A table of Laplace transforms can be found in many texts Here are a few f t 1 f t tn f t exp at f t sin at f t t a f t t f s 1 s f s n s n 1 f s 1 s a f s a s 2 a 2 f s exp sa f s 1 148 Application to ODE s If we take a Laplace transform of the equation of motion of a SDOF system m d2x dt2 c dx dt k x f t the right hand side becomes formally merely f s It can be calculated just by doing the defining integral f t exp st dt The third term of the left hand side may be replaced simply with k times x s For the other two terms we must use the above formulas for the Laplace transform of a derivative of x t in terms of s times the LT of x t The LT of the equation becomes m s 2 x s vo sxo c sx s xo kx s f s The differential equation for x t has become an algebraic equation for the transform x s The Laplace transformation has turned the differential equation into an algebraic equation Solving for x s we get x s f s cxo mvo msxo ms 2 cs k The part in curly brackets depends on the initial conditions the whole expression in the numerator is sometimes termed the generalized force because it includes all aspects of the excitation of the system forcing and initial conditions We define the other factor G s 1 ms 2 cs …