5.5 ConvolutionandtheLaplaceTrans- form

170 CHAPTER 5. laplace TRANSFORMS10. Solvey +2y +4y=f(t), y(0) = 0, y (0) = 0,wheref(t) is givenin the previous Graph the functionf(t) =t (2t 2)u(t 1) + (2t 4)u(t 2) (2t 6)u(t 3) +..12. Solvey +2y +4y=f(t), y(0) = 0, y (0) = 0,wheref(t) is givenin the previous Considerf(t) =e2tmade into a periodic function f(t) by takingfT(t) whereT= 1.(a) Plot f(t) for 0< t <4.(b) FindL[ f(t)](c)y + 2y + 3y= f(t), y(0) = 0, y (0) = 0,14. Use the differentiation theorem to verify thatL[t u(t a)] =e as1s215. Use appropriate theorems to computeL[tsintetu(t a)] Convolution and the laplace Trans-formWe introduce a new operation between two functions called of Two FunctionsLetf(t) andg(t) be two functions. Define a new function:f g(t) = t0f(t w)g(w)dwNote thatf gis itself a function oft. Moreover note that if wesubstitutev=t wthendv= dwand the integral becomes w=0w=tf(v)g(t v)( 1)dw= w=tw=0f(v)g(t v)dwwhich isg CONVOLUTION AND THE laplace TRANSFORM171 Solution:Sincef g=g f,we can compute 1 teasier, so letf(t) = 1 andg(t) = g= t0f(t w)g(w)dw= t0w dw=w22 t0=t22 We see that convolution is not the same as regular etSolution:We setf(t) =tandg(t) = g= t0f(t w)g(w)dw= t0(t w)ewdw=t t0tew wewdw= (tew wew+ew))|t0=et t 1 The following result shows why convolutions are important: laplace transform of the Convo

174 CHAPTER 5. LAPLACE TRANSFORMS Perhaps this was better done with PARTS, but we wanted to illustrate the power of the Laplace transform ¤ The advantage of …

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