5.5 ConvolutionandtheLaplaceTrans- form

1 170 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 ConvolutionLetf(t) andg(t) be two functions with laplace transformsF(s) andG(s),respectively.

2 Then:L[f g] =F(s)G(s)andL 1[F(s)G(s)] =f gExample [t2 et]172 CHAPTER 5. laplace TRANSFORMSS olution:SinceL[f g] =F(s)G(s),we haveL[f g] =2s31s 1 We can use convolution as an alternative to partial fractions asshown +y=e2t, y(0) = 0, y (0) = :After taking laplace transform of both sides we get:(s2+ 1)Y(s) =1s 2orY(s) =1s2+ 11s 2so settingF(s) =1s2+1andG(s) =1s 2we see thatY(s) =F(s)G(s) soy(t) =f gwheref(t) = sintandg(t) = convolution is t0e2(t w)sinw dwwhich is t0e2(t w)sinw is the same thing ase2t t0e 2wsinw this point, we could integrate by parts to get the solution, butwe wish to introduce a slick trick to avoid integration by PARTS,since the integrand looks like the definition of a laplace transform , CONVOLUTION AND THE laplace TRANSFORM173wheres= 2,and since 1 u(w t) is equal to zero forw > tandis equal to one forw < t(we viewtas fixed), we may rewritee2t t0e 2wsinw dw=e2t t0[1 u(w t)]e 2wsinw dw+e2t t[1 u(w t)]e 2wsinw dw=e2t 0[1 u(w t)]e 2wsinw dw=e2tL[(1 u(w t)) sinw] (2)(notes= 2 in the laplace transform definition).

3 =e2t(L[sinw](2) L[u(w t) sinw] (2))=e2t(L[sinw](2) L[u(w t) sin(w t+t)] (2))=e2t(122+ 1 L[u(w t) sin(w t) cost+u(w t) cos(w t) sint] (2))=e2t(122+ 1 L[u(w t) sin(w t) cost+u(w t) cos(w t) sint] (2))=e2t(15 costL[u(w t) sin(w t)](2) sintL[u(w t) cos(w t)](2))=e2t(15 cost(e 2t122+ 1) sint(e 2t222+ 1))=15e2t 15cost 25sint174 CHAPTER 5. laplace TRANSFORMSP erhaps this was better done with PARTS, but we wanted toillustrate the power of the laplace transform The advantage of convolution is that we can solve any spring masssystem without actually having the forcing function, as illustratedin the next +y=g(t), y(0) = 0, y (0) = 0for anyg(t).Solution:After taking laplace transform of both sides we get:(s2+ 1)Y(s) =G(s)orY(s) =1s2+ 1G(s)so settingF(s) =1s2+1and we see thatY(s) =F(s)G(s) soy(t) =f gwheref(t) = convolution (and hence the solution) isy(t) = t0sin(t w)g(w)dw Convolution and Second Order Linear with Constant CoefficientsConsideray +by +cy=g(t), y(0) = 0, y (0) = solution isy(t) =f gwhereF(s) =1as2+bs+c,which is called thetransfer functionand we callf(t)impulse response functionfor thissecond order superposition, we obtain the CONVOLUTION AND THE laplace TRANSFORM175 Convolution and Second Order Linear with Constant CoefficientsConsideray +by +cy=g(t), y(0) =c1, y (0) = we have the particular solution to the homogeneousyhomo part(t) that sat-isfied the initial conditionsy(0) =c1andy (0) =c2theny(t) =yhomo part(t) +f g(t)will solve the nonhomogeneous 1-5, find the e3t3.

4 Cost 14. cost (t 1) 16. Use convolution to solvey +y= 2, y(0) = 0, y (0) = 07. Use convolution to solvey 4y=t, y(0) = 0, y (0) = 08. For a fixed constant , use convolution to solvey + 4y=sin t, y(0) = 0, y (0) = 09. Use an integral approximation to estimatey(1) fory + 4y=et2, y(0) = 0, y (0) = 010. Find the impulse response function for an overdamped springmass systemmy +by +ky=g(t)11. Find the impulse response function for an underdamped springmass systemmy +by +ky=g(t)