Convolution solutions (Sect. 4.5).

1 Convolution solutions (Sect. ).IConvolution of two of Transform of a response decomposition solutions (Sect. ).IConvolution of two of Transform of a response decomposition of two piecewise continuous functionsf,g:R Risthe functionf g:R Rgiven by(f g)(t) = t0f( )g(t )d .Remarks:If gis also called the generalized product definition of Convolution of two functions also holds inthe case that one of the functions is a generalized function,like Dirac s of two the Convolution off(t) =e tandg(t) = sin(t).Solution:By definition: (f g)(t) = t0e sin(t )d .Integrate by parts twice: t0e sin(t )d =[e cos(t )] t0 [e sin(t )] t0 t0e sin(t )d ,2 t0e sin(t )d =[e cos(t )] t0 [e sin(t )] t0,2(f g)(t) =e t cos(t) 0 + sin(t).We conclude:(f g)(t) =12[e t+ sin(t) cos(t)].CConvolution solutions (Sect.)

2 IConvolution of two of Transform of a response decomposition of (Properties)For every piecewise continuous functions f , g , and h, hold:(i)Commutativity:f g=g f;(ii)Associativity:f (g h) = (f g) h;(iii)Distributivity:f (g+h) =f g+f h;(iv)Neutral element:f 0 = 0;(v)Identity element:f = :(v):(f )(t) = t0f( ) (t )d = t0f( ) ( t)d =f(t).Properties of :(1): Commutativity:f g=g definition of Convolution is,(f g)(t) = t0f( )g(t )d .Change the integration variable: =t ,henced = d ,(f g)(t) = 0tf(t )g( )( 1)d (f g)(t) = t0g( )f(t )d We conclude:(f g)(t) = (g f)(t). Convolution solutions (Sect. ).IConvolution of two of Transform of a response decomposition Transform of a (Laplace Transform)Iff,ghave well-defined Laplace TransformsL[f],L[g], thenL[f g] =L[f]L[g].Proof:The key step is to interchange two start wethe product of the Laplace transforms,L[f]L[g] =[ 0e stf(t)dt] [ 0e s tg( t)d t],L[f]L[g] = 0e s tg( t)( 0e stf(t)dt)d t,L[f]L[g] = 0g( t)( 0e s(t+ t)f(t)dt)d Transform of a :Recall:L[f]L[g] = 0g( t)( 0e s(t+ t)f(t)dt)d variables: =t+ t,henced =dt;L[f]L[g] = 0g( t)( te s f( t)d )d [f]L[g] = 0 te s g( t)f( t)d d key step: Switch the order of = tautaut0L[f]L[g] = 0 0e s g( t)f( t)d t d.

3 Laplace Transform of a :Recall:L[f]L[g] = 0 0e s g( t)f( t)d t d .Then, is straightforward to check thatL[f]L[g] = 0e s ( 0g( t)f( t)d t)d ,L[f]L[g] = 0e s (g f)( )d L[f]L[g] =L[g f]We conclude:L[f g] =L[f]L[g].Laplace Transform of a convolutions to find the inverse Laplace Transform ofF(s) =3s3(s2 3).Solution:We expressFas a product of two Laplace Transforms,F(s) = 31s31(s2 3)=321 3(2s3) ( 3s2 3)Recalling thatL[tn] =n!sn+1andL[sinh(at)] =as2 a2,F(s) = 32L[t2]L[sinh( 3t)]= 32L[t2 sin( 3t)].We conclude thatf(t) = 32 t0 2sinh[ 3(t ))]d .CLaplace Transform of a [f(t)] wheref(t) = t0e 3(t )cos(2 )d .Solution:The functionfis the Convolution of two functions,f(t) = (g h)(t),g(t) = cos(2t),h(t) =e [(g h)(t)] =L[g(t)]L[h(t)],then,F(s) =L[ t0e 3(t )cos(2 )d ]=L[e 3t]L[cos(2t)].We conclude thatF(s) =s(s+ 3)(s2+ 4).

4 CLaplace Transform of a the IVPy 5y + 6y=g(t),y(0) = 0,y (0) = :DenoteG(s) =L[g(t)]and compute LT of the equation,(s2 5s+ 6)L[y(t)] =L[g(t)] L[y(t)] =1(s2 5s+ 6)G(s).DenotingH(s) =1s2 5s+ 6,andh(t) =L 1[H(s)],thenL[y(t)] =H(s)G(s) y(t) = (h g)(t).Functionhis simple to compute:H(s) =1(s 2)(s 3)=a(s 2)+b(s 3)=a(s 3) +b(s 2)(s 2)(s 3)Laplace Transform of a the IVPy 5y + 6y=g(t),y(0) = 0,y (0) = :Then: 1 =a(s 3) +b(s 2).Evaluate ats= 2, 2 a= 3 b= (s) = 1(s 2)+1(s 3).Thenh(t) = e2t+ the formulay(t) = (h g)(t),we gety(t) = t0( e2 +e3 )g(t )d .CConvolution solutions (Sect. ).IConvolution of two of Transform of a response decomposition response response solutionis the solutiony to the IVPy +a1y +a0y = (t),y (0) = 0,y (0) = Laplace Transforms,(s2+a1s+a0)L[y ] = 1 y (t) =L 1[1s2+a1s+a0].Denoting the characteristic polynomial byp(s) =s2+a1s+a0,y =L 1[1p(s)].

5 Summary:The impulse reponse solution is the inverse LaplaceTransform of the reciprocal of the equation response :Theimpulse response solutionisy solution of the IVPy +a1y +a0y = (t),y (0) = 0,y (0) = the solution (impulse response att=c) of the IVPy c+ 2y c+ 2y c= (t c),y c(0) = 0,y c(0) = 0,c :L[y c] + 2L[y c] + 2L[y c] =L[ (t c)].(s2+ 2s+ 2)L[y c] =e cs L[y c] =e cs(s2+ 2s+ 2).Impulse response the solution (impulse response att=c) of the IVPy c+ 2y c+ 2y c= (t c),y c(0) = 0,y c(0) = 0,c :Recall:L[y c] =e cs(s2+ 2s+ 2).Find the roots of the denominator,s2+ 2s+ 2 = 0 s =12[ 2 4 8]Complex complete the square:s2+ 2s+ 2 =[s2+ 2(22)s+ 1] 1 + 2= (s+ 1)2+ ,L[y c] =e cs(s+ 1)2+ response the solution (impulse response att=c) of the IVPy c+ 2y c+ 2y c= (t c),y c(0) = 0,y c(0) = 0,c :Recall:L[y c] =e cs(s+ 1)2+ :L[sin(t)] =1s2+ 1,andL[f](s c) =L[ectf(t)].

6 1(s+ 1)2+ 1=L[e tsin(t)] L[y c] =e csL[e tsin(t)].Sincee csL[f](s) =L[u(t c)f(t c)],we concludey c(t) =u(t c)e (t c)sin(t c).CConvolution solutions (Sect. ).IConvolution of two of Transform of a response decomposition decomposition (Solution decomposition)The solution y to the IVPy +a1y +a0y=g(t),y(0) =y0,y (0) =y1,can be decomposed asy(t) =yh(t) + (y g)(t),whereyhis the solution of the homogeneous IVPy h+a1y h+a0yh= 0,yh(0) =y0,y h(0) =y1,andy is the impulse response solution, that is,y +a1y +a0y = (t),y (0) = 0,y (0) = decomposition the Solution Decomposition Theorem to express the solution ofy + 2y + 2y= sin(at),y(0) = 1,y (0) = :L[y ] + 2L[y ] + 2L[y] =L[sin(at)],and recall,L[y ] =s2L[y] s(1) ( 1),L[y ] =sL[y] 1.(s2+ 2s+ 2)L[y] s+ 1 2 =L[sin(at)].L[y] =(s+ 1)(s2+ 2s+ 2)+1(s2+ 2s+ 2)L[sin(at)].

7 Solution decomposition the Solution Decomposition Theorem to express the solution ofy + 2y + 2y= sin(at),y(0) = 1,y (0) = :Recall:L[y] =(s+ 1)(s2+ 2s+ 2)+1(s2+ 2s+ 2)L[sin(at)].But:L[yh] =(s+ 1)(s2+ 2s+ 2)=(s+ 1)(s+ 1)2+ 1=L[e tcos(t)],and:L[y ] =1(s2+ 2s+ 2)=1(s+ 1)2+ 1=L[e tsin(t)].So,L[y] =L[yh] +L[y ]L[g(t)] y(t) =yh(t) + (y g)(t),So:y(t) =e tcos(t) + t0e sin( ) sin[a(t )]d .CSolution decomposition :Compute:L[y ] +a1L[y ] +a0L[y] =L[g(t)],and recall,L[y ] =s2L[y] sy0 y1,L[y ] =sL[y] y0.(s2+a1s+a0)L[y] sy0 y1 a1y0=L[g(t)].L[y] =(s+a1)y0+y1(s2+a1s+a0)+1(s2+a1s+a0)L[g( t)].Recall:L[yh] =(s+a1)y0+y1(s2+a1s+a0),andL[y ] =1(s2+a1s+a0).Since,L[y] =L[yh] +L[y ]L[g(t)],soy(t) =yh(t) + (y g)(t).Equivalently:y(t) =yh(t) + t0y ( )g(t )d.