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Convolution solutions (Sect. 6.6). Convolution of two ...

Convolution 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. ).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 =f(t).

I Impulse response solution. I Solution decomposition theorem. Impulse response solution. Definition The impulse response solution is the function y δ solution of the IVP y00 δ + a 1 y 0 δ + a 0 y δ = δ(t − c), y δ(0) = 0, y δ 0(0) = 0, c ∈ R. Example Find the impulse response solution of the IVP y00 δ +2 y 0 δ +2 y δ = δ(t − ...

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Transcription of Convolution solutions (Sect. 6.6). Convolution of two ...

1 Convolution 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. ).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 =f(t).

2 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)( )dtL[f]L[g] =L[g f]We conclude:L[f g] =L[f]L[g]. Convolution solutions (Sect. ).IConvolution of two of Transform of a response decomposition response response solutionis the functiony solution of the IVPy +a1y +a0y = (t c),y (0) = 0,y (0) = 0,c the impulse response solution of the IVPy + 2y + 2y = (t c),y (0) = 0,y (0) = :L[y ] + 2L[y ] + 2L[y ] =L[ (t c)].(s2+ 2s+ 2)L[y ] =e cs L[y ] =e cs(s2+ 2s+ 2). impulse response the impulse response solution of the IVPy + 2y + 2y = (t c),y (0) = 0,y (0) = 0,.Solution:Recall:L[y ] =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 ] =e cs(s+ 1)2+ response the impulse response solution of the IVPy + 2y + 2y = (t c),y (0) = 0,y (0) = 0.

4 Solution:Recall:L[y ] =e cs(s+ 1)2+ :L[sin(t)] =1s2+ 1,andL[f](s c) =L[ectf(t)].1(s+ 1)2+ 1=L[e tsin(t)] L[y ] =e csL[e tsin(t)].Sincee csL[f](s) =L[u(t c)f(t c)],we concludey (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.

5 (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)].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.


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