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Boyce/DiPrima 10th ed, Ch 6.6: The Convolution Integral Elementary Differential Equations and Boundary Value Problems, 10th edition, by William E. Boyce and Richard C. DiPrima, ©2013 by John Wiley & Sons, Inc. • Sometimes it is possible to write a Laplace transform H(s) as H(s) = F(s)G(s), where F(s) and G(s) are the transforms of known functions f and g, respectively. • In this case we might expect H(s) to be the transform of the product of f and g. That is, does H(s) = F(s)G(s) = L{f }L{g} = L{f g}? • On the next slide we give an example that shows that this equality does not hold, and hence the Laplace transform cannot in general be commuted with ordinary multiplication. • In this section we examine the convolution of f and g, which can be viewed as a generalized product, and one for which the Laplace transform does commute.Observation • Let f (t) = 1 and g(t) = sin(t). Recall that the Laplace Transforms of f and g are • Thus and • Therefore for these functions it follows that 11sin)()(2stLtgtfL 11sin)(,11)(2stLtgLsLtfL )()()()( tgLtfLtgtfL 11)()(2sstgLtfLTheorem 6.6.1 • Suppose F(s) = L{f (t)} and G(s) = L{g(t)} both exist for s > a 0. Then H(s) = F(s)G(s) = L{h(t)} for s > a, where • The function h(t) is known as the convolution of f and g and the integrals above are known as convolution integrals. • Note that the equality of the two convolution integrals can be seen by making the substitution u = t - . • The convolution integral defines a “generalized product” and can be written as h(t) = ( f *g)(t). See text for more details. ttdtgtfdgtfth00)()()()()(Theorem 6.6.1 Proof Outline )()()()()()()()()()()()()()()()(000 0000 0)(0 0thLdtdgtfedtdgtfeddttfgeutdttfedgduufedgdgeduufesGsFtsttstststusssu Example 1: Find Inverse Transform (1 of 2) • Find the inverse Laplace Transform of H(s), given below. • Solution: Let F(s) = 1/s2 and G(s) = a/(s2 + a2), with • Thus by Theorem 6.6.1, )sin()()()()(11atsGLtgtsFLtf)()(222assasH tdatthsHL01)sin()()()(Example 1: Solution h(t) (2 of 2) • We can integrate to simplify h(t), as follows. 222000000)sin()sin(11)sin(1)][cos(11)cos(1)cos(1)cos(1)cos(1)sin()sin()sin()()(aatatatataataattaattadaaaaatadadatdatthtttttt tdatthsHL01)sin()()()(Example 2: Initial Value Problem (1 of 4) • Find the solution to the initial value problem • Solution: • or • Letting Y(s) = L{y}, and substituting in initial conditions, • Thus )}({}{4}{ tgLyLyL )(}{4)0()0(}{2sGyLysyyLs )(13)(42sGssYs 4)(413)(22ssGsssY1)0(,3)0(),(4 yytgyyExample 2: Solution (2 of 4) • We have • Thus • Note that if g(t) is given, then the convolution integral can be evaluated. dgttttyt)()(2sin212sin212cos3)(0)(42214221434)(413)(22222sGssssssGsssYExample 2: Laplace Transform of Solution (3 of 4) • Recall that the Laplace Transform of the solution y is • Note (s) depends only on system coefficients and initial conditions, while (s) depends only on system coefficients and forcing function g(t). • Further, (t) = L-1{ (s)} solves the homogeneous IVP while (t) = L-1{ (s)} solves the nonhomogeneous IVP )()(4)(413)(22sΨsΦssGsssY 1)0(,3)0(),(4 yytgyy1)0(,3)0(,04 yyyy0)0(,0)0(),(4 yytgyyExample 2: Transfer Function (4 of 4) • Examining (s) more closely, • The function H(s) is known as the transfer function, and depends only on system coefficients. • The function G(s) depends only on external excitation g(t) applied to system. • If G(s) = 1, then g(t) = (t) and hence h(t) = L-1{H(s)} solves the nonhomogeneous initial value problem • Thus h(t) is response of system to unit impulse applied at t = 0, and hence h(t) is called the impulse response of system. 41)( where),()(4)()(22ssHsGsHssGsΨ0)0(,0)0(),(4 yytyyInput-Output Problem (1 of 3) • Consider the general initial value problem • This IVP is often called an input-output problem. The coefficients a, b, c describe properties of physical system, and g(t) is the input to system. The values y0 and y0' describe initial state, and solution y is the output at time t. • Using the Laplace transform, we obtain or 00)0(,)0(),( yyyytgcyybya )()()0()()0()0()(2sGscYyssYbysysYsa )()()()()(2200sΨsΦcbsassGcbsasyaybassY Laplace Transform of Solution (2 of 3) • We have • As before, (s) depends only on system coefficients and initial conditions, while (s) depends only on system coefficients and forcing function g(t). • Further, (t) = L-1{ (s)} solves the homogeneous IVP while (t) = L-1{ (s)} solves the nonhomogeneous IVP 00)0(,)0(,0 yyyycyybya0)0(,0)0(),( yytgcyybya00)0(,)0(),( yyyytgcyybya)()()()()(2200sΨsΦcbsassGcbsasyaybassY Transfer Function (3 of 3) • Examining (s) more closely, • As before, H(s) is the transfer function, and depends only on system coefficients, while G(s) depends only on external excitation g(t) applied to system. • Thus if G(s) = 1, then g(t) = (t) and hence h(t) = L-1{H(s)} solves the nonhomogeneous IVP • Thus h(t) is response of system to unit impulse applied at t = 0, and hence h(t) is called the impulse response of system, with 0)0(,0)0(),( yytcyybyacbsassHsGsHcbsassGsΨ221)(

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