When a force f(t) is applied to a mass in a system, we recall that Newton s law says that. f(t) = ma = m d dt v,

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1 Impulse Functions In mny ppliction problems, n externl force f(t) is pplied over very short period of time. For exmple, if mss in spring nd dshpot system is struck by hmmer, the ppliction of the force to the mss hppens lmost instntneously. In such physicl contexts, it is cceptble to simplify the mthemticl model by ssuming tht the ppliction of the force is ctully instntneous; in other words, we think of the force s function tht tkes on nonzero vlue t only one point t in time. Mthemticins hve constructed resonble wy to model such phenomenon. We crete n object δ(t ) known s the Dirc delt function. The nme is rther misleding, s δ is not ctully function; insted, we will think of it is n d-hoc tool for modeling situtions in which force f(t) is pplied over n extremely short period of time. When force f(t) is pplied to mss in system, we recll tht Newton s lw sys tht f(t) = m = m d dt v, where m is the object s mss, its ccelertion, nd v its velocity. If the force is pplied from t = to t = b, we define f(t) dt to be the impulse of the force on the intervl t b. Of course, using the formul bove from Newton s lw, we cn rewrite the impulse s f(t) dt = m d dt v dt = mv b = m(v(b) v()). In other words, the impulse of function f(t) on the intervl t b is just the chnge in momentum of the mss in the system, so tht we cn think of the impulse s mesurement of the strength of the force. In order to describe n instntneous force t time t =, we wish to hve forcing function δ (t) whose only nonzero vlue occurs t t =, nd whose impulse t t = is 1; mthemticlly, this mens tht lim b δ (t) dt = 1. Thinking of the integrl bove s description of the re under δ (t), we see tht the sttement bove indictes tht the re under δ (t) from to b must pproch s b ; to stisfy ll of these requirements, δ (t) must hve form if t = δ (t) = if t, nd the property tht δ (t) dt = 1. 1

2 So for ny other function g(t), it cn be shown tht g(t)δ (t) dt = g(); in prticulr, we cn clculte the Lplce trnsform of δ (t): Lδ (t)} = e st δ (t) dt = e s. As with the unit step function u(t ), we rewrite δ (t) s δ(t ); then using the ides bove s motivtion, we define the Dirc delt function δ(t ) to be the object with the following properties: δ(t ) dt = 1 if t = δ(t ) = t Lδ(t )} = e s In prticulr, we interpret δ(t ) s description of force pplied (nerly) instntneously t time t = with impulse 1. Exmple. A 1 kg mss in spring nd dshpot system is ttched to spring with spring constnt k = 13 nd dshpot with dmping constnt c = 6. At time t =, the mss is relesed from its equilibrium position with n initil velocity of 1 m/s to the left. At time t = π nd t = π, the mss is struck by hmmer with impulse p = 3. Find the eqution x(t) modeling the position of the mss. Strting with the eqution mx + cx + kx = f(t), we hve m = 1, c = 6, nd k = 13; we know tht x() = nd x () = 1. In ddition, we cn pproximte the externl force f(t) by using the unit impulse functions δ(t π) nd δ(t π); of course, since the impulse in both cses is p = 3, we think of f(t) s f(t) = 3δ(t π) + 3δ(t π). Thus we need to solve the initil vlue problem x + 6x + 13x = 3δ(t π) + 3δ(t π), x() =, x () = 1. Applying the Lplce trnsform to the entire eqution, we hve s X(s) sx() x () + 6sX(s) 6x() + 13X(s) = πs + πs s X(s) sX(s) + 13X(s) = πs + πs s X(s) + 6sX(s) + 13X(s) = πs + πs 1 X(s)(s + 6s + 13) = πs + πs 1 X(s) = πs s + 6s πs s + 6s s + 6s + 13 X(s) = πs (s + 3) πs (s + 3) (s + 3) + 4.

3 Next, we need to pply the inverse trnsform to return to n eqution for x(t); we ll consider ech term seprtely. Beginning with πs (s + 3) + 4, we note tht we cn think of this function s ( 3 e πs (s + 3) + 4 whose inverse trnsform we cn find using Theorem 1 from section 7.5: If Lg(t)} = G(s), then L 1 e s G(s)} = u(t )g(t ). We simply need to find the inverse trnsform of G(s) = ), (s + 3) + 4 ; we recognize this s trnsltion of whose inverse trnsform is thus the inverse trnsform of G(s) is s + 4, sin t; g(t) = e 3t sin t. Thus the inverse Lplce trnsform of our originl frction is } L 1 πs (s + 3) = 3 u(t π)g(t π) + 4 = 3 u(t π)e 3(t π) sin((t π)). Next, we need to find the inverse trnsform of using the sme ides, we think of G(s) s πs (s + 3) + 4 = 3 e πs (s + 3) + 4 ; G(s) = (s + 3) + 4, whose inverse trnsform is thus the inverse trnsform of g(t) = e 3t sin t; πs (s + 3) + 4 3

4 is L 1 πs (s + 3) + 4 } = 3 u(t π)g(t π) = 3 u(t π)e 3(t π) sin((t π)). Finlly, the inverse trnsform of 1 (s + 3) + 4 = 1 (s + 3) + 4 is 1 e 3t sin t. Thus we hve } x(t) = L 1 πs (s + 3) πs (s + 3) (s + 3) + 4 } } } = L 1 πs (s + 3) + L 1 πs + 4 (s + 3) L (s + 3) + 4 = 3 u(t π)e 3(t π) sin((t π)) + 3 u(t π)e 3(t π) sin((t π)) 1 e 3t sin t. The eqution for x(t) is grphed below: 4

5 Exmple. A 1 kg mss is ttched to spring with k = 1. Initilly, the mss is t rest t equilibrium. At t =, the mss is struck with hmmer with impulse p = 1; the blow is repeted t ech integer multiple of π. Find the function x(t) modeling the mss s position. Beginning with the eqution mx + cx + kx = f(t), x() = x () =, we hve m = 1, c =, nd k = 1. We cn think of the externl force f(t) s n infinite sum of unit impulse functions: whose Lplce trnsform is f(t) = δ(t) + δ(t π) + δ(t π) δ( nπ) +... = δ(t nπ), Lf(t)} = Lδ(t)} + Lδ(t π)} + Lδ(t π)} Lδ(t nπ)} +... = 1 + e πs + e πs e nπs +... = e nπs. Thus our initil vlue problem is x + x = δ(t nπ), x() = x () =. Applying the Lplce trnsform to ech side of the eqution, we hve s X(s) sx() x () + X(s) = e nπs Since the inverse trnsform of the formul for x(t) is G(s) = e nπs s + 1 x(t) = s X(s) + X(s) = X(s) = e nπs e nπs s + 1. is g(t) = u(t nπ) sin(t nπ), u(t nπ) sin(t nπ). Of course, we cn use n identity to rewrite sin(t nπ): sin(t nπ) = sin t cos nπ cos t sin nπ = sin t cos nπ, = sin t sin t 5 n even n odd.

6 Rewriting the formul for x(t), we hve x(t) = = sin t sin t sin t sin t sin t + sin t. sin t t < π π t < π sin t. π t < 3π t < π π t < π π t < 3π or x(t) = sin t kπ t < (k + 1)π (k + 1)π t < (k + )π. The function x(t) is grphed below: 6