MIT18_03S10_c29

MIT18_03S10_c29 - 18.03 Class 29, April 16, 2010 Laplace...

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18.03 Class 29 , April 16, 2010 Laplace Transform IV: The pole diagram 1. Another example 2. t-shift rule 3. Poles 4. What the pole diagram of F(s) says about f(t) [1] We saw that if p(D)w = delta(t) , i.e. a_n w^(n) + . .. + a_1 w' + a_0 w = delta(t) with rest initial conditions, then p(s) W(s) = 1 , or W(s) = 1/p(s) , or w(t) = L^{-1}(1/p(s)) w(t) = weight function W(s) = transfer function This is another RULE. Other input signals? x" + 4x = cos(2t) rest initial conditions (No delta functions in this signal, so we know that x(0+) = 0 and x'(0+) = 0, but that will come out automatically.) X = W(s) LT[cos(2t)] = 2s/(s^2+4)^2 From our latest addition to the table, we find that x = u(t) 1/2 t sin(t) (Resonance!) GENERAL FACT: p(D) x = f(t) with rest initial conditions has Laplace transformed equation p(s) X(s) = F(s) with solution X(s) = W(s) F(s) In the s-domain, the system response is obtained by multiplying by the transfer function! [2] Another rule: Let a > 0 and define f_a(t) = 0 if t < a
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= f(t-a) if t > a (If f_s(t) contains c delta(t) , I also want f_a(t) to contain c delta(t-a) . So for example delta_a(t) = delta(t-a) .) If you know F(s) , what is the LT of f_a(t) ? integral_{0-}^infty f_a(t) e^{-st} dt = integral_{a-}^\infty f(t-a) e^{-st} dt Substitution: u = t-a , du = dt , ... = integral_{0-}^\infty f(u) e^{-s(u+a)} du = e^{-as} integral_{0-}^infty f(u) e^{-su} du = e^{-as} F(s) So we have the t-shift rule:
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MIT18_03S10_c29 - 18.03 Class 29, April 16, 2010 Laplace...

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