Differential Equations

E3t vs example solve du d2y dy 4u 3 2y 2 dt dt dt when

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Unformatted text preview: tion, ⇒ Vc (s + 1) = Vs The transfer function is Vc 1 = G (s ) = Vs s+1 The characteristic polynomial is a(s) = (s + 1) so the natural response is Vcn = A1e−t The forced response is −3t = − 1 e−3t Vcf = G(−3)e 2 The total response is −t − 1 e−3t Vc = Vcn + Vcf = A1e 2 The initial condition is Vc(0) = 6 1 ⇒ Vc(0) = A1 − = 6 2 13 1 ⇒ A1 = + 6 = 2 2 13 −t 1 −3t Vc = e − e 2 2 Check: dVc + Vc = Vs dt ￿ ￿ ￿ ￿ 13 −t 3 −3t 13 −t 1 −3t − e +e + e −e 2 2 2 2 = e−3t = Vs Example Solve du d2y dy + 4u + 3 + 2y = 2 dt dt dt when u(t) = 2e−3t and the initial dy (0) conditions are = −1 and y (0) = 1. dt In operator notation, ￿ ￿ s2 + 3s + 2 y = (s + 4) u so the transfer function is y s+4 = G (s ) = 2 u s + 3s + 2 s+4 = (s + 1)(s + 2) The poles are s = −1 and s = −2. The natural response is A1e−t + A2e−2 The forced response is −3t = −3 + 4 2e−3t = e−3t G(−3)2e 9−9+2 The total response is then y (t) = A1e−t + A2e−2 + e−3t The values of A1 and A2 are determined from the initial conditions. y (0) = A1 + A2 + 1 = 1 dy (t) = −A1e−t − 2A2e−2 − 3e−3t dt dy (0) = −A1 − 2A2 − 3 = −1 dt ⇒ A1 + A2 = 0 ⇒ A1 + 2A2 = −2 Solving these two equations gives A1 = 2, The solution is A2 = −2 y (t) = 2e−t − 2e−2 + e−3t Theorem 1 If u(t) = Aept, then one solution of eqn (1) is y (t) = G(p)u(t). Theorem 2 Systems described by n m ￿ dk y ￿ dk u ak k = bk k dt dt k =0 (1) k =0 are linear. Theorem 3 All solutions of eqn. (1) with u(t) = 0 are y (t) = n ￿ i=1 Aiepit Theorem 4 Suppose that u(t) is non-zero, and that y1(t) is any one solution of (1), and let y2(t) = n ￿ Aiepit i=1 Then y1 + y2 is also a solution, for all values of A1, . . . , An. dn−1y dny dy an n + an−1 n−1 + . . . + a1 + a0 y dt dt dt dm−1u du dmu + b0 u = bn m +bm−1 m−1 +. . .+b1 dt dt dt...
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This note was uploaded on 01/22/2014 for the course ENG 282IN taught by Professor Haines during the Spring '12 term at Pima CC.

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