Differential Equations

# 1 and assume a solution of the form y t bept n dk

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2 y 2 ) dt = k =0 m ￿ k =0 So dk b k k (c 1 u 1 + c 2 u 2 ) dt y3(t) = c1y1(t) + c2y2(t) is the output resulting from the input u3(t) = c1u1(t) + c2u2(t) since y3(t) and u3 obey (1), proving that such systems are linear. Natural Response Can there be a non-zero output when u(t) = 0 ? Set u(t) = 0 in eqn. (1), and assume a solution of the form y (t) = Bept. n ￿ dk y ak k = 0 dt k =0 n ￿ = k =0 = Bept ak pk Bept n ￿ k =0 ak p k = 0 The only way that this equation can be obeyed for all t is that B = 0 or n ￿ ak pk = 0 k =0 This says that p must be a pole of the system. The transfer function ￿m b k sk k G(s) = ￿n=0 ak sk k =0 has n poles. Let’s denote them by p1, p2, . . . , pn Consider y (t) = n ￿ Aiepit i=1 By linearity, this y (t) is also a solution. This shows that when u(t) = 0, y (t) = n ￿ Aiepit i=1 is a solution of eqn. (1), no matter what the values of the constants A1, . . . , An. Theorem 3 All solutions of eqn. (1) with u(t) = 0 are y (t) = n ￿ Aiepit i=1 Proof We showed above that such y ’s are solutions. These solutions are called the natural response. For this type of system, the natural response is a sum of complex exponentials, the frequencies of which coincide with the poles of the system. 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. Consequence To obtain a unique solution, need to specify the n initial conditions dy d2y dn−1y (0), 2 (0), . . . n−1 (0) y (o), dt dt dt which then determine the values of A1 , . . . , A n . Example ✟ ✟✟ ✟ ❡✟ ✟ ❡ 2k Ω 1mF The switch is closed at t = 0, and Vc(t) = 6V for t ≤ 0. Find Vc(t) for t &gt; 0. + Vc − dVc Vc C + =0 dt R Vc dVc Vc dVc + = + =0 ⇒ dt RC dt 2 In operator notation, ￿ ￿ 1 Vc s + =0 2 The characteristic polynomial is ￿ ￿ 1 a( s ) = s + 2 Its roots are the system’s poles. 1 There is one pole, at s = − . 2 So all solutions are Vc(t) = A1ep1t = A1e−1/2t Check: dVc Vc 1 −1/2t + 1 A e−1/2t = 0 + = − A1 e dt 2 2 21 The value of A1 is determined by the initial condition Vc(0) = 6 Vc(0) = A1e0 = A1 = 6 ⇒ Vc(t) = 6e−1/2t Example ✟ ✟✟ ✟ ❡✟ ✟ ❡ 1k Ω ✬✩ + Vs = e−3t ✫✪ − The switch is closed at t = 0, and Vc(t) = 6V for t ≤ 0. Find Vc(t) for t &gt; 0. 1mF + Vc − For t &gt; 0, dVc Vc − Vs + =0 C dt R dVc Vc dVc ⇒ + = + (Vc − Vs) = 0 dt RC dt In operator nota...
View Full Document

## 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.

Ask a homework question - tutors are online