differential-fall2011

differential-fall2011 - ECE 382 Review of Solutions of...

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Unformatted text preview: ECE 382 Review of Solutions of Linear Ordinary Differential Equations with Constant Coefficients We shall consider an n th-order, linear, ordinary differential equation with constant coefficients, and discuss some physical problems giving rise to such equation: a n d n x dt n + a n- 1 d n- 1 x dt n- 1 + + a 1 dx dt + a x ( t ) = q ( t ) , (1) where the coefficients a , a 1 ,..., a n are constants, a n 6 = 0, q ( t ) is the input variable and x ( t ) is the output variable of the differential equation. If we employ the differentiation operator symbols D , D 2 , D 3 ,..., etc. to denote d dt , d 2 dt 2 , d 3 dt 2 ,..., respectively, then Eq. (1) can be written as ( a n D n + a n- 1 D n- 1 + + a 1 D + a ) x ( t ) = q ( t ) . (2) Note that we are still in the time domain. We shall consider the solution for the homogeneous case first (i.e., when q ( t ) = 0) and then the in-homogeneous case (i.e., when q ( t ) 6 = 0). 1 The Homogeneous Case + _ q(t) t=0 PLANT y(t) + _ S Figure 1: Mathematical model of a physical plant. Consider that Eq. (1) is the mathematical model for some physical plant, which may be represented pictorially as in Figure 1. In Fig. 1, the plant has been operating for some time prior to t = 0. At t = 0, the switch S is opened and the input to the plant, the voltage q ( t ) , is zero for t . It is desired to find the output voltage y ( t ) for t 0. For simplicity we shall work with second-order- differential-equation examples. Figure 2 repre- sents an RLC circuit example of such a second- order plant. + _ t=0 y(t) E x(t) C R L S r Figure 2: A second-order plant example. The situation after t = 0 can be described by using Kirchhoffs voltage law around the closed circuit. If x ( t ) represents the inductor current, then L dx dt + Rx + 1 C Z t- xdt = . (3) If we differentiate this equation with respect to t , we have a second-order linear homogeneous ordinary differential equation with constant coefficients: L d 2 x dt 2 + R dx dt + 1 C x ( t ) = . (4) Copyright c 2011 C.S. George Lee 1 August 28, 2011 If we assign values to L , R and C of 1 henry, 3 ohms and 1/2 farad, respectively (not very realistic numbers, but convenient for illustrative purposes), we write Eq. (4) using operator notation as ( D 2 + 3 D + 2 ) | {z } a polynmial in D x ( t ) = ( D + 2 )( D + 1 ) x ( t ) | {z } z ( t ) = (5) By setting ( D + 1 ) x ( t ) = z ( t ) , we obtain ( D + 1 ) x ( t ) = z ( t ) , (6) ( D + 2 ) z ( t ) = . (7) Equations (6) and (7) show that we can decompose a second-order, linear differential equation into two first-order linear differential equations. One can generalize this statement to: An n th- order, linear differential equation can be decomposed into n first-order, linear differential equations....
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differential-fall2011 - ECE 382 Review of Solutions of...

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