differential-final

# differential-final - ECE 382 REVIEW OF SOLUTIONS OF LINEAR...

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ECE 382 REVIEW OF SOLUTIONS OF LINEAR ORDINARY DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS We shall consider ordinary differential equations of the form a n d n x dt n + a n - 1 d n - 1 x dt n - 1 + ··· + a 1 dx dt + a 0 x = q ( t ) (1) where the coefﬁcients a 0 , a 1 ,..., a n are constant, and discuss some physical problems giving rise to such equations. If we employ the symbols D , D 2 , D 3 ,..., etc. to denote d dt , d 2 dt 2 , d 3 dt 3 ,..., respec- tively, then Eq. (1) may be written as ( a n D n + a n - 1 D n - 1 + ··· + a 1 D + a 0 ) x = q ( t ) . (2) 1 The Homogeneous Case We shall suppose that Eq. (2) is the mathematical model for some physical plant which may be represented pictorially as in Figure 1. + _ q(t) t=0 PLANT y(t) + _ S Figure 1: Mathematical model of a physical plant. Here we suppose that it is desired to ﬁnd the output voltage y ( t ) . 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 0 . For simplicity we shall work with second-order examples. Figure 2 represents a concrete 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 Kirchhoff’s voltage law around the closed circuit. If x ( t ) represents the inductor current, then L dx dt + Rx + 1 C Z t - xdt = 0 . (3) If we differentiate this equation with respect to t , we have a second-order linear homogeneous ordi- nary differential equation with constant coefﬁcients: L d 2 x dt 2 + R dx dt + 1 C x = 0 . (4) 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 may write Eq. (4) using operator notation as ( D 2 + 3 D + 2 ) x = ( D + 2 )( D + 1 ) x = 0 (5) 1

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Setting ( D + 1 ) x = z , (6) we obtain ( D + 2 ) z = 0 . (7) From Eqs. (6) and (7), it appears that the building blocks of all linear differential equations with constant coefﬁcients are the ﬁrst-order differential equations of the form ˙ x + ax = p ( t ) . (8) By inspection, one can see that the function e at is an integrating factor of Eq. (8). This means that if one multiplies Eq. (8) by e at , then the left-hand side becomes derivative. We obtain ( ˙ x + ax ) e at = p ( t ) e at (9) or d dt ( xe at ) = e at p ( t ) . (10) Integrating this last equation with respect to t from t = 0 to some generic time t , we obtain x ( t ) e at - x ( 0 ) = Z t 0 e a τ p ( τ ) d τ (11) where τ is the dummy variable of integration. Thus, x ( t ) = e - at x ( 0 )+ e - at Z t 0 e a τ p ( τ ) d τ (12) where x ( 0 ) is an arbitrary constant. It is to be determined by the supposedly known initial value of the variable x ( t ) . As an example of the application of this formula, consider the circuit in Figure 3. + _ R
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## This note was uploaded on 02/04/2012 for the course ECE 382 taught by Professor Staff during the Spring '08 term at Purdue.

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differential-final - ECE 382 REVIEW OF SOLUTIONS OF LINEAR...

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