{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

differential-final

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

This preview shows pages 1–3. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 L x(t) p(t) Figure 3: An example circuit. If the voltage p ( t ) is given by p ( t ) = E sin ω t , then Kirchhoff’s law applied to this loop gives us L dx dt + Rx = E sin ω t , (13) or dx dt + R L x = E L sin ω t . (14) Equation (14) is now in the form of Eq. (8). An integrating factor is e R L t . Applying the formula in Eq. (12) we obtain x ( t ) = x ( 0 ) e - R L t + e - R L t Z t 0 ( E L sin ω s ) e R L s ds . (15) From a table of integrals, we obtain x ( t ) = x ( 0 ) e - R L t + E L e - R L t " e R L t ( R L sin ω t - ω cos ω t ) + ω ( R
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}