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we can rewrite equation (4.315), for this case, as F(2) yo(z +4) + Y1 1 (4.321) (z + 1)(2 + 3) + (z + 1)(z + 3) ' (z – 2)(z+ 1)(z+ 3)' The three terms on the right-hand side of this equation have the following partial-fraction expansions: 1 1 1 1 (4.322) (z + 1)(z + 3) z + 4 (z + 1)(z + 3) 2 z + 1 2 z + 3' 1 1 1 (4.323) 2 z + 1 2 z +3 1 1 1 1 1 1 1 (4.324) %3D (z – 2)(z+1)(z +3) 15 z - 2 6 z +1 10 z +3

we can rewrite equation (4.315), for this case, as F(2) yo(z +4) + Y1 1 (4.321) (z + 1)(2 + 3) + (z + 1)(z + 3) ' (z – 2)(z+ 1)(z+ 3)' The three terms on the right-hand side of this equation have the following partial-fraction expansions: 1 1 1 1 (4.322) (z + 1)(z + 3) z + 4 (z + 1)(z + 3) 2 z + 1 2 z + 3' 1 1 1 (4.323) 2 z + 1 2 z +3 1 1 1 1 1 1 1 (4.324) %3D (z – 2)(z+1)(z +3) 15 z - 2 6 z +1 10 z +3

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Example F Taking the z-transform of each term of the second-order equation Yk+2 + a1Yk+1+a2Yk = R¢ (4.313) gives (2?F(2) – 2²yo – zyı) + a1[zF(2) – zyo) + a2F(2) = G(z), (4.314) %3D or G(2) + yoz² + (a1Y0 + Y1)z F(2) (4.315) z2 + a1% + a2 where G(2) = Z(Rx). Let a1 = 0, a2 = 4, and R = 0. This corresponds to the difference equation Yk+2 - 4Yk = 0. (4.316) The z-transform is, from equation (4.315), given by the expression F(z) zYo + Y1 2y0 – Y1 1 2y0 + Y1 1 + (4.317) (z + 2)(2 – 2) 4 z + 2 4 Z – 2 From Table 4.1, we can obtain the inverse transform of F(z); doing this gives Yk = /¼(2y0 – Y1)(-2)* + '/¼(2yo + y1)2*. (4.318) Since yo and yı are arbitrary, this gives the general solution of equation (4.316). The equation Yk+2 + 4yk+1 + 3yk 2k (4.319) corresponds to a1 = 4, a2 = 3, and R = 2k. Using the fact that G(2) = Z(2*) (4.320) 2 - 2 we can rewrite equation (4.315), for this case, as F(2) (- E+1)(: + 3) * (z + 1)(z + 3) yo(z + 4) Y1 1 (4.321) (z – 2)(2 + 1)(z + 3)* The three terms on the right-hand side of this equation have the following partial-fraction expansions: 1 1 1 1 (4.322) (z + 1)(z + 3) z + 4 (z + 1)(z + 3) 2 z + 1 2 z + 3 3 1 1 1 (4.323) 2 z + 1 2 z + 3 1 1 1 1 1 1 1 (4.324) (z – 2)(z + 1)(z + 3) 15 z – 2 6 z + 1 10 z + 3
Transcribed Image Text:Example F Taking the z-transform of each term of the second-order equation Yk+2 + a1Yk+1+a2Yk = R¢ (4.313) gives (2?F(2) – 2²yo – zyı) + a1[zF(2) – zyo) + a2F(2) = G(z), (4.314) %3D or G(2) + yoz² + (a1Y0 + Y1)z F(2) (4.315) z2 + a1% + a2 where G(2) = Z(Rx). Let a1 = 0, a2 = 4, and R = 0. This corresponds to the difference equation Yk+2 - 4Yk = 0. (4.316) The z-transform is, from equation (4.315), given by the expression F(z) zYo + Y1 2y0 – Y1 1 2y0 + Y1 1 + (4.317) (z + 2)(2 – 2) 4 z + 2 4 Z – 2 From Table 4.1, we can obtain the inverse transform of F(z); doing this gives Yk = /¼(2y0 – Y1)(-2)* + '/¼(2yo + y1)2*. (4.318) Since yo and yı are arbitrary, this gives the general solution of equation (4.316). The equation Yk+2 + 4yk+1 + 3yk 2k (4.319) corresponds to a1 = 4, a2 = 3, and R = 2k. Using the fact that G(2) = Z(2*) (4.320) 2 - 2 we can rewrite equation (4.315), for this case, as F(2) (- E+1)(: + 3) * (z + 1)(z + 3) yo(z + 4) Y1 1 (4.321) (z – 2)(2 + 1)(z + 3)* The three terms on the right-hand side of this equation have the following partial-fraction expansions: 1 1 1 1 (4.322) (z + 1)(z + 3) z + 4 (z + 1)(z + 3) 2 z + 1 2 z + 3 3 1 1 1 (4.323) 2 z + 1 2 z + 3 1 1 1 1 1 1 1 (4.324) (z – 2)(z + 1)(z + 3) 15 z – 2 6 z + 1 10 z + 3
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