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)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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Explain the determine red

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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