Therefore, 1 3y1 + 9yo - 1, 11-5yo - 5yı 6. F(2) 1 1 (4.325) z +1 10 z+3 15 z – 2 and taking the inverse transform gives Yk = 16(3y1 + 9yo – 1)(-1)* +/10(1 – 5yo – 5y1)(-3)* + 1/152*. (4.326)

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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Difference Equations
Example F
Taking the z-transform of each term of the second-order equation
Yk+2 + a1Yk+1+a2Yk = Rµ
(4.313)
gives
[z?F(2) – 2?yo – zYyı] + a1[zF(z) – zyo] + azF(z) = G(2),
(4.314)
or
G(2) + yoz2
+ (a140 + Y1)z
z2 + a1z + a2
F(2) =
(4.315)
where G(z) = Z(Rk).
Let a1 = 0, a2 = 4, and R.
equation
= 0. This corresponds to the difference
Yk+2 – 4yk = 0.
(4.316)
The z-transform is, from equation (4.315), given by the expression
F(2)
zYo + Y1
(z + 2)(z – 2)
2y0 – Y1
1
2y0 + Y1
1
(4.317)
4
z + 2
4
z – 2
From Table 4.1, we can obtain the inverse transform of F(z); doing this gives
Yk = /¼(2yo – Y1)(-2)* +¼(2y0 + 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)
%3D
corresponds to a1 = 4, a2 = 3, and R = 2k. Using the fact that
G(2) = Z(2*)
(4.320)
z – 2
we can rewrite equation (4.315), for this case, as
F(2)
Yo(z + 4)
(z + 1)(z + 3)
Y1
1
(4.321)
(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
1
(4.322)
(z + 1)(z + 3)
2 z + 1
2 z +3
z + 4
3
1
1
1
(4.323)
(z + 1)(z + 3)
2 z + 1
2 z + 3
1
1
1
1
1
1
(4.324)
(z – 2)(z + 1)(z +3)
15 z – 2
6 z +1
10 z + 3
LINEAR DIFFERENCE EQUATIONS
157
Therefore,
1 1
+
15 z – 2
1 Зу1 + 9yo — 1
1 1- 5y0 – 5y1
(4.325)
6
z + 1
10
z + 3
and taking the inverse transform gives
Yk = /6(3y1 + 9y0 – 1)(-1)* + !/10(1 – 5y0 – 5y1)(-3)* + 1/152*. (4.326)
H IN
Transcribed Image Text:Difference Equations Example F Taking the z-transform of each term of the second-order equation Yk+2 + a1Yk+1+a2Yk = Rµ (4.313) gives [z?F(2) – 2?yo – zYyı] + a1[zF(z) – zyo] + azF(z) = G(2), (4.314) or G(2) + yoz2 + (a140 + Y1)z z2 + a1z + a2 F(2) = (4.315) where G(z) = Z(Rk). Let a1 = 0, a2 = 4, and R. equation = 0. This corresponds to the difference Yk+2 – 4yk = 0. (4.316) The z-transform is, from equation (4.315), given by the expression F(2) zYo + Y1 (z + 2)(z – 2) 2y0 – Y1 1 2y0 + Y1 1 (4.317) 4 z + 2 4 z – 2 From Table 4.1, we can obtain the inverse transform of F(z); doing this gives Yk = /¼(2yo – Y1)(-2)* +¼(2y0 + 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) %3D corresponds to a1 = 4, a2 = 3, and R = 2k. Using the fact that G(2) = Z(2*) (4.320) z – 2 we can rewrite equation (4.315), for this case, as F(2) Yo(z + 4) (z + 1)(z + 3) Y1 1 (4.321) (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 1 (4.322) (z + 1)(z + 3) 2 z + 1 2 z +3 z + 4 3 1 1 1 (4.323) (z + 1)(z + 3) 2 z + 1 2 z + 3 1 1 1 1 1 1 (4.324) (z – 2)(z + 1)(z +3) 15 z – 2 6 z +1 10 z + 3 LINEAR DIFFERENCE EQUATIONS 157 Therefore, 1 1 + 15 z – 2 1 Зу1 + 9yo — 1 1 1- 5y0 – 5y1 (4.325) 6 z + 1 10 z + 3 and taking the inverse transform gives Yk = /6(3y1 + 9y0 – 1)(-1)* + !/10(1 – 5y0 – 5y1)(-3)* + 1/152*. (4.326) H IN
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