Example F Taking the z-transform of each term of the second-order equation Yk+2 +a1Yk+1+a2Yk = Rk (4.313) gives [z?F(2) – 2*yo – 2yı] + a1[zF(2) – zyo] + a2F(z) = G(2), (4.314) or G(2) + yoz2 + (a1Y0 + Y1)z F(2) = (4.315) %3D 22 + a12 + a2 where G(z) = Z(Rx). 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(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 = 4(2y0 – y1)(-2)* + /¼(2yo + yı)2*. (4.318) Since yo and y1 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) %3D 2 we can rewrite equation (4.315), for this case, as yo(z + 4) (z + 1)(2 + 3) Y1 1 (4.321) (2+1)(2+3) (2 – 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) 2 z + 3' (z + 1)(z +3) z + 4 (z + 1)(z + 3) 2 z +1 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)(x +3) 15 z -2 6 z +1 10 z +3

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Example F
Taking the z-transform of each term of the second-order equation
Yk+2 +a1Yk+1+a2Yk =
(4.313)
gives
(2²F(2) – 2²yo – zyı) + a1[2F(2) – zyo) + azF(2) = G(2),
(4.314)
or
G(2)+ yoz² + (a1Y0 + Y1)z
22 + a1z + a2
F(2)
(4.315)
where G(2) = Z(Rk).
Let a1 = 0, a2 = 4, and Rk
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
2y0 – Y1
1
2y0 + Y1
1
(4.317)
(z + 2)(z – 2)
4
z + 2
4
2 – 2
From Table 4.1, we can obtain the inverse transform of F(z); doing this gives
Yk = /4(2y0 – Y1)(-2)* + /¼(2yo + yı)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 Rk = 2k. Using the fact that
G(2) = Z(2*)
(4.320)
2
we can rewrite equation (4.315), for this case, as
F(2)
yo(z + 4)
(z + 1)(2 + 3)
Y1
1
(4.321)
(z + 1)(2 + 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)
2 z + 3
(z + 1)(2 + 3)
z + 4
(z + 1)(z + 3)
2 z + 1
1
1
(4.323)
2 z + 1
2 z + 3
1
1
1
1
1
1
1
(4.324)
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 = (4.313) gives (2²F(2) – 2²yo – zyı) + a1[2F(2) – zyo) + azF(2) = G(2), (4.314) or G(2)+ yoz² + (a1Y0 + Y1)z 22 + a1z + a2 F(2) (4.315) where G(2) = Z(Rk). Let a1 = 0, a2 = 4, and Rk 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 2y0 – Y1 1 2y0 + Y1 1 (4.317) (z + 2)(z – 2) 4 z + 2 4 2 – 2 From Table 4.1, we can obtain the inverse transform of F(z); doing this gives Yk = /4(2y0 – Y1)(-2)* + /¼(2yo + yı)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 Rk = 2k. Using the fact that G(2) = Z(2*) (4.320) 2 we can rewrite equation (4.315), for this case, as F(2) yo(z + 4) (z + 1)(2 + 3) Y1 1 (4.321) (z + 1)(2 + 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) 2 z + 3 (z + 1)(2 + 3) z + 4 (z + 1)(z + 3) 2 z + 1 1 1 (4.323) 2 z + 1 2 z + 3 1 1 1 1 1 1 1 (4.324) 2)(z + 1)(z + 3) 15 z – 2 6 z + 1 10 z + 3
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