Taking the z-transform of each term of the second-order eq Yk+2 + a1Yk+1 + a2Yk = Rk gives

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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Difference Equations
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) – 2²yo – zyı] + a1[zF(z) – zyo] + azF(z) = G(2),
(4.314)
or
G(2) + Yoz2 + (a1Y0 + Y1)z
F(2) =
(4.315)
z2 + a1z + a2
where G(2) = Z(Rk).
0, а2
Let aj =
4, and Rk
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(2)
zyo + Y1
2yo – Y1
1
2yo + Y1
1
(4.317)
(z + 2)(z – 2)
4
z + 2
4
7 - z
From Table 4.1, we can obtain the inverse transform of F(z); doing this gives
Yk
14(2yo – Y1)(-2)* + 1¼(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'
we can rewrite equation (4.315), for this case, as
F(2)
yo(z + 4)
Y1
+
(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
(4.322)
(ɛ + 2)(I + 2)
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
1
(4.324)
(z – 2)(z + 1)(2 + 3)
15 z – 2
6 z +1
10 z + 3
Transcribed Image Text:Difference Equations 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) – 2²yo – zyı] + a1[zF(z) – zyo] + azF(z) = G(2), (4.314) or G(2) + Yoz2 + (a1Y0 + Y1)z F(2) = (4.315) z2 + a1z + a2 where G(2) = Z(Rk). 0, а2 Let aj = 4, and Rk 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(2) zyo + Y1 2yo – Y1 1 2yo + Y1 1 (4.317) (z + 2)(z – 2) 4 z + 2 4 7 - z From Table 4.1, we can obtain the inverse transform of F(z); doing this gives Yk 14(2yo – Y1)(-2)* + 1¼(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' we can rewrite equation (4.315), for this case, as F(2) yo(z + 4) Y1 + (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 (4.322) (ɛ + 2)(I + 2) 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 1 (4.324) (z – 2)(z + 1)(2 + 3) 15 z – 2 6 z +1 10 z + 3
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