Example E Consider the first-order equation Yk+1 - ayk = Rr, (4.294) where a is a constant. Taking the z-transform of each term in this equation gives zF(2) – zyo – aF(2) = G(2), (4.295) where G(z) = Z(R). Solving for F(2), we obtain G(2) + zyo F(2) = (4.296) Z - a Suppose R = 0; then G(2) 0 and equation (4.296) becomes yỗ F(2) = (4.297) a From Table 4.1, we see that the inverse transform of z/(z-a) is ak. Therefore, the general solution to the difference equation Yk+1 - ayk = 0 (4.298) is Yk = Yoak. (4.299) as can be readily checked by using some other method of solution. Note that there is one arbitrary constant, namely, yo. LINEAR DIFFERENCE EQUATIONS 155 Now suppose R = 1, with a + 1. Therefore, from equation (4.296), we have yoz2 + (1 - yo)z (n - 2)( 2/(z - 1) + zYo F(2) = (4.300) z - a where G(2) = Z(1) (4.301) 2 -1 Therefore, writing F(2) 2(A+ B) - (aA + B) (z – 1)(z - a) В (4.302) 2 - a and comparing with equation (4.300) gives the following two equations for A and B: A +B = yo, (4.303) aA + B = yo - 1. Solving gives (1- a)yo - 1 B = A = (4.304) 1- a 1- a and 1 (1 — а)уo — 1 F(Z) (4.305) a z - 1 1- a From Table 4.1, we determine the inverse transform of equation (4.305) to be 1 (1-a)yo -1ak. (4.306) 1- a 1- a This is the general solution to the difference equation
Example E Consider the first-order equation Yk+1 - ayk = Rr, (4.294) where a is a constant. Taking the z-transform of each term in this equation gives zF(2) – zyo – aF(2) = G(2), (4.295) where G(z) = Z(R). Solving for F(2), we obtain G(2) + zyo F(2) = (4.296) Z - a Suppose R = 0; then G(2) 0 and equation (4.296) becomes yỗ F(2) = (4.297) a From Table 4.1, we see that the inverse transform of z/(z-a) is ak. Therefore, the general solution to the difference equation Yk+1 - ayk = 0 (4.298) is Yk = Yoak. (4.299) as can be readily checked by using some other method of solution. Note that there is one arbitrary constant, namely, yo. LINEAR DIFFERENCE EQUATIONS 155 Now suppose R = 1, with a + 1. Therefore, from equation (4.296), we have yoz2 + (1 - yo)z (n - 2)( 2/(z - 1) + zYo F(2) = (4.300) z - a where G(2) = Z(1) (4.301) 2 -1 Therefore, writing F(2) 2(A+ B) - (aA + B) (z – 1)(z - a) В (4.302) 2 - a and comparing with equation (4.300) gives the following two equations for A and B: A +B = yo, (4.303) aA + B = yo - 1. Solving gives (1- a)yo - 1 B = A = (4.304) 1- a 1- a and 1 (1 — а)уo — 1 F(Z) (4.305) a z - 1 1- a From Table 4.1, we determine the inverse transform of equation (4.305) to be 1 (1-a)yo -1ak. (4.306) 1- a 1- a This is the general solution to the difference equation
Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Transcribed Image Text:Example E
Consider the first-order equation
Yk+1 - ayk = Rk,
(4.294)
where a is a constant. Taking the z-transform of each term in this equation
gives
zF(z) – zyo – aF(z) = G(z),
(4.295)
where G(2) = Z(Rx). Solving for F(2), we obtain
G(z) + zyo
F(2) =
(4.296)
2 - a
Suppose R = 0; then G(z) = 0 and equation (4.296) becomes
F(z) =
(4.297)
2 - a
From Table 4.1, we see that the inverse transform of z/(z-a) is a*. Therefore,
the general solution to the difference equation
Ук+1 — ау, —D 0
(4.298)
is
Yk = Yoa",
(4.299)
as can be readily checked by using some other method of solution. Note that
there is one arbitrary constant, namely, yo.
LINEAR DIFFERENCE EQUATIONS
155
Now suppose Rk = 1, with a + 1. Therefore, from equation (4.296), we
have
z/(z – 1) + zYO
Yoz2 + (1 – yo)z
(2 — 1) (2 — а)
F(z) =
(4.300)
z - a
where
G(2) = Z(1)
(4.301)
Z - 1
Therefore, writing
F(2)
A
z(A + B) – (aA + B)
(z – 1)(z – a)
В
(4.302)
Z - 1
z - a
and comparing with equation (4.300) gives the following two equations for A
and B:
A+B= y0;
(4.303)
аА + В %3D yо — 1.
Solving gives
1
A =
(1 — а)уо — 1
В —
(4.304)
1- a
and
(1 —а)уo
F(Z)
(4.305)
az - 1
1- a
2 - a
From Table 4.1, we determine the inverse transform of equation (4.305) to be
1
Yk =
(1 —а)yо — 1k
(4.306)
1- a
1- a
This is the general solution to the difference equation
(4 207)
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