082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

22.12 The z transform and difference equations 719
Example22.30 Solve the difference equationy[k +1] −3y[k] = 0,y[0] = 4.
Solution Taking theztransformof both sides of the equation we have
Z{y[k +1] −3y[k]} = Z{0} = 0
since Z{0} = 0.Usingthe properties of linearitywefind
Z{y[k +1]} −3Z{y[k]} = 0
Usingthe first shift theorem on the first of the terms on the l.h.s. weobtain
zZ{y[k]} −4z −3Z{y[k]} = 0
Writing Z{y[k]} =Y(z), thisbecomes
(z −3)Y(z) =4z
so that
Y(z)=
4z
z −3
Thefunctiononther.h.s.istheztransformoftherequiredsolution.Invertingthis,from
Table 22.2 we findy[k] = 4(3) k .
Higherorderequationsaretreatedinthe same way.
Example22.31 Solve the second-order difference equation
y[k +2] −5y[k +1] +6y[k] = 0 y[0] = 0 y[1] = 2
Solution Takingtheztransformofbothsidesoftheequationandusingthepropertiesoflinearity
wehave
Z{y[k +2]} −5Z{y[k +1]} +6Z{y[k]} = 0
Fromthe first shift theoremwehave
where
z 2 Y(z) −z 2 y[0] −zy[1] −5(zY(z) −zy[0]) +6Y(z) = 0
Y(z) = Z{y[k]}
Substitutingvalues forthe conditionsgives
Hence
so that
z 2 Y(z) −0z 2 −2z −5(zY(z) −0z) +6Y(z) = 0
(z 2 −5z +6)Y(z) =2z
Y(z) = Z{y[k]} =
z 2 Y(z) −5zY(z) +6Y(z) = 2z
2z
(z −2)(z −3)