9.2 • SECOND-ORDER HOMOGENEOUS DIFFERENCE EQUATIONS 367Calculate the residues for f(z) = Y(z)zn-l
poles2z3 n- 1 t th
(z+i)(z-i)(z+i)z a e2z^3Res[/(z), -1) = lim ( 2 ) zn-I = (- 1)",
z~-1 z ·+ lR es [/( z ) ,i ") = li m( l)( 2z3 .)z n-1 = (1 - i)·n d
2
+ -
2
i ,an
z-• Z + z +i
2 3 1.
Res[/(z), -i) = lim. ( ;( .) zn-I = (-
2
- 2
i )(- i)".
·-- · z + 1 z - iThus the solution isy [ n I = -( ) 1 n ( + -^1 + -i ) i ·n + ( -^1 - -i ) ( - i ")n
2 2 2 2 'which can be rewritten asy[n] = (-l)n + ~(i" + (-i)")-;i(in- (- i)")
= ein" + ~(eiT' + e=-i?) - ;i (el.¥ -e=-i?)
'Ir. 'Ir= cos(7rn)] + cos(
2
n) - sin( 2n).• EXAMPLE 9. 19 Solve y[n + 2) - yl2y[n + l ] + y[n) = 0 with y[O) = 2 and
y[l] = V2.
Solution Take the z-transforms of each term and getz^2 (Y(z) - 2-v'2z-^1 ) - v'2(z(Y(z) - 2)) + (Y(z)) = 0.
Solve for Y(z) and get Y(z) =^2 :2- 2z = 2•
2- 2•.
z2- 2 z+l z-!±! .,,.. (z-7z 1- •
Calculate the residues for /(z) = Y(z)zn-I at the poles 1ji and 1Tz:
R [/() l+i 1 _ 1. 2z
2
es z --- 1m -v'2zn. z _^1 _ 1 +- --i(l--+i)n-l_ ( l +- --i)"
' J2 ·-17. z - Ti J2 V2 J2Similarly, Res[f(z), lTzJ = Res[/(z), lTzJ = ( ~ r = ( ~ f. Therefore, the
solution is
y[n]= (_i^1 + ")n + (1 ~ ")"
J2 J2 '