9.2 • SECOND-ORDER HOMOGENEOUS DIFFERENCE EQUATIONS 365Solution
(a) Take the ~transform of each termz^2 (Y(z) - 1 - 5z-^1 ) - 4(z(Y(z) - 1)) + 5(Y(z)) = 0.Solve for Y(z) and get Y(z) = z'/.ifz"+s = (z-2.::rrz·-2-i)'
Calculate the residues of for f(z) = Y (z)zn-I = (z-z.::)t;_ 2 _i)zn-l
at the polesRes[f(z), 2 + i) = Jim. z
2
+ z. zn-l =^5 -^5 i (2 + i)n-1
z-+2+i(z - 2+t) 2=-1 --3i ( 2 +i ·}n an d
2 'R es [/( z )^2 -i·1 -- 1· im z2 + z z n- 1 -- --5 + 5i(2 -i ·)n-1
' • - 2 - i (z - 2 - i) 2
= 1 + 3i ( 2 _ i}n.
2Therefore, the solution isy[n] = Res[/(z), 2 + i] + Res(/(z), 2 - i]
yn [ 1 = 1 -- 3i ( 2 ·)n^1 + 3i ( 2 ·}n
2
- t + -
2
- t ,
- t + -
which agrees with the result of Example 9.14.Remark 9.9Observe that Res[f(z), 2-i] = Res[/(z), 2 + i] = Res[f(z) , 2 + i].
(b) Take the ~transform of each term2 z zz (Y(z) - 1 - oz-^1 ) - 4(z(Y(z) - 1)) + 5(Y(z)) =. +
1.
z - 1 - i z- + iS o ve 1 f or Y( z ) an d ge t Y( z } = (•- I+i)(z•^4 - 6z-l-i)(^3 + 1z-2+i)(2z^2 - 10z z-2-i) ·
Calculate the residues of f (z) = Y(z)zn-l =
z^4 - 6z^3 +12z^2 - 10z zn-1
(z- l+iHz-l-i)(z-2+i)(z-2-i)