612 ANSWERSresidues of f(z) = Y(z)zn- l at the poles Res[f(z), 3 +ii = limz ..... 3+i(z -
3 - ') i (z - J +2.2-Sz i)(z- 3-i) z n-l-1· - lmz-+3+ • .2»z-3-+i az z n-1=(2+ 4')(3i +t ')n-l _ (l + -
i)(3 + ir. At the conjugate pole we can use the computation Res[f(z), 3 -
i] = Res[f (z), 3 + ii = (1 + i)( 3 + i)" = (1 - i)(3 - i)n. Therefore, y[nl =
Res[f{z), 3 +ii+ Res[f(z), 3 - ii = (1 + i)(3 + i)" + (1 - i){3 - i)n.
3a. The characteristic equation r^2 -v'2r+ 1 = (r-^1 ,;i )(r-$) = 0 has complex
roots r 1 =^1 ,j';. The general solution is y[n) = c 1 ();) n + c 2 ( 1;i)"'. Solve
the linear system y[OI = c 1 + c2 = 2, y[ll = \l;;> c1 + <Al C2 = v'2 and
( )
n ( )n ~ - <n•
get c 1 = c 2 = 1. Therefore, y[n ] = ~ + 7! = 2 • +r •-
2 cos(in).Sa. The characteristic equation r^2 - r - 1 = (r - l-/^5 )(r -^1 + 2 ./5) = 0
has roots r 1 = i+/^5 and r 2 =^1 - /^5. The general solution is y[nl =
c 1 ( ¥f + C2 (^1 - 2 Af. Solve the linear system y[OI = c 1 + c 2 = 1,
y[l] = ¥c1^1 - 2 v'SC2 = 1 and get c1 = 4 and c2 = -4. Therefore, y[n] =
4 ( ¥ f - 4 ( l- 2 Af, and {Yn}:=o = {O, 1, 1, 2,3, 5,8, 13 , 21 , 34,. .. },
which is the sequence of Fibonacci numbers.7a. Take z-transforms and get z^2 (Y(z) - l -4z-^1 )-8z(Y(z)-1)+15Y(z) = ,.: 4.
S l /: Y( ) _ z3- sz2+ 17z _ z(^3) -8z't l 7.z _ z (^3) -Sz^2 tt7z C I
0 ve ior z - z•- 12z2+47z- 60 - (z- 4 )(z'- 8 ztl5) - (<-3)(•-4)(,z-5). a -
culate the residues f(z) = Y(z)zn-l at the poles Res[f(z), 31 = lim,_3(z -
3) z 3-8z^2 t17z zn- 1 = lim z^3 - Sz'tl7• zn-1 = 3. 3n- I = 3n and
(•-3)(z-4)(z-5) z-+3 (.z-4)(z-5) '
Re [f( ) 41 - Ii ( 4) .3-8.'~^17 .z n- 1 - r z(^3) - sz2+ 17z n - 1 -
s z ' - m,_ 4 z - (z-3)(z-4(z-5) z - un ...... 4 (z-JJ(z-5) z -
-4 · 4n-l = _4n ' and Res[f(z) 5] • = lim • - S (z -5) (z- 3•'-s)(zz'fj- 4^17 (z - 5) z zn - t =
I. z(^3) -s»+t7z n - 1 5 5n- l 5n Th t [ ] - R [f( ) 31 +
lillz-+ 5 (z-3)(z-4) z =. =. ere ore, y n - es z '
Res[f (z), 4] + Res[f(z), 51=3n - 4n + 5"'.
7c. Use the same method as 7a to get 3n - 4 (4 + ir - ~(4 -ir.
9a. Take z-transforms and get z^2 (Y(z) - 1 - l z -^1 ) - z(Y(z) - 1) +! Y(z) =
- Solve for Y(z) = 4 ,.:i~Z:z+l = ( 2 :.:.~)2 = (>~~)". Res[f(z), !J =
r 1mz-~ az a [( z - 2 1)2 (2z-l}'z 4 z2 n- (^11) = 1· im,_~ &.z a ( z z 2 n-1) = i· 1m,_1 (1 +n ) z n =
(1 + n) (!r' = (!r' + n (!t. T herefore, y[nl = Res[j(z), 41 = (!t +
n(!r.
lla. - l!i ( 3~4i r + l~i ( 354i r.
13a. e'r + e -•;· = 2cos{~n).