ANSWERS 6111 3a. Take the z-transform of both sides z(Y(z) - 1000) = (1 - ~)Y(z). Solve
for Y(z) and get Y(z) = ~~!', then find the inverse z-transform y [n] =
3 - 1 ( -;::-r lOOOz) = 10003 - 1 ( ;=r z ) = 1000 (l)n. 2
15a. Take the z-transform of both sides z(Y(z) - 1) - 3Y(z) = (z~~p · Solve
~ or Y( z ) an d ge t Y( z ) _ - (z-z^3 - IF(•-2z^1 +5z _ 2z z 2z d th
3 ) - ,_ 3 - •-l - ~· an en
find the inverse z-transform y(n] = 3 -^1 (z - 3 ..1!... - - •-z-1 - (z~- 1)' ) = 23 -^1 (-•-)-z-3
3 -^1 (..,: 1 )-23-^1 ( ( • -\);) = 2* 3n -1-2n.19a. Given x [n) = cos( ~n) and y(n) = -sin(~n) we have x (n + l] - y[n) =
cos( In+ I)+ sin( In) = -sin(~n) + sin( In) = 0, and y(n + l] + x[n) =
- sin(~n+ ~) +cos(~n) = -cos(~n) +cos(~n) = 0.
Section 9.2. Second-Order Homogeneous D ifference E q uations:
page 371la. Method 1. The characteristic equation r^2 -6r+ 8 = (r-2)(r-4) = 0 has
roots r 1 = 2 and r2 = 4. The general solution is y(n] = c 1 2n + c24n. Solve
the linear system y [O] = c1 + c2 = 3, y[l] = 2c1 + 4c2 = 4, and get c1 = 4
and c2 = -1. Therefore, y(n] = 4 · 2n - 4n.
Method 2. Take z-transforms and get z^2 (Y(z)- 3 -4z-^1 ) -6z(Y(z)-3)+8Y(z) = 0. Solve for Y(z) = ;;~6;!~ = (;1.:;~~~'.i)· Calculate residues of
f(z) = Y(z)zn- i at the poles Reslf(z), 2] = lim,,_ 2(z- 2) <z3,:;R!~" 4 )zn-I =
lim.- 2^3 ·::::~^4 • zn- l = 8. 2n-l = 4. 2", and Res[f(z), 4]
!Im. z - 4 (z _ 4 ) ( z -3z22)(z-4) -14z zn- 1 -- li'm z~4 3z2z - 2 - 14zzn-1 -_ 4. 4 n-I - _ 4 ., ·
l e. Method 1. The characteristic equation r^2 - 6r + 10 = (r - (3 - i))(r -(3 + i)) = 0 has complex roots r 1 = 3 ± i. The general solution is y(n] =
c 1 (3 + i )" + c2(3 - i)". Solve the linear system y(O] = c1 + c2 = 2, y[l] =(3 + i)c 1 + (3 - i)c2 = 4 and get c 1 = 1 + i and c2 = 1 - i. Therefore,
y(n) = (1 + i)(3 + i)" + (1-i)(3 - i)".
Method 2. Take z-transforms and get z^2 (Y(z) - 2 - 4z-^1 ) - 6(z(Y(z) -