364 CHAPTER 9 • Z-TRANSFORMS AND APPLICATIONS2Calculate the residue for f(z) = Y(z)zn- I = (~-'z) 1 zn-I at the pole
dR.€s[f(z), 2] = lim -d [(z^2 + z)zn-IJ = Jim ((2z + l )zn- I
z - 2 z z-2
+ (n -l)(z + z^2 )z"-^2 )= 5 · zn- I + 3(n - 1)2n-I = 2n + 3n2n-I.
Thus the solution isy[n) = 2n + 3n2n-^1 ,
which agrees with the result of Example 9.13.(b) Take the z-transforms of each termz2(Y(z) - 2 - 3z -^1 ) - 4(z(Y(z) - 2)) + 4(Y(z)) = z _^4
3.Solve for Y(z) and get Y(z) =^2 z3-11z
2
(z-2)2(z±- 3) !6z ·Calculat e the residue for f(z) = Y(z)zn-I =^2 »-(z-2)2(z-3)^11 »+^16 • zn-I at t he
polesR.€s(f(z),2] = lim[!£(2z3 - llz2 + l6z zn-1)]
z~2 dz (z - 3)I. 2(n + l)z^3 - (l7n + 12)z^2 + (49n + 17 )z - 48n n-I
= 1m z
z - 2 (z - 3)2= (2 - 2n)2n- I = 2n - 2nn, and
R.€8[!( ) 3] - r 2z
3- llz
2
+ 16z n-1 - 3. 3n- l - 3n
z ' - .. ~~ (z - 2)2 z - -.Thus the solut ion isy\n) = zn -nzn + 3n,
- EXAMPLE 9.17
(a) Use z-t ransform methods to solve y[n + 2] -4y[n + 1] + 5y[n] = 0 with
y[O) =Yo= 1 and y[l ] =YI = 5.
(b) Use z-transfor m methods to solve y [n + 2] - 4y[n + 1) + 5y[n)