1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

366 CHAPTER. 9 • Z-TR.ANSFOR.MS AND APPLICATIONS

at the poles

.. z^4 - 6z^3 + 12z^2 - lOz n - l

Res[f(z), 1 + il = z-1+• !rm. ( z - 1 +i ")( z - 2 +i ")( z - 2 - i ") z

• 1 + 3i. _ I..
  = (
  5

)(1+ir^1 =

10

(2+4i)(1+i)n,

R [! ( ) 1

.

1 1

. z^4 - 6z^3 + 12z^2 - lOz n - l

es z, - i = 1m... z

.z-1- i (z - l -i)(z-2+i)(z -2-i)

= (-1; 3 i)(1-ir - 1 = 110(2-4i)(l - iJ",

R [! ( ) 2 ·1 - 1· z

4

• 6z
  3
  
  
  • 12z
    2
    es z, +i- 1m.. - lOz .z n- 1
  
  
  
  

•-2+i (z-1 + i)(z-1-i) (z-2 +i)

2 + lli. 1

= (---io--)(2+ir-

1

= 10(3+4i)(2+ir, and

Reslf(z) 2 - ·1 = r z4 - 6z3 + 12z2 - lOz n- 1

' i z..!T-i (z - 1 + i}(z - 1 - i) (z -2 - i) z

= (

2

• ui)(2 -ir-^1 = 2...( 3 - 4i)(2 -i)n.
  10 10

Therefore, the solution is

y[n] = Reslf(z), 1 + i] + Res[f(z), 1 - i) + Reslf(z), 2 + i] + Reslf (z), 2 - i)

1

y[nl =

10

((2 + 4i)(1 + ir + (2 - 4i)(1 - i)" + (3 + 4i)(2 + ir

• (3 - 4 i)(2 - i)n).

Remark 9.10

Observe that Res[f(z), 1-i) = Reslf(z), 1 + i] = Res[/(z) , 1 + i] aud Res[/(z),

2 - i] = Reslf(z), 2 + i ) = Reslf(z), 2 + i). •

• EXAMPLE 9.18 Solve y[n + 3) + y(n + 2l + y[n + 1) + y[n] = 0 with y[O) = 2,

y(l] = -2, and y[2l = O.

Solution Take the z-transforms of each term

z^3 (Y(z) - 2 + 2z-^1 - oz-^2 ) + z^2 (Y (z) - 2 + 2z-^1 ) + z(Y(z} - 2) + Y(z) = O.