1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

360 CHAPTER 9 • z-TRANSFORMS AND APPLICATIONS

y[n) = c 1 (2 + i)" + c2(2 - i)". Use the initial conditions and form the linear
system

y[O] = C1 + C2 = 1 and y[l) = C1 (2 + i) + c2(2 - i) = 5,

then solve for the constants and get c 1 =^123 i and c 2 =^1 =t^3 i. Hence the solution

is

I l
yn = 1 - 3i - ( 2 ·)n^1 + 3i ( 2 ·)n
2

• 
  
  
  
  • i +-
    2
    
    
    • 
      
      
      
      • i.
      
      
      
      
    
    
    
    
  
  
  
  

We leave it for the reader to verify that this can be written as

y[n] = 5~ cos(arctan{!)n) + 3 · 5~ sin(arctan(4)n).

9.2.2 Higher-Order Difference Equations

The general form of a ?th-order linear constant coefficient difference equation

(LCCDE) is

y[n) + a1y[n - 1) + a2y[n - 2] +. .. + ap-1y[n - P + l] + apy[n - P]

= box[n) + b1x[n - 1) + ~x[n - 2) + ... + bQ-1x\n - Q + 1) + bQx[n - QI

(9-14)

where {a,,};= 1 and {bq}~=o· The sequence {xn = x[n]}::"=o is given and the

sequence {yn = y[n]} ::"=o is output. The integer P is the order of the difference

equation. The compact form of writing this difference equation is

p Q

y[n) + L apy[n - p] = L bqx[n - q). (9-15)

p=l q=O

The formula in (9-15} can be expressed in a recursive form:

Q p

y[n) = L b 9 x[n - q) -L apy[n -p). (9-16)

q=O p=l

This form of the LCCDE explicitly shows that the present output y [n ] is a

function of the past output values y[n - p), for p = 1, 2, ... , P; the present input

x[n); and the previous inputs x[n - q] for q = 1, 2, ... , Q.

Now we would like to emphasize the method of z-transforms for solving

difference equations. Applying the linearity and time delay shift property of the

z-transform to equation (9-15), we obtain

p Q

Y(z) + L:apY(z)z-P = L bqX(z)z-^9. (9-17)

p=I q=O