1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

370 CHAPTER 9 • z-TRANSFORMS AND APPLICATIONS

Solution

(b) The formula H(z) = z2-~~z+~ = 1/((z - ~^1 }i)(z - ~-'j1)) is the
transfer function. Using the z-transform

3[cos( ~n)] = z2 -cos (i) z = z2 - zv'Z

4 z2-2cos(f)z+l z2- v'zz+l

= (z2 __ 1 z)/((z _ 1 + i)(z _ 1 - i))

v'2 v'2 v'2

and the fact that H(z) = r~:\, we can write the z-transform Y(z)

using convolution

Y (z) = H(z)X(z) = (z - £ l+i) ( z - ~ 1-i) (z - l+i) (z - 1-i).

3 v'2 3 v'2 v'2 v'2

z^2 - zv'Z

Calculate the residues for f(z) = Y(z)z"-^1 at the poles. The residue

calculus can again be used to find the solution, but the details are

tedious. Let us announce that the following computations hold true:

[

Res f(z), 3 2 1 V2 + i] +Res [ f(z), 3 2 1 -V2 i]

= -^18 (2)n - cos(1f - n) - -^63 (2)- n. sm( 1f - n)

13 3 4 26 3 4 '

which is part of the homogeneous solution. The steady state or par-

t icular part of the solution is

Yp[n] =Res [!(z),

1

~i] +Res [!(z),

1

.Jzi]

18 (1f ) 27. (1f )

= - 13 cos 4n + 13 Sill 4n.

Therefore, the general solution to part (b) is

(

yin] = y,,[n] + Ypln] = c1 3 2)n cos( in) + c2 (2)n 3

Sill. (1f - n ) - -^18 cos (1f - n ) + -^27 sin. (1f - n )

3 13 4 13 4 '

and is illustrated in Figure 9.3.

For applications, it is useful to determine the limiting amplitude of y[n].

We need to simplify Yp[n] in a form that d isplays its amplitude, and to do this

we apply to the trigonometric identity (also known as the harmonic addition