1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

376 CHAPTER 9 • z-TRANSFORMS AND APPLICATIONS

x[n] y[n)

2

0.9

A

1 - --

'

' ,' v

'

' 10 0 4 0 60 n

" '

/

• 1 '' _,

-2 ~

Figure 9.5 The input x[n) = cos(O.lOn) + sin(0.77n) and output y(n].

(b) Calculate the amplitude responses A(0.10) and A(0.77):

A(0.10) = 11 - v'2 ( eo.10;)-1 + ( eo.10i)- 2 I

= I0.572918 - o.0574836il = o.575795,

a.nd

A(0.77) = 11 - v'2 (e^0 mr

1

+(e^0 mr

2

I= 10 .0155125- o.Ol50419il

= 0.0216078.

From these calculations we expect that components cos(O.lOn)

and sin(0.77n) of the signal are attenuated by the factors A(0.10) =

0.575795 and A(0.77) = 0.0216078, respectively. Hence the filter al-

most eliminates the signal component sin(0.77n) which is close to the

"zero-out" frequency 0. 785398 = i· This is illustrated in Figure 9.5.

(c) In part (b) we found A(0.10) = 0.575795, and now we make the

calculation

A(211' I ( .. )-1 ( .• )-2 I

3

)=1-v'2e'T +e¥

= l - v'2(--- - ) - -+ -

I

1 iJ3 1 iJ3

2 2 2 2

= 2.246507.

We expect that components cos(O.lOn) and 0.20sin(^23 " n) of the in-

put signal are attenuated by A(0.10) = 0.575795, and amplified by

A(^2 ;) = 2.246507, respectively. This is illustrated in Figure 9.6.