1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

9.3 • DIGITAL SIGNAL F ILTERS 377

x[n)

Figure 9.6 The input x(n] = cos(O.lOn) + 0.20sin(^23 " n) and output y [n).

• EXAMPLE 9. 22 Given the filter y[n] = x[n] + iv'2v[n - 1) - iv[n - 2).

(a) Show that it is a boosting up filter for the signals cos(fn) and sin(fn)
and calculate the amplitude response A(i)·
(b) Calculate the amplit ude responses A(i) and A(2.60) and investigate the
filtered signal for x[n] = cos(~n) + sin(fn) + sin(2.60n).

Solut ion In Section 9.2, Example 9.20(b), we found that the general solution

to y [n + 2] - ~v'2y[n + 1] + iv[n ] = cos(tn) is

y[n] = c1 (~) n cos (~n) + c 2 (~) n sin (~n) +~cos ( ~n+ arctan (~)).

Since limn-oo(c1 (~t cos( in)+ C2 (~)"' sin(tn)) = 0, t he output signal y[n)

tends to the steady state signal Jh cos(tn + arctan(~)) as n-+ oo.

Hence the signal cos(fn) is boosted up by an amplification factor of fi3 =

2.49615. A similar boost will be observed for the signal sin(tn).

For this filter, the amplitude response is A(9) = IH(e•^9 )1 where H(z) =

i-~J2z~'+a.-.-We compute the boost for sin(fn) by evaluating A(f):

(

A 'ff) I 2r,; ·•-I^4 ··)-21 I 2r,;(1-i) 4il

4

= 1/ 1 -

3

v2 (e«) + g (e« = 1/ 1 -

3

v2 v'2 - g

= 19

3

2

i 1 =

9

v'13 = 2 49615

13 13. '

which is the same value that was obtained in Example 9.20(b) in Section 9.2.
The amplitude response A(9) for an arbitrary frequency e for is given in Figure
9.7. Observe that a maximum occurs near e = T and there is amplification for
signals with 0 < e < 1.4944.