1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

9.3 • D IGITAL S IGNAL FILTERS 389

Imz

'--~~g~~g~~ 3 ~X~~gO

4 2 4

Figure 9.12 Amplitude response A(ll) and zero-pole plot for t he boosting up filter
y[n) = x[n) + ~ (3 + 4,/2) y [n - 1) - ~ (4 + 3VZ) y [n - 2) + ~y[n - 3].

• EXAMPLE 9.26 Design a combination filter using the zeros e*'"/^2 and ei"
  for zeroing out cos( mi), cos(~n), sin(~n) and poles ~e±hr/4, boosting up some
  of the low frequencies.

Solution The zeroing-out portion of t his filter design is similar to the filter in
Example 9.23(b), where we showed that

z - eif z - e~ z - e^1 '"
( )( )(--) = 1 + z-^1 + z-^2 + z-^3.
z z z

Of course, we could multiply by the constant ~ to give this filter a moving average
effect on the input signal x [n). However, for simplicity we use bo = b 1 = b 2 =

ba = 1 in the numerator of the transfer function.

For t he boosting up portion of this filter design we choose the one developed
in Example 9.22(a):
2 il! 2 !?!
(z-^3 e• )(z- 3e • )=l-~v'2z-1+~z- 2
z z 3 9 '

and we will use a1 = -~.;2 and a2 = ~ in the denominator of the transfer
function. Putting the two parts toget her we obtain

H (z) = bo + b1z -^1 + b2z -

2

+ b3z-^3 = 1 + z-^1 + z-^2 + z-^3 •

1 + a1z-^1 + a2z-^2 1 - ~ J2z-1 + ~z-2

The corresponding filter for this transfer function is

y(n] = box(n] + b1x[n - 1] + b2x[n - 2) + bJx(n - 3] -a1y[n - 1] -a2y[n - 2]

2 4

= x[n) + x[n - 1] + x(n - 2] + x [n - 3) +

3

v'2y[n - 1] - gY[n - 2).