386 CHAPTER 9 • Z..TRANSFORMS AND APPLICATIONS
. fil -•3w 4ff -•" in- in h ~
are z = e"', e • , e-·-, e•, e,-, e•, e•. There are no poles, so t e trans1er
function has the form
H(z) = 8 1 1 z (. ) ( ~) ( -<3•) ( ·~) ( -··)
7 z -e"' z -e • z -e-.- z -e• z - e- ·-
( z - e ilc) • ( z - e-·-••-)
Use b; = ~ for i = 0, ... , 7 to get
1
y[nJ =
8
(x[nJ + x[n - l} + x[n -2] + x[n -3} + x[n - 4)
- x(n -5) + x[n - 6} + x[n - 7]).
Remark 9.15
This is an extension of the filter in Example 9.23(b), and zeros out twice as
many frequencies. The function A(9) has additional zeros located ate = - f,
- I, -^3 ;. The transfer function can be written
1 + z + z^2 + z^3 + z^4 + z^5 + z^6 + z^7
H~)= &1.
The representation has a pole of order seven at the origin. Also, as in the
previous example the zeros are equally spaced points on the unit circle, and
their arguments correspond to frequencies that are zeroed out by the filter. The
situation is illustrated in Figure 9.10.
- EXAMPLE 9.25
(a) Design a filter with poles ~elf and ~e-lf for boosting up signals near
cos(fn) and sin(fn).
{b) Include the additional pole at! e^10 = ~ to the filter design in (a) so
that it also boosts up low-frequency signals.