1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

9 .3 • DIGITAL SIGNAL FILTERS 385

Imz

1t 1t

4 2

Figure 9.9 Ampli tude response A(8) and zero-pole plot for y[n) = l L; • x[n - k).

k=O

Remark 9. 14

The function A(B) can be proven to be even; i.e., A(-B} = A(B), which reinforces
the fa.ct that the z~ros come in conjugate pairs. Also, when expanded over a
common denominator, the transfer function H(z) = l+z!;;+z• actually has a
triple pole at t he origin. Finally, it can be shown that the zeros are all equally
spaced on the unit circle and that the arguments of the zeros correspond to
frequencies that are zeroed out by the filter. The situation is illustrated in
Figure 9.9. •

• EXAMP LE 9.24 The moving average filter

1

y[n) = S(x[n] + x [n - 1) + x[n - 2) + x[n -3} + x[n -4] + x [n - 5)

• x[n - 6} + x (n - 7})

is designed to zero out cos(mr}, cos(3;n), sin(3;n), cos(~n}, sin(~n}, cos( t n},

and sin(~n).

Solution We use the property (i) zeroing out filter. Recall that the solutions

to z^8 = 1 are the eighth roots of unity z = e ·~· for k = 0, 1, 2, 3, 4, 5, 6, 7 and

lie on the unit circle. Hence the roots of

1 z^8 - 1 1

---= -(z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z + 1) = 0

8 z-1 8