614 ANSWERS9. Recall that the solutions to z^6 = 1 are the sixth roots of unity z = e-'¥
for k = 0,1,2,3,4,5 and lie on the unit circle. Hence the roots of ~ ·:~ 11 =
~(z5 + z^4 + z^3 + z^2 + z + 1) = 0 are z = ei",e~,e~,e~,e-;·. We
now multiply the above expression by ~ to obtain a product of "zero-out"
factorsH(z) = ~(l+z-^1 +z-^2 +z-^3 +z-^4 +z-^5 ) = ~*(z^5 +z^4 +z^3 +z^2 +z+1),
Hz= ( ) 6 1 1 ( ,. z-e' '")( z - e• ... )( z - e-,--i2'1f )( z - e• <• )( z - e---r, _,.)
H(z) = ~( •-;'• )( z-~~ )( z-e~ )(•-:if)( z -e,=f'-).
Now use the property (i) for a zero-out filter. Use b; = ~ for i = 0, ... , 5 to
get the desired recursive formula y [n] = ~(x[n) + x [n - 1] + x[n - 2] + x[n -3] + x[n - 4] + x(n - 5]).
- We use the property (iii) designer specified filter. The solutions to z^8 = 1
are the sixth roots of unity z = e1¥ for k = 0,1,2,3,4,5,6,7 and lie on
the unit circle. Hence the roots of k(z^1 + z^6 + z^5 + z^4 + z3 + z^2 + z + 1) =
f(1- z^8 ) · i3·Jr f.3• -iw -hr t• -'n
i-z = 0 are z = e'",e.......-,e-.-,e> ,e--.--,e• ,e-.-. There are no poles(^1) (1-~} I
in the transfer function H(z) = • 1 _,. Use ai = - 1, bo = 8• bi= 0 for
i = 1, 2, ... ,7 and bs = -k and get y[n) = k(x[n] - x[n - 8)) + y[n - l].
13a. Use the conjugate pairs of zeros e±if and e±~ and calculate
(z-eif )(•-•,-if )(•-;if )(z-e,=¥) = (l -z-1 + z-2)(1+z-2)=1-z-1 +
2 z-~ - z-^3 + z-^4. There are no poles, so the transfer functiou has the form
H(z) = • ,,,,+ b ' ' _, + b •• -^2 b - s b -·
1 + " + •• , and we see that bo = b4 = 1, b2 =^2
and b 1 = b3 = -1. The filter isy[n] = box[n] + b1x[n - 1] + bzx(n - 2] + b3x[n - 3) + b,,x[n - 4],
y[n] = x[n] - x [n - l ] + 2x[n - 2] - x [n - 3] + x[n - 4] for "zeroing-out"
cos(~n), sin(~n), cos(~n), sin(~n).15a. Use the conjugate pairs of zeros e±;,r and e±~ and calculate
c·-;~ )(•-•;if )(•-;if)(•- •.=¥) = (1 - v'3z-l + z-2)(1 + z-2) = 1 -
v'3z-^1 +2z-^2 - v'3z-^3 + z-^4. There are no poles, so the transfer function has
the form H(z) = bo+biz-'+b·•-:±b.--^3 +b.z-• and we see that bo = b4 = 1,
~ = 2 and b1 = b3 = -v'3. The filter is
y[n] = box[n] + b1x[n - 1] + ~x[n - 2] + b3x(n - 3] + b4x[n - 4],
y[n] = x[n] - J3x[n - 1] + 2x[n - 2) - v'3x[n - 3) + x [n - 4)
for "zeroing out" 'cos( in), siu(in), cos( ~n), sin(~n).l 7 a. Use the conjugate pairs of zeros e±~ and ei" = -1 and calculate
(•-;if)(•-•,=¥-)(•-: '•)= (1 + JZz-^1 + z-^2 )(1 + z-^1 ) = 1 + (1 + v'z)z-^1 +
(1 + J2)z-^2 + z-^3. The transfer function for part (a) has the form H(z) =
bo+bi z - ' +'jaz-'+baz - • and we see that bo = b3 = 1 and bi = bz = 1 + J2.
The filter is y (n] = box[n) + b 1 x[n - 1] + b 2 x[n - 2] + b:ix[n -3),