The zeros of the transfer function
Hz
H
z
of a linear-phase filter lie in specific configurations.
We can write the symmetry condition
hn=hN-1-n
h
n
h
N
1
n
in the ZZ domain. Taking the
ZZ-transform of both sides gives
Hz=z-N-1H1z
H
z
z
N
1
H
1
z
(1)
Recall that we are assuming that
hn
h
n
is real-valued. If
z
0
z
0
is a zero of
Hz
H
z
,
H
z
0
=0
H
z
0
0
then
H
z
0
¯=0
H
z
0
0
(Because the roots of a polynomial with real coefficients
exist in complex-conjugate pairs.)
Using the symmetry condition, Equation 1, it follows that
H
z
0
=z-N-1H1
z
0
=0
H
z
0
z
N
1
H
1
z
0
0
and
H
z
0
¯=z-N-1H1
z
0
¯=0
H
z
0
z
N
1
H
1
z
0
0
or
H1
z
0
=H1
z
0
¯=0
H
1
z
0
H
1
z
0
0
If
z
0
z
0
is a zero of a (real-valued) linear-phase filter, then so
are
z
0
¯
z
0
,
1
z
0
1
z
0
, and
1
z
0
¯
1
z
0
.
It follows that
-
generic zeros of a linear-phase filter exist in sets of 4.
-
zeros on the unit circle (
z
0
=ⅇⅈ
ω
0
z
0
ω
0
) exist in sets of 2. (
z
0
≠±1
z
0
±
1
)
-
zeros on the real line (
z
0
=a
z
0
a
) exist in sets of 2. (
z
0
≠±1
z
0
±
1
)
-
zeros at 1 and -1 do not imply the existence of zeros at
other specific points.
The frequency response
H
f
ω
H
f
ω
of a Type II FIR filter always has a zero at
ω=π
ω
:
hn=
h
0
h
1
h
2
h
2
h
1
h
0
h
n
h
0
h
1
h
2
h
2
h
1
h
0
Hz=
h
0
+
h
1
z-1+
h
2
z-2+
h
2
z-3+
h
1
z-4+
h
0
z-5
H
z
h
0
h
1
z
-1
h
2
z
-2
h
2
z
-3
h
1
z
-4
h
0
z
-5
H-1=
h
0
-
h
1
+
h
2
-
h
2
+
h
1
-
h
0
=0
H
-1
h
0
h
1
h
2
h
2
h
1
h
0
0
H
f
π=Hⅇⅈπ=H-1=0
H
f
H
H
-1
0
H
f
π=0
H
f
0
always for Type II filters.
Similarly, we can derive the following rules for Type III and
Type IV FIR filters.
H
f
0=
H
f
π=0
H
f
0
H
f
0
always for Type III filters.
H
f
0=0
H
f
0
0
always for Type IV filters.
The automatic zeros can also be derived using the
characteristics of the amplitude response
Aω
A
ω
seen earlier.
| Type |
automatic zeros |
| I |
— |
| II |
ω=π
ω
|
| III |
ω=0∨π
ω
0
|
| IV |
ω=0
ω
0
|
The Matlab command zplane can be used to
plot the zero locations of FIR filters.
Note that the zero locations satisfy the properties noted
previously.