Laplace transform:
Hs=∫−∞∞
h
a
te−(st)d
t
H
s
t
h
a
t
s
t
Note that the continuous-time Fourier transform is
Hjλ
H
λ
(the Laplace transform evaluated on the imaginary
axis).

Since the early 1900's, there has been a lot of
research on designing analog filters of the form
Hs=
b
0
+
b
1
s+
b
2
s2+...+
b
M
sM1+
a
1
s+
a
2
s2+...+
a
M
sM
H
s
b
0
b
1
s
b
2
s
2
...
b
M
s
M
1
a
1
s
a
2
s
2
...
a
M
s
M
A causal IIR filter *cannot* have linear
phase (no possible symmetry point), and design work for analog
filters has concentrated on designing filters with equiriplle
(
∥L∥∞
L
) *magnitude* responses. These
design problems have been solved. We will not concern
ourselves here with the design of the analog prototype
filters, only with how these designs are mapped to
discrete-time while preserving optimality.

An analog filter with *real*
coefficients must have a magnitude response of the form
|Hλ|2=Bλ2
H
λ
2
B
λ
2

HjλHjλ¯=
b
0
+
b
1
jλ+
b
2
jλ2+
b
3
jλ3+...1+
a
1
jλ+
a
2
jλ2+...Hjλ¯=
b
0
−
b
2
λ2+
b
4
λ4+...+jλ(
b
1
−
b
3
λ2+
b
5
λ4+...)1−
a
2
λ2+
a
4
λ4+...+jλ(
a
1
−
a
3
λ2+
a
5
λ4+...)
b
0
−
b
2
λ2+
b
4
λ4+...+jλ(
b
1
−
b
3
λ2+
b
5
λ4+...)1−
a
2
λ2+
a
4
λ4+...+jλ(
a
1
−
a
3
λ2+
a
5
λ4+...)¯=
b
0
−
b
2
λ2+
b
4
λ4+...2+λ2
b
1
−
b
3
λ2+
b
5
λ4+...21−
a
2
λ2+
a
4
λ4+...2+λ2
a
1
−
a
3
λ2+
a
5
λ4+...2=Bλ2
H
λ
H
λ
b
0
b
1
λ
b
2
λ
2
b
3
λ
3
...
1
a
1
λ
a
2
λ
2
...
H
λ
b
0
b
2
λ
2
b
4
λ
4
...
λ
b
1
b
3
λ
2
b
5
λ
4
...
1
a
2
λ
2
a
4
λ
4
...
λ
a
1
a
3
λ
2
a
5
λ
4
...
b
0
b
2
λ
2
b
4
λ
4
...
λ
b
1
b
3
λ
2
b
5
λ
4
...
1
a
2
λ
2
a
4
λ
4
...
λ
a
1
a
3
λ
2
a
5
λ
4
...
b
0
b
2
λ
2
b
4
λ
4
...
2
λ
2
b
1
b
3
λ
2
b
5
λ
4
...
2
1
a
2
λ
2
a
4
λ
4
...
2
λ
2
a
1
a
3
λ
2
a
5
λ
4
...
2
B
λ
2

(1)
Let

s=jλ
s
λ
, note that the poles and zeros of

B−s2
B
s
2
are symmetric around

*both* the
real and imaginary axes: that is, a pole at

p
1
p
1 implies poles at

p
1
p
1,

p
1¯
p
1
,

−
p
1
p
1
, and

−
p
1¯
p
1
, as seen in

Figure 1.

Recall that an analog filter is stable and causal
if all the poles are in the left half-plane, LHP, and is
minimum phase if all zeros and poles are in the
LHP.

s=jλ
s
λ
:
Bλ2=B−s2=HsH−s=H(jλ)H(−(jλ))=H(jλ)Hjλ¯
B
λ
2
B
s
2
H
s
H
s
H
λ
H
λ
H
λ
H
λ
we can factor
B−s2
B
s
2
into
HsH−s
H
s
H
s
, where
Hs
H
s
has the left half plane poles and zeros, and
H−s
H
s
has the RHP poles and zeros.

|Hs|2=HsH−s
H
s
2
H
s
H
s
for
s=jλ
s
λ
, so
Hs
H
s
has the magnitude response
Bλ2
B
λ
2
. The trick to analog filter design is to design a
good
Bλ2
B
λ
2
, then factor this to obtain a filter with that
*magnitude* response.

The traditional analog filter designs all take the
form
Bλ2=|Hλ|2=11+Fλ2
B
λ
2
H
λ
2
1
1
F
λ
2
, where F
F is a rational function in
λ2
λ
2
.

Bλ2=2+λ21+λ4
B
λ
2
2
λ
2
1
λ
4
B−s2=2−s21+s4=(2−s)(2+s)(s+α)(s−α)(s+α¯)(s−α¯)
B
s
2
2
s
2
1
s
4
2
s
2
s
s
α
s
α
s
α
s
α
where
α=1+j2
α
1
2
.

Roots of

1+sN
1
s
N
are

N
N points equally spaced around the unit circle
(

Figure 2).

Take
Hs=LHP
H
s
LHP
factors:
Hs=2+s(s+α)(s+α¯)=2+ss2+2s+1
H
s
2
s
s
α
s
α
2
s
s
2
2
s
1