The Optimum "L" filter (also known as a Legendre–Papoulis filter ) was proposed by Athanasios Papoulis in 1958. It has the maximum roll off rate for a given filter order while maintaining a monotonic frequency response . It provides a compromise between the Butterworth filter which is monotonic but has a slower roll off and the Chebyshev filter which has a faster roll off but has ripple in either the passband or stopband . The filter design is based on Legendre polynomials which is the reason for its alternate name and the "L" in Optimum "L".
Comparison of filter magnitude between Butterworth-, Legendre- and Chebyshev-Type1-Filter Synthesizing the characteristic polynomials
edit
The solution to N order Optimum L filter characteristic polynomial synthesis emanates from solving for the characteristic polynomial,
L
N
(
ω
2
)
{\displaystyle L_{N}(\omega ^{2})}
[ 1]
L
N
(
0
)
=
0
L
(
1
)
=
1
d
L
N
(
ω
2
)
d
ω
≥
0 for 0
≤
ω
≤
1
d
L
N
(
ω
2
)
d
ω
|
ω
=
1
is maximum
{\displaystyle {\begin{aligned}&L_{N}(0)=0\\&L(1)=1\\&{dL_{N}(\omega ^{2}) \over d\omega }\geq {\text{ 0 for 0 }}\leq \omega \leq 1\\&{dL_{N}(\omega ^{2}) \over d\omega }{\biggr |}_{\omega =1}{\text{ is maximum}}\\\end{aligned}}}
The odd order case[ 2] [ 1] Legendre polynomials as follows.
N Odd:
L
N
(
ω
2
)
=
2
N
+
1
∫
−
1
2
ω
2
−
1
(
∑
i
=
0
i
=
k
a
i
P
i
(
x
)
)
2
d
x
Where
P
i
(
x
)
is the Legendre polynomial of the first kind of order i
k
=
N
−
1
2
a
i
=
2
i
+
1
2
(
k
+
1
)
N Even:
L
N
(
ω
2
)
=
∫
−
1
2
ω
2
−
1
(
x
+
1
)
(
∑
i
=
0
i
=
k
a
i
P
i
(
x
)
)
2
d
x
Where
k
=
N
−
2
2
a
i
=
{
2
i
+
1
(
k
+
2
)
(
k
+
1
)
,
if
i
is odd and
k
is odd OR
i
is even and
k
is even
0
,
otherwise
{\displaystyle {\begin{aligned}&{\text{N Odd:}}\\&L_{N}(\omega ^{2})={\frac {2}{N+1}}\int _{-1}^{2\omega ^{2}-1}{\bigg (}\sum _{i=0}^{i=k}a_{i}P_{i}(x){\bigg )}^{2}dx\\&{\text{Where}}\\&P_{i}(x){\text{ is the Legendre polynomial of the first kind of order i}}\\&k={\frac {N-1}{2}}\\&a_{i}={\frac {2i+1}{\sqrt {2(k+1)}}}\\&\\&\\&{\text{N Even:}}\\&L_{N}(\omega ^{2})=\int _{-1}^{2\omega ^{2}-1}(x+1){\bigg (}\sum _{i=0}^{i=k}a_{i}P_{i}(x){\bigg )}^{2}dx\\&{\text{Where}}\\&k={\frac {N-2}{2}}\\&a_{i}={\begin{cases}{\frac {2i+1}{\sqrt {(k+2)(k+1)}}},&{\text{if }}i{\text{ is odd and }}k{\text{ is odd OR }}i{\text{ is even and }}k{\text{ is even}}\\0,&{\text{otherwise}}\\\end{cases}}\\\end{aligned}}}
Frequency response and transfer function
edit
The magnitude frequency magnitude is created using the following formula. Since the Optimum "L" characteristic function is already in squared form, it should not be squared again as is done for other filter types such as Chebyshev filters and Butterworth filters .
T
(
ω
)
=
1
1
+
ϵ
2
L
N
(
ω
2
)
ϵ
2
=
10
|
δ
|
/
10
−
1.
δ
=
magnitude attenuation of the passband in dB, usually 3.0103
{\displaystyle {\begin{aligned}&T(\omega )={\sqrt {\frac {1}{1+\epsilon ^{2}L_{N}(\omega ^{2})}}}\\&\epsilon ^{2}=10^{|\delta |/10}-1.\\&\delta ={\text{magnitude attenuation of the passband in dB, usually 3.0103}}\\\end{aligned}}}
To obtain the transfer function,
T
(
j
ω
)
{\displaystyle T(j\omega )}
L
N
(
ω
2
)
{\displaystyle L_{N}(\omega ^{2})}
j
ω
{\displaystyle j\omega }
T
(
j
ω
)
{\displaystyle T(j\omega )}
L
N
(
(
j
ω
)
2
)
{\displaystyle L_{N}((j\omega )^{2})}
L
N
(
ω
2
)
{\displaystyle L_{N}(\omega ^{2})}
L
N
(
(
j
ω
)
2
)
{\displaystyle L_{N}((j\omega )^{2})}
T
(
j
ω
)
{\displaystyle T(j\omega )}
[ 3] [ 4]
T
(
j
ω
)
=
1
a
+
ϵ
2
L
N
(
(
j
ω
)
2
)
|
Left half plane
Where:
a
=
{
1
,
if
N
is even
−
1
,
if
N
is odd
ϵ
2
=
10
|
δ
|
/
10
−
1.
δ
=
magnitude attenuation of the passband in dB, usually 3.010
{\displaystyle {\begin{aligned}&T(j\omega )={\sqrt {\frac {1}{a+\epsilon ^{2}L_{N}((j\omega )^{2})}}}{\bigg |}_{\text{Left half plane}}\\&{\text{Where:}}\\&a={\begin{cases}1,&{\text{if }}N{\text{ is even}}\\-1,&{\text{if }}N{\text{ is odd}}\end{cases}}\\&\epsilon ^{2}=10^{|\delta |/10}-1.\\&\delta ={\text{magnitude attenuation of the passband in dB, usually 3.010}}\\\end{aligned}}}
The "Left Half Plane" constraint refers to finding the roots in all the polynomials contained in the brackets, selecting only roots in the left half plane, and recreating the polynomials from those roots.
Example: 4th order transfer function
edit
N = 4 (forth order), pass band attenuation = -3.010 at 1 r/s.
A forth order filter has a value for k of 1, which is odd, so the summation uses only odd values of i for
a
i
{\displaystyle a_{i}}
P
i
(
x
)
{\displaystyle P_{i}(x)}
The transfer function,
T
4
(
j
ω
)
{\displaystyle T_{4}(j\omega )}
k
=
N
−
2
2
=
1
(
k
is odd)
a
1
=
2
(
1
)
+
1
(
(
1
)
+
2
)
(
(
1
)
+
1
)
=
1.2247449
P
1
(
x
)
=
x
(
x
+
1
)
(
∑
i
=
0
i
=
k
a
i
P
i
(
x
)
)
2
=
(
x
+
1
)
(
1.2247449
(
x
)
)
2
=
1.5
x
3
+
1.5
x
2
L
4
(
x
2
)
=
∫
−
1
2
x
2
−
1
1.5
x
3
+
1.5
x
2
d
x
=
6
x
8
−
8
x
6
+
3
x
4
L
4
(
x
2
)
=
6
x
8
−
8
x
6
+
3
x
4
L
4
(
j
ω
2
)
=
6
(
j
ω
)
8
+
8
(
j
ω
)
6
+
3
(
j
ω
)
4
e
c
h
o
=
ϵ
3.0103
/
10
−
1
=
1
T
4
(
j
ω
)
=
[
1
1
+
1
2
(
6
(
j
ω
)
8
+
8
(
j
ω
)
6
+
3
(
j
ω
)
4
)
]
Left Half Plane
T
4
(
j
ω
)
=
1
2.4494897
(
j
ω
)
4
+
3.8282201
(
j
ω
)
3
+
4.6244874
(
j
ω
)
2
+
3.0412127
(
j
ω
)
+
1
{\displaystyle {\begin{aligned}&k={\frac {N-2}{2}}=1{\text{ (}}k{\text{ is odd)}}\\&a_{1}={\frac {2(1)+1}{\sqrt {((1)+2)((1)+1)}}}=1.2247449\\&P_{1}(x)=x\\&(x+1){\bigg (}\sum _{i=0}^{i=k}a_{i}P_{i}(x){\bigg )}^{2}=(x+1){\bigr (}1.2247449(x){\bigr )}^{2}=1.5x^{3}+1.5x^{2}\\&L_{4}(x^{2})=\int _{-1}^{2x^{2}-1}1.5x^{3}+1.5x^{2}{\text{ }}dx=6x^{8}-8x^{6}+3x^{4}\\&L_{4}(x^{2})=6x^{8}-8x^{6}+3x^{4}\\&L_{4}(j\omega ^{2})=6(j\omega )^{8}+8(j\omega )^{6}+3(j\omega )^{4}\\&echo={\sqrt {\epsilon ^{3.0103/10}-1}}=1\\&T_{4}(j\omega )={\bigg [}{\frac {1}{1+1^{2}(6(j\omega )^{8}+8(j\omega )^{6}+3(j\omega )^{4})}}{\bigg ]}_{\text{Left Half Plane}}\\&\\&T_{4}(j\omega )={\frac {1}{2.4494897(j\omega )^{4}+3.8282201(j\omega )^{3}+4.6244874(j\omega )^{2}+3.0412127(j\omega )+1}}\end{aligned}}}
A quick sanity check of
T
4
(
j
)
{\displaystyle T_{4}(j)}
Table of first 10 characteristic polynomials
edit
N
L
N
(
ω
2
)
{\displaystyle L_{N}(\omega ^{2})}
1
ω
2
{\textstyle \omega ^{2}}
2
ω
4
{\textstyle \omega ^{4}}
3
3
ω
6
−
3
ω
4
+
ω
2
{\textstyle 3\omega ^{6}-3\omega ^{4}+\omega ^{2}}
4
6
ω
8
−
8
ω
6
+
3
ω
4
{\textstyle 6\omega ^{8}-8\omega ^{6}+3\omega ^{4}}
5
20
ω
10
−
40
ω
8
+
28
ω
6
−
8
ω
4
+
ω
2
{\textstyle 20\omega ^{10}-40\omega ^{8}+28\omega ^{6}-8\omega ^{4}+\omega ^{2}}
6
50
ω
12
−
120
ω
10
+
105
ω
8
−
40
ω
6
+
6
ω
4
{\textstyle 50\omega ^{12}-120\omega ^{10}+105\omega ^{8}-40\omega ^{6}+6\omega ^{4}}
7
175
ω
14
−
525
ω
12
+
615
ω
10
−
355
ω
8
+
105
ω
6
−
15
ω
4
+
ω
2
{\textstyle 175\omega ^{14}-525\omega ^{12}+615\omega ^{10}-355\omega ^{8}+105\omega ^{6}-15\omega ^{4}+\omega ^{2}}
8
490
ω
16
−
1668
ω
14
+
2310
ω
12
−
1624
ω
10
+
615
ω
8
−
120
ω
6
+
10
ω
4
{\textstyle 490\omega ^{16}-1668\omega ^{14}+2310\omega ^{12}-1624\omega ^{10}+615\omega ^{8}-120\omega ^{6}+10\omega ^{4}}
9
1764
ω
18
−
7056
ω
16
+
11704
ω
14
−
10416
ω
12
+
5376
ω
10
−
1624
ω
8
+
276
ω
6
−
24
ω
4
+
ω
2
{\textstyle 1764\omega ^{18}-7056\omega ^{16}+11704\omega ^{14}-10416\omega ^{12}+5376\omega ^{10}-1624\omega ^{8}+276\omega ^{6}-24\omega ^{4}+\omega ^{2}}
10
5292
ω
20
−
23520
ω
18
+
44100
ω
16
−
45360
ω
14
+
27860
ω
12
−
10416
ω
10
+
2310
ω
8
−
280
ω
6
+
15
ω
4
{\textstyle 5292\omega ^{20}-23520\omega ^{18}+44100\omega ^{16}-45360\omega ^{14}+27860\omega ^{12}-10416\omega ^{10}+2310\omega ^{8}-280\omega ^{6}+15\omega ^{4}}
The table is calculated from the above equations for
L
N
(
ω
2
)
{\displaystyle L_{N}(\omega ^{2})}
^ a b Fukada, Minoru (September 1959). "Optimum Filters of Even Orders with Monotonic Response" . IRE Transactions on Circuit Theory . 6 (3): 277–281. doi :10.1109/TCT.1959.1086558 – via IEEE Xplore. ^ Papoulis, Athanasios (March 1958). "Optimum Filters with Monotonic Response" . Proceedings of the IRE . 46 (3): 606–609. doi :10.1109/JRPROC.1958.286876 – via IEEE Xplore. ^ Dr. Byron Bennett's filter design lecture notes, 1985, Montana State University , EE Department , Bozeman , Montana, US^ Sedra, Adel S.; Brackett, Peter O. (1978). Filter Theory and Design: Active and Passive ISBN 978-0916460143 {{cite book }}: CS1 maint: date and year (link )
![{\displaystyle {\begin{aligned}&k={\frac {N-2}{2}}=1{\text{ (}}k{\text{ is odd)}}\\&a_{1}={\frac {2(1)+1}{\sqrt {((1)+2)((1)+1)}}}=1.2247449\\&P_{1}(x)=x\\&(x+1){\bigg (}\sum _{i=0}^{i=k}a_{i}P_{i}(x){\bigg )}^{2}=(x+1){\bigr (}1.2247449(x){\bigr )}^{2}=1.5x^{3}+1.5x^{2}\\&L_{4}(x^{2})=\int _{-1}^{2x^{2}-1}1.5x^{3}+1.5x^{2}{\text{ }}dx=6x^{8}-8x^{6}+3x^{4}\\&L_{4}(x^{2})=6x^{8}-8x^{6}+3x^{4}\\&L_{4}(j\omega ^{2})=6(j\omega )^{8}+8(j\omega )^{6}+3(j\omega )^{4}\\&echo={\sqrt {\epsilon ^{3.0103/10}-1}}=1\\&T_{4}(j\omega )={\bigg [}{\frac {1}{1+1^{2}(6(j\omega )^{8}+8(j\omega )^{6}+3(j\omega )^{4})}}{\bigg ]}_{\text{Left Half Plane}}\\&\\&T_{4}(j\omega )={\frac {1}{2.4494897(j\omega )^{4}+3.8282201(j\omega )^{3}+4.6244874(j\omega )^{2}+3.0412127(j\omega )+1}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fdedd8a8062db23bfa01d29f2c505ad938f376eb)
