Comparison of filter magnitude between Butterworth-, Legendre- and Chebyshev-Type1-Filter The Optimum "L" filter (also known as a Legendre–Papoulis filter ) was proposed by
which is the reason for its alternate name and the "L" in Optimum "L".
Synthesizing the characteristic polynomials 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 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 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
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})}
See also References
![{\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)
