Legendre filter

Legendre filters , also known as Optimum L filters , are continuous frequency filters whose transfer function is based on the Legendre polynomials from which they are named . Legendre filters were introduced in 1958 by the Greek mathematician Athanasios Papoulis .

Legendre filters represent a compromise between the Butterworth filter and the Chebyshev filter : the absolute frequency curve is steeper than that of the Butterworth filter and, in contrast to the Chebyshev filter, has a monotonous curve in the stop and pass band.

Transfer function

Comparison of the absolute value curve between Butterworth, Legendre and Chebyshev type 1 filters

The squared magnitude frequency curve for the filter order is given by ${\displaystyle n}$

${\displaystyle M_{n}^{2}(\omega )={\frac {1}{1+L_{n}(\omega ^{2})}}}$

with the modified -th optimal polynomial , which is characterized by the fulfillment of several special criteria that ensure the desired properties of monotony of the transfer function and, at the same time, maximum steepness in the blocking range. These are the constraints ${\displaystyle n}$ ${\displaystyle L_{n}}$

${\displaystyle L_{n}(0)=0\quad {\text{(Gl. 1)}}}$

${\displaystyle L_{n}(1)=1\quad {\text{(Gl. 2)}}}$

and the demand for a monotonous increase

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} \omega }}L_{n}(\omega ^{2})\geq 0\quad {\text{(Gl. 3)}}}$

The main condition is the requirement for maximum steepness in the blocking range , e.g. B. from : ${\displaystyle \omega \geq 1}$

${\displaystyle \left.{\frac {\mathrm {d} }{\mathrm {d} \omega }}L_{n}(\omega ^{2})\right|_{\omega =1}={\text{Maximum}}\quad {\text{(Gl. 4)}}}$

Derivation

For linearly independent polynomials of degree , in the simplest case , an approach for the desired optimal polynomial can be formed with indirect fulfillment of (Eq. 3): ${\displaystyle k+1}$ ${\displaystyle Q_{i}(x)}$ ${\displaystyle 0\leq i\leq k}$ ${\displaystyle Q_{i}(x)=x^{i}}$

${\displaystyle L_{n}(\omega ^{2})=\int _{0}^{\omega ^{2}}\left[\sum _{i=0}^{k}a_{i}Q_{i}(x)\right]^{2}dx\quad {\text{(Gl. 5-1)}}}$

with unknown coefficients . Since the integrand is an even polynomial, is odd with . A straight with to get up offers: ${\displaystyle k+1}$ ${\displaystyle a_{i}}$ ${\displaystyle L_{n}(x)}$ ${\displaystyle n=2k+1}$ ${\displaystyle L_{n}(x)}$ ${\displaystyle n=2k+2}$

${\displaystyle L_{n}(\omega ^{2})=\int _{0}^{\omega ^{2}}x\left[\sum _{i=0}^{k}a_{i}Q_{i}(x)\right]^{2}dx\quad {\text{(Gl. 5-2)}}}$

Both approaches automatically fulfill the conditions from (Eq. 1) and (Eq. 3), since in (Eq. 5-2) is always positive. For the selected basic polynomials, for example, (Eq. 5-1) can be solved and converted into (Eq. 2) ${\displaystyle x}$

${\displaystyle L_{n}(1)=\int _{0}^{1}\left[\sum _{i=0}^{k}a_{i}Q_{i}(x)\right]^{2}dx=1\quad {\text{(Gl. 6)}}}$

This is a quadratic equation in the coefficients that can be solved for a coefficient, the easiest for . Inserted into (Eq. 5-1), there are still unknown coefficients that can be solved in non-linear equations from the partial derivatives of (Eq. 4). The straight-line approach in (Eq. 5-2) is to be used analogously. ${\displaystyle a_{i}}$ ${\displaystyle a_{0}}$ ${\displaystyle k}$ ${\displaystyle k}$

For general polynomials , the resulting system of equations is difficult to solve analytically for. However, the approach of (Eq. 5) suggests using the Legendre polynomials of the first kind as a basis, in the expectation that many partial integrals will vanish and the derivation will be simplified. Papoulis presented this in his first paper in 1958 for (Eq. 5-1). To do this, however, the integral limits must be adapted to the properties of the Legendre polynomials and scaled so that the following equation results: ${\displaystyle Q_{i}(x)}$ ${\displaystyle k>2}$ ${\displaystyle P_{i}(x)}$

${\displaystyle L_{n}(\omega ^{2})=\int _{-1}^{2\omega ^{2}-1}\left[\sum _{i=0}^{k}a_{i}P_{i}(x)\right]^{2}dx=\int _{-1}^{2\omega ^{2}-1}\sum _{i=0}^{k}a_{i}P_{i}(x)\sum _{j=0}^{k}a_{j}P_{j}(x)dx=\sum _{i=0}^{k}\sum _{j=0}^{k}a_{i}a_{j}\int _{-1}^{2\omega ^{2}-1}P_{i}(x)P_{j}(x)dx\quad {\text{(Gl. 7)}}}$

This simplifies (Eq. 2) or (Eq. 6) considerably to

${\displaystyle L_{n}(1)=\sum _{i=0}^{k}\sum _{j=0}^{k}a_{i}a_{j}\int _{-1}^{1}P_{i}(x)P_{j}(x)dx=\sum _{i=0}^{k}a_{i}^{2}\int _{-1}^{1}P_{i}^{2}(x)dx=2\sum _{i=0}^{k}{\frac {a_{i}^{2}}{2i+1}}=1}$

For you get so ${\displaystyle a_{0}}$

${\displaystyle a_{0}^{2}={\frac {1}{2}}-\sum _{i=1}^{k}{\frac {a_{i}^{2}}{2i+1}}\quad {\text{(Gl. 8)}}}$

(. Eq 4) for determining the maximum in the partial derivative is of after the still unknown coefficients with Needed: ${\displaystyle a_{0}}$ ${\displaystyle a_{j}}$ ${\displaystyle 0<j\leq k}$

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} a_{j}}}a_{0}={\frac {\mathrm {d} }{\mathrm {d} a_{j}}}{\sqrt {{\frac {1}{2}}-\sum _{i=1}^{k}{\frac {a_{i}^{2}}{2i+1}}}}={\frac {-a_{j}}{(2j+1)a_{0}}}\quad {\text{(Gl. 9)}}}$

Note: For the inner derivative, only the summand with the index makes a contribution, because all other summands are independent of. is identical to the root expression in (Eq. 9), but for the sake of simplicity it is included in the following like a constant parameter to which the solution of the unknown should refer. It is then determined that (Eq. 8) or (Eq. 2) are fulfilled. ${\displaystyle i=j}$ ${\displaystyle a_{j}}$ ${\displaystyle a_{0}}$ ${\displaystyle a_{j}}$ ${\displaystyle a_{0}}$

When forming the left side of (Eq. 4), the following insight is important. For everyone and the identity arises: ${\displaystyle P_{i}(x)}$ ${\displaystyle P_{j}(x)}$

${\displaystyle \left.{\frac {\mathrm {d} }{\mathrm {d} \omega }}\int _{-1}^{2\omega ^{2}-1}P_{i}(x)P_{j}(x)dx\right|_{\omega =1}=4\quad {\text{(Gl. 10)}}}$

Hence (Eq. 4) becomes

${\displaystyle 4\left[\sum _{i=0}^{k}a_{i}\right]^{2}=\operatorname {Maximum} (a_{j})\quad {\text{(Gl. 11)}}}$

A necessary condition for a maximum is that all partial derivatives of the left-hand side of (Eq. 11) with respect to the unknown coefficients are zero. It must be taken into account that it also depends on all according to (Eq. 8) and (Eq. 9) ${\displaystyle a_{j}}$ ${\displaystyle a_{0}}$ ${\displaystyle a_{j}}$

${\displaystyle 4{\frac {\mathrm {d} }{\mathrm {d} a_{j}}}\left[\sum _{i=0}^{k}a_{i}\right]^{2}=4\cdot 2\left(\sum _{i=0}^{k}a_{i}\right){\mathrm {d} \over \mathrm {d} a_{j}}\left(a_{0}+a_{j}\right)=8\left(1-{\frac {a_{j}}{(2j+1)a_{0}}}\right)\sum _{i=0}^{k}a_{i}=0_{j}\quad {\text{(Gl. 12)}}}$

Note: Only the two summands and are dependent on. ${\displaystyle a_{0}}$ ${\displaystyle a_{j}}$ ${\displaystyle a_{j}}$

The sum is only zero if and all are, but this is excluded because then and also (Eq. 8) would be violated. So the expression in brackets must be null and contain the solution ${\displaystyle a_{0}}$ ${\displaystyle a_{j}=0}$ ${\displaystyle L_{n}(x)={\text{konstant}}}$

${\displaystyle a_{j}=(2j+1)a_{0}\quad {\text{(Gl. 13)}}}$

Inserted into (Eq. 8) results in

${\displaystyle a_{0}^{2}={\frac {1}{2}}-\sum _{i=1}^{k}(2i+1)a_{0}^{2}={\frac {1}{2}}-k(k+2)a_{0}^{2}}$

or

${\displaystyle (k+1)^{2}a_{0}^{2}={\frac {1}{2}}}$

For

${\displaystyle a_{0}={\frac {1}{{\sqrt {2}}(k+1)}}\quad {\text{(Gl. 14)}}}$

With (Eq. 13) we get for all coefficients ${\displaystyle a_{i}={\frac {2i+1}{{\sqrt {2}}(k+1)}}\quad {\text{(Gl. 15)}}}$

For just after (Eq. 5-2) Papoulis published an analogous solution. After scaling to the more suitable interval limits, the following then applies ${\displaystyle n=2k+2}$

${\displaystyle L_{n}(\omega ^{2})=\int _{-1}^{2\omega ^{2}-1}(x+1)\left[\sum _{i=0}^{k}a_{i}P_{i}(x)\right]^{2}dx=\sum _{i=0}^{k}\sum _{j=0}^{k}a_{i}a_{j}\int _{-1}^{2\omega ^{2}-1}(x+1)P_{i}(x)P_{j}(x)\quad {\text{(Gl. 16)}}}$

Analogously to the helpful identity from (Eq. 10), we have for even ${\displaystyle n}$

${\displaystyle \left.{\frac {\mathrm {d} }{\mathrm {d} \omega }}\int _{-1}^{2\omega ^{2}-1}(x+1)P_{i}(x)P_{j}(x)dx\right|_{\omega =1}=8}$

The coefficients are:

${\displaystyle a_{i}={\begin{cases}{\frac {2i+1}{\sqrt {(k+1)(k+2)}}}&{\text{für }}k+i{\text{ gerade}}\\0&{\text{sonst}}\end{cases}}}$

Conclusion

As a basis for the optimal polynomial , the use of the eponymous Legendre polynomials is not absolutely necessary. Any other linearly independent, polynomial basis leads to the same result, but the analytical derivation is much more difficult, if not impossible. In order to simplify the already laborious and error-prone resolution of (Eq. 7) and (Eq. 16), the denominators of the respective factors can be placed in front of the integral. That leads to ${\displaystyle L_{n}(\omega ^{2})}$ ${\displaystyle Q_{i}(x)}$ ${\displaystyle a_{0}^{2}}$ ${\displaystyle a_{1}^{2}}$

${\displaystyle L_{n}(\omega ^{2})={\frac {1}{2(k+1)^{2}}}\int _{-1}^{2\omega ^{2}-1}\left[\sum _{i=0}^{k}(2i+1)P_{i}(x)\right]^{2}dx\quad {\text{(Gl. 17)}}}$

respectively

${\displaystyle L_{n}(\omega ^{2})={\frac {1}{(k+1)(k+2)}}\int _{-1}^{2\omega ^{2}-1}(x+1)\left[\sum _{i=0}^{k}b_{i}P_{i}(x)\right]^{2}dx\quad {\text{(Gl. 18)}}}$

With ${\displaystyle b_{i}={\begin{cases}2i+1&{\text{für }}k+i{\text{ gerade}}\\0&{\text{sonst}}\end{cases}}}$

Result

For the filter order from 1 to 6, the optimal polynomials of the filter are: ${\displaystyle n}$ ${\displaystyle L_{n}(\omega ^{2})}$

${\displaystyle n}$	${\displaystyle L_{n}(\omega ^{2})}$
1	${\displaystyle \omega ^{2}}$
2	${\displaystyle \omega ^{4}}$
3	${\displaystyle \omega ^{2}-3\omega ^{4}+3\omega ^{6}}$
4th	${\displaystyle 3\omega ^{4}-8\omega ^{6}+6\omega ^{8}}$
5	${\displaystyle \omega ^{2}-8\omega ^{4}+28\omega ^{6}-40\omega ^{8}+20\omega ^{10}}$
6th	${\displaystyle 6\omega ^{4}-40\omega ^{6}+105\omega ^{8}-120\omega ^{10}+50\omega ^{12}}$

Further polynomials up to the 10th order can be found in the sources mentioned.

literature

• Franklin F. Kuo: Network Analysis and Synthesis . 2nd Edition. Wiley, 1966, ISBN 0-471-51118-8 .

Individual evidence

1. ↑ ^{a b} Athanasios Papoulis: Optimum Filters with Monotonic Response . tape 46 , no. 3 . Proceedings to the IRE , March 1958, p. 606 to 609 .
2. ↑ ^{a b} Notes on "L" (Optimal) Filters by C. Bond. (PDF; 172 kB) 2011, accessed on August 31, 2012 .  
3. ↑ Athanasios Papoulis: On Monotonic Response Filters . tape 47 . Proceedings to the IRE , 1959, pp. 332 to 333 .
4. ↑ Optimum “L” Filters Polynomials, Poles and Circuit Elements. (PDF; 100 kB) 2004, accessed on August 31, 2012 .