I am trying to solve the Newton Coefficeint matrix for Newton sample rate converter please could you suggest for modifications?
11 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
So this coding is based on computation of coefficients of the
the filter structure formula as given in the img:
d is the bandwidth tuning factor :N is the polynomial order and M is subfilter order N =M-1
this is the fractional delay of Farrow Rate Converter d is fractional delay of Newton used for bandwidth tuning
My code :
function coeffMatrix = computeNewtonCoeff(N, d)
% Compute the Newton Coefficient matrix
% N: Order of the interpolator
% d: Bandwidth tuning parameter
N =M-1;
% Pre-allocate the coefficient matrix
coeffMatrix = zeros(N + 1, 1);
% Loop to compute coefficients using the equation
for k = 0:N
mu =linspace(-0.5,0.5);
d= compute_d_in_steady_state(mu, N);
% Compute the factorial term
numerator = prod(d:(d + k - 1)); % d * (d+1) * ... * (d+k-1)
denominator = factorial(k); % k!
coeffMatrix(k + 1) = numerator / denominator; % Coefficients
end
end
function d = compute_d_in_steady_state(~, N)
% Computes d in steady-state given mu and N
% mu: Fractional part of accum, should be in [-1, 0)
% N: Interpolator order
% Ensure mu is within the valid range
mu = linspace(-0.5,0.5);
% Compute d using the steady-state expression
d = mu - N / 2;
end
N= 4;
mu = linspace(-0.5,0.5);% Order of the interpolator
d =compute_d_in_steady_state(mu,N); % Example value of the bandwidth tuning parameter
coeffMatrix = computeNewtonCoeff(N, d);
disp('Newton Coefficient Matrix:');
disp(coeffMatrix);
Solution is showing after running :
Newton Coefficient Matrix:
1.0000
-2.5000
1.8750
-0.3125
-0.0391
The solution mentioned in the paper is different :
I have attached the pdf
- Design_of_Low_Complexity_Arbitrary_Pass-band_Filter_Hardware_using_Newton_Structure-based_Lagrange_Interpolators (3).pdf
for further reference you can read through .
0 commentaires
Réponses (0)
Voir également
Catégories
En savoir plus sur Multirate Signal Processing dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!