calculate with Twiddle factor
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Does anybody knows how to calculate the Twiddle Factors?
For example: for n=4 you have to do the calculate Wn= e^(-j*2*π)/n and ypu have to do it for a table 4x4. table looks like..
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the second is the table after calculate the 4x4
and i need this for calculate twiddle factors for n=8,16,24...
2 Commenti
Victor Cubedo
il 25 Dic 2023
cos(2pi*n*k/N)-j*sin(2pi*n*k/N) , N = 4,8,16, k and n = rows and colums
Risposte (2)
Sulaymon Eshkabilov
il 25 Dic 2023
This is one of the possible solutions:
N = 4;
Twiddle_Factors = zeros(N, N);
for k = 0:N-1
for n = 0:N-1
Twiddle_Factors(k+1, n+1) = exp(-1i * 2 * pi / N * k * n);
end
end
disp('Twiddle Factors of size 4-by-4: ');
disp(Twiddle_Factors);
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Pramita
il 19 Gen 2025
N = 8; % Number of points in DFT
w = zeros(1, N); % Pre-allocate array for twiddle factors
% Calculate the twiddle factors
for r = 0:N-1
w(r+1) = exp((-1j) * 2 * pi * r / N);
end
% Calculate Magnitude and Phase
Mag = abs(w);
Phase = (angle(w) / pi) * 180; % Convert phase to degrees
% Plotting
figure;
% Plot Magnitude
subplot(2,1,1);
stem(0:N-1, Mag, 'filled');
title('Magnitude of Twiddle Factors');
xlabel('Index');
ylabel('Magnitude');
grid on;
% Plot Phase
subplot(2,1,2);
stem(0:N-1, Phase, 'filled');
title('Phase of Twiddle Factors (degrees)');
xlabel('Index');
ylabel('Phase (degrees)');
grid on;
% Periodicity Check: W^(k+N) == W^k
k = 3; % Example index for periodicity check
disp(['Periodicity check: W^(', num2str(k), '+N) == W^(', num2str(k), '): ', num2str(w(k+1) == w(mod(k+N, N)+1))]);
% Symmetry Check: W^(-k) == conj(W^k)
disp(['Symmetry check: W^(-', num2str(k), ') == conj(W^', num2str(k), '): ', num2str(w(mod(-k, N)+1) == conj(w(k+1)))]);
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