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for generation of 10 i nput in each sequence for a sinusoidal signal

1 vue (au cours des 30 derniers jours)
Richa
Richa le 21 Mai 2023
Réponse apportée : VBBV le 21 Mai 2023
for k=4:1000
x(k) = 1.05 * sin(pi*k/45);
x(k-3:k+10-4)
end
In the above question, if I want to generate a input sequence in a batch of 10 for each iteration . Then how can we implement it?

Réponses (1)

VBBV
VBBV le 21 Mai 2023
clearvars
% size or range of iteration
k = 4:1000;
x = 1.05 * sin(pi*k/45)
x = 1×997
0.2894 0.3591 0.4271 0.4929 0.5564 0.6172 0.6749 0.7294 0.7803 0.8274 0.8705 0.9093 0.9437 0.9735 0.9986 1.0188 1.0340 1.0442 1.0494 1.0494 1.0442 1.0340 1.0188 0.9986 0.9735 0.9437 0.9093 0.8705 0.8274 0.7803
for J = 1:length(k)-9
X(J,:) = x(J:J+9);
end
% batch of 10 inputs per iteration
X
X = 988×10
0.2894 0.3591 0.4271 0.4929 0.5564 0.6172 0.6749 0.7294 0.7803 0.8274 0.3591 0.4271 0.4929 0.5564 0.6172 0.6749 0.7294 0.7803 0.8274 0.8705 0.4271 0.4929 0.5564 0.6172 0.6749 0.7294 0.7803 0.8274 0.8705 0.9093 0.4929 0.5564 0.6172 0.6749 0.7294 0.7803 0.8274 0.8705 0.9093 0.9437 0.5564 0.6172 0.6749 0.7294 0.7803 0.8274 0.8705 0.9093 0.9437 0.9735 0.6172 0.6749 0.7294 0.7803 0.8274 0.8705 0.9093 0.9437 0.9735 0.9986 0.6749 0.7294 0.7803 0.8274 0.8705 0.9093 0.9437 0.9735 0.9986 1.0188 0.7294 0.7803 0.8274 0.8705 0.9093 0.9437 0.9735 0.9986 1.0188 1.0340 0.7803 0.8274 0.8705 0.9093 0.9437 0.9735 0.9986 1.0188 1.0340 1.0442 0.8274 0.8705 0.9093 0.9437 0.9735 0.9986 1.0188 1.0340 1.0442 1.0494

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