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

1 Ansicht (letzte 30 Tage)
Richa
Richa am 21 Mai 2023
Beantwortet: VBBV am 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?

Antworten (1)

VBBV
VBBV am 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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