reduce values on a vector

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Miguel Albuquerque
编辑： Jan 2022-7-6
Hey guys,
I have this code. This two signal have a lot of values(millions) I want to use 1000 values(reference_signal_samples and Surveillance_signal_samples) with a step of 3000 samples. Im not able to do it with this code, I dont know where the error is.
Thank you
nRef = numel(Reference_signal)/Fs;
nSurv = numel(Surveillance_signal)/Fs;
pulse_size = 1000;
for k = 1:nRef
Reference_signal_samples =Reference_signal((k-1)*pulse_size+(1:pulse_size));
for kk =1:nSurv
Surveillance_signal_samples=Surveillance_signal((kk-1)*pulse_size+(1:pulse_size));
end
end

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采纳的回答

Jan
Jan 2022-7-5
You are overwriting the variable Surveillance_signal_samples repeatedly:
for kk =1:nSurv
Surveillance_signal_samples=Surveillance_signal((kk-1)*pulse_size+(1:pulse_size));
end
You want to use 1000 values with a stepwidth of 3000?
Reference_signal_samples = Reference_signal(1:3000:3000*1000);
Surveillance_signal_samples = Surveillance_signal(1:3000:3000*1000);
No loops.
A hint: Use more compact names for variablkes to improve the readability.
  2 个评论
Miguel Albuquerque
Miguel Albuquerque 2022-7-5
编辑：Miguel Albuquerque 2022-7-5
sorry, shoudnt it be?
And with that code, than I have to change the values, because next I want the iteration 4000 to 5000
I want first Reference_signal_samples(1:1000), plot the signals, next Reference_signal_samples(4000:5000), plot the signals, till the end of samples(millions)
Reference_signal_samples = Reference_signal(1:3000:3000);
Jan
Jan 2022-7-6
编辑：Jan 2022-7-6
Ran in:
Maybe I've misunderstood your question.
disp(1:3000:3000)
1
1:3000:3000 is the scalar 1. If you really want to get parts of the array, these indices are strange: 1:1000, 4000:5000 : The 2nd has 1001 elements, the first only 1000. Maybe you mean:
for k = 1:4000:numel(Reference_signal)
signal = Reference_signal(k:k+999);
... here your calculations...
end

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