im trying to chop my signal so it only plots 1 second but any second I have choosen as the length of the signal is 10 seconds

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zeshan chishti 2021-1-3

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评论： Star Strider 2021-1-5

t1=load('201710h1.lvm');  %test 1 
t2=load('201710h2.lvm');  %test 2
 
Fs = 100000; % frequency sample
 
x = t1 (:,1); % time for test 1 
sin = t1 (:,2); % test 1 sesnor 1
plot(sin)
title('Sensor1')

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Answer 1

Star Strider 2021-1-3

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First, please do not name the variable ‘sin’ since that makes the sin() function useless. This is called ‘overshadowing’ and is to be avoided.

Second, see if the buffer function will do what you want, with respect to segmenting your signal.

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zeshan chishti 2021-1-4

thank you will try the buffer

Star Strider 2021-1-5

My pleasure!

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Answer 2

Walter Roberson 2021-1-3

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在 MATLAB Online 中打开

x = t1(1:Fs,1);
sen1 = t1(1:Fs,2);
plot(x, sen1)
title('Sensor1')

or

idx = find(t1(:,1) <=1, 'last');
x = t1(1:idx,1);
sen1 = t1(1:idx,2);
plot(x, sen1)
title('Sensor1')

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zeshan chishti 2021-1-4

thanks but this does not chop the signal

Walter Roberson 2021-1-4

Yes it does chop the signal, showing two different ways of chopping the signal.

When your fs is accurate, then 1:Fs takes 1 second of data.

When your time vector is accurate, then finding the place where the time is last no more than 1 second and taking the samples to there takes the first 1 second of data, even if the data is sampled at irregular intervals. (Exception: if you have negative time, the code as written does not remove the samples corresponding to negative time.)

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