How can I get the maximum difference in the specific range?

2019 5 8
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更新时间：2019 5 8
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I am trying to find the value of the maximum difference of the wave, but I would like to eliminate some part of the graph.
In this case, I would like to delete the range from 0.0 to 0.5 and get the maximum value from 0.5 to 3.
I have no idea how to execute the above process.
t = 0:1/1000:3;
q1 = sin(2*pi*7*t).*exp(-t/2);
plot(t,q1)
[up,lo] = envelope(q1,100,'analytic');
hold on
plot(t,up,'-',t,lo,'--')
hold off
envelope(q1,300)
maxDiff = peak2peak(up-lo);
disp(maxDiff)

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KSSV
KSSV 2019-5-8
From what you want to delete the range from 0.0 to 0.5 ?
horizon
horizon 2019-5-8
I would like to delete data from 0.0 to 0.5.
horizon
horizon 2019-5-8
Thank you for your answer.
When I opened the array t, it has these consequtive data.
t = [0 0.00100000000000000 0.00200000000000000 0.00300000000000000 0.00400000000000000 0.00500000000000000 0.00600000000000000 0.00700000000000000 0.00800000000000000 0.00900000000000000 0.0100000000000000 0.0110000000000000 0.0120000000000000 0.0130000000000000 ....]
I can understand that your answer substitutes data range from 0 to 0.5 null, but in this case, what does it applied to data?
data(data >= 0 & data <= 0.5) = [];

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KSSV
KSSV 2019-5-8

0 个投票

Let t, data be your time and data. You have two options:
First option
% Remove the specified data, this will reduce the length of your arrays.
idx = data >= 0 & data <= 0.5 ;
t(idx) = [] ;
data(idx) = [] ;
Second option
% Make the unwanted data to NaN, this will not change the length of the array
idx = data >= 0 & data <= 0.5 ;
data(idx) = NaN ;

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horizon
horizon 2019-5-8
The result of the max value of differences were same both with and without elimination.
>> cutSentwave
1.8245
code without elimination
t = 0:1/1000:3;
q1 = sin(2*pi*7*t).*exp(-t/2);
% Make the unwanted data to NaN, this will not change the length of the array
%idx = t >= 0 & t <= 0.5 ;
%t(idx) = NaN ;
plot(t,q1)
[up,lo] = envelope(q1,100,'analytic');
hold on
plot(t,up,'-',t,lo,'--')
hold off
envelope(q1,300)
maxDiff = peak2peak(up-lo);
disp(maxDiff)
code wth elimination
t = 0:1/1000:3;
q1 = sin(2*pi*7*t).*exp(-t/2);
% Make the unwanted data to NaN, this will not change the length of the array
idx = t >= 0 & t <= 0.5 ;
t(idx) = NaN ;
plot(t,q1)
[up,lo] = envelope(q1,100,'analytic');
hold on
plot(t,up,'-',t,lo,'--')
hold off
envelope(q1,300)
maxDiff = peak2peak(up-lo);
disp(maxDiff)
horizon
horizon 2019-5-8
t = 0:1/1000:3;
q1 = sin(2*pi*7*t).*exp(-t/2);
% Make the unwanted data to NaN, this will not change the length of the array
for idx = (1:20)-1
t(idx) = nan;
q1(idx) = nan;
end
plot(t,q1)
[up,lo] = envelope(q1,100,'analytic');
hold on
plot(t,up,'-',t,lo,'--')
hold off
envelope(q1,300)
maxDiff = peak2peak(up-lo);
disp(maxDiff)
If I write the above, since idx is array index and it is required to reduce data for y-axis. However, I got an error message.
Array indices must be positive integers or logical values.
Error in cutSentwave (line 6)
t(idx) = nan;
Walter Roberson
Walter Roberson 2019-5-8
Where did you obtain the line
for idx = (1:20)-1
? I do not see anyone having suggested that.
MATLAB indices start with 1. Your 1:20 would be [1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20]. Then you subtract 1 to get [0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19] . Then you try to use that first value, 0, as an index, which is a problem.
Perhaps you are accustomed to Python, which indexes from 0 but has the odd range() construct that generates 0 to N-1 .

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