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

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)

[KSSV]

From what you want to delete the range from 0.0 to 0.5 ?

[horizon]

I would like to delete data from 0.0 to 0.5.

[Walter Roberson]

data(data >= 0 & data <= 0.5) = [];

[horizon]

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) = [];

Answer 1

[KSSV]

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 ; 

[horizon]

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]

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]

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 .