Putting random values in for loop

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SAZZAD HOSSAIN
SAZZAD HOSSAIN 2012년 10월 24일
Hi everyone
I am writing a program in which i am using the following loop.
for t = 1:1:81;
vv(t) = floor(+1*120*sin(2*pi*(t-1)/40));
end
The vales of vv are like 0,18,37...... However, i found that my program has some instability due to jumping directly from vv = 0 to vv = 18. Therefore, my intention is to keep the inputs for the loop as it is (otherwise it gets too long to run), but to put few more input values for vv. for example I want to start with 0, 1, 5, 9, 14, 18,37,....... Hence i need to put those 4 values in between 0 and 18 to bring stability into the result. Can anyone please help me.
Thanks in advance
Hossain

채택된 답변

Jonathan Epperl
Jonathan Epperl 2012년 10월 24일
Do what Matt suggested:
VV = floor(+1*120*sin(2*pi*(t-1)/40));
Then add your extra values into VV:
VV = [VV(1) 1 5 9 14 VV(2:end)];
Then run your loop
for i=1:numel(vv) % presumably 85
vv = VV(i);
% your other code
end
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SAZZAD HOSSAIN
SAZZAD HOSSAIN 2012년 10월 24일
Thanks Jonathan. Works perfect.

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추가 답변 (1개)

Matt Kindig
Matt Kindig 2012년 10월 24일
Hi Sazzad,
First of all, you don't need the loop: you can just do it like this:
vv = floor(+1*120*sin(2*pi*(t-1)/40));
Second, can you just use a finer resolution of t, such as:
t= 1:0.1:81;
vv = floor(+1*120*sin(2*pi*(t-1)/40));
Now vv changes from 0 to 1 to 3 to 5, etc.
  댓글 수: 1
SAZZAD HOSSAIN
SAZZAD HOSSAIN 2012년 10월 24일
Hi Matt
Thanks for the answer. The vv = floor (....) part is followed by a lot of other codes and it needs to be in a for loop. And i cannot use finer resolution. the current resolution i am using takes around 2 days to complete so i kind of want to stick to this resolution.
Thanks.

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