Solving difference equation with its initial conditions
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Hi,
Consider a difference equation:
8*y[n] - 6*y[n-1] + 2*y[n-2] = 1
with initial conditions
y[0]= 0 and y[-1]=2
How can I determine its plot y(n) in Matlab? Thank you in advance for your help!
2 个评论
John D'Errico
2017-2-19
Surely you can use a loop? Why not make an effort? You have the first two values, so a simple loop will suffice.
More importantly, you need to spend some time learning MATLAB. Read the getting started tutorials. It is apparent that you don't know how to even use indexing in MATLAB, nor how to use a for loop.
You will need to recognize that MATLAB does NOT allow zero or negative indices.
Walter Roberson
2017-2-19
I would call this a recurrence equation, not a difference equation.
采纳的回答
Resort the terms:
8*y[n] - 6*y[n-1] + 2*y[n-2] = 1
y[n] = (1 + 6*y[n-1] - 2*y[n-2]) / 8
or in Matlab:
y(n) = (1 + 6*y(n-1) - 2*y(n-2)) / 8;
Now the indices cannot start at -1, because in Matlab indices are greater than 0. This can be done by a simple translation:
y = zeros(1, 100); % Pre-allocate
y(1:2) = [2, 0];
for k = 3:100
y(k) = (1 + 6*y(k-1) - 2*y(k-2)) / 8;
end
Now you get the y[i] by y(i+2).
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Sindhuja Parimalarangan
2017-2-21
4 个投票
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