Calculate this matrix in a more general and shorter way

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Rowan Humphreys
Rowan Humphreys2021-11-25
评论: Rowan Humphreys ,2021-11-30
Is there a way of generating P in a way that doesn't involve writing sin/cos/sin/cos repeatedly? Perhaps like how you can generate an array by writing 'x = 1:0.1:10'?
the size of 'x' is 100by1 and 'w' is some constant.
P=[ones(size(x)),cos(w*x),sin(w*x),cos(2*w*x),sin(2*w*x),cos(3*w*x),sin(3*w*x),cos(4*w*x),sin(4*w*x),cos(5*w*x),sin(5*w*x)]

回答(2 个)

Stephen
Stephen 2021-11-25
x = rand(100,1);
w = pi;
m = (1:5).*w.*x;
m = [ones(100,1),reshape([cos(m);sin(m)],100,[])];
for comparison:
P = [ones(size(x)),cos(w*x),sin(w*x),cos(2*w*x),sin(2*w*x),cos(3*w*x),sin(3*w*x),cos(4*w*x),sin(4*w*x),cos(5*w*x),sin(5*w*x)];
isequal(P,m)
ans = logical
1

John D'Errico
John D'Errico 2021-11-25
x is 100x1. Now, suppose we want to generate multiple terms of the form sin(k*w*x), so k will be a vector. w is a fixed constant. Do you know how to do that?
x = (0:2:10)'; % arbitrary, your x is what you have. I've made x small so you can see what happens
w = 1; % again, my choice.
k = 1:4;
Ok, so what does this do in MATLAB?
x*w*k
ans = 6×4
0 0 0 0 2 4 6 8 4 8 12 16 6 12 18 24 8 16 24 32 10 20 30 40
This is just effectively a multiplication table. But do you see that it forms what you need? You can now compute sin(x*w*k), and cos(x*w*k).
In your case, x is a column vector, of size 100x1. Your code might look like:
x = ???? % whatever x is
w = ???? % again
n = 10; % how many terms?
P = [ones(size(x)),sin(x*w*k),cos(x*w*k)];
Now, the only problem is, all the sin terms are bundled together, then all the cos terms. If that bothers you, then reshuffle those columns. Something like this should work.
P(:,[2:2:end,3:2:end]) = P(:,[2:n+1,n+2:end]);
Now the sin and cos terms alternate.

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