Problem in using asin function
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Consider the followingj
t = 0 : 0.02 : 10;
nu = (t-5).^2 + 2;
omega = 2*pi*nu;%as a polynomial of degree 2
f = sin(omega);
Since f has defined as sin(omega), it should be possible to recalculate omega from f. That is:
Omega = asin(f);
plot(t,omega,'b',t,Omega,'r--')
Of course omega and Omega are no the same. But, is there any solution for this problem?
2 个评论
Sriram Tadavarty
2020-3-15
Should this use sind and asind functions?
Alex Dell
2021-3-30
You could try normalising your polynomial such that it fits within the first interval of the asin function and then multiply the final terms by this same factor.
f = sin(omega./max(omega));
Omega = asin(f).*max(omega);
This should then give a consistent output to your original polynomial.
采纳的回答
Ameer Hamza
2020-3-15
This is not a problem with MATLAB. This is the property of sin function. Sin is a periodic function, therefore, its inverse function asin can only the output value in a specific range. Consider this
sin(pi/2) = 1
sin(5*pi/2) = 1
sin(9*pi/2) = 1
so what should be the output of
asin(1)
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