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Symbolic solution by matching coefficients in trigonometric equation

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Gabriel Droguett
Gabriel Droguett 2021 年 4 月 6 日
コメント済み: Paul 2021 年 4 月 8 日
Hi, I am trying to solve the following symbolic equation:
id*cos(th) - iq*sin(th) == 2*x1*cos(th) - 2*y1*sin(th)
solving for x1 and y1 in terms of id and iq for all angles th. If considering all possible angles the solution should come from equating the coefficients of sines and cosines separetely. So I expected:
x1 = id/2
y1 = iq/2
However the solution I get is:
x1 = (id*cos(th) - iq*sin(th))/(2*cos(th))
y1 = 0
Is there an option that can be added to the solve function to handle this case?
  1 件のコメント
Paul
Paul 2021 年 4 月 8 日
I anticipated solve() returning three parametric solutions: one for th an odd mutiple of pi/2, one for th an even multiple of pi/2, one for th not a multiple of pi/2. However, it only returns one parametric solution:
>> sol = solve(eqn,[x1 y1],'ReturnConditions',true);
>> [sol.x1 sol.y1]
ans =
[ (id*cos(th) - iq*sin(th) + 2*z*sin(th))/(2*cos(th)), z]
>> sol.conditions
ans =
~in(th/pi - 1/2, 'integer')

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回答 (1 件)

David Goodmanson
David Goodmanson 2021 年 4 月 6 日
編集済み: David Goodmanson 2021 年 4 月 6 日
Hello Gabriel,
here is one way
syms id iq x1 y1 th1 th2
eq1 = id*cos(th1) - iq*sin(th1) == 2*x1*cos(th1) - 2*y1*sin(th1)
eq2 = id*cos(th2) - iq*sin(th2) == 2*x1*cos(th2) - 2*y1*sin(th2)
s = solve(eq1,eq2,x1,y1)
You have to persuade symbolics that the equation obtains for more than just one one angle.
  3 件のコメント
Paul
Paul 2021 年 4 月 7 日
David's solution yields:
>> [s.x1 s.y1]
ans =
[ id/2, iq/2]
What do you mean by substituting th1 for th2?

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