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Symbolic solution by matching coefficients in trigonometric equation

조회 수: 2(최근 30일)
Gabriel Droguett
Gabriel Droguett 2021년 4월 6일
댓글: Paul 2021년 4월 8일
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?
  댓글 수: 1
Paul
Paul 2021년 4월 8일
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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답변(1개)

David Goodmanson
David Goodmanson 2021년 4월 6일
편집: David Goodmanson 2021년 4월 6일
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.
  댓글 수: 3
Paul
Paul 2021년 4월 7일
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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