how to remove conj from equations? AND SOLVE THEM ?
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hii i am getting this kind of equations can someone please help me to solve these equations
here "t" is a variable i need values of A1,A2,A3,A4,A5,A6,Q1,Q2,Q3,Q4,Q5,Q6.
conj((A4*sin(Q4 + (2^(1/2)*t)/5))/10 == 0)
conj((A5*sin(Q5 + (2^(1/2)*t)/5))/10 == 0)
conj((A6*sin(Q6 + (2^(1/2)*t)/5))/10 == 0)
conj((10^(1/2)*A1*sin(Q1 + (2*2^(1/2)*5^(1/2)*t)/25))/100 == 1)
conj((10^(1/2)*A2*sin(Q2 + (2*2^(1/2)*5^(1/2)*t)/25))/100 == 1)
conj((10^(1/2)*A3*sin(Q3 + (2*2^(1/2)*5^(1/2)*t)/25))/100 == 0)
conj((2^(1/2)*A4*cos(Q4 + (2^(1/2)*t)/5))/50 == 0)
conj((2^(1/2)*A5*cos(Q5 + (2^(1/2)*t)/5))/50 == 0)
conj((2^(1/2)*A6*cos(Q6 + (2^(1/2)*t)/5))/50 == 0)
conj((2^(1/2)*5^(1/2)*10^(1/2)*A1*cos(Q1 + (2*2^(1/2)*5^(1/2)*t)/25))/1250 == 0)
conj((2^(1/2)*5^(1/2)*10^(1/2)*A2*cos(Q2 + (2*2^(1/2)*5^(1/2)*t)/25))/1250 == 0)
conj((2^(1/2)*5^(1/2)*10^(1/2)*A3*cos(Q3 + (2*2^(1/2)*5^(1/2)*t)/25))/1250 == 0)
THANKS IN ADVANCE
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採用された回答
Walter Roberson
2022 年 8 月 4 日
At some point in your code, you construct a row vector of equations, and then you use the ' operator thinking that it will transpose the row vector into a column vector. However, ' is not the transpose operator: it is the conjugate transpose operator. You need to switch from using ' to using .' which is the transpose operator
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その他の回答 (1 件)
Rahul Jangid
2022 年 8 月 4 日
1 件のコメント
Walter Roberson
2022 年 8 月 4 日
What difficulty are you encountering?
When I execute on my system, it thinks for a while, and eventually responds with 625 solutions.
The A* variables are all constant for the solutions, but there are 5 unique solutions for each of Q3, Q4, Q5, Q6, each of which appears in every combination, for a total of 5^4 = 625 solutions.
Some of the solutions can be simplified if you can assume that
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