How to solve equation
显示 更早的评论
m_c_theta1=(m_c0-1/M-1)*cos(theta1)+j_L0*sin(theta1)+1/M+1;
j_L_theta1=(-m_c0+1/M+1)*sin(theta1)+j_L0*cos(theta1);
m_c_gama=(m_c_theta1-1/M+1)*cos(gama-theta1)+j_L_theta1*sin(gama-theta1)+1/M-1;
j_L_gama=(-m_c_theta1+1/M-1)*sin(gama-theta1)+j_L_theta1*cos(gama-theta1);
Boundary Condition:
m_c_gama=-m_c0
j_L_gama=-j_L0
j_L_theta1=-(gamma*l)/2
Hi,
I need to solve these equations and these are my boundary conditions. I need to find m_c0,j_L0,theta1 and M.
Can anyone help me please?
7 个评论
Kaushik Lakshminarasimhan
2017-2-15
Do you want an exact or approximate solution?
safi58
2017-2-15
Walter Roberson
2017-2-16
Complicated trig functions rarely have exact solutions.
Are there constraints to consider? Real values? Only need to consider one period?
safi58
2017-2-16
safi58
2017-2-16
safi58
2017-2-16
Manuela Gräfe
2017-4-24
Hi, umme mumtahina.
I see you are working with the LLC converter and the IEEE document (Optimal design methodology for LLC Resonant Converter... by Zhijian Fang etc.).
I am looking for the same solution at the moment for my bachelor thesis and I was wondering if you could provide me your MATLAB code? So I 'don't have to annoy Walter Roberson with the same issues. Please contact me via private message.
采纳的回答
Walter Roberson
2017-2-16
m_c0 = 4*((m_c_theta1+1)*sin(gamma)^2+l*gamma*(ROOT-(1/2)*cos(gamma)-1/2)*sin(gamma)+(2*ROOT*m_c_theta1+2*ROOT-2)*cos(gamma)-2*ROOT*m_c_theta1+2*ROOT-2)*m_c_theta1/((gamma^2*l^2+4*m_c_theta1^2+8*m_c_theta1+4)*sin(gamma)^2-4*l*gamma*(m_c_theta1-1)*sin(gamma)-16*m_c_theta1)
j_L0 = (1/2)*(-l*gamma*(gamma^2*l^2+4*m_c_theta1^2+4*m_c_theta1+4)*sin(gamma)^3+(2*m_c_theta1*(gamma^2*l^2+4*m_c_theta1^2+12*m_c_theta1+8)*cos(gamma)+4*l^2*(-1+(ROOT+1/2)*m_c_theta1)*gamma^2+(16*ROOT+8)*m_c_theta1^3+(16*ROOT+8)*m_c_theta1^2+16*m_c_theta1)*sin(gamma)^2-2*l*gamma*((l^2*(ROOT-1)*gamma^2+(4*ROOT+4)*m_c_theta1^2+(4*ROOT-8)*m_c_theta1+4*ROOT-4)*cos(gamma)+l^2*(ROOT-1)*gamma^2+(4*ROOT+4)*m_c_theta1^2+(-4*ROOT-8)*m_c_theta1+4*ROOT-4)*sin(gamma)-8*(l^2*(ROOT-1)*gamma^2+4*m_c_theta1^2*(ROOT+1))*(cos(gamma)+1))/(((gamma^2*l^2+4*(m_c_theta1+1)^2)*sin(gamma)^2-4*l*gamma*(m_c_theta1-1)*sin(gamma)-16*m_c_theta1)*sin(gamma))
theta1 = arctan((-2*gamma*l*sin(gamma)^2*m_c_theta1+((-l^2*(ROOT-1)*gamma^2-4*(m_c_theta1+1)*(ROOT*m_c_theta1+ROOT-1))*cos(gamma)-l^2*(ROOT-1)*gamma^2+4*ROOT*m_c_theta1^2-4*ROOT-4*m_c_theta1+4)*sin(gamma)-4*l*gamma*(cos(gamma)+1)*(ROOT-1))/((gamma^2*l^2+4*(m_c_theta1+1)^2)*sin(gamma)^2-4*l*gamma*(m_c_theta1-1)*sin(gamma)-16*m_c_theta1), ROOT)
M = ((4*m_c_theta1+4)*sin(gamma)^2+4*l*gamma*(ROOT-(1/2)*cos(gamma)-1/2)*sin(gamma)+(8*ROOT*m_c_theta1+8*ROOT-8)*cos(gamma)-8*ROOT*m_c_theta1+8*ROOT-8)/((gamma^2*l^2+4*m_c_theta1^2+8*m_c_theta1+4)*sin(gamma)^2-4*l*gamma*(m_c_theta1-1)*sin(gamma)-16*m_c_theta1)
where
ROOT = RootOf((-4*gamma*cos(gamma)*l*sin(gamma)+4*sin(gamma)^2*m_c_theta1^2+gamma^2*l^2*sin(gamma)^2-4*sin(gamma)^2+8-8*cos(gamma)+4*gamma*l*sin(gamma))*z^2+(-cos(gamma)*l^2*gamma^2*sin(gamma)^2-gamma^2*l^2*sin(gamma)^2-2*l*sin(gamma)^3*m_c_theta1*gamma-4*l*sin(gamma)^3*gamma+4*cos(gamma)*m_c_theta1*sin(gamma)^2+4*cos(gamma)*sin(gamma)^2-4*sin(gamma)^2*m_c_theta1-4*sin(gamma)^2)*z+cos(gamma)*l^2*gamma^2*sin(gamma)^2+2*l*sin(gamma)^3*m_c_theta1*gamma+4*l*sin(gamma)^3*gamma-2*cos(gamma)*m_c_theta1^2*sin(gamma)^2+4*gamma*cos(gamma)*l*sin(gamma)-4*cos(gamma)*m_c_theta1*sin(gamma)^2-2*sin(gamma)^2*m_c_theta1^2-4*gamma*l*sin(gamma)-4*cos(gamma)*sin(gamma)^2+4*sin(gamma)^2*m_c_theta1+8*sin(gamma)^2+8*cos(gamma)-8, z)
and RootOf(f(z),z) means the values, z, such that f(z) = 0 -- the roots of the equation.
As ROOT is a quadratic, it has two exact solutions that can be substituted in to the other equations.
10 个评论
safi58
2017-2-16
Walter Roberson
2017-2-16
The MATLAB code would be
syms m_c0 j_L0 theta1 M m_c_gama j_L_gama j_L_theta1
syms m_c_theta1 gama l
eqn1 = m_c_theta1==(m_c0-1/M-1)*cos(theta1)+j_L0*sin(theta1)+1/M+1;
eqn2 = j_L_theta1==(-m_c0+1/M+1)*sin(theta1)+j_L0*cos(theta1);
eqn3 = m_c_gama==(m_c_theta1-1/M+1)*cos(gama-theta1)+j_L_theta1*sin(gama-theta1)+1/M-1;
eqn4 = j_L_gama==(-m_c_theta1+1/M-1)*sin(gama-theta1)+j_L_theta1*cos(gama-theta1);
m_c_gama=-m_c0;
j_L_gama=-j_L0;
j_L_theta1=-(gama*l)/2;
eqns = subs([eqn1, eqn2, eqn3, eqn4]);
sol = solve(eqns, m_c0, j_L0, theta1, M);
Note: if you try to add 'returnconditions', true to the solve() then it will take a long time.
The MATLAB results appear to be more compact than the ones I found earlier.
Note: I had to assume that the gamma in your boundary condition was the same as the gama in your equations.
Walter Roberson
2017-2-17
[m_c0 == sol.m_c0(1), j_L0 == sol.j_L0(1), theta1 == sol.theta1(1), M == sol.M(1)]
[m_c0 == sol.m_c0(2), j_L0 == sol.j_L0(2), theta1 == sol.theta1(2), M == sol.M(2)]
Walter Roberson
2017-2-17
sol = solve(eqns, m_c0, j_L0, theta1, M, 'returnconditions', true);
would take a long time, not worth doing.
You can read about ReturnConditions at https://www.mathworks.com/help/symbolic/solve.html#inputarg_ReturnConditions
Walter Roberson
2017-2-17
To verify the solution, go back to the equations in eqns and subs() in particular numeric values for m_c0, j_L0, theta1, M . Then vpasolve() to get numeric values for those. Now subs() the particular numeric values into the variables in the sol structure and vpa() that, and see whether the formula's numeric solutions agreed the vpasolve() numeric solutions.
Walter Roberson
2017-2-17
No, that is a duplicate. I am not going to answer duplicates.
safi58
2017-2-19
safi58
2017-2-20
Manuela Gräfe
2017-4-24
Hi, umme mumtahina.
I see you are working with the LLC converter and the IEEE document (Optimal design methodology for LLC Resonant Converter... by Zhijian Fang etc.).
I am looking for the same solution at the moment for my bachelor thesis and I was wondering if you could provide me your MATLAB code? So I 'don't have to annoy Walter Roberson with the same issues. Please contact me via private message.
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Programming 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
