How to solve equation

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safi58
safi58 on 15 Feb 2017
Commented: Manuela Gräfe on 24 Apr 2017
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 Comments
safi58
safi58 on 16 Feb 2017
Can anyone give any solution to this question please?
Manuela Gräfe
Manuela Gräfe on 24 Apr 2017
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.

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Accepted Answer

Walter Roberson
Walter Roberson on 16 Feb 2017
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 Comments
safi58
safi58 on 20 Feb 2017
i got it!!!!
Manuela Gräfe
Manuela Gräfe on 24 Apr 2017
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.

Sign in to comment.

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