What function should i use to solve the equation?

Q = 0.4; m = 6; K = (Fx.^2).*(m-1) / sqrt(((m.*Fx.^2 - 1).^2)+(Fx.^2.*(Fx.^2-1).^2.*(m-1).^2.*Q.^2));

After initialization, what should i set to use the ODE solver?

Answer 1

[Star Strider]

There is no ODE to solve. Set the ‘K(Q,m,Fx)’ equation up as a symbolic function (if you have R2012a or later), substitute in ‘Qmax=0.4’ and ‘m=6’ (when you declare them to be symbolic, this is automatic) to get:

syms Fx m Q
Q = sym(0.4); 
m = sym(6); 
K(Fx) = (Fx.^2).*(m-1) / sqrt(((m.*Fx.^2 - 1).^2)+(Fx.^2.*(Fx.^2-1).^2.*(m-1).^2.*Q.^2));
DK = simplify(diff(K(Fx), Fx), 'steps', 10)
Fx_min = vpasolve(DK == 0)

DK =

-(10*Fx*(2*Fx^6 + 4*Fx^2 - 1))/(4*Fx^6 + 28*Fx^4 - 8*Fx^2 + 1)^(3/2)

Fx_min =

0

[Joseph Chia]

Thanks for the respond. I realised too, not ode.

anyway, from the example, Q = 0.4, m =6.3 I'm suppose to get Fx_min = 0.489

i think i type the equation in the wrong manner?

[Torsten]

K(Q,m,F_x) >= 0 and K(Q,m,F_x) = 0 <=> F_x = 0.

So without any calculation, it's easy to see that F_x = 0 gives minimum K.

Maybe you try to maximize K ?

Best wishes

Torsten.

[Joseph Chia]

from the guide,

"The minimum normalized switching frequency occurs at the peak gain of the Qmax curve , so it can be found by solving Eq.2"

[Torsten]

Use

Fxmin=roots[1 0 0 0 2 0 -1/2]

Best wishes

Torsten.