Asked by swati tandon
on 20 Sep 2019 at 14:29

I want solve this equation to find Vcp value, and i am expecting Vcp value to be around 80 for the given constants.

options = optimoptions('fsolve');

options.MaxIterations = 1000;

options.MaxFunctionEvaluations = 500;

Po = 1000; % constants

Vin = 90;

Vo = 500;

n = 2;

fsw = 100e3;

Lr = 10e-6;

Cr = 35.29e-9;

Iin = Po/Vin;

Zr = sqrt(Lr/Cr);

Rfl= Vo^2/Po;

fr = 1/(2*pi*sqrt(Lr*Cr));

wr=2*pi*fr;

%% solve for Vcp

eqn = @(Vcp) -2*Vcp + ((Vo/(2*n)) +Vcp)* (1+ sqrt(1-(Iin*Zr/((Vo/(2*n))+Vcp))^2))+ (Iin/Cr)*( (0.5/fsw)- (1/wr)*(pi-asin((Iin*Zr)/((Vo/(2*n))+Vcp))) );

x = fsolve(@(Vcp) eqn(Vcp),80, options);

By using this code, without using options it says the iterations have reached maximum limit.

And if include "options" then NO SOLUTION FOUND.

( fsolve stopped because the relative size of the current step is less than the default value of the step size tolerance squared, but the vector of function values is not near zero as measured by the default value of the function tolerance.)

Could anyone suggest what can i chnage in my code to get the correct result. Any help would be deeply appreciated.

Thanks in advance!

Answer by Star Strider
on 20 Sep 2019 at 14:57

Accepted Answer

If you first vectorise your function:

eqn = @(Vcp) -2.*Vcp + ((Vo./(2.*n)) +Vcp) .* (1+ sqrt(1-(Iin*Zr./((Vo/(2*n))+Vcp)).^2))+ (Iin/Cr)*( (0.5/fsw)- (1/wr)*(pi-asin((Iin*Zr)./((Vo./(2*n))+Vcp))) );

then plot it:

v = linspace(0, 1000);

figure

plot(v, eqn(v))

grid

you will see that it is hyperbolic, has a positive y-asymptote, and never approaches 0 at all. (Note that fsolve is a root finder.) If you expect it to have a root, chack your constants and your function to be certain they are correct. It is also a good idea to plot your function first, so you know what it does.

swati tandon
on 20 Sep 2019 at 21:22

Thank You for answering to my query. So yes there was a sign mistake in the equation.

Star Strider
on 20 Sep 2019 at 21:27

As always, my pleasure.

(I have no idea which sign.)

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