Solution of implicit function - how to plot implicit functions?

Hi,

I'm trying to solve the following equation for (f,Cp):

I have an equation: Symmetric:

k= 2*pi/λ = ω/Cp,

ω=2*pi* f

Cp=f*λ

Cp:Phase velocity

cL = 5.850;% Longitudinal velocity in [mm/µseg]

cT = 3.184;% Shear velocity in [mm/µseg]

h = 1; %thickness [mm]

Thank you for the help.

This is my code, but it has errors. ¿what is my mistake?

cL = 5.850;% Longitudinal velocity in [mm/µseg]

cT = 3.184;% Shear velocity in [mm/µseg]

d = 1;%thickness [mm]

A=2*pi;

p = @(Cp,f)sqrt(A*f.^2/cL^2-A*f.^2/Cp^2);%p es una Función [Function Handle @], entrada Cp,f

q = @(Cp,f)sqrt(A*f.^2/cT^2-A*f.^2/Cp^2);%q es una Función [Function Handle @], entrada Cp,f

symmetric = @(f,Cp)tan(q(Cp,f)*d)./q(Cp,f)+4*A*f.^2/Cp^2*p(Cp,f).*tan(p(Cp,f)*d)./(q(Cp,f).^2-A*f.^2/Cp^2).^2;

figure

h1=fimplicit(symmetric,[0 3.5 0 22]);

Answer 1

[AndresVar]

Based on the equation you posted, the variable A=2*pi needs to be squared when it's inside the square root. But your code doesn't square A just f.

It's a messy formula so you should double check a couple of points by hand and then check for example the values

symmetric([0 3.5],[0 22])

Also you can put it in a separate function to avoid using 2 functions

fimplicit(@symmetric,[0 3.5 0 22])


function out = symmetric(cp,f)
CL = 5.850;% Longitudinal velocity in [mm/µseg]
CT = 3.184;% Shear velocity in [mm/µseg]
h=1;

om = 2*pi.*f;
k = om./cp;

p = sqrt(om.^2/CL^2-k.^2);
q = sqrt(om.^2/CT^2-k.^2);

out = tan(q*h)./q + 4*k.^2.*p.*tan(p*h)./(q.^2-k.^2).^2;
end

I noticed the plot changed a little when I use this method however.

[Carlos Andrés Galán Pinilla]

Thank you very much for your help and for your idea, is an easier way to solve it, however it is far from expected. What could it be?

I show what is expected:

Thank you