How do I get normal numerical answers using solve or vpasolve?
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I'm currently making a function within which I determine a Reflectivity value by solving the equation using solve().
However, the answers given are always values of pi, how do I solve this?
[Result, y] = lorentz(0, 0.725, 15)
function [Result, y] = lorentz(position, CavityLength, BaseRD)
%Enter position, cavity length in m and baserd in microseconds.
c = 299792458; %speed of light in m/s
BaseRD = BaseRD*10^-6;
syms Reflectivity
S = vpasolve(((pi*sqrt(Reflectivity))/(1-Reflectivity))/(c/(2*CavityLength)) == BaseRD, Reflectivity, [0.99 1]);
Result.Reflectivity = S;
Result.Finesse = (pi*sqrt(Result.Reflectivity)/(1-Result.Reflectivity));
Result.FSR = c/(2*CavityLength);
FWHM = Result.FSR./Result.Finesse;
Result.HWHM = FWHM/2;
x = linspace(-1e3,1e3,1e5);
y = (1/pi).*(Result.HWHM./(((x-position).^2)+Result.HWHM.^2));
%y = y.*(height/max(y));
%plot(x,y)
end
Reflectivity: 0.99898752118560808161741982765188
Finesse: 987.1748605317811539461548191278*pi
FSR: 206753419.310345
HWHM: 104719.75511965978378924223514237/pi
5 Comments
The equation you are solving does not have a solution.
BaseRD = 15;
BaseRD = BaseRD*10^-6;
c = 299792458; %speed of light in m/s
CavityLength = 0.725;
syms x
fun(x) = ((pi*sqrt(x))./(1-x))/(c/(2*CavityLength)) - BaseRD;
%Check the limit
limit(fun, x, Inf)
fplot(fun, [0 1e18])
Ben van Zon
on 30 Nov 2023
Dyuman Joshi
on 30 Nov 2023
I am aware that your problem was with the symbolic notation of pi.
But I was making another point, which I noticed for the given set of data.
Maybe you got that answer for a different set of data, but for the values noted above, there is no solution.
Dyuman Joshi
on 30 Nov 2023
The BaseRD value is different. And you get your answer.
Also, your comment was marked spam by the Auto Filter, idk how. I have removed the spam notice, and subsequently your flag as well.
Accepted Answer
More Answers (1)
vpa will turn those products involving pi into more normal looking numbers. For example...
x = 17*sym(pi)
So x has pi in it.
vpa(x)
vpa yields a symbolic high precision floating point number. But double will also work, and create a double as a result.
double(x)
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