清空筛选条件
清空筛选条件

how to solve integral in the defining the function command?

3 次查看（过去 30 天）
SAHIL SAHOO
SAHIL SAHOO 2022-7-19
编辑： Torsten 2022-7-20
clc
%defining constant
ti = 0; %inital time
tf = 10E-5;% final time
tspan=[ti tf];
o = 10E6; % detuning frequency
tc = 70E-9; %photon life time in cavity
tf = 240E-6; %flouroscence lifetime
a = 0.02; %round trip loss
P = 1; %pump strenght
k = 0.2; %critical coupling strength
l= 0.5;
% define function
%y(1) = I
%y(2) = G
%y(3) = phase difference
f = @(t,y) [
((y(2)-a-l.*(abs(cos(y(3)+ pi/4)))).*y(1) + k.*y(1).*cos(y(3)- pi/2)).*(2/tc);
(P - (y(2).*(y(1) + 1))) / tf;
o - (k / tc).*2.* sin(y(3));
];
%initial consitions
[T,Y] = ode45(f,tspan,[1;1;1]*10E-5);
%plotting the graphs
plot(T,Y(:,3));
ylim([0 30])
in this the program I represent y(1) as I(Φ), in the equation 2 intead of y(1) i want to use integral of I(Φ)dΦ , where the we assume I(Φ) is gaussian distribution with a variable mean Φ0 and constant rms width σ = 0.1
is it possible to do this, if yes, how ?

请先登录，再回答此问题。

回答（1 个）

Torsten
Torsten 2022-7-19
编辑：Torsten 2022-7-19
integral_{0}^(t) I(phi) dphi = normcdf(t,0,0.1) - 0.5
So you can work with
y(1) = normcdf(t,0,0.1) - 0.5
in your equations.
  2 个评论
SAHIL SAHOO
SAHIL SAHOO 2022-7-20
but how ill you write this in this?
f = @(t,y) [
((y(2)-a-l.*(abs(cos(y(3)+ pi/4)))).*y(1) + k.*y(1).*cos(y(3)- pi/2)).*(2/tc);
(P - (y(2).*(y(1) + 1))) / tf;
o - (k / tc).*2.* sin(y(3));
];
if you integrate outside it will give different result.
Torsten
Torsten 2022-7-20
编辑：Torsten 2022-7-20
I don't know the background of your equations.
If you want to take y1 as integral of I(Φ)dΦ , where we assume I(Φ) is gaussian distribution with a variable mean Φ0 and constant rms width σ = 0.1, the equations are
f = @(t,y) [ (P - (y(1).*(normcdf(t,0,0.1) - 0.5 + 1))) / tf; o - (k / tc).*2.* sin(y(2)) ];
Since the equation for y(1) is obsolete, y(2) became y(1) and y(3) became y(2) in the function handle.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Calculus 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by