Problem with plotting equation

2020 12 25
1 个回答
回答已采纳
更新时间：2020 12 26
5 次查看（30 天）

0 个投票

I have an implicit equation where x and y - variables. I want to make a graph, but ezplot function doesn't work. MATLAB writes: log(y/x)-STOI+(x/I_sat)*(y/x-1) cannot be plotted in the xy-plane. Also he offers to use fplot instead of ezplot, but it doesn't work correctly with fplot because, as far as I know, fplot doesn't work with implicit equations. So how can I solve this problem?
Also x should changes from 0 to 25. But I think it isn't a tough problem to solve.

6 个评论
显示 4更早的评论 隐藏 4更早的评论

George Bashkatov
George Bashkatov 2020-12-25
Maybe problems with other variables. Some of them are arrays. For example variable "STOI" is an array
Image Analyst
Image Analyst 2020-12-25
We can't run an image -- you'll need to give us the code. In the meantime, try using plot() instead of fplot().
George Bashkatov
George Bashkatov 2020-12-25
编辑：Image Analyst 2020-12-25
h_s=0.1; % step for integrals and graphs
n_0=1*10^19; %quantity of atoms
W=0.7; %wide (cm)
H=0.3; %height (cm)
L=0.78; %length (cm)
alpha=5; %absorption coefficient (1/cm)
I_0=0:h_s:25; %pump power (W)
I_s=3.5; %signal intensity (W)
I_in=5; %input power (W)
sigma_a=5*10^(-19); %absorption cross-section (cm^2)
sigma_e=1*10^(-18); %emission cross-section (cm^2)
h=6.62*10^(-27); %Plank's constant (cm^2*g/s)
tau=4*10^(-6); %lifetime of the exited state (s)
lambda_0=1.9*10^-4;%wavelength of the pump (sm)
c=3*10^10; %light velocity (sm/s)
w_s=80*1^-4; %spot radius (sm)
b=I_0*sigma_a*tau*lambda_0/(h*c); %W_p*tau, W_p - pump velocity
I_sat=h*c*tau/(lambda_0*(sigma_e+sigma_a));
STOI=(sigma_e/sigma_a)*(log(1+b)-log(1+b*exp(-alpha*L)));
syms x y;
ezplot('log(y/x)-STOI+(x/I_sat)*(y/x-1)');
Here is the code. When I try to use plot() MATLAB writes: "Invalid first data argument"
Image Analyst
Image Analyst 2020-12-25
编辑：Image Analyst 2020-12-25
Sorry, I don't have the symbolic toolbox but it runs up until the syms line just fine. Do you plan on doing anything with any of those variables, like plotting them?
For the final two lines, can you tell us what range you expect x to cover, and what is the y value over that range of x values? How about this for x:
x = linspace(0, 25, length(STOI));
??? Now, what is y? You never define it anywhere.
Star Strider
Star Strider 2020-12-25
The error is:
Error using ezplot (line 176)
log(y/x)-STOI+(x/I_sat)*(y/x-1) cannot be plotted in the xy-plane.
however using ezsurf also errors:
Error using ezgraph3>ezfixfun (line 682)
The expression log(y/x)-STOI+(x/I_sat)*(y/x-1) must only have 2 symbolic variables
Nopte that ‘STOI’ is a (1x251) double vector, and ‘I_sat’ is a scalar.
What do you want to do?
George Bashkatov
George Bashkatov 2020-12-25
I only want to plot the last equation. Other variables are only auxiliary.
"y" here is like a y=f(x). This is an inplicit equation, so I can't make equation like y=f(x) and I have to make and equation like f(x,y)=0.
I think x should cover the same range as I_0, but when I tryed to change x to I_0 it didn't work. The range of I_0 is 0..25.
Now I recognized that x and I_0 are the same variables. I substituted STOI with expression for this variable. After that in this formula I substituted b with expression for variable "b". And in expression for b I substituted the I_0 with x. So I had an expression with only two variables x and y without any vectors. But it also gives the same error.
syms x y;
ezplot('log(y/x)-(sigma_e/sigma_a)*(log(1+x*sigma_a*tau*lambda_0/(h*c))-log(1+(x*sigma_a*tau*lambda_0/(h*c))*exp(-alpha*L)))+(x/I_sat)*(y/x-1)');

请先登录，再进行评论。

请先登录，再回答此问题。

 采纳的回答

Mischa Kim
Mischa Kim 2020-12-26

0 个投票

Hi George, use fimplicit():
syms x y
f(x,y) = log(y/x)-(sigma_e/sigma_a)*(log(1+x*sigma_a*tau*lambda_0/(h*c))-log(1+(x*sigma_a*tau*lambda_0/(h*c))*exp(-alpha*L)))+(x/I_sat)*(y/x-1);
fimplicit(f)
Of course, you would have to assign values to the parameters.

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Calculus 的更多信息

产品

版本

R2020a

标签

Community Treasure Hunt

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

Start Hunting!

Translated by