Plotting a non linear equation.

Question

Sudhir Sahoo 2021-4-14

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/801681-plotting-a-non-linear-equation

评论： Sudhir Sahoo 2021-4-19

采纳的回答： Walter Roberson

How to plot the graph between

versus m from this non linear equation given as

where

and

also

and

are constants . Further

represents the upper incomplete gamma function

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Sudhir Sahoo 2021-4-18

在 MATLAB Online 中打开

@Walter Roberson my code is like

clc;
clear all;
close all;
%{
    This code is for phi vs rotation angle by considering Gamma_av 
    and R as constant
 %}
format long g
%% Initialization of the variables of the code;
syms theta_opt phi real
R = [1/2 1/4 1/8 1/16 1/32];
% phi = [pi/8 pi/6 pi/4 pi/3];
R = R(1);
% phi = pi/6;
snrdb = 20;
m = -1/(log(cos(phi)));
% Value of t_min and t_max and K;
t_min = (m + 2) * R/((1 + R)*(((1 + R)^(m +2))-1));
t_max = ((m+2)*R*((1 + R)^(m +2)))/((((1 + R)^(m +2))-1));
% Ke1 = 1/(R(i)*(m +3));
% Ke2 = (((m +2)*R(i)*((1 + R(i))^(m +2)))/((((1 + R(i))^(m +2))-1)))^(1/(m +3));
% K = Ke1*Ke2;
%% Main eauation optimization
% syms theta_opt m real
a = sin(theta_opt).^2;
b = (cos(theta_opt) - sin(theta_opt)).^2;
part1 = (4./(cot(theta_opt) - 1).^2).^((m+2)./(m+3));
part2a = igamma((m+1)/(2*m+6), b*t_min/2);
part2b = igamma((m+1)/(2*m+6), b*t_max/2);
part2c = igamma((m+1)/(2*m+6), a*t_min/2);
part2d = igamma((m+1)/(2*m+6), a*t_max/2);
part2 = (part2a - part2b) ./ (part2c - part2d);
part0 = part1 .* part2;
eqn = part0 - tan(2*theta_opt)
%% Equatoion solving
F = matlabFunction(eqn, 'Vars', [theta_opt, phi]);
% M = linspace(.1,0.5,30);
Phi = [pi/8 pi/6 pi/4 pi/3];
to0 = 1/2;
opt = optimoptions('fsolve','Display','off');
theta_vals = arrayfun(@(m) fsolve(@(t) F(t,phi), to0, opt), Phi)
plot(Phi, theta_vals)

Walter Roberson 2021-4-18

Looks plausible.

Sudhir Sahoo 2021-4-19

@Walter Roberson Yes I have done that too thanks for your code

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Answer 1

Walter Roberson 2021-4-14

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/801681-plotting-a-non-linear-equation#answer_674851

在 MATLAB Online 中打开

format long g

t_min = .3, t_max = .8

t_min =

0.3

t_max =

0.8

syms theta_opt m real

a = sin(theta_opt).^2;

b = (cos(theta_opt) - sin(theta_opt)).^2;

part1 = (4./(cot(theta_opt) - 1).^2).^((m+2)./(m+3));

part2a = igamma((m+1)/(2*m+6), b*t_min/2);

part2b = igamma((m+1)/(2*m+6), b*t_max/2);

part2c = igamma((m+1)/(2*m+6), a*t_min/2);

part2d = igamma((m+1)/(2*m+6), a*t_max/2);

part2 = (part2a - part2b) ./ (part2c - part2d);

part0 = part1 .* part2;

eqn = part0 - tan(2*theta_opt)

eqn =

F = matlabFunction(eqn, 'Vars', [theta_opt, m]);

M = linspace(.1,0.5,30);

to0 = 1/2;

opt = optimoptions('fsolve','Display','off');

theta_vals = arrayfun(@(m) fsolve(@(t) F(t,m), to0, opt), M)

theta_vals = 1×30

0.0069314830628348 0.00580857786032215 0.00476969071649739 0.00383107554970616 0.00300629942402464 0.00230365859998868 0.00172440452642074 0.00126245991980601 0.000905679893063352 0.000638173870072519 0.000442834585897566 0.000303403400001911 0.000205756788619826 0.00013842649511992 9.25693178041045e-05 6.16347434292645e-05 4.09168876360882e-05 2.71147042440712e-05 1.79533691833739e-05 1.18868228260001e-05 7.87482068082966e-06 5.22271731930821e-06 0.000261432834537425 0.000218383853943231 0.000182441265215855 0.000152451203889114 0.000127439225487863 0.000106583202718427 8.91937576217373e-05 7.46930184705892e-05

plot(M, theta_vals)

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Sudhir Sahoo 2021-4-14

Thanks for your answer @Walter Roberson @Matt J

Sudhir Sahoo 2021-4-18

@Walter Roberson Hello , here I one thing is that the value of m is discrete like only I have to take ten values of m in random so how to impliment that instead of using linspace function

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