How I can plot this equation ((1-e^-a)^m)

2 visualizaciones (últimos 30 días)
Ao Mohamed
Ao Mohamed el 28 de Mayo de 2022
Comentada: William Rose el 31 de Mayo de 2022
Selection combination outage probability

Respuesta aceptada

William Rose
William Rose el 29 de Mayo de 2022
The plot you provided has on the horizontal axis. Therefore define x, for use as the horizontal coordinate for plotting: . Then . Therefore
.
x=-10:40;
M=[1,2,3,4,10,20];
Pout=zeros(length(x),length(M));
for i=1:length(M)
Pout(:,i)=(1-exp(-1./(10.^(x/10)))).^M(i);
end
semilogy(x,Pout(:,1),'-r',x,Pout(:,2),'-g',x,Pout(:,3),'-b',...
x,Pout(:,4),'-c',x,Pout(:,5),'-m',x,Pout(:,6),'-y');
ylabel('Pout'); grid on;
xlabel('10log_{10}(\gamma_{bar}/\gamma_0)');
ylim([1e-4,1]);
legend('M=1','M=2','M=3','M=4','M=10','M=20')
The plot above matches the plot you provided.
  2 comentarios
Ao Mohamed
Ao Mohamed el 30 de Mayo de 2022
You are so Great ✨🌸
William Rose
William Rose el 31 de Mayo de 2022
@Aya Emam, you;r welcome, and thanks!

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Más respuestas (2)

William Rose
William Rose el 29 de Mayo de 2022
First you need to decide if you want a surface plot, with two independent variables, or a line plot, with one independent variable. If you want a line plot, then you must decide whether the variable for the horizontal axis is a or m.
Example 1: Assume a=1 and let m=0:.1:10.
Example 2: Assume m=1 and let a=0:.1:10.
a=1;
m=0:.1:10;
z=(1-exp(-a)).^m;
subplot(211), plot(m,z,'-r.');
xlabel('m'); ylabel('z'); title('z=(1-exp(-a))^m, a=1');
m=1;
a=0:.1:10;
z=(1-exp(-a)).^m;
subplot(212), plot(a,z,'-r.');
xlabel('a'); ylabel('z'); title('z=(1-exp(-a))^m, m=1')
Try it.
  3 comentarios
Ao Mohamed
Ao Mohamed el 29 de Mayo de 2022
I just want to plot this curve and this is the equation Po=((1-e^-gamma ratio)^m) And thanks for your efforts 🌸

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Ao Mohamed
Ao Mohamed el 29 de Mayo de 2022

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