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

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Selection combination outage probability

Risposta accettata

William Rose
William Rose il 29 Mag 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.

Più risposte (2)

William Rose
William Rose il 29 Mag 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 Commenti
Ao Mohamed
Ao Mohamed il 29 Mag 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 il 29 Mag 2022

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R2022a

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