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

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Aya Emam on 28 May 2022
Commented: William Rose on 31 May 2022
Selection combination outage probability

William Rose on 29 May 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 CommentsShowHide 1 older comment
William Rose on 31 May 2022
@Aya Emam, you;r welcome, and thanks!

William Rose on 29 May 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.
Aya Emam on 29 May 2022
This curve

Aya Emam on 29 May 2022
And this is the equation

R2022a

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