How I can plot this equation ((1-e^-a)^m)
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Aya Emam
am 28 Mai 2022
Kommentiert: William Rose
am 31 Mai 2022
Selection combination outage probability
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William Rose
am 29 Mai 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.
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Weitere Antworten (2)
William Rose
am 29 Mai 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.
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