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

3 次查看(过去 30 天)
Selection combination outage probability

采纳的回答

William Rose
William Rose 2022-5-29
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.

更多回答(2 个)

William Rose
William Rose 2022-5-29
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 个评论
Aya Emam
Aya Emam 2022-5-29
I just want to plot this curve and this is the equation Po=((1-e^-gamma ratio)^m) And thanks for your efforts 🌸

请先登录,再进行评论。


Aya Emam
Aya Emam 2022-5-29

类别

Help CenterFile Exchange 中查找有关 Signal Analysis 的更多信息

产品


版本

R2022a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by