how can eliminate negative data in erf function?

3 visualizaciones (últimos 30 días)
leuva
leuva el 10 de Mzo. de 2023
Comentada: leuva el 14 de Mzo. de 2023
clc;
clear all;
close all;
N = 30 ;
mu =N*pi/4;
M = N*(1-(pi/16));
lemdaD = 2;
avgsnr = 0.0001;
p = 100
n = 3.5;
deld =1e-3 * 10^(0/10); % 0 dBm in watts;
delr = 10^(-80/10)/1000; % -80 dBm in watts
delrd =deld+(p^2)*delr;
w = [0:1:40]
ps = zeros(size(w));
for i = 1:length(w)
%for j= 1:1:5
ps(i) = 10.^(w(i)./10)./1000;
dsr = 41.23;
drd = 60.82;
dsd =100;
d0 =1
avgsnrd(i) = (ps(i)./delrd);
b1(i) =sqrt(avgsnrd(i)).*p*(dsr*drd/d0^2)^(-0.5*n)
b2(i) = sqrt(avgsnrd(i)).*(dsd/d0)^(-0.5*n)
v1(i) = 1./b2(i)
v2(i) = b1(i)./b2(i)
e1(i) = sqrt(M.*v2(i)+lemdaD)
e2(i) = sqrt(2*M*lemdaD)
e3(i) = M.*v1(i).*v2(i)
e4(i) = (lemdaD.*v1(i))./v2(i)
e5 = lemdaD*mu
sqrt_avgsnr = sqrt(avgsnr)
d1(i)= 1./b1(i)
term1(i) = 0.5 .* (erf((d1(i).*sqrt_avgsnr - mu)./ sqrt(2* M)) + ...
erf(mu/sqrt(2 * M)))
term2(i) = (sqrt(lemdaD)./(2.*e1(i))).*exp(-(v1(i).*sqrt_avgsnr - v2(i).* mu).^2./(2.*e1(i).^2)).*....
erf((e4(i).*sqrt_avgsnr - e5)./(e1(i) .* e2(i)))
term3(i)= (sqrt(lemdaD)./(2.*e1(i))).*exp(-(v1(i).*sqrt_avgsnr - v2(i).* mu).^2./(2.*e1(i).^2)).* ....
erf((e3(i).*sqrt_avgsnr + e5)./(e1(i) .* e2(i)))
FyD1(i) = term1(i) - term2(i) - term3(i)
%FyD11(i) = abs(FyD1(i))
end
semilogy(w, max(FyD1, 0), '-or')
legend('p = 0 dB')
%,'p = 10 dB','p = 15 dB','p = 20 dB')
xlabel('ps(dBm)')
ylabel('Pout')
  2 comentarios
Arka
Arka el 10 de Mzo. de 2023
Editada: Arka el 10 de Mzo. de 2023
Do you want to remove the negative data points from the term1, term2, and term3 vectors?
leuva
leuva el 10 de Mzo. de 2023
yes

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Respuesta aceptada

Alan Stevens
Alan Stevens el 10 de Mzo. de 2023
Should
term2(i) = (sqrt(lemdaD)./(2.*e1(i))).*exp(-(v1(i).*sqrt_avgsnr - v2(i).* mu).^2./(2.*e1(i).^2)).*....
erf((e4(i).*sqrt_avgsnr - e5)./(e1(i) .* e2(i)));
be
term2(i) = (sqrt(lemdaD)./(2.*e1(i))).*exp(-(v1(i).*sqrt_avgsnr - v2(i).* mu).^2./(2.*e1(i).^2)).*....
erf((e4(i).*sqrt_avgsnr + e5)./(e1(i) .* e2(i)));
i.e. have a " + e5" instead of " - e5"?
This removes the negatives.
  3 comentarios
Alan Stevens
Alan Stevens el 11 de Mzo. de 2023
If the equation is correct then perhaps some part of your input data is incorrect; otherwise, why do you want to remove the negative values?
leuva
leuva el 14 de Mzo. de 2023
yes sir ..i have to check it and thankyou so much for replying

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