# how can eliminate negative data in erf function?

4 views (last 30 days)
leuva on 10 Mar 2023
Commented: leuva on 14 Mar 2023
clc;
clear all;
close all;
N = 30 ;
mu =N*pi/4;
M = N*(1-(pi/16));
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)
e3(i) = M.*v1(i).*v2(i)
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 CommentsShow NoneHide None
Arka on 10 Mar 2023
Edited: Arka on 10 Mar 2023
Do you want to remove the negative data points from the term1, term2, and term3 vectors?
leuva on 10 Mar 2023
yes

Alan Stevens on 10 Mar 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 CommentsShow 1 older commentHide 1 older comment
Alan Stevens on 11 Mar 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 on 14 Mar 2023
yes sir ..i have to check it and thankyou so much for replying

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