rho1=1;
h=1;
sigma1=sqrt(10^-3);
alpha=2;
beta=(2500);
d=40; % assumption
r=[50 ,55,60,65,70];
n=100; % number of data points
x_axis=linspace(0,35,8)
Pt_values=(10^-3)*10.^(x_axis/10);
%n - no of values for z
z0=1; % cdf meh z = 1 daal rahe hai
startpow=10;
endpow=1000;
noofpow=30;
out1=zeros(1,8);
k=10^3.7;
cindex = 1;
for index =1:8
Pt=Pt_values(index);
c= (rho1 * Pt)/sigma1^2;
sum=0;
for i=1:5
r0=sqrt(r(i)^2+d^2);
z=z0;
%const j
%(exp(-((r0^alpha)*z0)/(c*beta))/(r0^alpha))
%
j=(1/(r0^alpha));
sum=sum +((exp(-1*(((r0^alpha)*z)/(c *beta))))/(r0^alpha) - j) ;
end
index;
out1(index)=sum*(-0.2)*k ;
end
out1
plot(x_axis,out1);
semilogy(x_axis,out1);
ylabel("Outage Probability")
xlabel("c in decibel, where c is gamma1 dash")
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
r=[50 ,55 ,60 ,65 ,70];
alpha1=3;
alpha2=4;
lambda1=1.12201
lambda2=1.12201
omega1=1
omega2=1
q=1;
% Calculations for SNR2 |
%Expected value of higher order moments of h.g where h and g are rician
%random variables
%qth order moment
alpha11=2.33
alpha12=1.4
M=mu_k(1,lambda1,lambda2,omega1,omega2,alpha1,alpha2);
%defining phis
no_of_irs_ele=4
% lets call 2 paramters , alpha 11 , alpha 12
phi2=mu_k(2,lambda1,lambda2,omega1,omega2,alpha11,alpha12) /mu_k(1,lambda1,lambda2,omega1,omega2,alpha11,alpha12)
phi3=mu_k(3,lambda1,lambda2,omega1,omega2,alpha11,alpha12) /mu_k(2,lambda1,lambda2,omega1,omega2,alpha11,alpha12)
phi4=mu_k(4,lambda1,lambda2,omega1,omega2,alpha11,alpha12) /mu_k(3,lambda1,lambda2,omega1,omega2,alpha11,alpha12)
mu_1=no_of_irs_ele*mu_k(1,lambda1,lambda2,omega1,omega2,alpha11,alpha12)
%defining 'a ' constants
a3 = ( 4*phi4 - 9*phi3 +6*phi2 - mu_1) /(-phi4 + 3*phi3 - 3*phi2 + mu_1 )
a2 = ((a3/2) * ( phi4 - 2*phi3 + phi2) ) + 2*phi4 - 3*phi3 + phi2
%a6 = ((a3*(phi2 - mu_1) + 2*phi2 - mu_1 )/a2 ) -3
a6 = (1/a2)*(a3*(phi2 - mu_1) + 2*phi2 - mu_1) -3
a7 = sqrt((a6+2)^2 - ((4*mu_1*(a3+1))/a2))
a5= (a6-a7)/2
a4 = ( a6+ a7 )/2
a1 = gamma(abs(a3+1)) / ( a2 * gamma(abs(a4+1))*gamma(abs(a5+1)))
%defining the CDF function of meijer G ; consider it to be 5 different
%independent distributions and act accordingly
% make it into a summation format
a_vec=[1];
b_vec=[a3+1];
c_vec=[a4+1,a5+1]
d_vec=[0];
dist2=(500)^alpha2;
sigma2=((10)^(-7.5));
no_of_points=8;
maxpow=35;
minpow=0;
out2=zeros(1,no_of_points);
%x_axis=linspace(minpow,maxpow,no_of_points);
Pt2_values=10.^(x_axis/10)*(10^-3);
rho2=1;
for index=1:no_of_points
curr_pow=Pt2_values(index);
gamma2=curr_pow/(sigma2^2);
sum=0;
for i=1:5
dist1=r(i)^alpha1
dist2
rho2*gamma2
z=(1/a2)*sqrt((dist1*dist2)/(rho2*gamma2))
% ai ,a2 * value
value=a1*a2*meijerG(a_vec,b_vec,c_vec,d_vec,z);
sum=sum + real(value);
end
%(10^(-(index-1)/3))*
out2(index)=0.2*sum;
end
out2
plot(x_axis,out2)
title('outage prob 2 ')
semilogy(x_axis,out2)
title('outage prob 2 semilog')
out4=1.-out1;
out1
out2
out10=out1+out2
out11=out1.*out2
out3=out1+out2-(out1.*out2)
plot(x_axis,out3)
title('final outage ')
semilogy(x_axis,out3)
title('final outage semilog ')
function expected_moment = mu_k(q,lambda1,lambda2,omega1,omega2,alpha11,alpha12)
e1=laguerreL(q,0,- (((lambda1)^2)/omega1));
e2=laguerreL(q,0,-(((lambda2)^2)/omega2));
Exp =((factorial(q))^2 / (alpha11*alpha12)^q)*e1*e2;
expected_moment = Exp; % Mean value of the elements in q
end