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function Y = admittance_matrix(bus_n,branch_n,Admittance,lined)
Bus_Number = bus_n;
Branch_Number = branch_n;
Y = zeros(Bus_Number,Bus_Number);
line_val = lined;
% Formation of the Off Diagonal Elements...
for k = 1:Branch_Number
Y(line_val(k,1),line_val(k,2)) = Y(line_val(k,1),line_val(k,2)) - Admittance(k);%互导纳的求解
Y(line_val(k,2),line_val(k,1)) = Y(line_val(k,1),line_val(k,2));%互导纳对称性
end
% Formation of Diagonal Elements....
for m = 1:Bus_Number%自导纳的求解(所有和节点有链接的支路的导纳之和)
for n = 1:Branch_Number
if line_val(n,1) == m
Y(m,m) = Y(m,m) + Admittance(n);
elseif line_val(n,2) == m
Y(m,m) = Y(m,m) + Admittance(n);
end
end
end |