极坐标下的雅可比矩阵形成
for m=1:Npqfor n=1:Nbus
pt(n)=u(m)*u(n)*(G(m,n)*cos(delt(m)-delt(n))+B(m,n)*sin(delt(m)-delt(n))); %由节点电压求得的PQ节点注入有功功率
qt(n)=u(m)*u(n)*(G(m,n)*sin(delt(m)-delt(n))-B(m,n)*cos(delt(m)-delt(n))); %由节点电压求得的PQ节点注入无功功率
end
Unbalance(2*m-1)=p(m)-sum(pt); %计算PQ节点有功功率不平衡量
Unbalance(2*m)=q(m)-sum(qt); %计算PQ节点无功功率不平衡量
end %是节点不平衡量矩阵
for m=1:Npq
for n=1:Nbus
h0(n)= u(m)*u(n)*(G(m,n)*sin(delt(m)-delt(n))-B(m,n)*cos(delt(m)-delt(n)));
n0(n)=-u(m)*u(n)*(G(m,n)*cos(delt(m)-delt(n))+B(m,n)*sin(delt(m)-delt(n)));
j0(n)=-u(m)*u(n)*(G(m,n)*cos(delt(m)-delt(n))+B(m,n)*sin(delt(m)-delt(n)));
l0(n)=-u(m)*u(n)*(G(m,n)*sin(delt(m)-delt(n))-B(m,n)*cos(delt(m)-delt(n)));
end
H(m,m)=sum(h0)-u(m)^2*(G(m,m)*sin(delt(m)-delt(m))-B(m,m)*cos(delt(m)-delt(m)));
N(m,m)=sum(n0)+u(m)^2*(G(m,m)*cos(delt(m)-delt(m))+B(m,m)*sin(delt(m)-delt(m)))-2*u(m)^2*G(m,m);
J(m,m)=sum(j0)+u(m)^2*(G(m,m)*cos(delt(m)-delt(m))+B(m,m)*sin(delt(m)-delt(m)));
L(m,m)=sum(l0)+u(m)^2*(G(m,m)*sin(delt(m)-delt(m))-B(m,m)*cos(delt(m)-delt(m)))+2*u(m)^2*B(m,m);
Jacobi(2*m-1,2*m-1)=H(m,m);
Jacobi(2*m-1,2*m)=N(m,m);
Jacobi(2*m,2*m-1)=J(m,m);
Jacobi(2*m,2*m)=L(m,m);
end %计算m=n情况下的Jacobi矩阵中的子矩阵元素
for m=1:Npq
for n=1:Npq
if m==n
else
H(m,n)=-u(m)*u(n)*(G(m,n)*sin(delt(m)-delt(n))-B(m,n)*cos(delt(m)-delt(n)));
J(m,n)= u(m)*u(n)*(G(m,n)*cos(delt(m)-delt(n))+B(m,n)*sin(delt(m)-delt(n)));
N(m,n)=-J(m,n);
L(m,n)= H(m,n);
Jacobi(2*m-1,2*n-1)=H(m,n);
Jacobi(2*m-1,2*n)=N(m,n);
Jacobi(2*m,2*n-1)=J(m,n);
Jacobi(2*m,2*n)=L(m,n);
end
end
end %计算m≠n情况下的Jacobi矩阵中的子矩阵元素
问题：１、在算对角元素时候，为什么不能用前面pt(n)和qt(n)代呢
２、对角元素H(m,m),N(m,m),J(m,m),L(m,m)中还有u(m)^2*(G(m,m)*sin(delt(m)-delt(m))-B(m,m)*cos(delt(m)-delt(m)));等，书上不是就u(m)^2*G(m,m)＋Ｑ（i）?
觉得很疑问？谢谢大家回答我！
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