該樓層疑似違規已被系統折疊?隱藏此樓查看此樓

求大神幫忙看看這個程序，解釋一下！

%形成節點導納矩陣

%考慮參考節點

%考慮互感支路

function [d_g,d_b,u_g,u_b]=calyy(itoi,noels_hgln,noelsnum_hgln,bs_noels,bs_cp,gb_hgln,gb_noels,gb_cp,lv_bs,lv_hgln,lv_noels,lv_cp,type_bs,iy,jy,remv_noels)

%lv_bs

d_g=zeros(1,lv_bs);

d_b=zeros(1,lv_bs);

u_g=zeros(1,length(jy));

u_b=zeros(1,length(jy));

type_noels=zeros(1,lv_noels);

for i=1:lv_hgln

%%支路自阻抗

lnnum=noelsnum_hgln(i);

lnnum;

gb_noels;

for ii1=1:lnnum

iln=noels_hgln(i,ii1);

[d_g,d_b,u_g,u_b]=getlnselfgb(iln,bs_noels,type_bs,itoi,gb_noels,iy,jy,d_g,d_b,u_g,u_b,lv_bs);

type_noels(iln)=1;

end

for ii1=1:lnnum

iln=noels_hgln(i,ii1);

for ii2=ii1+1:lnnum

zln=noels_hgln(i,ii2);

b=gb_hgln(i,ii1,ii2);

[d_g,d_b,u_g,u_b]=getlnhugb(iln,zln,bs_noels,type_bs,itoi,0,b,iy,jy,d_g,d_b,u_g,u_b,lv_bs);

end

end

end

for i=1:lv_noels

if remv_noels(i)==1%被除去線路

continue;

end

ibs=bs_noels(i,1);

zbs=bs_noels(i,2);

if type_noels(i)==1

continue;

end

type_noels(i)=1;

iibs=itoi(ibs);

zzbs=itoi(zbs);

g=gb_noels(i,1);

b=gb_noels(i,2);

j=0;

if iibs~=zzbs

j=getlocal(iy,jy,iibs,zzbs,lv_bs,1);

end

bch=gb_noels(i,3);

%分接頭所在端的自導納要+=(g+jb)/t^2

if bch<0

bch=-bch;

if gb_noels(i,4)==1%變比位于末端

ibs=iibs;

zbs=zzbs;

else %變比位于首段

ibs=zzbs;

zbs=iibs;

end

if ibs<=lv_bs

d_b(ibs)=d_b(ibs)+b;

d_g(ibs)=d_g(ibs)+g;

end

if zbs<=lv_bs

d_b(zbs)=d_b(zbs)+b/bch/bch;

d_g(zbs)=d_g(zbs)+g/bch/bch;

end

if j~=0

u_g(j)=u_g(j)-g/bch;

u_b(j)=u_b(j)-b/bch;

end

continue;

end

%disp('線路支路')；

if iibs<=lv_bs

d_b(iibs)=d_b(iibs)+b;

d_b(iibs)=d_b(iibs)+bch;

d_g(iibs)=d_g(iibs)+g;

end

if zzbs<=lv_bs

d_b(zzbs)=d_b(zzbs)+b;

d_b(zzbs)=d_b(zzbs)+bch;

d_g(zzbs)=d_g(zzbs)+g;

end

if j~=0

u_g(j)=u_g(j)-g;

u_b(j)=u_b(j)-b;

end

end

for i=1:lv_cp

ibs=bs_cp(i,1);

if ibs==0

continue;

end

headbs=itoi(ibs);

if headbs>lv_bs

continue;

end

d_b(headbs)=d_b(headbs)+gb_cp(i,1);

end

k=iy(lv_bs);

u_g=u_g(1:k-1);

u_b=u_b(1:k-1);

function [d_g,d_b,u_g,u_b]=getlnselfgb(iln,bs_noels,type_bs,itoi,gb_noels,iy,jy,d_g,d_b,u_g,u_b,lv_bs)

g=gb_noels(iln,1);

b=gb_noels(iln,2);

bch=gb_noels(iln,3);

if bch<0

disp('error!!!');

return;

end

ibs=bs_noels(iln,1);

zbs=bs_noels(iln,2);

iibs=itoi(ibs);

zzbs=itoi(zbs);

if iibs~=lv_bs+1

d_b(iibs)=d_b(iibs)+b+bch;

d_g(iibs)=d_g(iibs)+g;

end

if zzbs~=lv_bs+1

d_b(zzbs)=d_b(zzbs)+b+bch;

d_g(zzbs)=d_g(zzbs)+g;

end

j=0;

if iibs~=zzbs

j=getlocal(iy,jy,iibs,zzbs,lv_bs,1);

end

if j==0

disp('找不到支路位置');

%exit；

return;

end

u_g(j)=u_g(j)-g;

u_b(j)=u_b(j)-b;

function [d_g,d_b,u_g,u_b]=getlnhugb(iln,zln,bs_noels,type_bs,itoi,g,b,iy,jy,d_g,d_b,u_g,u_b,lv_bs)

ibsi=bs_noels(iln,1);%首

zbsi=bs_noels(iln,2);%末

ibsz=bs_noels(zln,1);%首

zbsz=bs_noels(zln,2);%末

first_head=itoi(ibsi);

first_tail=itoi(zbsi);

second_head=itoi(ibsz);

second_tail=itoi(zbsz);

if first_head==second_head&&first_head~=lv_bs+1

disp('add');

d_b(first_head)=d_b(first_head)+2*b;

d_g(first_head)=d_g(first_head)+2*g;

end

if first_tail==second_tail&&first_tail~=lv_bs+1

disp('add');

d_b(first_tail)=d_b(first_tail)+2*b;

d_g(first_tail)=d_g(first_tail)+2*g;

end

if first_head==second_tail&&first_head~=lv_bs+1

disp('add');

d_b(first_head)=d_b(first_head)-2*b;

d_g(first_head)=d_g(first_head)-2*g;

end

if first_tail==second_head&&first_tail~=lv_bs+1

disp('add');

d_b(first_tail)=d_b(first_tail)-2*b;

d_g(first_tail)=d_g(first_tail)-2*g;

end

[d_g,d_b,u_g,u_b]=addlnval(first_head,second_head,g,b,iy,jy,d_g,d_b,u_g,u_b,lv_bs);

[d_g,d_b,u_g,u_b]=addlnval(first_head,second_tail,-g,-b,iy,jy,d_g,d_b,u_g,u_b,lv_bs);

[d_g,d_b,u_g,u_b]=addlnval(second_head,first_tail,-g,-b,iy,jy,d_g,d_b,u_g,u_b,lv_bs);

[d_g,d_b,u_g,u_b]=addlnval(second_tail,first_tail,g,b,iy,jy,d_g,d_b,u_g,u_b,lv_bs);

function [d_g,d_b,u_g,u_b]=addlnval(first_head,second_head,g,b,iy,jy,d_g,d_b,u_g,u_b,lv_bs)

if first_head==second_head

return;

end

j=getlocal(iy,jy,first_head,second_head,lv_bs,1);

if j==0

return;

end

u_b(j)=u_b(j)+b;

u_g(j)=u_g(j)+g;

總結

以上是生活随笔為你收集整理的matlab节点导纳阵求逆,关于利用矩阵稀疏技术求解节点导纳矩阵的MATLAB编程的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。