clear all;
clc;
syms rbe ro gm rc re
y11=1/rbe
y13=0
y31=gm
y33=1/ro
y12=-(y11+y13)
y21=-(y11+y31)
y23=-(y13+y33)
y22=-(y21+y23)
y32=-(y31+y33)
Y=[y11,y12,y13;y21,y22,y23;y31,y32,y33]
% CC
yce=Y
yce(2,2)=yce(2,2)+1/re
yce(3,:)=[]
yce(:,3)=[]
% CB
% yce=Y
% yce(2,2)=yce(2,2)+1/re
% yce(3,3)=yce(3,3)+1/rc
% yce(1,:)=[]
% yce(:,1)=[]
% CE
% yce=Y
% yce(3,3)=yce(3,3)+1/rc
% YY=yce
% yce(2,:)=[]
% yce(:,2)=[]
Y2T=@(y) [-y(2,2)/y(2,1),-1/y(2,1);(y(1,2)*y(2,1)-y(1,1)*y(2,2))/y(2,1),-y(1,1)/y(2,1)]
T=Y2T(yce)
hs=1/T(1,1)
hs=subs(hs,rbe,1500)
hs=subs(hs,ro,50e3)
hs=subs(hs,gm,0.5)
hs=subs(hs,rc,10e3)
hs=subs(hs,re,1e3)
@叶春勇 我不确定第一步是否正确,Ro与gm项并联,结果是什么? 1-2的步骤和2的计算没错。