§173 Loop analysis 8 17-4 Analysis of networks containing dependent sources A typical branch with dependent current source k KCL bk k b sk k OrI6(s)-(s)=l2(s)-(S) Ye(sve(s)-I(s) bk y(s川[Vb(S)+(S)-,(s) bbs)-l(s)=e(svb(s)+vs(s)(s)
§17-3 Loop analysis §17-4 Analysis of networks containing dependent sources A typical branch with dependent current source. bk dk ek s k KCL: i − i = i − i or I (s) I (s) I (s) I (s) b d e s • • • • − = − Y (s)V (s) I (s) e e s • • • = − Y (s)[V (s) V (s)] I (s) e b s s • • • • = + − I (s) I (s) b d • • − Y (s)[V (s) V (s)] I (s) e b s s • • • • = + − bk i ek i sk i +ek − + bk − sk Zek + dk i
b(s)-Ia(s)=y2(川1()+V()]-I() 1.If L, (s)=BJ1(sy=IBkJY(s V,(s)+V(s) IBiki represent the strengths of current sources in branchi as a function of the element current in branch k. For one such source the matrix is zero;except for the element j,k.(the j, k-and the general branch have the same orientation, Pik->+) Let r(s)=BkYe(s)=BikJk(s It is a matrix of zeros except the jk-th element, whose value is BikJk(s) a(s)=Y(s川b(s)+V( 6(s)=[y2(s)+Y(s)Vb6(s)+V,(s)]_I,(s)
I (s) I (s) b d • • − Y (s)[V (s) V (s)] I (s) e b s s • • • • = + − [ ]Y (s)[V (s) V (s)] j k e b s • • • = + [ ] jk represent the strengths of current sources in branch j as a function of the element current in branch k. For one such source the matrix is zero except for the element j,k. (the j,k-and the general branch have the same orientation, → +) jk Let Y (s) [ ]Y (s) [ y (s)] = j k e = j k kk • • It is a matrix of zeros except the jk-th element, whose value is y (s). jk kk I (s) Y (s)[V (s) V (s)] d b s • • • • = + I (s) [Y (s) Y (s)][V (s) V (s)] I (s) b e b s s • • • • • • = + + − 1. If I (s) [ ]I (s) d j k e • • =
6(s)=[y2(s)+Y"(s)[Vb(s)+V(s)-I(s) Al(s)=0 and Vb(s)=A V(s) AY (S)+Y(SJA V,(S)=Als(S)-AlYe(s)+r(s)Is(s) or A(SAT, (s)=A,(s)-Y,(s)V(s) b(s)=le(s)+r(s)--branch admit tance matrix Let Y,(s)=AY(S)A J,(s)=-Alb(s)Vs(s)+Als(s) The node voltage equations are Y (SV(s)=J,(s)
I (s) [Y (s) Y (s)][V (s) V (s)] I (s) b e b s s • • • • • • = + + − AI (s) 0 and V (s) A V (s) n T b b • • • • • • = = AI (s) A[Y (s) Y (s)]V (s)] s e s • • • • • • A[Y (s) Y (s)]A V (s) = − + n T e • • • • • + A[I (s) Y (s)V (s)] s b s • • • • or AY (s) A V (s) = − n T b • • • • Yb (s) = Ye (s)+Y (s)− −branch admit tance matrix • • • • • • • = T Let Yn (s) AYb (s) A J (s) AY (s)V (s) AI (s) n b s s • • • • • • = − + The node voltage equations are Y (s)V (s) J (s) n n n • • • =
2. If I,(s)=I8iklVe(s)=g(sve(s)=g(s)vb(s)+s(sI G() is matrix of zeros except 8体k≠0 b()=ⅣY2(s)+G(s川Vb(s)+(S)-1(s) 'Al(S=0 and V(s=A V(s) IYe (s)+G(s) A V(s)=Als(s)-alle(s)+G(sVs(s) or AY(S)ATV, (s)=All,(s)-Yb(s)V(s)
2. If I (s) [g ]V (s) d j k e • • = G(s)V (s) G(s)[V (s) V (s)] e b s • • • • • = = + G(s) • is matrix of zeros except g jk 0. I (s) [Y (s) G(s)][V (s) V (s)] I (s) b e b s s • • • • • • = + + − AI (s) 0 and V (s) A V (s) n T b b • • • • • • = = AI (s) A[Y (s) G(s)]V (s)] s e s • • • • • • A[Y (s) G(s)]A Vn (s) = − + T e • • • • • + A[I (s) Y (s)V (s)] s b s • • • • or AY (s) A V (s) = − n T b • • • •
A(SA VN(S=A,(S)-Yb(S)Vss) Let I,()=Al,S)A JuS=-Al,(sv(s)+al(s) The node voltage equations are Y(SVs)=J(s) (s=Ye(s)+Y(s)+G(s/--branch admit tance matrix
A[I (s) Y (s)V (s)] s b s • • • • AY (s)A V (s) = − n T b • • • • • • • • = T Let Yn (s) AYb (s) A J (s) AY (s)V (s) AI (s) n b s s • • • • • • = − + The node voltage equations are Y (s)V (s) J (s) n n n • • • = Y (s) [Y (s) Y (s) G(s)] branch admit tance matrix b = e + + − − • • • •
