《电路》（英文版）17-2 Nodal analysis

817-2 Nodal analysis k The node-to-datum voltage vector. bk The branch voltage vector: 十 bk b10b2 bB D-D=A D Ai1=0 b i一i∴.Ai1=Ai-AL.=0 b Ai=Ai

§17-2 Nodal analysis The node-to-datum voltage vector: T n n n nN [ ]  =  1  2  • The branch voltage vector: T b b b bB [ ]  =  1  2  • • • • b = e − s • • • = n T  b A  • • • •  − = n T e  s A  • • •  Ai b = 0 • • • b = e − s i i i • • • • • • •  Ai b = Ai e − Ai s = 0 • • • • e = s Ai Ai bk i ek i sk i +ek − + bk −  sk Zek

bk +Uek -lek sk k k 十 bk Ie(s=Y(sve(s)--complex frequence domain Ay2(s)V2(s)=Al2(3)=A(s) V(S)=Vb(S)+V(s=AV,(s)+V(s) AYe(s)A V(S)+ASV(s=l(s) or AY(S)A V(S)=-AY(s)V(s)+Al(s)

• • • • e = s Ai Ai Ie s = Ye s Ve s − −complex frequence domain • • • ( ) ( ) ( ) AY (s)V (s) AI (s) AI (s) e e e s • • • • • • •  = = V (s) V (s) V (s) A V (s) V (s) n s T e b s • • • • • •  = + = + AY (s) A V (s) AY (s)V (s) AI (s) n e s s T e • • • • • • • • • + = or AY (s) A V (s) AY (s)V (s) AI (s) n e s s T e • • • • • • • • • = − + bk i ek i sk i +ek − + bk −  sk Zek

AY(S)A V(S=-AY(S(S)+Al(S) Let Y(s=AY(s)A node admittance matrix Jn(s)=-Ay2(s)V,(s)+A/,(s) -equivalent current source vector The node voltage equation: Y,(S)Vn(s)=Jn(s) or V(s)=n (s)J,(s)

AY (s)A V (s) AY (s)V (s) AI (s) n e s s T e • • • • • • • • • = − + J (s) AY (s)V (s) AI (s) n e s s • • • • • • = − + The node voltage equation: Y (s)V (s) J (s) n n n • • • = ( ) ( ) ( ) 1 or V s Y s J s n n n • • − • = ----node admittance matrix ----equivalent current source vector • • • • = T Let Yn (s) AYe (s) A

Let us now summarize the steps required for a complete nodal analysis of given network. Choose an arbitrary node as the datum and write down the incidence matrix A from the network graph. 2. Find the element admittance matrix Y, (s)from the branch Vcr. 3. Use A, obtained in step l, to calculate yn (s)from: Y, (S)=AY, (S)A 4. Obtain the voltage and current source vectors v (s)and I,(s) from the network 5. Use Y, (s)and Jn(s)=-AYe(Ss)V,(s)+Als(s)to compute In(s) from Y(SV(S=J,(s) 6. Use Vb(s)=A Vn(s) to compute branch voltage,(s)

2. Find the element admittance matrix from the branch VCR. Y (s) e • 1. Choose an arbitrary node as the datum and write down the incidence matrix from the network graph. • A 4. Obtain the voltage and current source vectors and from the network. V (s) s • I (s) s • 5. Use and to compute from Y (s) n • V (s) n • Y (s)V (s) J (s) n n n • • • = J (s) AY (s)V (s) AI (s) n e s s • • • • • • = − + 6. Use to compute branch voltage V (s) A Vn (s) . T b • • • = V (s) b • Let us now summarize the steps required for a complete nodal analysis of given network. 3. Use , obtained in step 1, to calculate from: • • • • = T Yn (s) Yn (s) AYe (s) A • • A

DP: Write node voltage equations 11100000 -10010100 x∏S入1S 3 00-10-100 2 00000-1-1-1 4k 10000000 IS 01000000 7A8 00100000 IS 00010000 00001000 1-10 00000100 130 00000010 Y=AY A 103-1 0000000 =104000000 J =a -v 0 2000000 0

DP: Write node voltage equations.             − − − − − − = • 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 A                           = • 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 Ye             − − − − − − − − = = • • • • 0 1 1 3 1 0 3 1 1 3 0 1 3 1 1 0 T Yn AYe A     T s T s I V 1 2 0 0 0 0 0 0 0 4 0 0 0 0 0 0 = = • •             − − = − = • • • • • 0 0 1 1 ( ) n s Ye Vs J A I • • • Yn Vn = J n 1S 2A 1S 1S 1S 1S 1S 1S 1S 1A 4V + − n1 n2 n3 n4 n1 n2 n3 n4 0 • 1 2 3 4 5 6 7 8