第三章多端口网络3

1、本科时所列的节点电压方程，是以选定网络,N,内,某一节点,为,参,考,点而列写的,1,全节点方程,n,n,n,I,U,Y,称为,定,导纳阵,1,n,Y,其,方程数,等于,独立节点数,n-1,非奇异,存在,n,Y,3-6,全节点电压,方程与,不定,导纳阵,P131,若把电位,参考,点改在电路,N,的,外部,则得到电路的,全部节点,电压,为变量的,n,个方程,n,n,i,I,U,Y,称为,全节点电压方程,求法与原来相同,1,1,n,n,n,n,T,a,b,a,n,n,i,I,U,A,Y,A,Y,显然,Y,i,的,行,是线性,相关的,det,Y,i,0,Y,i,是,奇异,的,1,Y,i,不存在，称为

2、,不定导纳阵,此不定意指参考点任意选定,由于,参考点,选在电网络,N,外,全节点电压方程和不定导纳阵,非常灵活,可用于,不同网络,的,连接,和,同一网络,的,变换,2,意义（应用,I,n,中去掉,I,nk,就得到以网络,N,中,k,节点,为参考点的,节点,电压方程,n,n,n,I,U,Y,例如，划去,i,Y,的,k,行,k,列得,n,Y,从,U,n,中去掉,U,k,各端,U,1,U,2,U,3,U,n,作为激励,数值上等于各节点电压,，各端,电流作为零状态响应,I,1,I,2,In,按线性叠加，可以得到,n,个端,子电流表达式,N,I,1,I,2,I,n,U,1,U,2,U,n,设,N,中,不

3、含独立源且零状态,n,n,i,I,U,Y,1,1,n,n,n,n,I,U,N,I,1,I,2,I,n,U,1,U,2,U,n,n,k,j,U,I,y,k,i,U,k,j,jk,i,2,1,0,nn,n,n,n,n,i,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,2,1,2,22,21,1,12,11,称为,n,端网络的不定导纳参数，其物理意义,j=k,短路驱动点导纳,j,k,短路转移导纳,与多口网络的短路导纳相似，但又有不同。一个是,端口,电流,端口,电压,这里是,端电压、端电流,如,是除第,k,端外，其余各端都与参考点短接,0,i,U,i,k,N,I,1,I,2,I,n,U,1,U,2,U,n,

4、Y,i,称为,零和,矩阵,1,0,n,k,k,I,11,12,1,21,22,2,1,2,n,n,i,i,n,n,nn,Y,Y,Y,Y,Y,Y,Y,YU,I,Y,Y,Y,M,M,M,3,不定导纳阵,Y,i,的,基本性质,n,个端子可对应原网络的,n,个节点，各端子可与网络,外,的参考点,之间施加电压源（参考点可能与,N,相连，也可能不与,N,相连,1,Y,i,的零和特性,1,1,1,0,2,0,n,kj,k,n,jk,k,Y,Y,每列之和,每行之和,1,2,1,1,1,0,n,n,n,k,k,kn,k,k,k,Y,Y,Y,U,0,0,0,1,0,n,k,k,I,1,2,U,n,u,u,u,M,

5、1,0,n,kj,k,Y,n,n,nn,n,n,n,n,I,I,I,U,U,U,Y,Y,Y,Y,Y,Y,Y,Y,Y,2,1,2,1,2,1,2,22,21,1,12,11,具有任意性,每列之和为零,1,0,n,jk,j,k,Y,U,1,2,0,n,jk,k,Y,每行之和,让,j,端子,与,外,部,参考点,之间接入,U,j,相当于,电压源,其余,支,路,开路,由于,N,中无独立源,I,j,0,1,0,n,jk,k,Y,N,I,1,I,j,I,n,U,1,U,j,U,n,0,j,I,无电流,通路,U,j,1,2,3,j,U,U,U,U,L,L,1,2,3,0,j,I,I,I,I,L,L,1,1,2

6、,2,1,0,j,j,j,jn,n,n,jk,k,k,I,Y,U,Y,U,Y,U,Y,U,因为,U,j,具有任意性,每行之和为零,Y,i,为,零和矩阵,因此,所有一阶,代数,余子式相等,称为,等余因子,特性,2,Y,i,为奇异阵，所以,det,Y,i,0,Y,i,1,不存在，所以称不定导纳矩阵,3,Y,i,的等余因子特性：任意两个元素的代数余子式相等,1,1,det,0,i,j,j,jk,jk,jn,jn,Y,Y,Y,Y,1,2,n,j,jk,k,Y,Y,由前式得，带入上式得,2,2,1,1,det,i,j,j,j,jk,jk,j,Y,Y,Y,1,0,jn,jn,j,Y,ij,11,12,nn

7、,i,j,1,2,n,由于,Y,jk,无论取何值，上式都成立，所以,1,1,jk,j,k,n,任一行的一阶代数余子式皆相等；同样，按照列展开，可,得任一列的一阶代数余子式相等，所以有,4,不定导纳阵的,运算,端子接地与浮地,短路收缩,开路抑制,网络并联,求端口的短路导纳阵,1,端子接地与浮地,端子接地,将,n,端网络的第,k,个端子选为参考点,所以,U,k,0,可以划去,Y,i,中,第,k,列,另外，第,k,个端子电流,I,k,仅仅是其余端子电流之和的负值,所以可不在变量中出现,Y,i,中删去第,k,行。如此将,n,端网络的第,k,个,端子选为参考点，相当于把对应的不定导纳矩阵,Y,i,的,第

8、,k,行和第,k,列都删除,得到一个,n,1,阶方阵,Y,det,Y,不再为“零”，故称,为,定导纳矩阵,33,32,31,23,22,21,13,12,11,i,y,y,y,y,y,y,y,y,y,Y,以端子,3,为公共端构成双口网络,n,sc,Y,y,y,y,y,Y,22,21,12,11,N,3,1,2,N,3,1,2,例,三端网络的不定导纳矩阵为,N,1,2,3,N,1,2,3,11,12,21,22,n,y,y,Y,y,y,22,21,12,11,22,12,21,11,22,21,22,21,12,11,12,11,3,2,1,y,y,y,y,y,y,y,y,y,y,y,y,y,y

9、,y,y,Y,i,端子浮地,现以,2,端作为公共端,并在,3,端和公共端之间跨接,2,电阻,1,端施,以单位阶跃电压源,试求此时端子,1,和,3,电流的单位阶跃响应,例,某非含源线性三端网络,N,以,3,端作为公共接地端。当,2,端短路,1,端施以单位冲激电压源时,1,端和,2,端电流的冲激响,应分别为,1,2,A,0.75,A,t,t,i,t,e,t,i,t,e,t,1,2,0.25,1,A,0.0625,4,3,A,t,t,i,t,e,t,i,t,e,t,而当,1,端短路,2,端施以单位阶跃电压源时,两个端子电流的阶,跃响应分别为,图释,1,i,2,i,N,V,t,1,i,2,i,N,V,

10、t,求,1,i,3,i,3,i,1,i,N,V,t,2,解,1,2,1,1,0.25,0.25,1,1,4,3,4,16,16,1,16,1,I,s,s,s,s,s,s,I,s,s,s,s,s,1,2,1,0,U,s,U,s,时,1,2,1,0.75,1,1,I,s,I,s,s,11,21,1,0.75,1,1,Y,Y,s,s,1,2,0,1,U,s,U,s,s,时,1,2,1,i,2,i,N,V,t,1,i,2,i,N,V,t,1,A,t,i,t,e,t,1,0.25,1,A,t,i,t,e,t,2,0.75,A,t,i,t,e,t,2,0.0625,4,3,A,t,i,t,e,t,12,2

11、2,1,4,4,1,16,1,s,Y,Y,s,s,则三端网络的不定导纳矩阵为,3,1,1,1,4,1,3,4,4,1,16,1,S,S,S,S,S,Y,则,3,端为公共端时的,Y,参数矩阵为,1,1,3,1,4,1,4,1,3,4,8,4,1,16,1,16,1,1,4,4,1,16,1,16,1,i,s,s,s,s,s,s,s,s,s,s,s,s,s,Y,以,2,端为公共端时的,Y,参数矩阵为,2,1,3,1,4,1,1,4,4,1,16,1,s,s,s,s,s,Y,当,3,端与公共端之间跨接,2,电阻时,3,3,0.5,I,s,U,s,1,1,2,3,3,I,s,U,s,I,s,U,s,Y

12、,以,2,端为公共端时的,Y,参数方程为,1,1,U,s,s,并且,求,1,i,3,i,3,i,1,i,N,V,t,2,三端网络的不定导纳矩阵为,1,1,3,1,4,1,4,1,3,4,8,4,1,16,1,16,1,1,4,4,1,16,1,16,1,i,s,s,s,s,s,s,s,s,s,s,s,s,s,Y,3,31,1,33,1,2,1,9,4,3,I,s,Y,s,U,s,Y,s,s,s,联立解得,1,11,1,13,3,1,2,4,3,I,s,Y,s,U,s,Y,s,I,s,s,s,4,3,1,4,3,3,0.75(1,A,1,1,A,6,t,t,i,t,e,t,i,t,e,t,1,1

13、1,1,13,3,1,2,4,3,I,s,Y,s,U,s,Y,s,I,s,s,s,3,31,1,33,1,2,1,9,4,3,I,s,Y,s,U,s,Y,s,s,s,取拉斯反变换可得零状态响应分别为,解得,1,1,1,6,4,3,s,s,3,1,1,4,4,3,s,s,这种方法能从一种共点组态过渡到另外一种共点,组态,例：共集电极、共基极三极管,由不定导纳阵,短路导纳阵：划去相应于公共,端的行、列，成为共点,n-1,口网络,若由短路导纳阵,不定导纳阵：根据零和特性,加上与公共端对应的行、列后，即得将公共端,浮地”后所形成得,n,端网络的不定导纳阵,小结,33,32,31,23,22,21,13

14、,12,11,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,i,2,短路收缩,同一网络,又称端子缩并,将两个或多个端子,连接起来形成一个新的端子。这时,N,的外接端子减少，不定导纳,矩阵的阶次也要相应降低。两个以上端子的短路收缩可以看成是,两个端子短路收缩的组合,相应的,行和列相加,N,1,3,2,1,3,1,1,3,1,1,U,U,U,I,I,I,33,32,33,31,23,22,23,21,33,13,32,12,33,13,31,11,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,i,33,32,31,23,22,21,33,13,32,12,31,11,Y,Y,Y,Y

15、,Y,Y,Y,Y,Y,Y,Y,Y,Y,i,22,23,21,32,12,33,13,31,11,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,i,降阶了,假定把端子,1,和,2,短接,1,2,1,U,U,U,1,1,2,I,I,I,N,1,2,n,3,1,I,2,I,3,I,n,I,1,I,1,U,KVL,规则,将原网络的不定导纳矩阵,第,2,列加到第,1,列，再将第,2,列划去,或者将第,1,列加到第,2,列,再将第,1,列划去,规则,将原网络不定导纳矩阵的第,2,行加到第,1,行,再将第,2,行划去,或者将第,1,行加到第,2,行，再将第,1,行划去,KCL,对应行相加,对应列相加,Jump,J

16、ump,Jump,应用：电力系统双母线,母联开关合上,两节点合并为一个节点,这种情况相当于令两节点电压相等，新节点注入电流等于原两个,节点地注入电流之和,n,nn,n,n,n,n,n,n,n,U,y,U,y,U,y,I,U,y,U,y,U,y,I,U,y,U,y,U,y,I,2,2,1,1,2,2,22,1,21,2,1,2,12,1,11,1,1,2,1,U,U,U,代入上式,n,nn,n,n,n,n,n,n,n,n,U,y,U,y,U,y,y,I,U,y,U,y,U,y,y,I,U,y,U,y,U,y,y,I,3,3,1,2,1,2,3,23,1,22,21,2,1,3,13,1,12,1

17、1,1,Jump,1,1,2,I,I,I,代入,n,nn,n,n,n,n,n,n,n,n,U,y,U,y,U,y,y,I,U,y,U,y,U,y,y,I,U,y,U,y,y,U,y,y,y,y,I,3,3,1,2,1,2,3,23,1,32,31,3,1,3,23,13,1,22,21,12,11,1,得到,1,2,短接后所形成的,n-1,个端子的不定导纳矩阵,n,nn,n,n,n,n,n,n,n,n,U,y,U,y,U,y,y,I,U,y,U,y,U,y,y,I,U,y,U,y,U,y,y,I,3,3,1,2,1,2,3,23,1,22,21,2,1,3,13,1,12,11,1,Jump,

18、例,四端网络的不定导纳矩阵为,44,43,42,41,34,33,32,31,24,23,22,21,14,13,12,11,i,y,y,y,y,y,y,y,y,y,y,y,y,y,y,y,y,Y,如果将端子,2,和,4,短路收缩为新端子,2,则,33,34,32,31,43,23,42,44,24,22,41,21,13,14,12,11,i,y,y,y,y,y,y,y,y,y,y,y,y,y,y,y,y,Y,I,2,N,1,3,2,I,3,I,1,U,3,I,3,0,I,2,N,1,3,2,I,1,U,3,3,开路抑制,开路扼制、端子的删减、端子的封禁,将一个或多个端子开路,其对应的端电流

19、限定为零,端子变成,不,可及节点,抑制的端钮压进网络内，切断端钮与外部电路的联系,1,11,12,1,2,21,22,2,I,Y,Y,U,I,Y,Y,U,1,21,1,22,2,1,21,2,22,2,22,1,21,2,12,1,11,1,0,U,Y,Y,U,U,Y,U,Y,U,Y,U,Y,U,Y,U,Y,I,仍为不定导纳阵,21,1,22,12,11,1,21,1,22,12,11,1,Y,Y,Y,Y,Y,U,Y,Y,Y,Y,I,k,n,i,要开路抑制的端子电,流、电压列向量分别,记为,I,2,和,U,2,其它端,子电流、电压列向量,分别,记为,I,1,和,U,1,将原来的不定导纳矩阵也做

20、相应的分块,设要删除的节点为,K,个,在开路抑制端子,k,的情况下,划去原网络不定导纳矩阵,Y,i,的第,k,行和第,k,列,其他行和列的元素作如下变化,1,ij,ik,ik,kj,ij,ij,kj,kk,kk,kk,y,y,y,y,y,y,y,y,y,y,11,12,21,22,1,i,y,Y,Y,Y,Y,如果开路抑制网络的一个端子,21,1,22,12,11,Y,Y,Y,Y,Y,k,n,i,开路抑制（抑制了,k,个端子）后网络的不定导纳矩阵为,若某节点无注入电流（浮游节点），可将其消去；或网,络化简也需要消去某些节点。节点消去，导纳矩阵将降阶,但奇异性不变,P137,例,对于图示四端网络，

21、将端子,4,接地，端子,1,和,2,分,别与端子,4,构成一个端口。这样就改造成了一个共地,双口网络，求该双口网络的,Y,参数矩阵,1,C,2,C,2,G,1,G,1,I,2,I,3,I,4,I,1,C,2,C,2,G,1,G,1,I,2,I,4,I,解,图中四端网络的不定导纳矩阵为,2,2,2,2,2,1,2,1,2,1,2,1,1,1,1,1,i,sC,sC,0,0,sC,sC,G,G,G,G,0,G,sC,G,sC,0,G,sC,sC,G,Y,1,1,1,1,1,2,1,2,1,2,1,2,2,G,sC,sC,G,sC,G,sC,G,G,G,G,G,sG,Y,划去,Y,i,的第,4,行和

22、第,4,列，得端子,4,接地的定导纳矩阵,1,C,2,C,2,G,1,G,1,I,2,I,3,I,4,I,为了得到双口网络，必须开路抑制端子,3,将,分块,11,1,1,1,11,12,1,2,1,2,21,1,2,22,1,2,2,G,sC,sC,G,sC,G,sC,G,G,G,y,G,G,sC,Y,Y,Y,1,1,1,1,1,2,1,2,1,2,1,2,2,1,G,sC,sC,G,G,G,sC,G,sC,G,G,G,sC,Y,则双口网络的,Y,参数矩阵为,11,12,21,22,y,Y,Y,Y,Y,1,1,1,1,1,2,1,2,1,2,1,2,2,G,sC,sC,G,sC,G,sC,G,

23、G,G,G,G,sG,Y,2,1,2,2,1,2,2,2,1,1,2,1,2,2,1,2,2,1,2,1,1,2,1,2,2,1,2,2,1,2,1,1,2,1,2,2,1,2,2,1,1,2,2,1,1,2,1,2,G,G,sC,G,G,G,C,G,G,C,s,C,C,s,G,G,sC,G,G,G,G,sC,C,C,s,G,G,sC,G,G,G,G,sC,C,C,s,G,G,sC,G,G,G,C,G,G,C,s,C,C,s,1,C,2,C,2,G,1,G,1,I,2,I,4,I,定义,指网络,N,1,和,N,2,具有,公共的参考点,且,N,1,和,N,2,的,对应端子相连接,1,2,n,21

24、,I,22,I,12,I,2,n,I,11,I,1,n,I,1,I,2,I,n,I,1,N,2,N,4,网络并联,1,1,1,2,2,2,1,2,1,2,1,1,2,2,1,1,2,2,i,i,i,i,i,i,i,i,i,I,Y,U,I,Y,U,U,U,U,I,Y,Y,I,I,Y,Y,U,Y,U,Y,Y,U,并后,33,33,32,32,31,31,23,23,22,22,21,21,13,13,12,12,11,11,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,i,两,n,端网络并联,I,2,N,1,1,3,2,I,3,I,1,U,3,N,2,1,3,2,33

25、,32,31,23,22,21,13,12,11,2,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,i,33,32,31,23,22,21,13,12,11,1,Y,Y,Y,Y,Y,Y,Y,Y,Y,Y,i,KCL,1,11,21,2,12,22,1,2,n,n,n,I,I,I,I,I,I,I,I,I,M,M,M,1,2,I,I,I,简记为,KVL,1,11,21,2,12,22,1,2,n,n,n,U,U,U,U,U,U,U,U,U,M,M,M,1,2,U,U,U,简记为,N,1,和,N,2,的不定导纳矩阵方,程分别为,1,2,i,i,I,Y,U,Y,U,则,由,KVL,得,1,2,U,U,U,1,

26、1,1,i,I,Y,U,2,2,2,i,I,Y,U,1,2,I,I,I,1,1,2,2,i,i,Y,U,Y,U,1,2,i,i,i,Y,Y,U,YU,1,2,i,i,i,Y,Y,Y,总网络的不定导纳矩阵等,于各个并联网络的不定导,纳矩之和,两,n,端网络并联,n,端,网络与,m,端,网络并联,n,m,当相并联的两个网络的端子数目不等时，需先对端子数目,少的网络补充孤立节点，在原不定导纳矩阵中插入零行和零,列，形成增广不定导纳矩阵,然后再相加,增广网络,1,2,3,1,N,2,N,补零法,将,m,阶增广到,n,阶后相加,SC,SC,SC,SC,Y,i,2,3,3,3,3,2,1,2,1,2,2,

27、1,1,1,0,0,0,0,0,0,G,G,G,G,G,G,G,G,G,G,G,G,Y,i,2,1,4,3,G,1,G,2,G,3,1,2,C,4,3,0,0,0,0,0,0,0,0,0,0,0,0,2,SC,SC,SC,SC,Y,i,1,C,2,G,1,G,1,I,2,I,3,I,4,I,3,G,网络都是由元件连接而成的，因此，一个网络可以看成是由元件,并联连接而成，网络的不定导纳矩阵就等于各个元件的增广不定,导纳矩阵之和。这样就可以用添加元件的方法形成网络的不定导,纳矩阵。也就是说,可以把一个复杂的,n,端网络写成由,n,个二端网,络,n,次并联生成的,3,3,3,3,2,1,2,1,1,

28、2,1,1,2,1,0,0,0,0,G,G,G,G,G,G,G,G,G,SC,G,SC,G,SC,SC,G,Y,Y,Y,i,i,i,5,几种,常用元件,的不定导纳阵,P139-P141,作为作业验证几个,3,3,3,3,2,1,2,1,2,2,1,1,1,0,0,0,0,0,0,G,G,G,G,G,G,G,G,G,G,G,G,Y,i,0,0,0,0,0,0,0,0,0,0,0,0,2,SC,SC,SC,SC,Y,i,6,由不定导纳阵求多端口网络的,短路导纳矩阵,P141P144,短路导纳矩阵（多口网络的赋定关系或约束,端口电流与电压的关系为,N,1,u,1,i,n,u,n,i,1,11,12,

29、1,1,2,21,22,2,2,1,2,n,n,n,n,n,nn,n,I,y,y,y,U,I,y,y,y,U,I,y,y,y,U,L,K,M,M,M,O,M,M,L,j,k,U,j,i,ij,SC,k,U,I,Y,U,Y,I,0,直接按定义求，并不好求,下面介绍,借助,于,不定导纳阵,求短路导纳阵的方法,为,j,端口加电压源,U,j,其余端口短路,设多端口网络,N,的不定导纳阵为,Y,i,n,i,U,Y,J,则相应的,全节点方程,为,这里用,J,表示,I,n,设网络,N,可以构成,m,个端口,对应原来的,2,m,个端,子,相当于,2,m,个节点,如图所示,N,无,独,立,源,1,U,1,I,m

30、,U,m,I,1,2,1,2,m,m,2,2,1,1,n,n,U,U,U,4,3,2,n,n,U,U,U,2,1,2,m,n,m,n,m,U,U,U,3-6-10,该,m,端口网络（短路）导纳参数为,Y,sc,相应的方程为,U,Y,I,sc,所谓由不定导纳阵求短路导纳阵就是把,n,i,U,Y,J,U,Y,I,sc,3-6-8,3-6-9,端口电压与端钮电压的关系为,2,1,2,1,1,2,1,1,1,J,J,I,J,I,J,I,2,1,4,3,2,4,2,3,2,J,J,I,J,I,J,I,2,1,2,1,2,2,1,2,m,m,m,m,m,m,m,J,J,I,J,I,J,I,3-6-11,由

31、,3-6-11,还可得,0,2,1,2,1,J,J,0,2,1,4,3,J,J,0,2,1,2,1,2,m,m,J,J,3-6-12,端口电流与端钮电流的关系为,N,无,独,立,源,1,U,1,I,m,U,m,I,1,2,1,2,m,m,2,把,3-6-11,3-6-12,写成矩阵形式,m,m,m,m,m,J,J,J,J,J,J,I,I,I,2,2,1,2,1,2,1,1,1,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,1,1,2,1,0,0,0,该矩阵记

32、为,K,1,上式简写为,记为,I,记为,J,记为,0,J,K,1,0,I,3-6-14,1,1,1,0,K,K,T,1,1,1,2,K,K,人为引入,下列变量,2,1,1,n,n,U,U,H,4,3,2,n,n,U,U,H,2,1,2,m,n,m,n,m,U,U,H,3-6-13,把,3-6-10,3-6-13,写成矩阵形式,2,2,1,2,1,2,1,2,1,1,1,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,1,1,m,n,m,n,m,n,nm,n,n

33、,m,m,U,U,U,U,U,U,H,H,H,U,U,U,2,2,1,2,1,2,1,2,1,1,1,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,1,1,m,n,m,n,m,n,nm,n,n,m,m,U,U,U,U,U,U,H,H,H,U,U,U,记为,U,n,记为,U,记为,H,该矩阵记为,K,2,2 K,1,上式简写为,n,n,H,U,U,K,U,K,1,2,2,3-6-15,J,K,1,0,I,3-6-14,n,n,H,U,U,K,U,K,1,2,2

34、,3-6-15,把,3-6-8,式,代入,3-6-14,得,n,i,U,Y,J,n,i,I,U,Y,K,1,0,由,3-6-15,式得,H,U,H,U,2,H,U,H,U,T,T,n,1,1,1,1,2,K,K,K,K,U,2,1,2,1,1,3-6-18,把,3-6-18,式代入上式得,H,U,K,Y,K,I,1,1,2,0,i,3-6-17,H,U,K,Y,K,I,T,i,1,1,0,U,Y,I,sc,利用,0,列向量,回代消去,H,例：图,a,为一个有公共端的三端口网络，它的,Y,参数方程是,3,2,1,3,2,1,4,1,2,1,6,3,2,1,5,U,U,U,I,I,I,如果将,1F

35、,电容器接在端子,1,与,2,之间，如图,b,所,示，求这个二端口网络的短路导纳矩阵,N,U,2,I,2,U,1,I,1,1,2,4,U,3,3,I,3,a,N,1F,b,I,a,U,a,U,b,I,b,1,4,2,3,3,2,1,3,2,1,4,1,2,1,6,3,2,1,5,U,U,U,I,I,I,N,U,2,I,2,U,1,I,1,1,2,4,U,3,3,I,3,a,解,(1,分析：这是一个共地的三端口网络,1,1,2,2,3,3,n,n,n,U,U,U,U,U,U,则,原方程,可以当做节点电压方程，其系数矩阵为,定导纳阵,Y,n,因为,改接后,的二端口网络,没有公共端,显然不能通过,3

36、,端子的,开路抑制来实现。下面应用,P143,公式,3-6-17,的方法处理,设,4,节点为电位参考点，则,N,1F,b,I,a,U,a,U,b,I,b,1,4,2,3,S,N,I,2,I,1,1,2,4,3,I,3,I,4,a,图变为,N,1,I,1,I,2 2 1 S 2,I3 3,3,4,4,4,3,2,1,3,2,1,4,1,2,1,6,3,2,1,5,n,n,n,U,U,U,J,J,J,2,求解,4,3,2,1,4,3,2,1,5,1,4,0,1,4,1,2,2,1,6,3,2,2,1,5,n,n,n,n,U,U,U,U,J,J,J,J,做,4,端子浮地,运算（由,Y,i,的零和特性

37、,0,0,0,0,0,0,0,0,0,0,0,0,s,s,s,s,节点电压方程,全节点电压方程,把电容该画为,4,端网络，其不,定导纳阵为,然后将二网并联,4,3,2,1,4,3,2,1,5,1,4,0,1,4,1,2,2,1,6,3,2,2,1,5,n,n,n,n,U,U,U,U,s,s,s,s,J,J,J,J,可根据式,3-6-17,求解,H,U,K,Y,K,I,T,i,1,1,0,先写出,K,1,N,1F,b,I,a,U,a,U,b,I,b,S,N,I,2,I,1,1,2,4,3,I,3,I,4,要把该电路改造成,b,电路，必须满足相应的端口条件,要注意重新排列端子顺序,令,2,4,3,

38、1,4,3,2,1,J,J,J,J,J,J,J,J,S,N,I,2,I,1,1,2,4,3,I,3,I,4,2,4,3,1,4,3,2,1,n,n,n,n,n,n,n,n,U,U,U,U,U,U,U,U,1,1,2,2,3,3,4,4,n,n,n,n,U,J,U,J,U,J,U,J,K,1,1,2,2,3,3,4,4,n,n,n,n,U,J,U,J,U,J,U,J,K,N,1F,b,I,a,U,a,U,b,I,b,把第四行放到第二行的位置,1,2,3,4,5,1,2,2,0,4,1,5,3,6,1,2,2,1,4,1,n,n,n,n,U,s,s,U,U,s,s,U,1,2,3,4,5,2,1,

39、2,0,5,4,1,3,2,6,1,2,1,1,4,n,n,n,n,U,s,s,U,U,s,s,U,把第四列移到第二列的位置,先化简,T,i,1,1,K,Y,K,1,1,0,0,0,0,1,1,1,1,0,0,0,0,1,1,2,1,1,K,1,0,1,0,1,0,1,0,0,1,0,1,0,1,0,1,2,1,T,1,K,T,i,1,1,K,Y,K,1,1,0,0,0,0,1,1,1,1,0,0,0,0,1,1,2,1,T,1,K,4,1,1,2,1,6,2,3,1,4,5,0,2,1,2,5,s,s,s,s,3,5,3,5,3,5,3,5,5,7,1,1,1,3,7,5,2,1,s,s,s

40、,s,s,s,s,s,1,0,1,0,1,0,1,0,0,1,0,1,0,1,0,1,2,1,8,8,2,2,8,8,2,2,2,2,12,2,2,4,12,4,1,s,s,s,s,s,s,s,s,s,s,s,s,s,s,s,s,把上式代入,H,U,K,Y,K,I,T,i,1,1,0,得,2,1,2,1,2,1,8,8,2,2,8,8,2,2,2,2,12,2,2,4,12,4,1,0,0,H,H,U,U,s,s,s,s,s,s,s,s,s,s,s,s,s,s,s,s,I,I,与,H,1,H,2,对应块,H,1,H,2,对应,块,上式第三行与第四行是线性相关的，任取一行，有,8,2,2,1,2

41、,1,U,U,s,s,H,H,代入前式化简得,2,1,2,1,8,23,4,4,8,1,4,8,7,4,8,5,20,4,1,U,U,s,s,s,s,s,S,s,s,I,I,2,1,2,1,23,4,1,7,5,5,8,1,U,U,s,s,S,s,s,I,I,1,1,1,2,2,2,1,1,2,2,1,2,12,4,2,2,1,1,12,2,2,4,4,12,4,1,2,12,4,4,2,I,U,H,s,s,s,s,I,U,H,s,s,s,s,U,H,H,s,s,s,U,H,H,s,s,s,8,2,2,1,2,1,U,U,s,s,H,H,1,1,1,2,2,2,1,2,1,1,2,2,1,2,

42、12,4,1,2,12,4,4,2,12,4,1,2,12,4,4,8,I,U,H,H,s,s,s,I,U,H,H,s,s,s,U,U,U,s,s,s,U,U,U,s,s,s,大家可以验证，上面的求解过程是,较简单,的，因为有固定的公,式。还有校验功能，即如果在计算中,H1,H2,对应块,不一致,或,后,一半行,不是线性相关的，则计算有问题,解法二,4,3,2,1,4,3,2,1,5,1,4,0,1,4,1,2,2,1,6,3,2,2,1,5,n,n,n,n,U,U,U,U,s,s,s,s,J,J,J,J,0,1,1,0,1,0,0,1,0,1,1,0,1,0,0,1,2,1,1,K,0,1,

43、0,1,1,0,1,0,1,0,1,0,0,1,0,1,2,1,T,1,K,S,N,I,2,I,1,1,2,4,3,I,3,I,4,T,1,1,1,2,K,K,1,2,2,0,1,1,0,1,0,0,1,0,1,1,0,1,0,0,1,K,K,先化简,T,i,1,1,K,Y,K,1,2,1,K,K,T,T,i,1,1,K,Y,K,5,1,4,0,1,4,1,2,2,1,6,3,2,2,1,5,s,s,s,s,T,1,K,3,3,5,5,3,3,5,5,1,5,7,1,7,1,3,5,2,1,s,s,s,s,s,s,s,s,0,1,1,0,1,0,0,1,0,1,1,0,1,0,0,1,2,1,

44、0,1,0,1,1,0,1,0,1,0,1,0,0,1,0,1,2,1,12,4,2,2,12,2,2,1,2,2,8,8,4,2,2,8,8,s,s,s,s,s,s,s,s,s,s,s,s,s,s,s,s,T,i,1,1,K,Y,K,12,4,2,2,12,2,2,1,2,2,8,8,4,2,2,8,8,s,s,s,s,s,s,s,s,s,s,s,s,s,s,s,s,把上式代入,H,U,K,Y,K,I,T,i,1,1,0,得,1,1,2,2,1,2,12,4,2,2,12,2,2,1,0,2,2,8,8,4,0,2,2,8,8,U,I,s,s,s,s,U,I,s,s,s,s,H,s,s,s,

45、s,H,s,s,s,s,与,H,1,H,2,对应块,H,1,H,2,对应块,2,1,2,1,8,23,4,4,8,1,4,8,7,4,8,5,20,4,1,U,U,s,s,s,s,s,S,s,s,I,I,2,1,2,1,23,4,1,7,5,5,8,1,U,U,s,s,S,s,s,I,I,上式第三行与第四行是线性相关的，任取一行，有,8,2,2,1,2,1,U,U,s,s,H,H,代入前式化简得,2,端子数为,奇数,或端口间,有公共端,的处理,4,1,2,1,0,0,2,1,0,1,2,1,4,1,0,1,0,1,3,1,1,0,0,1,2,i,Y,N,U,2,I,2,U,1,I,1,1,2,

46、4,U,3,3,I,3,5,例,图示网络的不定导纳阵为,求,三端口,网络的导纳参数（矩阵,下面我们用一道例题就是说明其原理,这时的处理方法与,1,不同，较,1,复杂,解：该题与前面讲的不同，有一个,公共端子,下面我们一起处理,N,U,2,I,2,U,1,I,1,1,2,4,U,3,3,I,3,5,1,1,J,I,3,3,3,2,2,1,2,1,J,J,J,I,4,3,J,I,0,3,2,1,J,J,J,2,1,3,J,J,J,3,2,1,3,3,3,2,2,1,2,1,2,1,2,1,2,1,J,J,J,J,J,J,I,1,分析并构造,K,1,和,K,2,5,4,J,J,0,5,4,J,J,3

47、,2,1,2,2,1,2,1,2,1,J,J,J,I,5,4,3,2,1,3,2,1,1,1,0,0,0,0,0,1,1,1,0,2,0,0,0,0,0,1,1,1,0,0,0,0,2,2,1,0,0,J,J,J,J,J,I,I,I,5,4,3,2,1,2,1,3,2,1,2,0,0,0,0,0,0,1,1,0,1,1,0,0,0,0,0,1,1,0,0,0,0,1,1,n,n,n,n,n,U,U,U,U,U,H,H,U,U,U,K,1,K,2,1,1,0,0,0,0,0,1,1,1,0,2,0,0,0,0,0,1,1,1,0,0,0,0,2,2,1,1,K,1,0,0,0,0,1,0,2,0

48、,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,2,2,1,1,T,K,2,0,0,0,0,0,0,1,1,0,1,1,0,0,0,0,0,1,1,0,0,0,0,1,1,2,K,2,0,1,0,0,0,0,1,0,0,0,1,0,1,0,0,1,0,1,1,0,0,0,0,1,2,T,K,2,0,0,0,0,0,0,1,1,0,1,1,0,0,0,0,0,1,1,0,0,0,0,1,1,1,0,0,0,0,1,0,2,0,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,2,2,1,2,0,0,0,0,0,2,0,0,0,0,0,2,0,0,0,0,0,2,0,0,0

49、,0,0,2,2,1,T,1,2,K,K,所以有,1,2,1,K,K,T,2,代入公式求解,H,U,K,Y,K,I,1,1,2,0,i,1,1,2,K,Y,K,i,1,1,0,0,0,0,0,1,1,1,0,2,0,0,0,0,0,1,1,1,0,0,0,0,2,2,1,T,1,4,1,2,1,0,0,2,1,0,1,2,1,4,1,0,1,0,1,3,1,1,0,0,1,2,K,5,5,2,1,2,5,5,2,1,2,4,4,8,0,4,1,1,2,9,2,2,2,0,6,8,4,1,2,1,H,H,U,U,U,I,I,I,3,2,1,3,2,1,5,5,2,1,2,5,5,2,1,2,4,

50、4,8,0,4,1,1,2,9,2,2,2,0,6,8,4,1,0,0,上式,后两行是线性相关,的,任取一行得,2,2,5,1,3,2,1,2,1,U,U,U,H,H,代入前式,得,3,2,1,3,2,1,3,2,1,5,48,5,4,5,12,5,8,5,44,5,12,5,4,5,32,5,36,4,1,5,8,8,5,4,5,8,4,5,2,2,5,1,9,5,2,2,5,4,5,2,6,5,4,8,4,1,U,U,U,U,U,U,I,I,I,3,2,1,3,2,1,12,1,3,2,11,3,1,8,9,5,1,U,U,U,I,I,I,由此列可以看出当,端子,数为,奇数,或端口间,有公

51、共端,的处理方法。此例虽然较繁，但如果,只要求写出,K,1,和,K,2,并不难,3,小结,端口电流,除满足端口条件外，还必须满足,相,应的约束,端口电压,满足,端口条件,人为变量,应尽量使,易求,1,2,K,2,含（独）源多口网络表征方程,1,非含源（独立源）多口网络的常见矩阵表示法,3,含源多口网络的等效电路,含源多口,罗森定理,4,散射（参数）矩阵（不要求,本章小结,5,不定导纳阵,1,定义,n,n,n,I,U,Y,定,导纳阵,方程数,等于,独立节点数,n-1,1,n,Y,n,Y,非,奇,异,若把电位,参考点,改在电路,N,的外部,则得到的电路的,全,部节点,为变量的,n,个,方程,称为,

52、全节点电压方程,节点电压,方程,n,n,i,I,U,Y,求法与原来,相同,T,i,n,n,a,b,a,Y,A,Y,A,1,Y,i,不存在，称为,不定,导纳阵,显然,Y,i,的,行,是线性,相关的,det,Y,i,0,1,n,n,U,1,n,n,I,2,基本性质,5,不定导纳阵,1,Y,i,的零和特性,1,1,1,0,2,0,n,kj,k,n,jk,k,Y,Y,每列之和,每行之和,2,Y,i,的等余因子特性：任意两个元素的代数余子式相等,Y,i,为,零和矩阵,由于,参考点,选在电网络,N,外,全节点方程和不定导纳阵,非常,灵活,可用于,不同,的,网络,的,连接,和,同一网络,的,变换,3,引入不

53、定导纳阵的意义,从,U,n,中去掉,U,k,I,n,中去掉,I,nk,就得到以网络,N,中,k,节点,为参考点的,节点,电压,方程,的,k,行,k,列得定导纳矩阵,i,Y,例如，划去,n,Y,n,n,n,I,U,Y,4,不定导纳阵的运算,1,端子,接地,与,浮地,划去,j,行,j,列,得到以,j,节点为参考点,的定导纳阵,Y,n,增加,j,行,j,列,和为零）得到不定导纳阵,Y,i,2,短路收缩,同一网络,将,两个,或多个,端子连接,起来形成,一个新的端子,相应的,行和列,相加,3,开路抑制,端口删减,使,删减后,的,某些端子,变成,不可及,节点,两,n,端网络并联,4,网络,并联,两个具有,

54、相同参考点,的网络,N,1,N,2,对应端子相连,n,端,网络与,m,端,网络并联,n,m,补,零,法,将,m,阶增广到,n,阶后相加,与生成的,Y,n,的送值表类似，即可以,把一个复杂的,n,端,网络写成,由,n,个,二端网络,n,次并联,生成的，几种,常用元件,的,不定导纳,阵,P139-P141,5,由不定导纳矩阵求,短路导纳,矩阵,P141P144,短路导纳,矩阵,多口网络,的赋定,关系,或约束）端口,电流,与,电压,的关系为,N,1,u,1,i,n,u,n,i,j,k,U,j,i,ij,SC,k,U,I,Y,U,Y,I,0,n,i,U,Y,J,n,i,U,Y,J,U,Y,I,sc,令

55、,每两个端子,构成,一个端口,按,端口,条件，有,2,1,1,n,n,U,U,U,4,3,2,n,n,U,U,U,2,1,2,m,n,m,n,m,U,U,U,2,1,2,1,1,2,1,1,1,J,J,I,J,I,J,I,2,1,4,3,2,4,2,3,2,J,J,I,J,I,J,I,2,1,2,1,2,2,1,2,m,m,m,m,m,m,m,J,J,I,J,I,J,I,0,2,1,2,1,J,J,0,2,1,4,3,J,J,0,2,1,2,1,2,m,m,J,J,人为引入,下列,变量,2,1,1,n,n,U,U,H,4,3,2,n,n,U,U,H,2,1,2,m,n,m,n,m,U,U,H,

56、J,K,1,0,I,1,1,1,0,K,K,T,1,1,1,2,K,K,n,n,H,U,U,K,U,K,1,2,2,当,端子,数为,奇数,或端口间,有公共端,的处理方法较繁,端口电流,除满足端口条件外，还必须满足,相应的约束,端口电压,满足,端口条件,人为变量,应尽量使,易求,1,2,K,由不定导纳,阵求多端口网络,短路导纳,矩,阵的步骤,1,T,n,2,1,U,U,U,K,K,H,H,1,T,2,1,K,K,T,1,1,i,n,1,i,1,I,U,K,J,K,Y,U,K,YK,0,H,不定导纳,最重要的应用,n,端,网,络,与,m,端,网络的,并联,5,6,3,4,7,1,2,1,5,6,3

57、,4,7,2,1,例,列出图示电路的节点方程,该题前面用矩阵,分块,处理,过，下面用不定导纳阵,并,联方法,处理处理,解,从节点处把原网络分开,网络,N,1,网络,N,2,C,6,M,L,2,L,1,u,S5,R,5,C,7,R,3,i,S7,R,4,2,1,5,6,3,4,7,1,C,6,M,L,2,L,1,E,5,R,5,C,7,R,3,i,S7,R,4,0,0,0,0,0,0,0,0,0,0,7,7,5,4,3,2,1,3,7,7,6,4,7,7,6,6,7,5,1,S,I,S,I,s,E,G,U,U,U,U,G,SC,SC,SC,G,SC,SC,SC,SC,SC,G,N,s,s,s,n

58、,n,n,n,对网路,N,1,列,节点,电压方程,网络,N,1,2,1,网络,N,2,2,1,0,0,2,1,0,3,1,1,4,0,1,N,A,对网,N,2,列,全节点,电压方程（方法一,1,2,1,2,2,2,1,2,1,2,1,I,SL,SM,U,I,SM,SL,U,S,L,L,M,C,6,M,L,2,L,1,E,5,R,5,C,7,R,3,i,S7,R,4,1,1,1,2,1,1,1,2,2,2,1,2,2,2,U,SL,I,SMI,U,SL,SM,I,U,SL,I,SMI,U,SM,SL,I,2,1,1,L,M,Y,M,L,2,1,2,S,L,L,M,令,T,i,n,n,a,b,a,

59、Y,A,Y,A,2,1,网络,N,2,2,1,0,0,2,1,0,3,1,1,4,0,1,N,A,C,6,M,L,2,L,1,E,5,R,5,C,7,R,3,i,S7,R,4,2,2,2,2,1,2,1,1,1,1,2,T,i,L,M,L,M,Y,AY,A,M,L,L,L,M,M,L,M,M,L,L,2,0,0,0,T,n,I,2,1,1,L,M,Y,M,L,2,1,2,S,L,L,M,令,2,1,网络,N,2,N,2,的,全节点,方程（方法二,1,2,1,2,2,2,1,2,1,2,1,I,SL,SM,U,I,SM,SL,U,S,L,L,M,1,1,1,2,2,2,2,1,U,SL,I,SMI,U,SL,I,SMI,2,1,2,S,L,L,M,令,1,1,1,2,2,2,U,SL,SM,I,U,SM,SL,I,2,3,2,2,2,3,4,1,1,2,1,n,n,n,n,U,U,SL,SM,U,U,SM,SL,S,L,L,M,2,3,2,3,4,1,1,n,n,n,n,U,U,L,M,U,U,M,L,2,2,3,1,4,1,1,0,1,0,1,1,n,n,n,U,L,M,U,M,L,U,2,2,2,3,1,1,4,1,n,n,n,U,L,M,L,M,U,M,M,L,L,U,直接利用支路