answer to foster's open question

2
126 IEEE TR.ANSACTIONS ON CIRCUIT THEORY March strained by only one operational amplifier (as defined by Nathan). Let us call the constrained network N, and its unconstrained form N. Then the admittance matrix of N is given by [2] Y = APA’ (1) where A is the incidence matrix of N, A’ is the transpose of A and Y is the branch admittance matrix of N. The admittance matrix Y, of the net- work N, is then [l] Y, = A+PALi (2) where the matrices A-< and A-j are obtained by deleting the ith and jth rows, respectively, from A. Al-i is the transpose of A-i. So, the determinant of Y, is [2] det. Y, = (-l)i+‘W,i,,(Y) where (3) Wij, r(Y) = C 2-tree products of the network without the constraint over all the 2-trees containing i and j in one part and T in the other. Next let us say we want to evaluate the cofactor Amn of the element of the matrix Y, that lies in the mth row and the nth column of the matrix Y. It can be easily shown that it is given by ernn -= (- 1) i+i+u(m--i)+u(n--i)+k . [Wij, mn, T + Win, jm, r] (4) where K = c u(x - p) (5) z over z = those of i, j, m and n that are not equal, p = value of those of i, j, m and n that are equal. Example : Find the input impedance and the voltage transfer function of the network shown in Fig. 1. ,NPYT T c, OUTPUT Fig. 1. In this case, i = 4, j = 3 (the node numberings are arbitrary). The input impedance = All/A where A = det. Y, and the transfer function F(& = AM/AII. From (3) A = (- 1)3+4W,,,, = -G, [S2C1C, f SG,C, + SC&C, + G,G,I. For A ,,, m = n = 1. Thus, K = -2 [from (5)] and from (4) Al1 = (- l)4+3-2 [c 3 - treeproducts] = - [S”C,C, + SC&5 + G, + GJ + G,G,]. For AXa,m = 1, n = 4, K = u(3 - 4) + U(1 - 4) = 0 and 44 = (-l>“+““[G~G~] = GIG,. Finally, we have y 11 = S2C,C, + SG(Gl +. G, + G,) + G,G, G,[S2C,C, + SG,C, + SG,C, + GzG,] and F(S) = - 2 ~ &JlG, -. S ClCz + SCdG, -t G, + G,) + G2G, U’S are step functions defined by u(a - b) = -1 for a > b (6) = 0 for a < b and W’s are the sums of the 3-tree products of the network [2]. The suflixes indicate the node groupings. Thus, Wij, mn, r runs over the 3-trees that contain i and j in one part, m and n in another and r in another. Eqs. ;( 1) through (6) may now be used to obtain the transfer function and the self- and transfer-impedances of the type of network under discussion. The procedure is perhaps best explained through an example. V. P. SINHA Dept. of Elec. Engrg. Northampton College of Adv. Tech. London, England REFERENCES [l] A. Nathan, “Matrix analysis of networks having infinite gain operationd amplifiers,” PROC. IRE (~orrespondace). vol. 49, pp. 1577-1578; October, 1961. [2] 5. Seshu and M. B. Reed,, “Linear Graphs and Electrical Networks,” Ad&son-Wesley Publishing Co., Inc., Reading, Masc.; 1961. [3] L. Weinberg, “Kirchhoff’s third and fourth laws,” IRE TRANS. ON CIRCUIT THEORY, vol. CT-5, pp. S-30; March, 1958. [4] L. Weinberg, “Fundamentals of linear graphs and their applications to networks without transformers.” Estmtta da Alta Bequenza, vol. 30, no. 10; 1961. [5] L. K. Wadhwa, “Simulation of third order systems with one operational amplifier.” PROC. IRE (Correspondence), vol. 50, pp. 201-202; February, 1962. [6] S. J. Mason, “Topological analysis of linear non- reciprocal networks,” PROC. IRE, vol. 45, PP. 829-838; June, 1957. [7] W. Mayede, “Topological Formulae for Active Networks.” University of Illinois, Urbana, Int. Tech. Rept. No. 8; January, 1958. Answer to Foster’s Open Question* The problem considered here is one of the open questions connected with the “n-port problem.” (Readers who are interested in the n-port problem are referred to the recent panel discussion in these TRANSACTIONS.~) The question posed by Foster is the following? Is it possible to realize the matrix 7 -1 -2 -2 -2 -1 7 -3 -2 -1 -2 -3 9 -1 -3 -2 -2 -1 8 -3 --2 -1 -3 -3 9_ as the open-circuit resistance matrix of a network (without using transformers)? Foster’s contention is that no such network exists. The following is a proof of this conjecture. Note that Foster’s matrix is an indefinite matrix of order five and rank four. (A matrix is said to be indefinite if the sum of elements. in each row and each column is zero.) Instead of specifically considering Foster’s matrix, the procedure described here is general and will decide the realiz- ability of any nth order indefinite resistance matrix of rank n - 1. It is a well-known fact that if the n ports of a network form an oriented circuit as shown in Fig. 1, then the resistance matrix of the network is an indefinite matrix.3 In the following development a slight general- ization of this idea is needed and it is now stated as a theorem without proof: Theorem 1 If (rii) is the resistance matrix of an n-port network, and if the ports 311 3, ‘2 * . . jk(k < n) form an oriented circuit, then i$rii = 0, for i = I,2 ... n. * Received February 28, 1962; revised manuscript received, October, 1962. 1 “The realization of n-port networks without transformers-A panel discussion.” IRE TRANS. ON CIRCUIT THEORY, vol. CT-S, pp. 202-214; Sep- terober. 1962. 2 R. M. Foster, “An open question,” IRE TRANS. ON CIRC~T THEORY (Correspondence), vol. CT-8, p. 175; June, 1961. 2 T. H. Puck&t. “A note on the admittitnee and impedance matrices of ILL n-terminal network,” IRE TRANS. ON CIRCUIT THEORY, vol. CT-3, pp. 70-75; March, 1956.

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Page 1: Answer to Foster's Open Question

126 IEEE TR.ANSACTIONS ON CIRCUIT THEORY March strained by only one operational amplifier (as defined by Nathan).

Let us call the constrained network N, and its unconstrained form N. Then the admittance matrix of N is given by [2]

Y = APA’ (1) where A is the incidence matrix of N, A’ is the transpose of A and Y is the branch admittance matrix of N.

The admittance matrix Y, of the net- work N, is then [l]

Y, = A+PALi (2) where the matrices A-< and A-j are obtained by deleting the ith and jth rows, respectively, from A.

Al-i is the transpose of A-i. So, the determinant of Y, is [2]

det. Y, = (-l)i+‘W,i,,(Y) where

(3)

Wij, r(Y) = C 2-tree products of the network without the constraint over all the 2-trees containing i and j in one part and T in the other.

Next let us say we want to evaluate the cofactor Amn of the element of the matrix Y, that lies in the mth row and the nth column of the matrix Y. It can be easily shown that it is given by

ernn -= (- 1) i+i+u(m--i)+u(n--i)+k

. [Wij, mn, T + Win, jm, r] (4) where

K = c u(x - p) (5) z

over z = those of i, j, m and n that are not

equal, p = value of those of i, j, m and n that

are equal.

Example : Find the input impedance and the voltage

transfer function of the network shown in Fig. 1.

,NPYT T c, OUTPUT Fig. 1.

In this case, i = 4, j = 3 (the node numberings are arbitrary).

The input impedance = All/A where A = det. Y, and the transfer function F(& = AM/AII.

From (3)

A = (- 1)3+4W,,,,

= -G, [S2C1C, f SG,C,

+ SC&C, + G,G,I. For A ,,, m = n = 1. Thus, K = -2 [from (5)] and from (4)

Al1 = (- l)4+3-2 [c 3 - treeproducts]

= - [S”C,C,

+ SC&5 + G, + GJ + G,G,]. For AXa, m = 1, n = 4,

K = u(3 - 4) + U(1 - 4) = 0

and

44 = (-l>“+““[G~G~] = GIG,. Finally, we have

y 11 = S2C,C, + SG(Gl +. G, + G,) + G,G, G,[S2C,C, + SG,C, + SG,C, + GzG,]

and

F(S) = - 2 ~ &JlG, -. S ClCz + SCdG, -t G, + G,) + G2G,

U’S are step functions defined by

u(a - b) = -1 for a > b (6)

= 0 for a < b

and W’s are the sums of the 3-tree products of the network [2]. The suflixes indicate the node groupings. Thus, Wij, mn, r runs over the 3-trees that contain i and j in one part, m and n in another and r in another.

Eqs. ;( 1) through (6) may now be used to obtain the transfer function and the self- and transfer-impedances of the type of network under discussion. The procedure is perhaps best explained through an example.

V. P. SINHA Dept. of Elec. Engrg.

Northampton College of Adv. Tech. London, England

REFERENCES

[l] A. Nathan, “Matrix analysis of networks having infinite gain operationd amplifiers,” PROC. IRE (~orrespondace). vol. 49, pp. 1577-1578; October, 1961.

[2] 5. Seshu and M. B. Reed,, “Linear Graphs and Electrical Networks,” Ad&son-Wesley Publishing Co., Inc., Reading, Masc.; 1961.

[3] L. Weinberg, “Kirchhoff’s third and fourth laws,” IRE TRANS. ON CIRCUIT THEORY, vol. CT-5, pp. S-30; March, 1958.

[4] L. Weinberg, “Fundamentals of linear graphs and their applications to networks without transformers.” Estmtta da Alta Bequenza, vol. 30, no. 10; 1961.

[5] L. K. Wadhwa, “Simulation of third order systems with one operational amplifier.” PROC. IRE (Correspondence), vol. 50, pp. 201-202; February, 1962.

[6] S. J. Mason, “Topological analysis of linear non- reciprocal networks,” PROC. IRE, vol. 45, PP. 829-838; June, 1957.

[7] W. Mayede, “Topological Formulae for Active Networks.” University of Illinois, Urbana, Int. Tech. Rept. No. 8; January, 1958.

Answer to Foster’s Open Question* The problem considered here is one of

the open questions connected with the “n-port problem.” (Readers who are interested in the n-port problem are referred to the recent panel discussion in these TRANSACTIONS.~) The question posed by Foster is the following? Is it possible to realize the matrix

7 -1 -2 -2 -2

-1 7 -3 -2 -1

-2 -3 9 -1 -3

-2 -2 -1 8 -3

--2 -1 -3 -3 9_ as the open-circuit resistance matrix of a network (without using transformers)? Foster’s contention is that no such network exists. The following is a proof of this conjecture.

Note that Foster’s matrix is an indefinite matrix of order five and rank four. (A matrix is said to be indefinite if the sum of elements. in each row and each column is zero.) Instead of specifically considering Foster’s matrix, the procedure described here is general and will decide the realiz- ability of any nth order indefinite resistance matrix of rank n - 1.

It is a well-known fact that if the n ports of a network form an oriented circuit as shown in Fig. 1, then the resistance matrix of the network is an indefinite matrix.3 In the following development a slight general- ization of this idea is needed and it is now stated as a theorem without proof:

Theorem 1 If (rii) is the resistance matrix of an

n-port network, and if the ports

311 3, ‘2 * . . jk(k < n) form an oriented circuit, then

i$rii = 0, for i = I,2 ... n.

* Received February 28, 1962; revised manuscript received, October, 1962.

1 “The realization of n-port networks without transformers-A panel discussion.” IRE TRANS. ON CIRCUIT THEORY, vol. CT-S, pp. 202-214; Sep- terober. 1962.

2 R. M. Foster, “An open question,” IRE TRANS. ON CIRC~T THEORY (Correspondence), vol. CT-8, p. 175; June, 1961.

2 T. H. Puck&t. “A note on the admittitnee and impedance matrices of ILL n-terminal network,” IRE TRANS. ON CIRCUIT THEORY, vol. CT-3, pp. 70-75; March, 1956.

Page 2: Answer to Foster's Open Question

1963

Fig. I-An n-port network in which the ports form an oriented circuit.

Now consider the following question: If it is given that the resistance matrix R = (rdi) of an n-port network is indefinite and of rank n - 1, can it be concluded that the ports of the network form an oriented circuit? The answer to this question is in the affirmative and the following arguments are given to substantiate it.

Suppose unit current generators are con- nected to the ports. The total power dissipated in the network is:

[1 1 ‘I’ l]

r11 r12 .-* ?-I?2 1-i

f-21 7.22 . . * rPn

i I!

l 0 . . . ,.. . . . . . . = t . . .

r nl r,, . . . r,, 1 since

g rii = 0.

This means that the current through any resistor is zero and, hence, at any node current generators are totally responsible for delivering and taking away current. Since all current generators are of unit magnitude, it can be inforred that the number of current generators deiivering current to any node is exactly equal to the number of current generators taking current away from the same node. Hence the con- clusion is that a network with an indefinite matrix as its resistance matrix must have its ports forming an oriented circuit or a disjoint union of oriented circuits. It can be shown easily that the ports must form an oriented circuit and cannot form a disjoint union of circuits if the rank of R is n - 1. To see this assume that the ports j,, jz, . . . jk with k < n, form an oriented circuit. Then

ig rii = 0

by Theorem 1. Also R is an indefinite matrix, so that

g rii = 0.

Since this contradicts the hypothesis that the rank of R is n - 1, the conclusion is:

Theorem 2 A network with an indefinite matrix R

of order n and rank n - 1 as its resistance matrix must have its ports forming an oriented circuit.

If R is to be realized as a resistance matrix of a network, it has been shown above that the port structure must be as shown in Fig. 1. If one of the ports is left open-circuited, the port structure becomes a “linear tree.“4 It is known that the conductance matrix of a network with ports forming a linear tree can be arranged to be in the “uniformly tapered form.“4 The following theorem is stated without proof, since the proof is implicit in the above statements.

Theorem 3 Let R’ be a nonsingular matrix obtained

by deleting one row and corresponding column .of the given indefinite matrix R of order n and rank n - 1. Let G’ be the inverse of R’. The necessary and sufficient condition that R be realizable as a resistance matrix is that G’, or G’ after a proper rearrangement of columns and rows, is in the uniformly tapered form.

The above theorem gives a finite proce- dure to decide the realizability of Foster’s matrix. Deleting the last column and last row of Foster’s matrix gives

r 7 -1 -2 -2

1 R’= --I 7 -3 -2

-2 -3 9 -1

-2 -2 -1 8 1 Inversion of the above matrix gives

l- 377 165 156 155 1

G' & 165 421 196 171 =

156 196 320 128

1155 171 128 329_)

4 E. A. Guillemin, “On the analysis and synthesis of single-element kind networks,” IRE TRANS. ON CIRCUIT THEORY. vol. CT-7, pp. 303-312; September, 1960.

127

Interchanging rows 1 and 3 and then columns 1 and 3 gives

r 320 196 156 128 1 ,’ G” 1 196 421 165 171 =

1852 1.

156 165 377 155

Ll28 171 155 329-l In the above matrix, the minimum element is at the top right position and the elements of the first row decrease from left to right. In this manner the elements of G” are arranged in such a way that G” will appear in.tapered form whenever that is possible. Inspection of G” shows that it is not in the tapered form and, therefore, as conjectured by Foster, R is not realizable.

The author is indebted to Dr. Laveen Kanal for helpful discussions.

K. K. NAMBIAR Philco Scientific Lab.

Blue Bell, Pa.

Enumeration of Trees* This brief note describes an efficient and

algebraic method of finding the sum of tree products of a graph and as a corollary constitutes the algebraic proof of Fujisawa’s enumeration of trees.*

Let the graph have b branches and n nodes.

Let A, be the fundamental cut set matrix with respect to a tree whose branches are ytl, yt2, . . . , yth-1). Since Af is related to the vertex matrix by a nonsingular transformation, the determinant A/Y A/, where A/ is the transpose of A,, and Y is a diagonal matrix whose diagonal elements are ye, . . * , Yt(n-I), Ycl, Yc2, . ’ ’ , ?ic(b-n+l),

gives the sum of all tree products.

The following is an efficient expansion applicable to the proof of Fujisawa’s enumeration. Let Al be partitioned as

where the columns of the unit matrix correspond to the tree branches and those of Afll to chords. Let Y be partitioned a.3

in which Y t is the diagonal matrix diag lYt1 YlZ . . . yt cn-l,] and Y C is the diagonal matrix diag [y,l yC2 .. * yc(b-,+~)l. The product A,Y A/ is

* Received August 15, 1962. This work was supported by the National Science Foundation under grant NSF G-15078. The publication of this note is supported by the Marcellus Hartley Fund, School of Engineering and Applied Science, Columbia University.

1 T. Fujisawa, “On & problem of network topol- ogy,” IRE TRANFJ. ON CIRC~T THEORY, vol. CT-6. pp. 261466; September, 1959.