Transient analysis of symmetrical networks by the method of symmetrical components

  • View

  • Download

Embed Size (px)


  • Transient Analysis of Symmetrical Networks by the Method of Symmetrical Components

    Discussion and author's closure of paper 40-2 by Louis A . Pipes, presented at the AIEE winter convention, New York, . Y., January 22-26 , 1940, and published in AIEE TRANSACTIONS, 1940 (August section) pases 457-9 .

    W. V. Lyon (Massachusetts Institute of Technology, Cambridge): A method, such as Doctor Pipes proposes in his paper, which avoids solving a high-order determinantal equation, is welcome. I should like to point out two things. First, the condition of symmetry which Doctor Pipes imposes is more severe than necessary. Second, the resolution of the determinantal equation into factors can be accomplished by means of symmetrical components alone without recourse to matrix theory. For example, consider six equal circuits, each having the same operational impedance. Let each of these circuits be coupled to one of six equal coils equally spaced around a uniform magnetic ring, so that the mutual inductances between adjacent coils are equal, between alternate coils are equal, and between diametrical coils are equal. Let Zab be the operational coupling impedance between adjacent coils, Zac between alternate coils, and Zad between diametrical coils ; and let be the resulting operational self-impedance of each of the six loops. Since the self-impedances of the six loops are equal, and because of the symmetry imposed on the coupling, the potentials of each sequence are proportional to the currents of tha t sequence alone. For simplicity, assume that an electromotive force, E, is impressed on one loop a and all other loops, b, c, d, e, f are short-circuited, there being no initial currents or charges. The symmetrical components of the applied electromotive force are each E/6. Consequently the zero-sequence equation is

    Eo = /6 = Zi0+Zabi:Q-f-Zaci0+Zadi0+ -\-

    The last two terms represent the mutual effect of the zero-sequence currents in coils e and / . Hence,

    0 = / 6 = (Z+2Zab+2Zac+Zad)i0 (1)


    E1 = E/6 =

    where a operator a + ab = l,

    E, = E/6 =


    E, = E/6 =

    Ei = E/6 =

    4 = / =

    Eb = E/6 =

    = Zii +Zab(ai1) +Zac(aHi) - f

    = e-J2w/(> j s the characteristic for a six-phase circuit. Since and 2 + 4 = 1 , and 3 = 1 ,

    = (Z-\-Zab Zac Zad)ii (2)

    it is readily shown that

    = (Z Za0 Zac+Zad)i2 (3)

    --(Z-2Zab+2Zac-Zad)iz (4) Z(Z ZabZac-\-Zad)t4 (5)

    = (Z+ZabZacZad)ib (6)

    Equations 2 and 6 show tha t ii = u. Also equations 3 and 5 show that i2 U.

    The impedance operators in equations 1, 2, 3, and 4 are the factors of the determinantal equation. Having determined the component currents, to, iif i2, from equations 1, 2, 3, and 4, the currents in the loops a,b,c, etc., are

    * = 5 !*0 + 2 + 2 2 + * 8 (7)

    i io+

  • Here

    (5 ) =

    Za %b z c z b

    Zb Za Zb Zc _

    Zc 2ft Za Zft Zf, Zc Zf, ,Za

    ( z a - ze) 2(za+zc+2zb) (za+zc- 2zb) (5 )

    These factors can be readily found by: (1) transforming the determinant into a factor times a determinant having the first element of the first row unity, and all other elements of this row zero; (2) reduction of this determinant to a third-order determinant; and (3) transformation of this third-order determinant into a factor times a determinant having the first element of its first column unity, and all other elements of this column zero. A similar but briefer procedure reduces Njjc(s) to a product of factors.

    To complete the example, the current in loop 1 will be found, considering 2 , 3 , and 4 are zero, and Ti = y/s, y being the net initial condenser voltage in loop 1. Here

    US): Nn(s)_

    " Ms) ' (za-zc) (za2 +ZaZc ~ 2s&2)

    (za-Zc)2(Za+Zc+2zb)(za+Zc-2zb) J

    { /v 2 { [_za + (zc+2zb) za-zc za-\-{zc-2zb) J s

    (6) in which

    = A 3 = 0.25 and K2 = 0.5

    I t will be noted tha t the partial fraction expansion has been made considering za as the variable.

    In the terminology of symmetrical components, {za-\-zc-\-2zb) will be recognized as the zero-sequence impedance, (zazc) as the first- and third-sequence impedances, and (za-\-zc2zb) as the second-sequence impedance. Taken with the factor y/s, the first, second, and third terms are the Laplace transforms, respectively, of the zero-sequence component, the combined first- and third-sequence components, and the second-sequence component of the current of loop 1.

    Replacing za, zb, and zc by their respective functions of s,

    Ids) = ( A l M A z \ (7)

    in which

    teristic equation for its roots in the usual manner. I t was not necessary to employ matrices, nor the aid of symmetrical-component theory. Furthermore the procedure was the usual one throughout, being shortened simply by the ease with which the determinant could be factored.

    In this discussion the direct Laplace transformation used is the ordinary form, that is,

    A ! = 0.25(L+2Mi + M2) ~1

    A2 = 0ML-M2)-1

    .4 3 = 0 . 2 5 ( L - 2 M 1 - r - M 2 ) - 1

    / 3 1 2 = [ C ( L + 2 M 1 + M 2 ) ] - 1

    ^ 3 2 = [ C ( L - 2 M 1 - f M 2 ) ] - 1

    Finally, the inverse Laplace transformation of Ii(s) gives

    t\(t) = [ sin -\ sin sin ) \ 2 / Out ( 8 )

    1 / ( 0 1 = 8 f f ( t ) e ~ S t d t = F ( s ) ( 9 )

  • Figure 1

    L = 1.0 henry C = 10~ 4 farad /W = 0.1 henry = 1 0 0 volts

    Capacitors are initially uncharged

    These can be written

    \ii + Mpi2 = E

    The roots of the determinantal equation (a fourth-degree equation) are


    100 V 9


    pA = -j


    100 /

    By examination of the circuit, the forms for the two currents are :

    /, = / I 'ePlt+h'e^+h'eW+IiW h = Il''ePlt+l2''ep*t+I3',ep*t+I4,'ep*t

    Determining the coefficients of the different terms on the right-hand side and substituting them in the equations, the expressions obtained for i\ and i2 are the same as those given by Pipes' method. By substitution, the currents satisfy the two Kirchhoff's electromotive-force equations.

    Louis A. Pipes: The method of obtaining the transient solution of networks presented in this paper is a special case of a more general method which may be briefly described as follows:

    Following the notation of the paper, let

    (e)=[Z](i) (1) represent the canonical equations of the general -mesh. circuit under consideration. As in the paper, let the equation (1) be subjected to a Laplacian transformation of the ^-multiplied type. The equation is transformed to:

    \Z{p)](D = (E)+p[L](i*)-[S]W) (2)

    Now let the transformed current and voltage matrices be subjected to a linear transformation of the type:

    [A ]() = (),



    where [B]~x and [ 4] are nonsingular matrices whose elements are complex numbers. If we substitute for (I) and (E) the expressions given in (3) and (4) into (2), we obtain :

    [Zip) ]\B ] ( / ) ,= (E) +p[L](i') - [S](q*) (5)

    If (5) is premultiplied by [A ], we obtain:

    [A ) \Z{p) ] [B ](/), = [A ]() +p[A ] [L](io) -[A}[S](q) (6)

    Now, if we let:


    we have :


    (I)t=[Z}rH(E)t+p[A}[L](io)-[A][S](q)} (8)

    If we choose the elements of |yl ] and [B ] so tha t [Z]t has the diagonal form, then (I)t may be readily computed from (8) and the inverse transform (i)t is then obtained by the usual operational methods. The actual currents are obtained from (i)t by the equation:

    (i) = [B](i)t ()

    In the method of the paper, the matrices [A ] and [B] were denned by the equations:





    where [C] is the fundamental matrix of the theory of symmetrical components. If [A ] and [B ] are specified in this manner, the matrix [Z] must have a certain symmetry for it to be diagonalized. The reduction of a given matrix to the diagonal form by a transformation of the type (7) is one of the fundamental problems of matrix algebra.

    In his discussion, Professor Lyon points out that the assumed symmetry of the matrix [Z] which was considered in the paper is more severe than necessary. He shows tha t the matrices of circuits tha t are coupled in a certain manner so tha t they have "ring symmetry" are diagonalized by the symmetrical-component transformation. The fact tha t the symmetrical-component transformation diagonalizes the matrices of networks having "ring symmetry" makes the method applicable to a much larger class of problems.

    I t would appear from this tha t an investigation of the sufficient conditions tha t must be satisfied by the elements of the impedance matrix \Z] of the general w-mesh network in order tha t this matrix may be diagonalized by a symmetrical-component transformation is in order. A brief t reatment of this question will now be given.

    The general transformation matrix, [C], of the theory of symmetrical components may be written in the form :

    [C] = 1 1 1 a-


    1 a ~ ( n _ 1 ) . . . a-^-\){n-i)


    where a = e

    Now, from the basic equation connecting the sequence currents and the sequence voltages, we have:

    [C](E)S=[Z][C](I)S (13)

    [Z]s is diagonalized by the transformation

    [Z].= [C]-MZ][C] (14)

    If we introduce a column matrix defined by :

    " l a-r

    Let us assume tha t only the r th sequence voltage is impressed on the system and tha t this produces only r th sequence currents. This is equivalent to stating tha t the matrix



    Then the assumption tha t the r th sequence voltage produces only r th sequence current leads to the matrix equation:

    )=[]() (16)

    These equations may be solved for the ratios of ET to IT and we h