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# Simplified Mutual Impedance of Nonplanar Skew Dipoles

K.E. Schmidt
Department of Physics and Astronomy
Arizona State University
Tempe, AZ 85287

This note was published in IEEE Trans. on Antennas and Propagation, 44, 1298 (1996).

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### Abstract:

By including the point charges at the ends of the nonplanar skew monopoles with sinusoidal current distributions, the expressions given by Richmond and Geary are simplified. The simplified expressions satisfy reciprocity.

# 1 Introduction

Richmond and Geary[1] worked out the general mutual impedance of nonplanar skew dipoles with piecewise sinusoidal currents in terms of exponential integrals. They calculated the mutual impedance between two monopoles'' each with a filamentary oscillatory current given by

 I(x) = , (1)

where = j and xa and xb are the ends of the monopole with the current zero at xa and one at xb .

Since the current at the xb end is not zero, there is necessarily a point charge there which contributes to the electric field. Richmond and Geary neglected this contribution because it cancels with a similar term of opposite sign when monopoles are connected together to produce the continuous physical current. In this note, I show that adding these point charge contributions cancels one of the terms in the expression of Richmond and Geary. The result is that the Fik terms in Eq. 13 of reference 1 may be dropped since they cancel when summed over the terms making up the mutual impedance of the dipole. An added advantage is that the resulting monopole-monopole mutual impedance satisfies reciprocity so no additional bookkeeping is needed to keep track of source and observation monopoles. To demonstrate these statements I will follow Richmond and Geary and set up the system as they do, but keep the point charge terms in the fields.

# 2 Geometry

The two monopoles are defined to have ends at , , and , , where the current as in Eq. 1 is zero at the a ends and one at the b ends. The and axes are defined by the unit vectors along the monopoles,

 = , = , (2)
with the angle between these axes,

 cos() = , (3)

and d the perpendicular coordinate between the elements,

 = , d = ( - ) . (4)
The origin of the integration coordinates is where the z and t axes are separated by d,

 = + (5)

and z1 , z2 , t1 , t2 are defined similarly to Richmond and Geary as the coordinates along the and directions of elements 1 and 2 respectively.

 z1 = ( - ) , z2 = ( - ) , t1 = ( - ) , t2 = ( - ) . (6)
The signed lengths of the elements are

 d1 = z2 - z1, d2 = t2 - t1. (7)

# 3 Mutual Impedance

Richmond and Geary calculate the mutual impedance in the usual way by integrating the dot product of the current density and the electric field. The point charge contributions can be calculated similarly. An alternative method that displays reciprocity immediately and also simplifies the derivation of the expressions is given by writing the mutual impedance in terms of the scalar and vector potentials. The mutual impedance is

 Z = dtdz(- R)/R, (8)

where R is the distance between the source (z) and observation (t) points and q(z) is the time derivative of the charge density corresponding to I(z). That is

 q(z) = - + (z - z2), (9)

and similarly for q(t) .

Eq. 8 can be written in terms of exponential integrals. As in Richmond and Geary an E function is defined to be

 E(( + j)) = exp(j)dw (10)

which can be evaluated using the exponential integral E1 ; efficient and accurate algorithms are available for this function[2,3,4].

The mutual impedance expression Eq. 8 is evaluated as combinations of two kinds of integrals,

 - st = exp(stz2cos())E((R2 - st(t - z2cos()))) (11)
and

 dtdzexp(-(R+stt+szz))/R=[ - exp(- (st + szcos())t)E((R + sz(z - tcos()))) - exp(- (sz + stcos())z)E((R + st(t - zcos()))) + E((R + stt + szz + j)) + E((R + stt + szz - j))], (12)
where st and sz take values 1 .

R2 in Eq. 11 is the distance between the z2 point and the t point. The right hand side of Eq. 12 is evaluated at the four limits of the integration. The E function is defined above as being the integral from t1 to t2 , and this definition can be used for the last three terms of Eq. 12. In the first term, the roles of z and t must be interchanged; the first E function is the integral from z1 to z2 and the result at t1 is subtracted from the result at t2 . This reversal of roles comes from an integration by parts in deriving Eq. 12.

The mutual impedance of two monopoles is,

 Z = [ + stszexp((stt1 + szz1))(- 1)iE((Ri + stt + szzi + jsb))], (13)
where

 = d (cos() + szst)/sin() (14)

The first term in Eq. 13 is the interaction between the two point charges at the ends of the monopoles. Since it obeys reciprocity and separately cancels when the current is made continuous it can be dropped if desired. If analytic integrations are combined with monopole impedances calculated by numerically integrating the electric field over the current on a monopole element, the same terms in both the analytic and numerical integrations must be included so that the point charge contributions will properly cancel.

The Richmond-Geary expression, with the additional Fik terms, is obtained by dropping the function term in q(z) .

As an aside, the expression for the mutual impedance between an electric current monopole and a magnetic current monopole[5] can also be derived using just Eq. 12. This impedance is proportional to the volume integral of the magnetic current dotted with the magnetic field from the electric current. Writing the magnetic field as the curl of the vector potential and using the identity

 ( x ) = sin() (15)

the mutual impedance is proportional to

 sin()dtdzIm(t)I(z)exp(- R)/R, (16)

where Im is the magnetic current of the same sinusoidal form as the electric current term. The calculation of the derivative of Eq. 12 is straightforward and the result of ref. [5] is obtained.

# 4 Conclusion

By dropping the Fik terms of Richmond and Geary, the mutual impedance of two sinusoidal monopoles is simplified. The expressions now satisfy reciprocity. An explicit numerical example is given in Table I. The mutual impedance of two v-dipoles is calculated. The monopole mutual impedances are different for each of the four expressions used, but the total mutual impedance of the dipoles is independent of the method as required.

The efficiency improvement can be estimated by assuming that the computation is dominated by the calculation of the exponential integrals. Dropping the Fik terms will reduce the number of exponential integrals needed to calculate the mutual impedance of an isolated dipole from 36 to 32 per monopole pair, giving roughly a 10 percent increase in efficiency. For a large structure, each monopole needs to be evaluated with the current equal to one at each end in turn except for those monopoles at the ends of wires. In this case, the number of exponential integrals required per monopole pair is reduced from 40 to 32 or a 20 percent savings. For the special case where the monopoles are skew coplanar, d is zero and the number of exponential integrals required drops from 24 to 16 for about a 30 percent reduction in computational cost.

# Acknowledgments

I wish to thank Prof. Constantine Balanis for a helpful discussion. This work was supported by a computer time grant from the Cornell Theory Center.

## References

1
J. H. Richmond and N. H. Geary,Mutual Impedance of Nonplanar-Skew Sinusoidal Dipoles,'' IEEE Trans. Antennas Propagat. vol. AP-18, pp. 412-414, 1970.

2
F. L. Whetten, K. Liu, and C. A. Balanis, An Efficient Numerical Integral in Three-Dimensional Electromagnetic Field Computations,'' IEEE Trans. Antennas Propagat. vol. AP-38, pp. 1512-1514, 1990.

3
D. E. Amos, Computation of Exponential Integrals of a Complex Argument'', ACM Trans. Math. Softw. vol. 16, pp. 169-177, 1990.

4
D. E. Amos, Algorithm 683, A Portable FORTRAN Subroutine for Exponential Integrals of a Complex Argument'', ACM Trans. Math. Softw. vol. 16, pp. 178-182, 1990.

5
K. Liu, C. A. Balanis, and G. C. Barber,Exact Mutual Impedance Between Sinusoidal Electric and Magnetic Dipoles'', IEEE Trans. Antennas Propagat. vol. AP-39, pp. 684-686, 1991.

Table I. The mutual impedance in ohms calculated for two v-dipoles in free space. The planes of the dipoles are parallel and the midpoints of the dipoles are separated by .01 wavelength. Each dipole consists of two monopoles of equal length (.1 wavelength for one set and .2 wavelength for the other) at an angle of 90 degrees. The four monopole mutual impedances (labeled 11, 12, 21, and 22) are shown, as well as the final dipole-dipole result Ztot . Four calculations are compared. These use Eq. 13 without the first term point-charge point charge interaction; Eq. 13; and lastly the Richmond and Geary expressions. RG1 is the Richmond and Geary result from integrating the E field from z with the current from t. RG2 corresponds to integrating the field from t with the current from z.

 Z Eq. 13 w/o charge Eq. 13 RG1 RG2 11 34.60 , 116.18 4.64, -360.02 5.32, -15.33 7.69, 61.32 12 -29.92 , -153.15 0.04, 323.04 -0.63, -21.64 -3.01, -98.30 21 -29.92 , -153.15 0.04, 323.04 -0.63, -21.64 -3.01, -98.30 22 34.60 , 116.18 4.64, -360.02 5.32, -15.33 7.69, 61.32 Ztot 9.36 ,-73.95 9.36, -73.95 9.36, -73.95 9.36, -73.95

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