7. Appendix: High-frequency impedance of the rectangular strip

At the end of the section on poles and zeros, we succeeded in obtaining a good approximation of the asymptotic impedance of the ellipse, without using its closed-form expression (17). The result was (24) but, instead of the correct coefficient 1.088... of (25), we obtained 1 and 1.044... by successive approximations. In this Appendix, we apply the same method to the rectangle. The analysis is, however, much more difficult, essentially because the matrix (28) is a full matrix whereas the one (56) for the ellipse was tridiagonal, so that only the first approximation will now be worked out. Consequently, the accuracy of the result cannot be assessed. Also, Silvester's numerical results cover too narrow a frequency range to provide an adequate verification. In spite of its limited significance, our result is the only one presently available; it can be improved by further research, and has been obtained by a method having its own mathematical interest.

As for the ellipse, the following analysis is based on the equivalent circuit, but could be translated into purely mathematical terms. Because the matrix (28) is not tridiagonal, the circuit is not a ladder network, but a general RL network with an infinite number of nodes. Since, however, terms in k-1 only occur on the diagonal, inductances only connect each node to ground and the interconnections between non-ground nodes are purely resistive. Finally, for k = , the sum of all elements in any row of (28) is zero, on account of (63) for u = 0. This means that the direct conductance from any node to ground is zero, so that the branch connecting node n to ground is the inductance of value 1/2n alone, as in fig. 4.

When the network is truncated at n nodes, by opening all resistances leading to further nodes, what remains is an n-node resistive network with an inductance from each node to ground, but with no resistive path from the first node to ground. The impedance from node 0 to ground is thus infinite at high frequency and (62) its principal value is due to the inductances alone, so that all resistances can equally be short-circuited. The resulting impedance is the parallel combination of the first n inductances, and this yields (36), as in the elliptic case.

In the complementary method of truncation at n nodes, all further nodes are shorted to ground. The resistances leading to further nodes then produce a resistive path from node 0 to ground, so that the impedance is resistive at high frequency, as in (35), and can be evaluated by open-circuiting all the inductances. This amounts to computing (27) for the matrix (28) truncated at order n and for k = . Note that the sum of the elements in each row is no longer zero, owing to the truncation, so that the matrix is non-singular. Finally, the evaluation of the truncated impedance is equivalent to solving (64) for z with

(65)

Note that the truncation of the Fourier series of coefficients Bi corresponds to the shorting of the higher nodes, whereas the Fourier series (63) appearing in (64) is not truncated, for this corresponds to the preserved resistive connections to higher nodes. By (63), the form of (64) modified by (65) is thus

(66)

In the original (non-truncated) form (64), originating from the boundary condition (45), the coefficients Bi and the impedance z were determined by making the latter independent of u. Because of the truncation, this becomes impossible rigorously, and zs can only be made constant at n points (the number of undetermined coefficients) and, owing to the quadrantal symmetry of (66), these may be all chosen in one quadrant, say the first. The quadrantal symmetry is preserved by choosing n equidistant points with intervals front each other and half that interval from the ends. The interpolation points are thus

For this classical trigonometric interpolation, the "d.c. component" zs of (66) is simply the mean value of the second term of (66) at the interpolation points:

(67)

Since the above derivation of (67) is rather indirect, we check that the same derivation yields the known value (35) in the elliptic case. Equation (55) with the reductions (65) then yields

and (67) is replaced by

(68)

which is indeed (35), by a known identity.

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