6. Mathematical derivation

The mathematical formulation of the lateral skin-effect problem for a thin conductor will now be derived simultaneously for the elliptic section (fig. la) and for the rectangular section (fig. lb) with the common notation 2h for the thickness, h being given by (9) in the first case and equal to the constant b in the second. The linear current density is the discontinuity of the tangential component Hx of the magnetic field. By symmetry we have for a right-handed coordinate system:

(40)

On the other hand, the true current density (A/cm2) is

(41)

Finally, Lenz's law yields:

(42)

It is convenient to introduce the vector potential A which has only a z-component. We then have:

 ; 

(43)

whereas the voltage drop ZI along the conductor is

(44)

In (44), Ez is expressed in terms of Az by (41), (40) and the first equation (43); this yields:

(45)

On the other hand, by elimination from (40)-(4l)-(42), we obtain the Biot boundary condition

(46)

When the magnetic fields are eliminated from (46) by (43), the resulting relation shows that the derivative of (45) with respect to x is zero. Biot's condition is thus equivalent to saying that the impedance computed by (45) is independent of x. The form (45) of the boundary condition is to be preferred to (46), since it yields the impedance without additional effort. Finally, the problem amounts to solving the Laplace equation for Az, with the condition (45) on the conductor (i.e. for y = 0, x < a), and the prescription of the value

(47)

at large distance, corresponding to the vector potential of a filament of current I at the origin, with a coaxial return of large radius D.

We introduce the conformal transformation

(48)

yielding

(49)

The transformation is one-to-one with the restrictions

In the (x,y)-plane, the curves of constant v are homofocal ellipses (fig. 7); the segment y = 0, is the infinitely flat ellipse v = 0, and v increases outwards to infinity. The curves of constant u are the hyperbolae of fig. 7 but there is a cut along the segment v = 0, so that u is positive in the upper half-plane Re y > 0 and negative in the lower half-plane. The semi-infinite segment y = 0, corresponds to u = 0, whereas the segment y = 0, corresponds to .


Fig. 7. The conformal representation .

The expression of Az is of the form

(50)

where the first two terms give the principal value (47), because (48) yields

whereas the sum (extending from n = 1 to ) with undetermined coefficients An is the general harmonic function, vanishing at infinity, and having the appropriate quadrantal symmetry.

For y = + 0, and hence v = 0, u > 0, then by (49):

so that (45) becomes:

(51)

The coefficients An must now be determined so that (51) takes a constant value for all u (). For the elliptic section, the first relation (49) reduces to (18) for v = 0, and (9) becomes

,

(52)

so that the denominator in (51) simplifies to

(53)

With the notations (15)-(16) and (12), and with the substitution

Bn = -k An

(54)

(51) multiplied by (53) becomes

(55)

Replacing the product of cosines occurring in the left-hand side of (55) by cosines of sums and differences, one obtains a Fourier series, and its identification with 1 yields the infinite system

(56)

of linear equations in z and the unknown coefficients Bn.

Disregarding temporarily the first two equations contained in (56), one obtains the three-term recurrence relation

Bn-1 + Bn+1 = 2(1+n/k)Bn(n = 2, 3, ...)

(57)

which is very similar to the recurrence relation

(58)

for Bessel functions. For v = n + k, s = k, (58) shows that

Bn = C Jn+k(k)

(59)

satisfies (57), with C an arbitrary constant. In fact (59) is the solution of the second-order difference equation (57) in our case, because the other linearly independent (51) solution (involving a Bessel function of the second kind) is excluded on physical grounds since it makes all coefficients Bn infinite at d.c. Since (57) holds down to n = 2 and thus involves B1, (59) holds down to n = 1 and only two unknowns remain: the common factor C of (59), and the impedance z. These are determined by the first two equations (56) which have been disregarded. By solving these equations, and making use of the known expression for the derivative of a Bessel function:

(60)

(used for v = s = k), we obtain

(61)

and (17).

From the known expression (50) for Az, where An is deduced from (54), (59) and (61), we can compute Hz, by (43) and i by (40). This gives (19). The other form (20) is obtained when i is computed by (41) with the value of Ez, deduced from (44) where is ZI. This completes the proof of all the basic results, (17) to (21), for the ellipse.

The expansion (22) of (17) can be deduced from the three-term recurrence relations (57). In the equivalent network of fig. 4, they correspond to the Kirchhoff relations between the currents in branches incident to a common node. and the following electrical proof is therefore equivalent to a mathematical discussion of (57). Consider the ladder network of fig. 8, where the series admittances are denoted Y1, Y2, ... and the shunt admittances Ya, Yb, ... Denote by Vi the node potentials with respect to ground as indicated. If a unit current is injected at the input, elementary node analysis yields the linear system

(62)

for the node voltages, and the solution V0 is the input impedance. From a comparison of (56) and (62), it is found that the solution z of (56) is the input impedance of the ladder network of fig. 4 where the elements are normalized (k is taken as the complex frequency). Since the input impedance of fig. 4 is (22), we have indirectly obtained the continued-fraction expansion of the function (17).


Fig. 8. A ladder network.

For the rectangular section, h is constant in the denominator of (51). After multiplication by the known Fourier series

(63)

and substitution of (54), (51) becomes

(64)

The linear system resulting from (64) is analogous to (56) except that the matrix is now (28). This establishes (27).

[ Back to index ]