SURFACE IMPEDANCE OF AN INFINITE PARALLEL-WIRE GRID AT OBLIQUEANGLES OF INCIDENCE*

By G. G. MACFARLANE, B.Sc, Dr.Ing., Associate Member, f

(The paper was first received 10th May, 1946, and in revised form 26th March, 1947.)

SUMMARYA study is made of the scattering of a plane wave incident obliquely

upon an infinite plane grid of parallel equidistant circular wires. Asimple means is described of picturing the mechanism by which sidewaves are produced as the angle of incidence is varied. Exact formulaeand curves for the shunt reactance of the grid are given for the mostuseful case in which the spacing is less than a wavelength. Thus thereactance of the grid referred to the component of the wave impedanceof the incident plane wave which is normal to the grid, viz. 376-5sec 0 ohms, is

376-5d[lo

376-5dlog.

+ F(d/X,6)]

for a <^ d <^ A,

where d — spacing of wires

a = radius of each wire

6 — angle of incidence.

The correction term is plotted in Fig. 6.

The reflection coefficient is

r ••-• ~ 1/(1 + 2XgIZsQC 0), Zo = 376-5 ohms

(1) INTRODUCTIONThe contents of this paper were first described in a T.R.E.

report dated November, 1942. The theory of reflection of aplane wave at a grating of parallel wires had been discussed bya number of authors for the case of normal incidence, notablyby Wessel {Hochfrequenztechnik, 1939, 54, p. 62) and Ignatowsky(Annalen der Physik, 1939, 44, p. 369). The extension tooblique angles of incidence does not seem to have been published.

The grating gives rise to an inductive storage field in its imme-diate vicinity. It is useful to express this reactive field as aninductance shunted across a transmission line. A convenienttransmission line is obtained by inserting two parallel, infinite,perfectly conducting sheets 1 metre apart perpendicular to theaxes of the wires and by considering 1 metre width of sheet.Since the electric field is assumed to be everywhere normal tothese sheets, their presence does not alter the field. If the angleof incidence is 0, the wave impedance of the line is Zo sec d,%where Zo is the specific impedance of the medium.

For the case of normal incidence the grid is inductively reactivewhen the spacing of wires is less than a wavelength. At spacingof a wavelength the inductance is infinite; the grid transmits theincident wave without reflection. At spacing greater than awavelength part of the energy of the incident wave is scatteredin side waves; the grid is then equivalent to a shunt impedancewith resistive as well as reactive component.

The discussion will now be extended to the case in which theplane wave is incident at any angle, with the electric field vibratingparallel to the wires.

* Radio Section paper.t Telecommunications Research Establishment.t Cf. ScHfcLKUNOn-: "Electromagnetic Waves" (McGraw-Hill Book Co.), p. 254

(2) THEORYThe system under discussion is indicated in Fig. 1 where a

small section of the infinite grid is shown. The radius of wireis a and the spacing between wires is d. Let us assume

a < 0)and that a plane wave is incident at angle 9 to the normal tothe plane of the grid.

In the first instance we shall consider the production of sidewaves by the grid without reference to their amplitudes, and weshall show how these side waves arise out of the storage field ofthe grid as the angle of incidence is changed. We shall thengive the formula for the equivalent parallel impedance of thegrid, and discuss its properties when the spacing is less than awavelength.

(2.1) Side WavesThe difference of phase of the incident field at adjacent wires

is (see Fig. 1)kds\n9

where k = 2TT/A.

Fig. 1.—Grating of infinite parallel wires.

Consequently currents equal in magnitude but differing inphase by

kdsind (2)

are induced in adjacent wires. The waves, re-radiated by thewires, arrive therefore in the same phase in any plane whosenormal makes an angle tf/n with the normal to the grid, where

sin iftn = sin 6 ± nX/d n ~ 0,1, 2 (3)

Eqn. (3) clearly defines the angles of scatter when

Isin 9 ± n\ld\ < 1 (4)[ 1523 ]

1524 MACFARLANE: SURFACE IMPEDANCE OF AN INFINITE PARALLEL-WIRE

for then the distances of two adjacent wires from the planedefined by tftn differ by exactly n wavelengths. Thus the gridre-radiates waves only in certain discrete directions whichdepend on spacing and angle of incident according to eqn. (3).These discrete directions are analogous to the discrete frequencylines in the spectrum of a regularly repeated pulse. The infinitearray of wires is analogous to the infinite train of pulses. To

Fig. 2.—Directions of propagation of incident and scattered waves.

illustrate this refer to Fig. 2 in which scattered waves are shownfor the case

A/d - 1 < sin 9 < 1

The incident wave Uo excites the grid to scatter the followingwaves.

(a) Direct Waves.

U^ at angle 9n = 0 in eqn. (3)

U^Q at angle n — 9(b) Side Waves.

UsX at angle tjtx — arc sin (sin 9 — X/d)n = 1 in eqn. (3)

Ual at angle IT — fa

[Note: From eqn. (3) side waves always appear in pairssymmetrical about the plane of the grid.]

These are the only waves which are re-radiated by the grid assimple plane waves. Other waves, however, are produced andthey too are contained in eqn. (3). They are defined by thecomplex values of tpn which satisfy eqn. (3) and there is aninfinite number of them.

The complex scattering angles are found to be

tfin = TT/2 ± j arc cosh (wA/rf + sin 0) = TT/2 ± jVn

ifi'n = 37r/2 ± j arc cosh (n\jd — sin 9) = TT/2 ± jV'n

The meaning to be attached to a complex angle of scatter andthe reason why the angles occur in conjugate pairs will be clear if

is substituted for ift in the wave function

exp [jcot — jk(y sin ifj + z cos if/)]

of a scattered plane wave making angle ip with the normal, OZ,to the plane of the grid. It becomes

exp. (=p kz sinh Vn) exp (Jcot — jky cosh Vn)

which represents a plane wave travelling in the positive directionalong the grid normal to the wires with amplitude falling off

exponentially with distance from the plane of the grid providedthe positive/negative sign is taken for z ^ 0. Thus the posi-tive/negative sign of Vn in the scattering angle ipn corresponds to awave on the positive/negative side of the grid. These waves alwaysoccur in pairs in this case. For them surfaces of constant phase arenormal to the grid and parallel to the wires but surfaces of con-stant amplitude are parallel to the plane of the grid. These sur-face or evanescent waves are therefore confined to the immediateneighbourhood of the grid. They constitute a storage fieldvibrating 90° out of phase with the incident field.

We have thus seen how eqn. (3) gives not only the re-radiatedsimple plane waves but also the evanescent plane waves pro-duced by the grid. To picture the transformation of a pair ofevanescent waves into two symmetrical side waves it is helpfulto plot the points ijjn in the complex plane. This has been donein Figs. 3 and 4 for the case

d<X<2d

Points on the real axis OPLRM correspond to simple planescattered waves, while points on the imaginary axes Q'PQ andS'RS through TT/2 and 3TT/2 correspond respectively to surfacewaves travelling in opposite directions along the grid normal tothe wires.

For small angles of incidence only two points 9 and -n — 9lie on the real axis. The points TT/2 ± }Vn all lie on Q'PQ andthe points 3TT/2 ± JV'H on S'RS. This is shown in Fig. 3. As

Fig. 3.—Complex angles of scatter for 0 < 6 < arc sin (\Jd — 1).

6 is increased the points 6 and TT — 6 converge on P, the pointsTT/2 ± jVn recede from P along PQ and PQ' and the points3TT/2 ± }V'n approach R along SR and S'R. This general move-ment is indicated by the arrows in Fig. 3(a). When 6 hasreached the value arc sin (kid — 1) the points 3TT/2 ± jVn havereached the real axis at R. Upon further increasing B thepoints 0i = 3TT/2 ± jV\ diverge symmetrically from R along thereal axis. Thus the first pair of evanescent waves has becomea symmetrical pair of side waves at the critical angle

9\ = arc sin (Xfd - 1)

This case is shown in Fig. 4, the diagram of scattered wavescorresponding to Fig. 4(a) being shown in Fig. 4(6).

In the particular case under consideration a further increaseof 9 will not transform any other evanescent waves into sidewaves. However, if we had chosen a greater spacing some ofthe points ifjn would have been on the real axis for small valuesof 9. The general movement of points discussed above would

GRID AT OBLIQUE ANGLES OF INCIDENCE 1525

o1

1\-Q

TO

%+M

s'

90"

Zfl

Fig. 4.—Complex angles or scatter for arc sin (A/rf — 1)< 0 < w/2.

still take place as 9 is increased. The critical angle at whicha pair of evanescent waves 3TT/2 ± jV'n on S'RS becomes twoside waves on LRM is

9 = arc sin — 1)

and at which two side waves ifjn on OPL become a pair ofevanescent waves on Q'PQ is

9 = arc sin (1 — ii\jd)

One specially interesting case arises when

d<\<2d

and 9 = arc sin (A/2*/)

For then the two pairs of points

and = IT + 9, 2<IT - $

are equally spaced about P and R respectively. The side wavesU# and l/al are re-radiated symmetrically from the grid and,since they are equal in amplitude, they interfere to produce anHQX wave-guide wave. Be it noted, however, that although theresultant reflected wave is of the form H0l the resultant trans-mitted wave is not, since the latter is the sum of the incidentand the scattered transmitted waves.

(2.2) Equivalent Shunt Impedance of GridWhen a planar grid is to be used as a reflector, loss of power

in side waves must be avoided. Only points «/r0 = 9, v — 9must appear on the real axis. This will be the case for all anglesof incidence provided.

2d<\

and for angles of incidence in the range

0 < 9 < arc sin (Xfd — 1)

when A < 2d < 2A

This limiting value of 9 is plotted in Fig. 5.For a plane wave incident at angle 9 on the grid with the

electric field vibrating parallel to the wires the field impedancelooking normal to the grid is

ZQ sec 9

70°

a 60

$5Cf—

4 0 -

I 30°

I

.—_V_

\\

\

\

N

—_—

20°

lCf

005 06 07 08 09 10

Ratio of wire spacing to wavelength

Fig. 5.—Condition for the occurrence of side-waves. No side-wavesoccur for operating points which lie below this curve.

for incident, reflected, and transmitted waves. It will be shownin Section 3 that the shunt reactance of the grid referred to thetransmission line of surge impedance Zo sec 9 is

whereand

and

AS) =

sin 0+ = sin 9 -f n\jd

The term F(d/\,9) does not involve the radius of the wireand for spacing much less than A/2 is only a small correction.It has been plotted in Fig. 6 for values of d/X from 0-1 to 0-8.Thus the shunt reactance of the grid is substantially independentof the angle of incidence when the radius of the wire is smallcompared with the spacing and the spacing is small comparedwith the wavelength.

(3) APPENDIXDerivation of Formula for Shunt Reactance of Grid

The formula for shunt reactance could be derived by anextension of the method used by Wessel, but it would involveextensive use of Bessel functions. It can, however, be obtainedmuch more simply by means of the quasi-stationary* method,as we shall show.

The y and z components of a plane wave that is incident on• See Journal I.E.E., 1946, 93, Part IIIA, p. 703.

1526 MACFARLANE: SURFACE IMPEDANCE OF AN INFINITE PARALLEL-WIRE

0 10° 20° 30" 40° 50" 60° 70° 80° 90"Angle of incidence *-

Fig. 6.—Correction term F(d/A, 0).The parameter is <//>>.

the grid of parallel wires shown in Fig. 1, and which is polarizedwith <f parallel to the wires can be written:—

H'~ A exp. ( - jky sin 9 — jkz cos 9)

e'x = AZ0 sec

The field scattered by the grid may be expressed in the form:—+ 00

Hy =

+ 00

n exp. ( - jky sin tfjn ~ jkz cos iftn)

SeC i n C0S

for z > 0, wheresin j/rw = sin 9 + nXjd

(6)

(7)

The constants An are determined by the boundary condition thatthe resultant electric field component, &x, is zero at the surfaceof each wire.

If we assume that the radius of the wire, a, is small comparedwith the spacing of wires, d, and the wavelength, A, then thefield near the surface of a wire is nearly symmetrical about theaxis of the wire. Moreover, the boundary condition is satisfiedat every point on the wire if it is satisfied at any one point.

If we choose this point to be (y = 0, z = a) the boundarycondition is:—

(A + AQ) sec 9 exp. (— jka cos 9)

An sec i/»w exp. ( - jka cos ^ ) = 0 (8)

n = — oo

where the dash means that the term with n ~ 0 is omitted.For large values of n

sin j/fn ~ wA/rf

cos «/rfl ~ — i\n\X/d (9)Therefore

sec \jia exp. ( - jka cos if,n) ~ ( ^ ) exp. ( -

and eqn. (10) can be written

(A +

( 1 0 )

+ 0 ° ' r7 ^ 4 , (sin2 ijjn

- l ) - i exp. ( - /£« cos </rn

In the quasi-stationary method it is assumed that the con-stants An are independent of A. In this case it can be shownby a fuller analysis that this is true. We can therefore putA = oo in the formula for the scattered field Hy and obtain thevalues of An by solving the resulting magneto-static problem.The incident field becomes constant and drives equal currentsalong each wire. The magneto-static field is well known andis given in terms of the magnetic potential V and the streamfunction U by the formulae

U + jV = log sin iT(y + jz)ld*

(12)

(13)

From eqns. (6) and (7) we can express the magneto-static fieldas a sum of evanescent waves

00

Hy ( / = 0) = AQ + 2 V A n cos (2nnyfd) exp. (~ (14)

smce> f r o m symmetry, A_n — An.T h e c o n s t a n t s ^ can be found most easily by noting that,

since the magneto-static field of (11) represents an array ofcurrent filaments at y = d, ± Id, . . . ± nd, . . ., the ^-com-ponent of H is zero in the plane z = 0, except at the pointsy = ± nd. Therefore from eqn. (14)

= ,lim "bV, . flimy— d y , s ince c o s (—r—~by \ d

-> 1 as v -> 0

* Cf. SMYTHE: "Static and Dynamic Electricity" (McGraw-Hill Book Co.),1st Edn., p. 84.

GRID AT OBLIQUE ANGLES OF INCIDENCE 1527

Therefore= An = (15)

Although we now know the values of all the amplitudes Anand can find A from eqn. (11) we still require to know the sumof the series in eqn. (11) before we can use the results. Thefirst sum can be obtained very simply from eqn. (14) on puttingV - 0 and integrating with respect to z from a to /. Thus:—

exp.27Tfl/

-> ]But in the plane y — 0, V — TT/2 and

U — logE sinh

Therefore for large values, /, of z

U(l) ~ irl/d - log.2

r ^ AQI — lOg£2

(16)

(17)

• • (18)

On substituting from eqns. (17; and (18) into eqn. (16) andletting / -> oo we get

2\^An(d/nX) exp. ( - 2-nna\d) = k logs (d/lira)1

(19)n -•- 1

In eqn. (19) we have neglected the term AQa — ira/d ande sinh (jra/d) by (jrald), since aid is assumed to be small.

But the first sum from n = — oo to n — oo, omitting the termn — 0, of eqn. (11) is equal to twice the sum from n = 1 ton oo, which is eqn. (19). Moreover, since a < d, A, we canput a = 0 in the first term and in the second sum of eqn. (11)without introducing appreciable error. Eqn. (11) thus becomes:

(A 4- Ao) sec 9 + jk logg + jS'n = 0 (20)

where

S' =

andsin «/r± — sin 9 i nXfd

S'n can be rewritten more neatly as

where

and

m

sin sin

fl - 1)-* -

nXld

(20B)

On substituting the expression (20A) for S'n into eqn. (20) thereflection coefficient r is found to be

+ j(2d/X) cos 9[F(dl\, 9) + logE

where X is the shunt reactance and Z the equivalent line impe-dance.

If we refer the reactance to a line of impedance Zo sec 9 we getX = Z0(d/\)[\ogz(dl27ra) + FW\,9)] . . (22)

which is the result given in Section 2.2.The power transmission coefficient is:—

t = (1 - |r|2)lwhich for a < d < A reduces to the simple form

t=(2d/X)cos9\ogz(dl27ra) . . . (23)