Samuel N Karp.

Diffraction by an infinite grating of arbitrary cylinders online

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NEW YORK UNIVERSITY

INSTITUTE OF IV ' TICAL SCIENCES

AFCRC-TN-57-974 L

ASTI A DOCUMENT AD 1 33799 M Wr^y PU«, New York 3, N. V,



/O^k. ^x r^ NEW YORK UNIVERSITY

tu I ^ I T ^ Institute of Mathematical Sciences



"^A^^^wvH-^ Division of Electromagnetic Research

^ C C C X A



RESEARCH REPORT No. EM-108



Diffraction of a Plane Wave

by a Unidirectionally Conducting Half-plane



S. N. KARP



CONTRACT No. AF 19(604)1717



^ AUGUST, 1957



^ ^

^



NEW YORK UNIVERSITI

Institute of Mathematical Sciences
Division of Electromagnetic Research

Research Report No. EM-3D8



Diffraction of a Plane Wave by a Unidirectionally
Conducting Half -plane

S.N, Karp



^■//'^^Ho



S. N. Karp



l^^l^TY^ /j(\^^^ - i*1 fZ,
where

(1) £; = X cos a + y sin oi,

(2) 1 = -X sin a + y cos a.

The screen is conducting in the ? direction only. The incident wave is specified

^-
by its magnetic field in the form



(3) ft„ =Ae



13^.



r



inc



where k is s prescribed propagation vector, whose x,y,z components are k, , kp, k_

2 2 2 2

respectively, with k, + kp + k_ = k . We assume the number k to have a small

positive imaginary part. The components of the vector r in the xyz system are
X, y, z respectively. The vector A is a prescribed constant vector. Of course
the vectors k and !\ are not independent^ from the divergence condition we have

(1|) k ' t = 0.

(The incident electric vector can be determined from the incident magnetic vector
via Maxwell's equations.) Our total electric and magnetic fields are denoted by

E, H respectively.

*, -icot .
A factor e is to be understood as multiplying all our fields.



- 3 -



We now turn to the formulation of the boundary conditions which are
intended to express the unidirectional conductivity of the screen. /According
to Toraldo di Francia (private communication) these are expressible as follows:



(5) E^ =



(6) Fh^I s H^(x,y,0*) - H^(x,y,0") =



(7)



[\]



z = 0,x >



J



Here the notation Ew means the C component of the total vector E. That these
conditions are reasonable is suggested by thEor successful employment in the
theory of a traveling wave tube constmcted of a tightly wound helical wire.
(See [2], for example).

In addition to the above conditions Toraldo proposes the edge condition



(8)



N ->°



at the edge of the screen. This condition, which is necessary in order to have

a unique solution, represents the vanishing of the currents at the ends of the

wires composing the screen.

To complete the specification of the fields to be determined, we have

to impose the conditions that (5f - E. ) and (H - H. ) are to be outgoing at
^ xnc mc

infinity. In view of our use of a complex k, it follows that these vectors
vanish at infinity.

In addition to the above conditions we have found it necessary, in

analogy with the theory of perfectly conducting screens, to require (cf. Meixner l_2j )

1 -/I 2

that our fields are less singular at the edge than — , where p = / x + z .

P

Of course, in the exterior of the screen, E and H must satisfy Maxwell's
equations



- u -



(9)



a) V / E - i^H
c



b) VXH = -^E
c



c) div E =

d) div H - 0,



3. Reduction to a scalar problem

We now reduce our problem to a scalar problem. To this end we note
that, from 9a)






_c_ a



aE

z



a7



while we know that



a^E,



az



a^E.



^'



^\ 2

arc *•



Using 9c) we infer that



(11)



an



c

ioi



eE,



aE,



_a, / 3 ^ -"t
avj. I ac §Y



a^,



a^E



^.k^E



34



n.



Since [e^] = by (5) and JE ] = by (?), this must also be true of
their derivatives in directions tangential to the screen, whence we infer that



(11a)



aH^
az^



0.



We recall, in addition, that [H^I = 0, u:f, equation (6)).

By the above arguments we have established that the wave function Hw
is continuous together with its derivatives through the screen. It follows that

it is a i*egiilar wave function everyw! ere except possibly in the neighborhood of

inc
the edge. The same is therefore true of h^ = PL. - H^ . The function h^. is also



- 5 -

outgoing at infinity. Such a function can be developed, by separation of
variables, in the form e multiplied by a series of Hankel functions of
integer and zero order. Our restriction on the order of magnitude of edge singu-
larity excludes the non-zero orders. Therefore h^. must be either a multiple of

^2^ (1)
e H^ (Kp) or it must vanish. We adopt the latter ansatz, in order to avoid

.^
the possibility of too large E fields which would otherwise be generated by

equation (9b). We shall show that a solution may be constructed on the assump-
tion h», = which will thus be justified. It is to be noted that the simpler
problem of reflection ard transmission of plane waves by a fully infinite
unidirectional screen can be shown to have this behavior.
Since we have h^ = 0, we deduce from (9d) that



(12)



ah.



ah.



= 0,



ai^ az

so that we can introduce a scalar wave function u(x,y,z) such that



(13a)



h^ = ^



8u

ez



(13b)



V, _ au



We can then express the quantity



e - E - E



mc



in terms of u via (9b), (13a) j and (13b). '/.'e find



(lU)



s



c a^u a^u
c f a^ul

La^azJ •



c

123



c



Li?



+ k'



^1



This font! is valid for a straip;ht edge, but the ensuing result may be plausibly
extended to any smooth edged screen, since it is a statement about behavior in
the small (p -> O).



- 6 -



Since our incident fields are certainly regular at the screen, the conditions
(5), (6), (7), and (8), become



(15)



(15a)



(16)



(17)



2
3 u . , 2 aro „inc



i?



+ k



. [ul = 0; i.e., [u] = M\)e^^ + B(»|^)e"^



^ for z =
X >



J



ecan



[u] =



6u
dz



at the edge



' for z = and x >



The natural procedure is to obtain the general solution for the values
of u on the screen from (l5), in the form of a particular solution plus an un-
determined complementary function. Using these values of u as boundary conditionSj
we determine the outgoing wave function u in all space, and compute \-r-\ . Then
we impose condition (17) and thereby determine the complementary function, where-
upon u is determined. This procedure will be followed in the ensuing section.



U. Determination of function u(x,y,z)

The concluding remarks of Section 3 constitute a program which is
applicable to any screen whatever, but of course we have not shown that a suit-
able complementary function, allowing the fulfillment of (17), always exists. We
shall now call in to play the fact that we are dealing with a half-plane in
order to simplify the matter further. To this end we assume a special form of
solution, based on the fact that the conditions of the problem depend on y only
via the incident wave. Accordingly, we assume that (as is the case for the
incident wave), all our field quantities have the form e times a function of



- 7 -



A ^2^
X and 2 alone. Thus we set u = u(x,z)e . Since all our fields are wave



functions, this implies that



(18) L| . i^ . (k2 . k^) ^ = 0.

dx dz



Now from (l5a) we have



(19) [u] = A(n)e^ + B(a)e-^.



Differentiating (19) with respect to K aixl *[ , and applying (16), we conclude
that

(20) A'(t ) = B^"^) = 0,

in view of the absurd result (e~ ^ = constant) which follows from the contrary-
hypothesis.



Now we note that if k is in the C direction, we have E =0 (since



,inc

E^"° . Ic = 0), and then the solution of our problem is, trivially, E = E ,
H = H^"*^. V/e exclude this case from what follows. Then, from (l5), we have

i(k,x + k„y) . .

(21) u^^ + k^u = De ^ ^ J D = -^E^°(0,0,0).



The general solution is
(22)



u(x,iO,y) = /-(^) e"^^ + B^(ri) e ^ + D^e



where the superscripts denote the value on upper and lower sides of the screen,

and where

(22a) D^ = D/(k^- (k^ cos a + kg sin a)^) = - ^ E^"''(0,0,0)/[k2-(k-LCOs a



* kpsin



a)'].



- 8 -



ik



2^

Now for a half plane as noted above, we have u = u(x,z)e so that we must

have

^ + i(k4-k y) ^ -i(k4+k y) ilc x

(23) u(x,+0) = A^(l|) e + B^(^) e ^ + Ilj_e ^ ,

where the right side of (23) must be independent of y. We differentiate with
respect to y and set the result equal to 0, and obtain



(2U) /^(^)co3 a + i A^(k sin a - k^) e^^ +



b' ("7)003 a - iB^(k sin a + kg) e~^^=0



ik£ -ik£
which implies that the coefficients of e ^ and e - must vanish. It follows

from (2U) that



(25)



± t r -I

U('T) = aT exp[i(-k tan a + kp sec a) "tj



+ -r

B"(*1) = b~ exp[i(k tan a + kp sec a)"^"]



where aZ and >:'. are constants. Substituting in the expression (23) for u(x,0)
we find

ik,x



(26)

where
(27)



^ + ic-,x + xCpX ik^x

u(x,10) = a' e + b' e + D^e ^ x >



c, = k sec a - k^tan a



(28) Cg = -k sec a - kp tan a.

Ifow our incident wave has a real angle of incidence, whence it follows
that k_ = k • Y where the direction-cosine y is less than 1 in absolute value.
Since k has a positive imaginary part, we can now infer that c, has a positive
imaginary part while c_ has a negative imaginarj'' part. The wave function u
should be finite at infinity, and consequently we must discard the term in C-



- 9 -



which is exponentially large at x = +00.

From the resulting simplification of (26) and from (20), we can now
conclude that [u] =0, unless k sin a = k^. We shall postulate that [u] ■
in any case. The E and II field so obtained is verifiable by direct calculation.

In view of equations (17), (16) and (26), and in view of the arguments
of the last paragraph, our boundary value problem for u(x,z)£u(x,y,z)e
can therefore be expressed as follows:

(29) ( -^ + j! + K^ K = 0; K^ = k^- k^ > .

(30) u -> at 00.

ic,x ik,x

(31) u(x,0) o a^e + D^e J x>0



(32) [llj ->o as X ->0* .



These equations suggest that we solve for u in the form



(33) u(x,z) = a,v(x,z) + D-|_w(x,z)

v/here

ic-x ik-x
(3U) V = e ; w = e for z = 0, x > 0,

and both v and w vanish at 00. V/e shall see that a solution can actually be
achieved by this ansatz.

After solving for v and w, we must compute ] v [ and [w 1 , and choose

Actually, if k sin a = kp the foregoing analysis merely shows that
[G] = ra-.lexp(ic^x) = [a."lexp(ii: cos ax). But, at a later stage of the
argviment, after applying the edge condition on jH 1 one could show the
necessity of Fa-l » 0, i.e. of [u] =0.



- 10 -



the constant a, so as to enforce the condition [u 1 = at the edge. Now it is

apparent that the function w is nothing but the scattered field produced by the

diffraction of the plane wave exp i(k,x + /Y. -k,z ) at the half plane z = 0,

X > 0. The solution of this two-dimensional problem is well known. On the

other hand, the function v cannot be interpreted in this way, since we cannot

form a two-dimensional plane wave (with real angle of incidence) which adopts

ic,x
the value e at z = 0, It is however possible to solve for v by a simple

modification of a method due to Lamb'- -^ . For the sake of completeness we also

give the determination of the function w by this method.

Let us first determine v. We define a fvinction T(x,z) by the equation

(35) T(x,z) = g - ic^v = e'"'^'' 4 (ve"''''^^).

x/7 ?

The function T must vanish at z =0, x > 0. Let p = /x + z ,
© = arctan — , < 6 i; 2n, Then a solution T obeying the boundary conditions
is

iKp

(36) Y = Yq . sin 9/2 j Y^ = constant.

We now solve the ordinary differential equations (35) for v and obtain



ic,x



(37) v(x,z) = e ■^'



I exp[-ic^x^l Y(x^,z) dx^ + f(z)



X =-00

o



Here f(z) is a constant of integration. Using the fact that Y is a wave function
in order to calculate



■M- 2 2 2 2

Note K - c, = -k (sin a - sin p) < 0, if we define sin p = k^A.



- 11 -






-00 -00



by partial integration, we find that the first term in our expression for v is
a wave function. Thus the function f(z) must be of the form



f^ exp iz f^- c^ + fg exp -iz )^- c^ ,



(where f, and fp are constants), if the whole expi^ssion (37) for v is to be

a wave function. However, one of these terms is exponentially large as

z -> +00, and the other is large as z -> -oo, provided k_ f^ k sin a. (If k^ = k sin a,

then the terms are constants). On the other hand, the integral tenn occurring in

(37) goes to zero as |z| -> oo, as can be seen by substituting the formula for

Y in the integral. Since we require that v -> at infinity, the above remarks

show that f- = f p B (whether or not k- ■= k sin a).

We have yet to determine Y . This can be done by evaluating v(x,0) for

° ic,x

x positive, from equation (37) j and setting it equal to e .To do so we first

note that for z = 0, x > 0, Y(x ,0) =0, while for z = 0, x^ < 0, we have



(38) Y(x^,0) = Y^



expFiKlx^lj



since 9 « n/2 in this range. Consequently we obtain

^ -ic^x^ exp|iK|xj]



v(x,0) = Ye ^



ICtX

0~'



1C,X

dx = e
o



X =-00

o



whence it follows that



- 12 -



(39)

and

(39a)



^0= 1'



{ exp -i(K+c, )x 1
I o



dx



/^



'1 -inA
— e '



-oo



v(x,z) = exp



ic-,x



X

J exp(-ic^x^)Y(xQ,z)dx J



X =-CXD
O



We determine w similarly, by introducing the function






The function w must represent outgoing waves at oo. Of course, as is well
known, in certain regions of space, these outgoing waves will be partly plane
waves. In fact, w is the negative of that scattered wave which corresponds to

) ik-x + iz / K - k^



the incident wave exp ik->



, and therefore w will contain



shadow- forming waves or reflected waves, in the appropriate regions of space



■K-



indicated by geometrical optics. However, such waves depend on x in the form

ik-x
e . Consequently ^ = w - ik,w -> at oo. Thus



(ID)



j25=0.



^iKp
1^



3xn ^ , 0^



constant.



Solving for w we obtain



(Ul)



ik^x



w " e



^ J exp[-iJc^xJ 0(x^,z)dx^

-00



where the homogeneous exponential solutions which are otherwise possible are



The shadow-forming wave is the ne gativ e of the incident wave while the reflected
wave has the form exp(
as remarked above.



ive is the ne gativ e of the incident wave while the reflected
)(ik-x - iz )iC -k^ ). Thus both are proportional to exp(ik^x),



- 13 -



rejected as violating the outgoing condition. The constant is determined
by the same procedure that we employed in the case of the function v. We find

o
(U2)




-^1^0



exp



-00



iK|x 1
' o'



/K^



o' 4



dx



^



!i g-inA .



We have now determined the functions v and w, and we have
u(x,0) = a-v + D,w.
The constant a, is now to be determined by calculating U— and requiring it



to vanish as x



0^.



is useful to notice that the



Faul

In order to study the behavior of ^ it i;
wave functions (v - exp i(c^x + /K -c, z) ) and (w - exp i(k,x + jr K -kZ^ z) )
vanish on the screen, and are finite at the edge. Consequently they can be
developed near the edge in Fourier series of the form



^ c^ J^/2 (Kp) si^ § 9.
n=l '



In these series the terms of odd index have discontinous z derivatives at the

screen. The leading terms are 0( ^'r sin 9/2) and give rise to a discontinuity

in r- and ■5— of order of magnitude r~ . The remaining terms of L— tend to
oz oz ^ '=' JjBzJ

zero. Therefore we must choose the constant a^ so as to cause the leading terms

3v Sw

of a- ^ and D, ^ to cancel each other. This is achieved by selecting



ih3)



¥0 , /^"^



as is evident from the fact that



a^*¥ = ^l(i - i^l)^^^l(i-^l)^



- m -



then vanishes, whence a-. ^ + D^ ^ must be finite, whence in turn the odd
index terms in a.v + D, w must vanish more rapidly than y'r sin 9/2.

We have now detern.ined the constant a, and the functions v and w. (These
functions can be expressed as Fresnel integrals, but this is not necessary for
our purposes.) ^fe now observe that via equations (UO) to (U2), (39) and (39a),

and (22a), and (33) and (U3) we have a determination of u(x,z), and of

^2^
u(x,y,z) = u(x, z) e . We then obtain h^ and h from equations (13) and

(13a), Now, using (9b) we can determine the scattered field e, and our solution
is complete. This solution is collected and given explicitly in Section 6. Inspec-
tion shows that it is valid whether or not k sin a = kp. Thus that restriction
is irrelevant, as noted earlier.



$, Determination of the far field

It is of interest to supply a determination of the far field. In order

A ^2^

to do this we note that, since u(x,y, z) = u(x, z)e , we expect a representa-
tion of the far field to have the standard form,

(hU) M ^ e ^ i U(9) + D expTik^x + ik |z| m(9)

vrhere m(9) is r step function (whose form we indicate subsequently) which
represents the shadow- forming and reflecting properties of the screen. Our
ob.ject is now to determine U(&). Once U(9) is deterrained, we can obtain the
corresponding representations of the cartesian components of e and h. Now we
note that

(U5) u = a,v + D^w,

so that to obtain U(9) we only need to know the amplitude functions V(9), W(©)



- 15 -



such that



(U6)



^



iKp
2 Y{»)

iKp



w



/P



W(©) + in(e) exp ik,x + ik^lxl



' as p ->oo with fixed &.



\-Ie now must specify the function m(©), and then determine V(€)) and W(B). The
function m(9) is readily obtained from geometric considerations since the term
it affects represents the value of w according to geometrical optics. We have

/7^



(U8) m(9) =1, < 9 arctan r^ = arctan

^1



^



= 9
- o



(the shadow region).



m(9) =0, 9 < 9 < 2n - 9^



m(6) =1, 2n - 9^ < 9 < 2n.



(the reflection region).



The lines 9 = 9,G = 2tt-9 are celled 'shadow lines' in the literature. On
o o

these lines the function w(6) becomes infinite, as we shall see, (cf. (5l)),
and the asymptotic representation (U6) fails.

These results follow from the fact that at infinity, in the successive
angular regions indicated in (U8),w must successively c?.ncel the incident wave
to form a shadow, leave the incident wave unperturbed, or provide a reflected
wave. The functions V(9), W(&) are now obtained as follows. >fe have



(U9)



dv J

^ - ic^v =



iKp



° y?



in 9/2



sm



^ iKp

|2 _ il, „ = Y e g^^ ^/2.

6x 1 o



/P



- 16 -



We express the differentiations in polar coordinates, apply then to the

asjTnptotic representations given in equation (U6), and re.iect terrr^s of higher

-1/2
order than p . Tlien vre obtain

i(K cos 9 - c^) V(&) = Y^ sin I

i(K cos & - k ) W(») = 0Q sin •^ .
Consequently

($)) v(e)



Y sin 9/2





(51)



W(&) =



iCl^ cos & - c, j



sin &/2
i(K cos e - k^)



From equation (U5) we deduce that U(&) = a^V(9) + D^W(e), whence

^^0



(52)



*



U(&) =



in 9/2 ,



sm



^li^o



K cos & - C-, K cos 9 - k.



for any angle © which is bounded away from the angles 9 , 2n - 9 , where
9 = arc tan k_/^ . The far field fonn of hy. and h^ can be obtained from the
above equations by the asymptotic differentiation process already employed.
The relevant formulas for differentiating the terms involving u(9) and v(9)are



J. = iK sin ej ^ = JJ + D^in(e)ik- -^ exp ik^x + k |z| .



/P
These are the results we aimed at in this section.

6, Conclusion

A solution has been obtained for the diffraction of the electromagnetic

plane wave

ik«r



-inc ^ f gXk.r ^ - exp[i(k^x + l:^ + k^z)]



by a screen which occupies the semi-infinite plane x > 0, z =0, and which is
perfectly conducting in the direction of the 4 axis, where

K = X cos a + y sin a

>1 = -X sin a + y cos Qj

a being a prescribed angle, < a < 5, under the restriction that k2-k sin a
and k- are not simultaneously zero. The field obtained is a solution of the
boundary value problem formulated in Section 2, pp. 2, 3. We conclude that
the idealization embodied in this boundary value problem makes sense. The
solution we obtained is defined by



H. . , = H. + h
total inc



E, .T =-^VXH
total 10)



whei>e



with



h = (hg, h^, h^)






For u we have the formula
u = uCxjz; e



- 18 -



^2^ r 1



-0



If 1 o

e I e

-00



-ic^x^ iKp

" ® sin 9/2 dx^



/P



where



ik,x r -xk^x iKp



-00



sin &/2 dx

yp



c^ 5 k sec a - kp tan a



1 -



=

^0 -



ICO

c



E^"yp - (k^cos a + kgsin a)^j



+k




1 ^-inA



Simple asymptotic representations of the solution which are valid for large p

are indicated in Section 5. Note that if H is in the ? direction then

Ev = 0> whence D, = 0, u = 0, and h = 0, so that the screen is transparent to

a plane wave whose incident magnetic vector is parallel to the direction of

conductivity.

In our analysis we have excluded the case where k cos a ~ ^ -i » ^ ^^ '^ " kp>

k_ = 0, i.e. k in the direction of the C axis. However, in this excluded esse

"^inc -inc
we have E^ =0 and IC = Oj since the incident S and H vectors are orthogonal

to the direction of propagation. Then we again have the trivial solution

E = E , H = H , since the boundary conditions are then trivially fulfilled.



Our solution is therefore complete.



- ]9 -
References



fl] G. Toraldo dl Francia - Electromagnetic cross section of a small circxJ.ar

disc with unidirectional conductiTityj II Nuovo
Cimento, Series X, 3, 1276-8U (1956).



[2] J. R. Pierce
[3] J. Meixner



[U] H. Lamb


1 3