Some relations between the radiation patterns in the two main planes and the whole radiation pattern. *

Transcription

1 Some relations between the radiation patterns in the two main planes and the whole radiation pattern. * abstract: In theoretical studies, the far fields radiated by antennas are derived by using a radiation integral. In many cases, only a cut in the two main planes called -Plane and H- Plane, is of interest. Outside these two main planes, the behaviour of the field is generally not known. It seems that no detailed study has been made in order to determine whether a relation exists between the,h Planes and the whole radiation pattern. The aim of this paper is to try to link the far field in any direction of space to the far field which is present in the two main planes, and to analyse the conditions under which this pattern reconstruction is possible. I INTRODUCTION The far field radiation from current distributions is obtained by using a well known radiation integral. Following the notations of FIG, a particular form of this integral may be given as followed: Figure : Representation of a current element on a plane rectangular surface, in polar coordinates. *P. VAUDON B. CKO P. BRACHAT : annals of telecommunications,5,n 5-6,996

2 jkη (P Ψ(r ((M U U.e 4 Λ Λ π S jk.om ds jkr with Ψ (r e and η the wave impedance. r P represents the observation point, which is assumed to be far away from the radiation sources. In the general case, the current distribution is a volume distribution. However, we shall restrict this study to the case of a current distribution on a rectangular plane surface. M is a point on this plane surface, and (M represents the vector of current density. Analytical formulations of eq( are obtained in a very classical manner. The details are not given, since it may be found in most of antenna books. If we refer to FIG( where it has been assumed that the currents where lying in the x,y plane, we obtain: ( jkη (P Ψ(r ((x,y U U.e 4 Λ Λ π S jk(sin.cos.+sin.sin.y dx.dy ( The current vector is decomposed into its two vector components: and : (x, y (x, y + (x, y (x, y.u + (x, y. U (3 where U and U represent the unit vectors in the x and y directions. It remains to evaluate the double vector product, and one more time, since it constitutes a very classical result, we omit further details. As a last simplification we can see that the term ahead of the integral ( or ( is a constant for a fixed frequency and a fixed distance r. This term doesn't play any role in the next sections and it will be omitted. Then, we rewrite eq( under the form: (, + S S (x,y.e (x,y.e jksin(cos.x+ sin.y jksin(cos.x+ sin.y dx.dy.( cos.cos.u dx.dy.( cos.sin.u + sin.u cos.u where and are the polar co-ordinates of the observation point P and vectors of the polar co-ordinate representation. U, U (4 the unit At this stage, it seems difficult to further progress on the analysis of this relation without introducing a new assumption: the x and y currents can be separated with respect to the x and y variables, so that we are authorised to put:

3 3 (x, y (x. (y and (x, y (x. (y (5 q(5 may seems to be a very restrictive case. However, most of antenna radiation mechanisms are based on stationary phenomena, and to solve these problems, implicitly or explicitly, the condition of variable separation (5 is assumed, then we think that it still applies in a lot of cases. This property allows to completely separate the surface integral of eq(4 and to rewrite it as: (, + (x.e (x.e jksin.cos.x jksin.cos.x dx. dx. (y.e (y.e jksin.sin.y jksin.sin.y dy.( cos.cos.u dy.( cos.sin.u + sin.u cos.u (6 part. This formulation will be at the basis of the analysis which will be given in the next II TH FOUR MAIN RADIATION PATTRNS By observing eq(6, we see that the component along U is obtained in summing the contribution of two terms: the first term represents the radiation of the currents, and the second term, respectively, the radiation of the currents. The same remark can be made for the component along U. Let us consider a cut in the plane 0, the angle varying from corresponding field is given by: (,0 π (, (x.e jksin.x dx. A second cut in the plane (xdx. (y.e (ydy.( cos.u + (x.e jksin.x dx. π allows to obtain a second relation: jk sin.y dy.(u + (xdx. (y.e jksin.y π to (ydy.( U dy.( cos.u π +. The (7 (8 We are now trying to determine whether the only knowledge of the radiation patterns (7 and (8 is a sufficient condition to retrieve the whole radiation pattern (6.

4 4 To answer to this question, we shall suppose that the four patterns which are presents in eq(7 and (8 are known, and they will be named in the following way: dx.a.( cos.u jksin.x.0( (x.e with A (y dy (9.0, π / ( ( (x.e (y.e jk sin.x jksin.y dx.b.( U dy.c.(u with with B (y dy (0 C (x dx (. π / ( (y.e jk sin.y dy.d.( cos.u with D (x dx ( For a given current distribution, A,B,C,D are constant. The subscript,,0,π / have been introduced with the following meaning: -, : to recall that the or the axis or the axis. - 0,π/ : to recall that the 0 or π / π /. field is due to the current flowing along field is taken in the plane 0 or III TH RCONSTRUCTION OF TH WHOL RADIATION PATTRN. It remains to link the four relations (9,(0,(,( with the representation of the field in any direction (, given by eq(6. A detailed observation of these equations suggests to put in (9 and (0 : sin sin. cos (3 cos sin.cos (4 The positive square root has been retained for cos, since is assumed to vary in the interval [ π/, + π/ ]. A similar observation leads to a corresponding relation for ( and (:

5 5 sin sin. sin (5 cos sin.sin (6 By using this variable change, it becomes possible to retrieve the four integrals of eq(6 which are necessary to recompose the total field. From eq(9 to (, we obtain a new set of four equations: jk sin.cos.x (x.e dx A..0 sin (.cos (7 (x.e jk sin.cos.x dx.0 B ( (8 (y.e jk sin (.sin.y. π / dy C (9 jksin.cos.y (y.e dy D.. π / sin (.sin (0 A particular care must be exercised in deriving the expressions (7..(0, because the four constants A,B,C,D must be different from 0. This physically means that the sum of the elementary currents x and y along any direction or must not be nil. We may note that if one of these constants is equal to 0, this implies that one of the four radiation patterns (9..( is nil for all values of. This condition forbid the use of the expressions (7..(0 in the case of a symmetrical distribution of opposite currents for example. This point brought some restrictions to the general formulation which comes from (6,(7..(0: (, ( (0. (. sin (. sin. π / (.cos. π / (.sin.( cos.cos.u.(cos.sin.u + sin.u + cos.u ( with

6 6 Arc sin(sin.cos Arc sin(sin.sin ( To use this reconstruction formula, it is supposed that the four radiation patterns (9..( are exactly known in there real and imaginary components, or in their amplitude and phase components. This is adequate for theoretical applications. However, for practical applications, and particularly for measurements, generally only the modulus of the field is known, and we may ask what possibilities are offered by eq( under this new condition. It becomes clear that no reconstruction is possible from a general surface current distribution. But, if we restrict our analysis to a rectangular sheet of currents which are altogether flowing in the same direction along the axis for example, then we obtain a simple relation for each far field component: ( (. (.0. π/,.cos. cos.0(0. sin.cos (3 ( (. (.0. π/,. sin.0(0. sin.cos (4 These two expressions shows that the knowledge of the two main patterns in the plane 0 and π/ allows to determine the field in the whole half space. In this case, in any direction (,, except the plane 0, a simple relation links the and the component; namely: cos.cot g (6 IV SOM NUMRICAL AMPLS No numerical justification is needed for eq( since it has been established in a rigorous manner. However, it would be interesting to observe how this formula works on a practical example. We have chosen a current distribution such as:

7 7 (x, y C (x, y C m. π nπ.sin x.cos y a b m. π nπ.cos x.sin y a b (5 (6 where Cx, Cy, a, b, are constants and m,n are integers. These currents appear on a rectangular patch antenna with εr and with a height h<<λ. a and b are the patch dimensions, and m,n characterise the mode on which the antenna works. One corner of the patch is put at he origin, and the currents spread from x0 to xa in the x direction, and from y0 to yb in the y direction. As a first remark, we can see that there are only a limited number of modes which authorise a reconstruction pattern using eq(, namely: m odd, n0 or n odd, m0. In all other cases, the symmetrical and opposite distribution of currents doesn't allow a correct pattern reconstruction. From eq(5 and eq(6, we have derived the analytical far field expression. Omitting further details, we give the result in the following formula: n.v. β m.w. α (, D cos(.sin( + cos(.cos( U b a n.v. β m.w. α + D cos( sin( U b a (7 with : j.v.a m j.w.b n [ e.( ][ e.( ] D j (8 [ α² V² ][ β² W² ] and : V k.sin(.cos(, W k.sin(.sin(, m. π α, a n. π β (9 b Then we have constructed four files containing the four radiation patterns which must be known (Only two were necessary in this particular case. These data files were sampled each degree.

8 8 Next, we have compared the results obtained in any plane using two methods: eq(7 - the analytical expressions derived from eq(5 and eq(6, that is to say - The field obtained by the reconstruction formula ( using the four data files. The resulting curves are plotted on FIG for 35. We can see that the curve obtained with eq( is very close to the analytical curve. If we don't recover exactly the same result, it's because the exact angle computed from eq( or eq(3 is not available in the data files, and rather than making an interpolation, we have taken the nearest point, which corresponds to the nearest degree. *0-3 4, , D egrés -4, , ,0-0 Figure : Comparison between the exact radiation pattern and the reconstructed pattern with 35, M 3, N 0, in the case of a patch antenna. (exact :, reconstructed : As previously mentioned, since we can only consider eq(5 and eq(6 for m0 or n0, this implies that only one direction of current is possible: x or y, and the test has been presented under this condition. If we want to test the complete formula (, we have to put in eq(5 and eq(6 current distributions which have not a destructive symmetry. This can be made for example by giving to m and n non integer values.

9 9 ven if we may ask on the physical meaning of these non integer values, there are no mathematical problems in their introduction in eq(, and this is the point which interests us. The results obtained for m. and n.5 are plotted on FIG 3. They confirm validity of eq( in the general case. *0-3 3,00-03,00-03, , ,00-03 D egrés -, ,00-03 Figure 3 : Comparison between the exact radiation pattern and the reconstructed pattern with 40, M., N.5, in the case of a patch antenna. (exact :, reconstructed : V TWO APPLICATIONS OF TH DSCRIBD MTHOD The reconstruction formula ( may seem to be restrictive because of the choice of equation (. However, more than the result itself, we want to show that the derivation which has been described applies in some other cases. As a first example, we shall consider the use of the Huygens - Fresnel principle. This principle is often the only manner allowing to obtain analytical formulations for the radiated field. When it is applied to a plane surface, with the exact knowledge of the fields on this surface, it can be shown that the far field may be established using three equivalent representation:

10 a by using the equivalent electric current NΛ H and the equivalent magnetic current M ( NΛ b by using twice the equivalent electric current ( NΛH c by using twice the equivalent magnetic current M ( NΛ In practical cases, only approximated fields are known, and the plane surface is always limited; then we must keep in mind that the obtain results are not exact. The previous relation ( applies readily to the case b where only the electric currents are considered, but it do not apply to the cases a or c. If the reader want to obtain a reconstruction formula in theses cases, and also in other different situations, he must use step by step the derivation which has been made, and introduce himself the modifications which are imposed by the studied problem. 0 To illustrate these considerations, we are going to present an example which refers to the situation a. Let us consider a rectangular aperture equal to the plane surface which is represented on FIG. Also, let us consider, in this aperture, an electromagnetic field with the following components: xy (, xy (,. U x Hxy (, xy (,. U y (30 η For a practical purpose, this field configuration may be observed in an open wave guide working on its fundamental mode. The equivalent electric and magnetic currents are then given by: NΛH x y U x η (,. and M ( NΛ ( x, y. U y (3 and the corresponding radiated far field is: jkη (P Ψ(r ((ΛU ΛU + (MΛU.e 4π η which leads to: S jk.om ds (3 jk jk.om (, ψ(r (x,y.e ds.( + cos(.(cos( U sin( U (33 4π s Then, omitting the constant factor and taking again step by step the relations which have been presented in the sections I and II, we obtain the corresponding reconstruction formula:

11 .0 (.. π / ( (, ( + cos.(cos.u sin.u (0.( + sin.cos ( + sin.sin.0 with : (34 Arc sin(sin.cos Arc sin(sin.sin (35 As a second example, we shall consider a sheet of currents which radiates over a perfect ground plane: this is the case for patch antennas with the air as substrate. Thus we have tried to extend the formulation ( to take into account the ground plane effect. To solve this classical problem, we use the image principle. This leads us to introduce in the radiation integral ( a multiplicative factor equal to: SIN(kh.cos (36 where h represents the height of the current sheet over the x,y plane. Then, we derive the new relation corresponding to this configuration, in a similar manner than in the previous case. The final result is given by eq( multiplied by the following factor: sin(kh sin(kh.sin(kh cos sin ².cos ².sin(kh sin ².sin ² (37 We have made several computational verifications of eq(37. All confirm the validity of this formulation which takes into account the presence of a perfectly conducting ground plane. These two examples show that the reconstruction formula depends on the jk.om function of q and f which follows the radiation integral x.or.y (x,y.e ds in the expression of the far field (Cf eq(4 and eq(33. Then, for each particular problem, it is necessary to derive a new relation. S

12 VI CONCLUSION We have studied the far field obtained from a rectangular current distribution. Assuming that these currents were separable in the x and y directions, we have shown that the only knowledge of four radiation patterns was a sufficient condition to determine the field in any direction of space. However, no one of these patterns must be nil, elsewhere the reconstruction is not possible. We have proved that this method was still applicable in connection with the HUGNS-FRSNL principle and also when the sheet of current was located at a height h above a perfectly conducting ground plane. In these two particular cases, we have given the corresponding formulations.