APERTURE ANTENNA

Love’s Equivalence Principle of Figure 3(a) produces a null field within the imaginary surface S. Since the value of the E = H = 0 within S cannot be disturbed if the properties of the medium with in it are changed, let us assume that it is replaced by a perfect electric conductor (σ =∞).As the electric conductor takes its place, as shown in Fig 3(b), the electric current density Js , which is tangent to the surface S, is short-circuited by the electric conductor.Thus the equivalent problem of Fig3(a) reduces to that of Figure 3(b). There exists only a magnetic current density Ms over S, and it radiates in the presence of the electric conductor producing outside S the original fields E1, H1.

Fig:3 Equivalence principle models.

The steps that must be used to form an equivalent and solve an aperture problem are as follows:

1. Select an imaginary surface that encloses the actual sources (the aperture). The surface must be judiciously chosen so that the tangential components of the electric and/or the magnetic field are known, exactly or approximately, over its entire span. In many cases this surface is a flat plane extending to infinity.

2. Over the imaginary surface form equivalent current densities Js ,Ms which take one of the following forms:a. Js and Ms over S assuming that the E- and H-fields within S are not zero.b. or Js and Ms over S assuming that the E- and H-fields within S are zero (Love’s theorem)c. or Ms over S (Js = 0) assuming that within S the medium is a perfect electric conductor

a1

a2

auxiliary potential functions :A and F

EA can be found usingMaxwell’s equation of with J = 0.HF can be found usingMaxwell’s equation with M = 0.

a3

a4

a5

a6

d. or Js over S (Ms = 0) assuming that within S the medium is a perfect magneticconductor.3. Solve the equivalent problem. For forms (a) and (b), above equations can be used. For form (c), the problem of a magnetic current source next to a perfect electric conductor must be solved above equations cannot be used directly, because the current density does not radiate into an unbounded medium]. If the electric conductor is an infinite flat plane the problem can be solved exactly by image theory.For form (d), the problem of an electric current source next to a perfect magnetic conductor must be solved.

RADIATION EQUATIONS

it was stated that the fields radiated by sourcesJs and Ms in an unbounded medium can be computed by using (a1)–(a6) where the integration must be performed over the entire surface occupied by Js and Ms.

for far-field observations R can most commonly beapproximated by

5a

5b

6

6a

7

7a

In the far-field only the θ and φ components ofthe E- and H-fields are dominant

8a

8b

8c

8d

Combining (8a)–(8d) with (9a)–(9d), and making use of (6)– (7a) the total E- and H-fields can be written as

9a

9b

9c

9d

10a

10b

10c

10d

10e

10f

The Nθ, Nφ, Lθ, and Lφ can be obtained from (6a) and (7a).

11a

11b

Using the rectangular-to-spherical component transformation (11a) and (11b) reduce for the

12a

12b

12c

12d

Figures 2(a) and 2(b) are used to indicate the geometry.1.Select a closed surface over which the total electric and magnetic fields Ea and Ha are known.

2. Form the equivalent current densities Js and Ms over S using (3) and (4) with H1 = Ha and E1 = Ea.

3. Determine the A and F potentials using (6)–(7a) where the integration is over the closed surface S.

4. Determine the radiated E- and H-fields using

3’. Determine Nθ, Nφ, Lθ and Lφ using (12a)–(12d).4’. Determine the radiated E- and H-fields using (10a)–(10f).