Antenna

1. APERTURE ANTENNA

2. 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.

3. Fig:3
Equivalence
principle
models.

4. 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

5. a1
a2
auxiliary potential functions :A and F

6. EA can be
found using
Maxwell’s
equation of
with J = 0.
HF can be
found using
Maxwell’s
equation with
M = 0.
a3
a4
a5
a6

7. d. or Js over S (Ms = 0) assuming that within S the medium is
a perfect magnetic
conductor.
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.

8. RADIATION EQUATIONS
it was stated that the fields radiated by sources
Js 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 be
approximated by
5a
5b

10. 6
6a
7
7a

11. In the far-field only the θ and φ components of
the E- and H-fields are dominant
8a
8b
8c
8d

12. 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

13. 10a
10b
10c
10d
10e
10f

14. 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

15. 12a
12b
12c
12d

16. 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

17. 3’. Determine Nθ, Nφ, Lθ and Lφ using (12a)–(12d).
4’. Determine the radiated E- and H-fields using
(10a)–(10f).