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Electromagnetic Theorems and Principles
Duality Theorem:
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Ex:
A very thin linear magnetic current element of very small length (l<<λ) is often used to represent the fields of a very small loop radiator. It can be shown that the fields radiated by a small linear magnetic current element are identical to those radiated by a small loop whose area is perpendicular to the length of the dipole. Assume that the magnetic dipole is placed at the origin and is symmetric along the z axis with a constant magnetic current of
Im=az Im
Find the fields radiated by the dipole.
Ans:
Since the magnetic dipole is the dual of the electric dipole,
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Uniqueness Theorem:
Whenever a problem is solved, it is important to know that the solution is unique.
Under what condition or what information are needed for a unique solution:
The important cases for this equation is satisfied:
Or
Image Theory:
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(Vertical dipole and its image)
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In summary,
(Electric and Magnetic Sources and Their Images)
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Ex: Vertical Electric Dipole
Refer to the following geometry
Find the far zone fields. Plot the power patterns for h=2λ and 5λ
Ans:
The far zone component of the electric field of the infinitesimal dipole of length l, constant current I0, and observation point P1 is given according to the dominant terms (βr>>1)
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The reflected components can be accounted for by the images
(since the reflection coeff. is unity)
The total field above the interface is sum of the incident and reflected components.
To simplify the expression for the total field,
For h<<r
And
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Or
Then the total field is the product of a field from a single source and a factor, that is a function of h and θ. This product is known as a pattern multiplication rule, and the factor is the array factor.
Power patterns for h=2λ and 5λ:
![Page 10: Çankaya Üniversitesiece311.cankaya.edu.tr/uploads/files/Lecture_Notes_9(1… · Web viewElectromagnetic Theorems and Principles Duality Theorem: Ex: A very thin linear magnetic](https://reader033.vdocuments.net/reader033/viewer/2022053014/5f137fdbd4345f178758c7ae/html5/thumbnails/10.jpg)
Reciprocity Theorem:
The positions of the electric and magnetic sources can be interchanged without affecting the fields.
Reaction Theorem:
The coupling effect in the previous equation is the basis for the reaction theorem.
Volume Equivalence Theorem:
The sources Ji and Mi generates the fields E0 and H0 in free space.
The same sources generates fields E and H in the dielectric with ϵ, µ.
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Substracting these equations from the former ones
Define scattered fields (disturbance fields) as
and
Define volume equivalent electric and magnetic sources as
which exists only in the region ϵ≠ϵ0 and µ≠µ0.
Then,
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Surface Equivalence Theorem (Huygen’s Principle):
Actual sources are replaced by equivalent sources.
a) Actual, b)Equivalent problem
Equivalent sources
The equivalent problem of b) is
(Love’s equivalence principle)
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Equivalence principle models, a) Love’s equivalent, b) Electric conductor equivalent, c) Magnetic conductor
equivalent.
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(Equivalent models for magnetic source radiation)
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Induction Theorem:
Where E, H = total fields in the presence of the obstacle.
E1, H1 = total fields in the absence of the obstacle.
Es, Hs = Scattered fields owing to the obstacle.
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On the boundary
or
The equivalent currents are
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(Induction Theorem for PEC)
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(Induction equivalent for Scattering by flat PEC of infinite extend.)
Physical Equivalent and Physical Optics Equivalent:
Considering the problem
(Physical equivalence for scattering by a PEC, a) Actual problem, b) Equivalence)
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or
and
or
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(Physical Equivalent of a flat PEC, infinite extend)
Induction and Physical Optic Approximations:
(Approximate Induction Equivalence)
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(Approximate Physical Equivalence)
Ex: A parallel polarized uniform plane wave on the xy plane, in a free space medium, is obliquely incident on a rectangular flat PEC as shown below. The dimensions of the plate are a and b.
Find the far zone fields scattered by the plate. Solve the problem using Induction and Physical equivalents.
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a) Actual problem, b) Induction Equivalent, c) Physical Equivalent
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where
In summary
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For θs=π/2, θs=θi
Physical Equivalent:
where
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In summary
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For θs=π/2, θs=θi