prof. david r. jackson dept. of ece fall 2013 notes 26 ece 6340 intermediate em waves 1

Prof. David R. JacksonDept. of ECE

Fall 2013

Notes 26

ECE 6340 Intermediate EM Waves

1

Equivalence Principle

Basic idea:We can replace the actual sources in a region by equivalent sources at the boundary of a closed surface.

Keep original fields E , H outside S.

Put zero fields (and no sources) inside S.

E , H

a

b

ANT

S

E , H

2

Equivalence Principle (cont.)

Note: (Ea , Ha) and (Eb , Hb) both satisfy Maxwell’s equations.

Eb = 0 Hb = 0

No sources

Ea = E Ha = H

3

The B.C.’s on S are violated.

Introduce equivalent sources on the boundary to make B.C.’s valid.

n̂

a

b

esJ

esM

ˆe a bsJ n H H

ˆe a bsM n E E

Equivalence Principle (cont.)

4

Zero fields

Original fields

Outside S, these sources radiate the same fields as the original antenna,

and produce zero fields inside S.

This is justified by the uniqueness theorem:

Hence ˆ

ˆ

es

es

J n H

M n E

Equivalent sources:

Zerofields

esJ

esMZero

sources

S

Equivalence Principle (cont.)

Maxwell's equations are satisfied along with boundary conditions at the interface.

5

Note about materials:

If there are zero fields throughout a region, it doesn’t matter what material is placed there (or removed).

Equivalence Principle (cont.)

This object can be added

Zerofields

esJ

esM

r

(E , H)

S

6

Scattering by a PEC

sE

(E , H)

Source

iE

0tE

PEC

S

sJ

i sE E E i sH H H

7

Scattering by a PEC (cont.)

ˆ ˆ 0

ˆ

es t

es s

M n E n E

J n H J

S

Source

0 0,

esJesM

(E , H)

(0 ,0)

Equivalent Sources:

n̂

sJ (E , H)

SourcePEC

S

We put zero fields and sources inside of S, and remove the PEC object.

8

Conclusion: The conductor can be removed.

Equivalent problem:

Source

0 0, sJ

(E , H)

(0 ,0)

S

(E , H)

Source PEC

S

sJ

Original problem:

Scattering by a PEC (cont.)

9

Source

0 0, sJ

(E , H)

(0 ,0)

0,it tsE J E r S

it tsE J E

This integral equation has to be solved numerically.

Integral equation for the unknown current

so

0s it tE E

“Electric Field Integral Equation” (EFIE)

Note: The bracket notation means

“field due to a source.”

Scattering by a PEC (cont.)

10

Scattering by Dielectric Body

sE

(E , H)Source

,r r iE

i sE E E

i

s

E

E

incident field

scattered field

(E , H)

Note: The body is assumed to be homogeneous.

11

Exterior Equivalence

Replace body by free space(The material doesn’t matter in the zero-field region.)

S

Source

,r r

Ea = E

Ha = H

Eb = 0 Hb = 0

ˆ 0

ˆ 0

e as

e as

J n H

M n E

S

Source

0 0,

esJ

esM

(E , H)(0 ,0)

n̂12

Summary for Exterior

ˆ

ˆ

es

es

J n H

M n E

,r r

S

Original problem:

Free-space problem:

S

0 0,

esJ

esM

(0 ,0)

n̂

(E , H)

13

Interior Equivalence

e es s

e es s

J J

M M

S

No sources or fields outside S

,r r

Ha, Ea

Eb = 0

Hb = 0

Place dielectric material in the "dead region" (region b).

ˆ ˆ ˆ0

ˆ ˆ ˆ0

e a es s

e a es s

J n H n H n H J

M n E n E n E M

ˆ ˆn n

Ea = E

Ha = H

14

Interior Equivalence (cont.)

S

e es sJ J

e es sM M

(0 ,0)

E, H

,r r

,r r n̂

When we calculate the fields from these currents, we let them radiate in an infinite dielectric medium.

15

Summary for Interior

,r r

S

Original problem:

S

esJ esM

(0 ,0)

E, H

,r r

,r r

Homogeneous-medium problem:

n̂

16

Integral Equation

,r r

S

, ,

, ,

e e i e et s t t ss s

e e i e et s t t ss s

E J M E E J M

H J M H H J M

, ,

, ,

e e e e it s t s ts s

e e e e it s t s ts s

E J M E J M E

H J M H J M H

Boundary conditions:

Hence:

The “–” means calculate the fields just inside the surface, assuming an infinite dielectric region.

Note: The “+” means calculate the fields just outside the surface,radiated by the sources in free space.

“PMCHWT" Integral Equation*

* Poggio-Miller-Chang-Harrington-Wu-Tsai 17

e es s

e es s

J J

M M

Recall:

Fields in a Half Space

Equivalent sources:

ˆ

ˆ

es

es

J z H

M z E

(0 ,0)es

es

J

M

(E , H)

Sources, structures,

etc.

z

(E , H)Region of interest

(z > 0)

Equivalence surface S (closed at infinity in the z < 0 region)

18

Hence, we have:

Put PEC in zero-field region:

The electric surface current on the PEC does not radiate.

Fields in a Half Space (cont.)

Note: The fields are only correct for z > 0.

19

z

(E , H)es

es

J

MPEC

ˆesM z E

(E , H)

esMPEC z

Note: The fields are still correct for z > 0.

Now use image theory:

ˆ2sM z E

Fields in a Half Space (cont.)

This is useful whenever the electric field on the z = 0 plane is known.

20

Incorrect fields

(E , H)

sMCorrect fields

0 0( , )

2 es sM M z

Fields in a Half Space: Summary

Sources + objects

z

(E , H)Region of interest

(z > 0)

21

Incorrect fields

(E , H)

sMCorrect fields

0 0( , )

ˆ2sM z E

z

Alternative (better when H is known on the interface):

Mse does not radiate on PMC,

and is therefore not included.

Image theory:

Incorrect fields

(E , H)

sJCorrect fields0 0( , )

ˆ2 2es sJ J z H

Fields in a Half Space (cont.)

22

esJPMC

z

Example: Radiation from Waveguide

zb

y

0ˆ( , ,0) cosx

E x y y Ea

b

a

x

y

23

Example (cont.)

Step #1

0ˆ2 coss

xM x E

a

Step #2

0

ˆ

ˆ cos

esM z E

xx E

a

PEC

z

z

24

The feeding waveguide was removed from the dead region that was created, and then a continuous PEC plane was introduced.

Image theory is applied to remove the ground plane and double the

magnetic surface current.

Apply equivalence principle

Apply image theory

0ˆ2 coss

xM x E

a

x

a

b

y

z

Example (cont.)

25

A three-dimensional view

0ˆ2 coss

xJ x E

a

a

b

x

y

z

ssJ M

E H

H E

Solve for the far field of this problem first.

(This is the “A” problem in the notation of the duality notes.)

Then use:

Example (cont.)

Use the theory of Notes 22 to find the far field from this rectangular strip of electric surface current.

0 0

0 0

26

Step #3

Apply duality

A radiating electric current can be replaced by a magnetic current, and vice versa.

Volume Equivalence Principle

27

y

x

z

J

y

x

z

M

1 1( , )E H 2 2( , )E H

P. E. Mayes, “The equivalence of electric and magnetic sources,” IEEE Trans. Antennas Propag., vol. 6, pp. 295–29, 1958.

We wish to have the same set of radiated fields.

Volume Equivalence Principle (Cont.)

28

1 0 1

0 0 1

E j H

j J j E

1 0 1

1 0 1

E j H

H J j E

21 0 1 0E k E j J

Set 1 (electric current source):

Hence

Therefore, we have

Volume Equivalence Principle (Cont.)

29

2 0 2

0 0 2

E j H M

j j E M

2 0 2

2 0 2

E j H M

H j E

22 0 2E k E M

Set 2 (magnetic current source):

Hence

Therefore, we have

Volume Equivalence Principle (Cont.)

30

22 0 2E k E M

21 0 1 0E k E j J

0j J M

0

1J M

j

Set

Compare:

Hence

Volume Equivalence Principle (Cont.)

31

1 2 0 1 0 1E E j H j H M

1 20

1H H M

j

Hence, the two electric fields are equal everywhere, but the magnetic fields are only the same outside the source region.

Next, examine the difference in the two Faraday laws:

1 0 1

2 0 2

E j H

E j H M

so

This gives us

Summary

32

y

x

z

J

y

x

z

M

0

1J M

j

0

1H J H M M

j

E J E M

Volume Equivalence Principle (Cont.)

Apply duality on the two curl equations to get two new equations:

33

22 0 2H k H J

21 0 1 0H k H j M

y

x

z

M

1 1( , )E H

y

x

z

J

2 2( , )E H

Volume Equivalence Principle (Cont.)

34

22 0 2H k H J

21 0 1 0H k H j M

0

1M J

j

1 0 1

2 0 2

H j E

H j E J

1 2 0 1 2H H j E E J

Similarly, from duality we have

Volume Equivalence Principle (Cont.)

35

y

x

z

M

y

x

z

J

0

1M J

j

0

1E M E J J

j

H M H J

Summary

Volume Equivalence Principle (Cont.)