1

Prof. David R. JacksonECE Dept.

Spring 2014

Notes 3

ECE 6341

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Grounded Dielectric Slab

Goal: Determine the modes of propagation and their wavenumbers.

Assumption: There is no variation of the fields in the y direction,

and propagation is along the z direction.

x

z

,r r h

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Dielectric Slab

TMx & TEx modes:x

z

H ,r r E

TMx

z E ,r r

H TEx

x

4

Surface Wave

1 c

The internal angle is greater than the critical angle, so there is exponential decay in the air region.

z

,r r 1

x

Exponential decay

1 1 0sinzk k k

The surface wave is a “slow wave”.

2 20 0 0x z xk k k j

ch

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Surface Wave

The wave must also satisfy a “consistency condition”:

1 1 12 sec 2zk h k z n

This forces the angle 1 to be a discrete value, depending on n.

1 1 1 1 1 12 sec sin 2 tan 2k h k h n or

z

,r r 1

x

z

h

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TMx Solution

Assume TMxˆ ( , )xA x A x z

B.C. :(see TMx-TEx tables in Appendix) 00 ( 0)x

x y z

AE E

x

zjk zxA e f xAssume

2 2 22

2 2 20x x x

x

A A Ak A

x y z

2

2 22

0xz x

Ak k A

x

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Hence

22 2

20z

f xk k f x

x

Denote 1/ 22 2

0 0x zk k k

1/ 22 21 1x zk k k

2

202

0x

f xk f x

x

2

212

0x

f xk f x

x

x h

x h

TMx Solution (cont.)

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x h

x h

1 1cos( )zjk zx xA e k x

00

xz jk xjk zxA Ae e

Applying boundary conditions at the ground plane,

Note: Since the surface wave is a slow wave, we have:

0 0x xk j2 2 1/ 2

0 0( ) 0x zk k 1/ 22 2

0 0x zk k k

TMx Solution (cont.)

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Boundary Conditions

BC 1) 0 1 @y yH H x h

0 1

0 1

1 1x xA A

z z

0 1

0 0 1 1

1 1x xA A

x x

0 1 @z zE E x h BC 2)

0 1x xr r

A A

x x

0 1r x xA A

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Boundary Conditions (cont.)

These two equations yield:

0

0

1

0 1 1

cos( )

( ) sin( )

x

x

hr x

hr r x x x

Ae k h

Ae k k h

Divide second by first:

0 1 1( ) tan( )x r x xk k h

0 1 1tan( )x r x xk k h or

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Final Result: TMx

This may be written as:

2 2 1/2 2 2 1/2 2 2 1/20 1 1( ) ( ) tan ( )r z z zk k k k k k h

This is a transcendental equation for the unknown wavenumber kz.

Note: The choice of square root for kx1 is not important, but it is for kx0:

2 2 1/20( ) 0zk k

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AppendixTMx

Note: There is a factor difference with the Harrington text.

21yE

j x y

0xH

21zE

j x z

1yH

z

22

2

1xE k

j x

1zH

y

xA

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Appendix (cont.)TEx

Note: There is a factor difference with the Harrington text.

xF

0xE

21yH

j x y

1

yEz

21zH

j x z

1

zEy

22

2

1xH k

j x