notes 6 - waveguides part 3 attenuation

8/13/2019 Notes 6 - Waveguides Part 3 Attenuation

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Prof. David R. Jackson

Dept. of ECE

Notes 6

ECE 5317-6351

Microwave Engineering

Fall 2011

Waveguides Part 3:

Attenuation

1



For most practical waveguides and transmission lines the loss associated with

dielectric loss and conductor loss is relatively small.

To account for these losses we will assume

z k j attenuation constantPhase constant for lossless

wave guide

c d Attenuation

constant due to

conductor loss

Attenuation constant

due to dielectric loss

Attenuation on Waveguiding Structures

2



Attenuation due to Dielectric Loss: d

Lossy dielectric complex permittivity

1

(1 tan )

c

c c

cc

c

c

j

j

j

j

2 2 2 (1 tan )c ck j

tan c

c

0

0

rc

rc

complex wavenumber k

ck Note:

k k jk

3



Thus

2 2

z d ck j k k

Attenuation due to Dielectric Loss (cont.)

ck

Remember: The value k c is always real, regardless of whether the

waveguide filling material is lossy or not.

Note: The radical sign denotes the principal square root :

Arg z

4

This is an exact formula for

attenuation due to dielectric

loss. It works for both

waveguides and TEM

transmission lines (k c = 0).



2 (1 tan )

(1 tan )

c

c

k j

j

tan 1

1 1 / 2 1 z z z for

Small dielectric loss in medium:

Approximate Dielectric Attenuation

Use

1 tan / 2ck j

tan / 2

ck

k k

5



2 2

2 2

2 2 2

(1 tan )

tan

z d c

c c

c c c

k j k k

j k

k j

2 2 2tan ( )c c ck

1 1

1 / 12 2

z z a z a z a a a z a

a a

for

Small dielectric loss:

Approximate Dielectric Attenuation (cont.)

Use

6



2

22

2

2

2

2

2

tan

an

2

t

c

c

cc

z c c c

c

d

k k j

j

k jk

2 tan

2

cd


2 2

c ck 2 2

ck k

2 tan

2d

k

We assume

here that weare above

cutoff.

7



d

k

k

z k j k k jk

For TEM mode


tan

2d

k

ck

k jk

8

We can simply put k c = 0 in the previous formulas.

Or, we can start with the following:



Assuming a small amount of conductor loss:

We can assume fields of the lossy guide are

approximately the same as those for lossless guide,

except with a small amount of attenuation.

We can use a perturbation method to determine c.

Attenuation due to Conductor Loss

Note: Dielectric loss does not change the shape of the fields at all, since the boundary

conditions remain the same (PEC). Conductor loss does disturb the fields slightly.

9



Surface Resistance (cont.)

Assume

1/2 1/21

(1 )22

j

k j j j

Note that

1

2k k jk j

2

k k

The we have

Hence

1

11




Denote

Then we have

2k k

1

" p

d k

“skin depth” = “depth of penetration”

2

1' "

pd

k k

At 3 GHz, theskin depth for

copper is about

1.2 microns.

12




2

Example: copper

7

0 4 10 [H/m]

75.8 10 [S/m]

Frequency

1 [Hz] 6.6 [cm]

10 [Hz] 2.1 [cm]

100 [Hz] 6.6 [mm]

1 [kHz] 2.1 [mm]

10 [kHz] 0.66 [mm]

100 [kHz] 21 [mm]

1 [MHz] 66 [m]

10 [MHz] 20.1 [m]

100 [MHz] 6.6 [m]1 [GHz] 2.1 [m]

10 [GHz] 0.66 [m]

100 [GHz] 0.21 [m]

13




where

x y E H

1

2

(1 )2

(1 )

c

s

s

j

j j

j

j

j R

Z

Inside conductor:

2 s R

“Surface resistance ()”

“Surface impedance ()”

1 s s Z j R

15

ˆ

t t E n H

Note: To be more general:

n̂ outward normal




(1 )

1

2

s

s s

s

Z

Z j R

R

Summary for a Good Conductor

16

ˆt s t

E Z n H



17

ˆt s t

E Z n H

ˆ

t

t

E

H

n

tangential electric field at surface

tangential magnetic field at surface

outward unit normal to conductor surface

n̂

conductor

ˆ

eff

s t J n H

eff t s s

E Z J

“Effective surface current”

The surface impedance gives us the ratio of the

tangential electric field at the surface to the

effective surface current flowing on the object.

Hence we have


For the “effective”

surface current density

we imagine the actual

volume current density

to be collapsed into a

planar surface current.




1

2 s R

Example: copper

7

0 4 10 [H/m]

75.8 10 [S/m]

Frequency Rs

1 [Hz] 2.6110-7 []

10 [Hz] 8.2510-7 []

100 [Hz] 2.6110-6 []

1 [kHz] 8.2510-6 []

10 [kHz] 2.6110-5 []

100 [kHz] 8.2510-5 []

1 [MHz] 2.6110-4 []

10 [MHz] 8.2510-4 []

100 [MHz] 0.00261 []6.6

1 [GHz] 0.00825 []

10 [GHz] 0.0261 []

100 [GHz] 0.0825 []

18




2

0

1

2d s t

R H PIn general,

2 2

*0 0 01 1 1Re ( ) Re

2 2 2d x y z y s y E H H R H P

We then have

For a good conductor, 0

ˆeff

s t J n H

Hence 2

12

eff d s s R J P

This gives us the power dissipated per

square meter of conductor surface, ifwe know the effective surface current

density flowing on the surface.

19

PEC limit:eff

s s J J eff

s s J J Perturbation method : Assume that



Perturbation Method for c

Power flow along the guide:

2

0( ) z

P z P e

0 (0) P P

Power @ z = 0 is calculated

from the lossless case.

Power loss (dissipated) per unit length:l

dP P

dz

2

0( ) 2 2 ( ) z

l P z P e P z

0

( ) (0)

2 ( ) 2

l l P z P

P z P

20



There is a single

conductingboundary.0

(0)

2

l c

P

P

*

0

0

1ˆ

Re 2S z

P E H z dS

2

0

(0)

2

sl s

C z

R P J d

Surface resistance of

metal conductors:2

s R

ˆ s J n H Note :

On PEC conductor

Perturbation Method: Waveguide Mode

C

S

21

For these calculations, we

neglect dielectric loss.



0

(0)

2

l

c

P

P

*

0

0

2

0

1ˆRe

2

1

2

S z

lossless

P E H z dS

Z I

1 2

2

0

(0) 2 sl s

C C z

R P J d

Surface resistance of

metal conductors:

2

s R

ˆ s J n H Note :

On PEC conductor

2C

S

1C

Perturbation Method: TEM Mode

There are two

conductingboundaries.

22

For these calculations, we

remove dielectric loss.

0 0

lossless Z Z

h l l d l



0

02

lossless s

c lossless lossless

R dZ

Z d

2C

S

1C

Wheeler Incremental Inductance Rule

23

The Wheeler incremental inductance

rule gives an alternative method for

calculating the conductor attenuation

on a transmission line (TEM mode): It is

useful when Z 0 is already known.

In this formula, d l (for a given conductor) is the distance by which the

conducting boundary is receded away from the field region.

The top plate of a

PPW line is shown

being receded.

The formula is applied for each conductor

and the conductor attenuation from each

of the two conductors is then added.

d



Example: TEM Mode Parallel-Plate Waveguide

Previously, we showed

2

0

0 * 2

0 0

2

0 *

2

0

1ˆ ˆRe

2

1 1Re

2

1

2

w d

lossless

V P z z dydxd

wV

d

wV

d

0

0

ˆ

ˆ

jkz

jkz

V

E y ed

V H x e

d

0

ˆ

ˆ

top

s

jkz

J y H

V z e

d

0ˆ ˆ

bot jkz

s

V J y H z e

d

y

z w

d , ,

x

On the top plate:

On the bottom plate:

24

j

lossless

c



s

c lossless

R

d

Example: TEM Mode PPW (cont.)

26

Let’s try the same calculation using the Wheeler incremental inductance rule.

0

02

lossless s

c lossless lossless

R Z

Z

0

0

lossless lossless

d Z

w

d Z

w

From previous calculations:top bot

c c c

0 0lossless lossless Z dZ

d

d

w

, ,

0 02 2 22

top lossless s s s sc lossless lossless lossless lossless

lossless

R R R Rd

d Z d w wZ d w

w

0 02 2 2

2

bot lossless s s s sc lossless lossless lossless lossless

lossless

R R R Rd

d Z d w wZ d

w w



Results for TM/TE Modes (above cutoff): (derivation omitted)

TMn modes of parallel-plate

2, 0 s

cn lossless

k Rn

d

TEn modes of parallel-plate

22

, 0c s

cn lossless

k Rn

k d

c

nk

d

Example: TM z /TE

z Modes of PPW

y

z w

d , ,

x

27

notes 6 - waveguides part 3 attenuation

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