Waveguides part 3 attenuation

Report
ECE 5317-6351
Microwave Engineering
Fall 2011
Prof. David R. Jackson
Dept. of ECE
Notes 6
Waveguides Part 3:
Attenuation
1
Attenuation on Waveguiding Structures
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
k z    j
Phase constant for lossless
wave guide
attenuation constant
  c  d
Attenuation
constant due to
conductor loss
Attenuation constant
due to dielectric loss
2
Attenuation due to Dielectric Loss: d
Lossy dielectric  complex permittivity
 complex wavenumber k
k 2   2  c   2  c (1  j tan  )
k  k   jk 
Note: k     c


  c  j c
c    j
 c 


  c 1  j 
 c 

  c (1  j tan  )
 c
tan  
 c
    0 rc
    0 rc
3
Attenuation due to Dielectric Loss (cont.)
Thus
k z    j d  k  k
2
2
c
k    c
This is an exact formula for
attenuation due to dielectric
loss. It works for both
waveguides and TEM
transmission lines (kc = 0).
Remember: The value kc 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
Approximate Dielectric Attenuation
k   2  c (1  j tan  )
   c (1  j tan  )
Small dielectric loss in medium:
 tan   1
Use
1  z  1  z / 2 for z  1
 k   c 1  j  tan   / 2 
k    c
k   k   tan   / 2
5
Approximate Dielectric Attenuation (cont.)
k z    j d  k 2  kc2
  2  c (1  j tan  )  kc2
  2  c  kc2  j 2  c tan 
Small dielectric loss:
  2  c  tan   ( 2  c  kc2 )
Use
 1  z 
1 z 
a  z  a 1   z / a   a 1      a  
for z  a

2 a 
 2  a 
6
Attenuation due to Dielectric Loss (cont.)
 k z   2  c  kc2  j 2  c tan 
  2  c  kc2  j

 2  c tan 
2
2
2
k


 c  c 
We assume
here that we
are above
cutoff.

   j d
   2  c  kc2
  k 2  kc2
  c tan 
d 
2
2

k tan 
d 
2
2
7
Attenuation due to Dielectric Loss (cont.)
For TEM mode
We can simply put kc = 0 in the previous formulas.
Or, we can start with the following:
k z    j  k  k   jk 
  k

 d  k 
k    c
 k   jk 
k  tan 
 d 
2
8
Attenuation due to Conductor Loss
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 .
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
This is a very important concept for calculating loss at a metal surface.
x
 0 , 0
 ,  ,
z
S
C
Plane wave in a good conductor
Note: In this figure, z is the direction normal to the metal surface,
not the axis of the waveguide. Also, the electric field is assumed to
be in the x direction for simplicity.
10
Surface Resistance (cont.)
Assume


1
The we have


k      j 


1/2
 
   j 
 
Note that
k  k   jk  
Hence
k   k  
1/2

2
  1  j 



(1  j )


  2 
2
1  j 

2
11
Surface Resistance (cont.)

k   k  
Denote
  dp 
1
k"
2
“skin depth” = “depth of penetration”
Then we have
2
  dp 
k '  k"

1
At 3 GHz, the
skin depth for
copper is about
1.2 microns.

12
Surface Resistance (cont.)
Frequency

1 [Hz]
6.6 [cm]
2
10 [Hz]
2.1 [cm]

100 [Hz]
6.6 [mm]
1 [kHz]
2.1 [mm]
10 [kHz]
0.66 [mm]
100 [kHz]
21 [mm]
  0  4 107 [H/m]
1 [MHz]
66 [m]
  5.8 107 [S/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]

Example: copper
13
Surface Resistance (cont.)
x
S
z
fields evaluated on this plane
Pd
= time-average power dissipated / m2 on S
Pd 
1
1
*
Re ( E  H )  zˆ  Re ( Ex H y* ) z 0
2
2
14
Surface Resistance (cont.)
Inside conductor:
Ex   H y
Note: To be more general:
Et    nˆ  Ht 
nˆ  outward normal
where








c
j
j


1  j 

2 

 (1  j )

2
 (1  j ) Rs
 Zs
j


“Surface resistance ()”

Rs 
2
“Surface impedance ()”
Zs  1  j  Rs
15
Surface Resistance (cont.)
Summary for a Good Conductor
Et  Zs  nˆ  Ht 
Zs  
Z s  (1  j ) Rs

1
Rs 

2 
16
Surface Resistance (cont.)
nˆ
conductor
Et  Zs  nˆ  Ht 
Et  tangential electric field at surface
H t  tangential magnetic field at surface
nˆ  outward unit normal to conductor surface
“Effective surface current”
J
eff
s
  nˆ  Ht 
For the “effective”
surface current density
we imagine the actual
volume current density
to be collapsed into a
planar surface current.
Hence we have
Et  Zs J seff
The surface impedance gives us the ratio of the
tangential electric field at the surface to the
effective surface current flowing on the object.
17
Surface Resistance (cont.)
Frequency

1
Rs 

2 
Example: copper
  0  4 10 [H/m]
7
  5.8 107 [S/m]
Rs
1 [Hz]
2.6110-7 []
10 [Hz]
8.2510-7 []
100 [Hz]
2.6110-6 []
1 [kHz]
8.2510-6 []
10 [kHz]
2.6110-5 []
100 [kHz]
8.2510-5 []
1 [MHz]
2.6110-4 []
10 [MHz]
8.2510-4 []
100 [MHz]
0.00261 []6.6
1 [GHz]
0.00825 []
10 [GHz]
0.0261 []
100 [GHz]
0.0825 []
18
Surface Resistance (cont.)
We then have
Pd 
2
2
1
1
1
Re ( Ex H y* ) z 0  Re   H y 0  Rs H y 0
2
2
2
In general,
For a good conductor,
Hence
1
2
Pd  Rs H t 0
2
J seff  nˆ  Ht 0
1
Pd  Rs J seff
2
PEC limit: J seff  J s
2
This gives us the power dissipated per
square meter of conductor surface, if
we know the effective surface current
density flowing on the surface.
Perturbation method : Assume that
J seff  J s
19
Perturbation Method for c
P0  P (0)
Power flow along the guide:
P( z )  P0 e2 z
Power @ z = 0 is calculated
from the lossless case.
dP
Power loss (dissipated) per unit length: Pl  
dz
 Pl ( z )  2 P0e2 z  2 P( z )
Pl ( z ) Pl (0)

 
2 P ( z ) 2 P0
20
Perturbation Method: Waveguide Mode
Pl (0)
c 
2 P0
There is a single
conducting
boundary.
S
C
  1

*
P0  Re   E  H   zˆ dS 

 S  2
z 0 

Rs
Pl (0) 
2
J
2
d
s
C
Surface resistance of
metal conductors:
For these calculations, we
neglect dielectric loss.
Note : J s  nˆ  H
z 0
Rs 

2
On PEC conductor
21
Perturbation Method: TEM Mode
Pl (0)
c 
2 P0
There are two
conducting
boundaries.
  1

*
P0  Re    E  H   zˆ dS 

 S  2
z 0 

1 lossless 2
 Z0
I
2
Rs
Pl (0) 
Js

2 C1 C2
2
Surface resistance of
metal conductors:
C2
C1
For these calculations, we
remove dielectric loss.
Z
d
0
 Z 0lossless 
Note : J s  nˆ  H
z 0
Rs 
S
On PEC conductor

2
22
Wheeler Incremental Inductance Rule
The Wheeler incremental inductance
rule gives an alternative method for
calculating the conductor attenuation
on a transmission line (TEM mode): It is
useful when Z0 is already known.
S
C2
C1

 dZ 0lossless
Rs
 c   lossless lossless 
 2Z 0 
 d
The formula is applied for each conductor
and the conductor attenuation from each
of the two conductors is then added.
In this formula, dl (for a given conductor) is the distance by which the
conducting boundary is receded away from the field region.
d
The top plate of a
PPW line is shown
being receded.
23
Example: TEM Mode Parallel-Plate Waveguide
Previously, we showed
2
wd
1   V0
P0  Re     * 2
2  0 0   d


y


zˆ   zˆ dydx 



1 2 w  1 
V0   Re  * 
2
 d   
1 2 w 
 V0  lossless 
2
d

On the top plate:
, ,
d
x
     j 


c
 lossless 
J stop   yˆ  H
 zˆ
On the bottom plate:
J sbot  yˆ  H   zˆ
V0  jkz
e
d
V0  jkz
e
d
z
w


V0  jkz
e
d
V
H   xˆ 0 e  jkz
d
E   yˆ
24
Example: TEM Mode PPW (cont.)
w
Rs  top
 Pl (0)    J s
2 0
w
2
y  d , z 0
V0
 Rs 
dx
d
0
V0
(
0

dx 
y 0, z 0

2
w
 Rs
dx   J
bot 2
s
(equal contributions from both plates)
2
lossless
d)
2
w


w
Rs V0  lossless 2 
(
d) 
Pl (0)

 c 

2 P0
1 2 w 
2   V0  lossless 
d
2

2
The final result is then
c 
Rs
 lossless d
25
Example: TEM Mode PPW (cont.)
Let’s try the same calculation using the Wheeler incremental inductance rule.

 Z 0lossless
Rs
 c   lossless lossless 
 2Z 0 
 
Z 0lossless dZ 0lossless


d
 c   ctop   cbot
From previous calculations:
d
Z0    
 w
d
Z 0lossless   lossless  
 w

  
R
Rs
Rs
Rs
d
 ctop   losslesss lossless   lossless  


w  2wZ 0lossless
 lossless d  2 lossless d
 2Z 0 
 d 
2 w 

w


  
R
Rs
Rs
Rs
d
 cbot   losslesss lossless   lossless  

 lossless
lossless
d  2
w  2wZ 0
d

 2Z 0 
 d 
2w  lossless 
w

w
d
 ,  ,
c 
Rs
 lossless d
26
Example: TMz/TEz Modes of PPW
y
, ,
d
x
z
w
Results for TM/TE Modes (above cutoff): (derivation omitted)
TMn modes of parallel-plate  cn 
TEn modes of parallel-plate
kc 
 cn 
2k Rs

lossless
d
, n0
2kc2 Rs
k 
lossless
d
, n0
n
d
27

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