prof. david r. jackson dept. of ece notes 6 ece 5317-6351 microwave engineering fall 2011 waveguides...
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Prof. David R. JacksonDept. of ECE
Notes 6
ECE 5317-6351 Microwave Engineering
Fall 2011
Waveguides Part 3:Attenuation
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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
zk j
attenuation constant Phase constant for lossless wave guide
c d
Attenuation constant due to conductor loss
Attenuation constant due to dielectric loss
Attenuation on Waveguiding Structures
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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
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Thus
2 2z d ck j k k
Attenuation due to Dielectric Loss (cont.)
ck
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
This is an exact formula for attenuation due to dielectric loss. It works for both waveguides and TEM transmission lines (kc = 0).
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2 (1 tan )
(1 tan )
c
c
k j
j
tan 1
1 1 / 2 1z z z for
Small dielectric loss in medium:
Approximate Dielectric Attenuation
Use
1 tan / 2ck j
tan / 2
ck
k k
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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 11 / 1
2 2
z za z a z a a a z a
a a
for
Small dielectric loss:
Approximate Dielectric Attenuation (cont.)
Use
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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
2c
d
Attenuation due to Dielectric Loss (cont.)
2 2c ck 2 2
ck k
2 tan
2d
k
We assume here that we are above cutoff.
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d
k
k
zk j k k jk
For TEM mode
Attenuation due to Dielectric Loss (cont.)
tan
2d
k
ck
k jk
We can simply put kc = 0 in the previous formulas. Or, we can start with the following:
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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.
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This is a very important concept for calculating loss at a metal surface.
Surface Resistance
z
0 0, , , x
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.
C
S
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Surface Resistance (cont.)
Assume
1/2 1/21
(1 )22
jk j j j
Note that
12
k k jk j
2k k
The we have
Hence
1
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Surface Resistance (cont.)
Denote
Then we have
2k k
1
"pdk
“skin depth” = “depth of penetration”
2
1' "
pd
k k
At 3 GHz, the skin depth for copper is about 1.2 microns.
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Surface Resistance (cont.)
2
Example: copper7
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]
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Surface Resistance (cont.)
dP = time-average power dissipated / m2 on S
* *0
1 1ˆRe( ) Re( )
2 2d x y zE H z E H P
z
x
S
fields evaluated on this plane
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Surface Resistance (cont.)
where
x yE H
1
2
(1 )2
(1 )
c
s
s
jj j
j
j
j R
Z
Inside conductor:
2sR
“Surface resistance ()”
“Surface impedance ()” 1s sZ j R
ˆt tE n H Note: To be more general:
n̂ outward normal
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Surface Resistance (cont.)
(1 )
1
2
s
s s
s
Z
Z j R
R
Summary for a Good Conductor
ˆt s tE Z n H
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ˆt s tE 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
n̂conductor
ˆeffs tJ n H
efft s sE 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
Surface Resistance (cont.)
For the “effective” surface current density we imagine the actual volume current density to be collapsed into a planar surface current.
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Surface Resistance (cont.)
1
2sR
Example: copper7
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 []
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Surface Resistance (cont.)
2
0
1
2d s tR HPIn general,
2 2*0 0 0
1 1 1Re( ) Re
2 2 2d x y z y s yE H H R H P
We then have
For a good conductor, 0ˆeffs tJ n H
Hence21
2eff
d s sR JPThis gives us the power dissipated per square meter of conductor surface, if we know the effective surface current density flowing on the surface.
PEC limit: effs sJ J eff
s sJ JPerturbation method : Assume that
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Perturbation Method for c
Power flow along the guide: 20( ) zP z P e
0 (0)P P
Power @ z = 0 is calculated from the lossless case.
Power loss (dissipated) per unit length: l
dPP
dz
20( ) 2 2 ( )z
lP z P e P z
0
( ) (0)
2 ( ) 2l lP z P
P z P
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There is a single conducting boundary.0
(0)
2l
c
P
P
*0
0
1ˆRe
2S z
P E H z dS
2
0
(0)2
sl s
C z
RP J d
Surface resistance of metal conductors:
2sR
ˆsJ n H Note :
On PEC conductor
Perturbation Method: Waveguide Mode
C
S
For these calculations, we neglect dielectric loss.
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0
(0)
2l
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
RP J d
Surface resistance of metal conductors: 2sR
ˆsJ n H Note :
On PEC conductor
2C
S
1C
Perturbation Method: TEM Mode
There are two conducting boundaries.
For these calculations, we remove dielectric loss.
0 0losslessZ Z
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0
02
losslesss
c lossless lossless
R dZ
Z d
2C
S
1C
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.
In this formula, dl (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
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Example: TEM Mode Parallel-Plate Waveguide Previously, we showed
2
00 * 2
0 0
2
0 *
2
0
1ˆ ˆRe
2
1 1Re
2
1
2
w d
lossless
VP z z dydx
d
wV
d
wV
d
0
0
ˆ
ˆ
jkz
jkz
VE y e
dV
H x ed
0
ˆ
ˆ
tops
jkz
J y H
Vz e
d
0ˆ ˆbot jkzs
VJ y H z e
d
y
z w
d , ,
x
On the top plate:
On the bottom plate:
j
lossless
c
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2 2
, 0 0, 00 0
(0)2
w wtop bots
l s sy d z y z
RP J dx J dx
2
0
0
w
s
VR dx
d
2
02( )s lossless
VR w
d
2
0 2
200
( )(0)
2 12
2
s losslessl
c
lossless
wR V
dP
P wV
d
sc lossless
R
d
Example: TEM Mode PPW (cont.)
(equal contributions from both plates)
The final result is then
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sc lossless
R
d
Example: TEM Mode PPW (cont.)
Let’s try the same calculation using the Wheeler incremental inductance rule.
0
02
losslesss
c lossless lossless
R Z
Z
0
0lossless lossless
dZ
w
dZ
w
From previous calculations:top bot
c c c
0 0lossless losslessZ dZ
d
d
w
, ,
0 02 2 22
top losslesss s s sc lossless lossless lossless lossless
lossless
R R R RddZ d w wZ dww
0 02 2 22
bot losslesss s s sc lossless lossless lossless lossless
lossless
R R R RddZ d w wZ dww
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Results for TM/TE Modes (above cutoff): (derivation omitted)
TMn modes of parallel-plate2
, 0scn lossless
k Rn
d
TEn modes of parallel-plate22
, 0c scn lossless
k Rn
k d
c
nk
d
Example: TMz/TEz Modes of PPWy
z w
d , ,
x