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10/26/2017
1
Lecture 5b Slide 1
EE 4347
Applied Electromagnetics
Topic 5b
Parallel Plate Waveguide
These notes may contain copyrighted material obtained under fair use rules. Distribution of these materials is strictly prohibited
Lecture Outline
Lecture 5b Slide 2
• What is a parallel plate waveguide?
• TEM Analysis
• TM Analysis
• TE Analysis
• Conclusions
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Lecture 5b Slide 3
What is a Parallel Plate Waveguide?
Lecture 5b Slide 4
Geometry of Parallel Plate Waveguide
x
y
z
d,
w
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Notes on the Parallel Plate Waveguide
• Becoming very popular for transmitting differential signals around a circuit board.
• Simply analysis and demonstrates most of the concepts of waveguides.
• Differential lines have confined fields for reduced interference with other devices in close proximity.
• Differential lines exhibit common mode rejection for noise immunity.
Lecture 5b Slide 5
Lecture 5b Slide 6
Vision for 3D High‐Frequency Interconnects
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Lecture 5b Slide 7
TEM Analysis
Lecture 5b Slide 8
Starting Point for TEM Analysis
Assuming the parallel plate waveguide has an LHI dielectric between the plates, we start with the homogeneous Laplaces’ equation.
2 2 2
22 2 2
, , 0V V V
V x y zx y z
The parallel plate waveguide is uniform in the x and z directions so our governing equation reduces to
2
2
V
x
2 2
2 2
V V
y z
2
2
2
2
0
0
0
V
y
d V
dy
Note, by assuming the field is uniform in the xdirection, we are ignoring the fringing fields at the edges.
The derivative becomes ordinary because y is the only independent variable left.
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Lecture 5b Slide 9
How to Interpret Governing Equation
y
d, 0V y y d
2
20
d V
dy
Our governing equation is now
The solution to this will give us V(y).
Lecture 5b Slide 10
Boundary Conditions
d,
0 ?
?
V
V d
We need boundary conditions to solve our differential equation.
Apply a voltage V0 across the plates and the boundary conditions will be
0
0 0V
V d V
+‐0V
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Lecture 5b Slide 11
General Solution to Differential Equation
Our differential equation with boundary conditions is
00 0 and V V d V 2
20 0
d Vy d
dy
This is solved by integrating by y twice.
2
20
d V
dy
dVA
dy
V y Ay B
Lecture 5b Slide 12
Apply Boundary Conditions
Our general solution is now
We apply the boundary condition at y = 0.
V y Ay B
0 0
0 0
0
V
A B
B
We apply the boundary condition at y = d.
0V d V
A d B
0
0
0
V
A d V
A V d
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Lecture 5b Slide 13
The Solution (1 of 2)
The final solution to the governing equation is
The electric field is calculated from the electric potential as
0VV y yd
0 0 0
0
ˆ ˆ ˆ
ˆ ˆ ˆ
ˆ
x y z
x y z
x
V V VE V a a a
x y z
V V VE a y a y a y
x d y d z d
VE a y
x d
0 0ˆ ˆy z
V Va a yd z d
0ˆyV
E ad
We are still not done because we still do not know much about the waveguide.
Lecture 5b Slide 14
The Solution (2 of 2)
If we ignore the fringing fields outside of the waveguide, the electric field is expressed as
0ˆ for 0 and 0
,0 otherwise
y
Va x w y d
E x y d
x
y
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Lecture 5b Slide 15
The Wave Solution
We derived the form of the TEM wave by way of an electrostatic analysis.
This ignores the wave nature of a TEM wave.
To account for the wave nature, we must incorporate a term that accumulates phase in the z direction.
0ˆ for 0 and 0
, , 0 otherwise
j zy
Va e x w y d
E x y z d
It follows that the magnetic field component is
0
0 0
TEM
0
ˆ ˆˆ
ˆ ˆ ˆ, ,
ˆ for 0 and 0, ,
0 otherwise
j zz y
j z j zzz y x
j zx
Va a e
V Va E dH x y z a a e a e
Z d d
Va e x w y d
dH x y z
Lecture 5b Slide 16
Impedance from Wave Solution (1 of 2)
The impedance of the TEM wave is defined as
0TEM
VZ
I
1 sheet
ˆ
2
K nH
We must determine the current term I. Recall the magnetic field above an infinite current sheet is
It follows that the field between two current sheets (i.e. in our parallel plate waveguide) is
2 sheets ˆH K n
surface current density A mK
Using this equation for the parallel plate waveguide ignores fringing fields at the edges.
Solving this for the surface current term yields
ˆ ˆyn a
ˆ ˆ ˆy yK n H a H H a
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Lecture 5b Slide 17
Impedance from Wave Solution (2 of 2)
We can find the total current I by integrating the surface current across the plate.
0 0 0
ˆ ˆ ˆw w w
z y z xI K a dx H a a dx H dx
00x
VH z
d
Let z = 0 and our magnetic field solution reduces to
Substituting this into our equation for I yields
0 0 0 0
0 0
w wV V V VwI dx dx w
d d d d
The characteristic impedance is found by substituting this into the original definition.
0 0TEM
0
V V dZ
VwI wd
TEM
dZ
w
Lecture 5b Slide 18
Propagation Constant We cannot calculate the phase constant from our solution because we analyzed it using an electrostatic approximation.
TEM waves propagate with nearly the same speed as a plane wave in an infinite medium composed of the dielectric that is between the plates.
TEM
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Lecture 5b Slide 19
Distributed Inductance and Capacitance
The characteristic impedance of the parallel plate transmission line is
TEM
dZ
w
The distributed capacitance C can be estimated as
wC
d
It follows that the distributed inductance L is
TEM
d L LZ
w C w d
d
Lw
Lecture 5b Slide 20
Example #1 (1 of 3)
Suppose we have the following parallel plate waveguide.
w = 2.0 mm
d = 0.5 mmr 2.3
What is the characteristic impedance?
What value of w would make this transmission line 50 ?
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Lecture 5b Slide 21
Example #1 (2 of 3)
The equation for characteristic impedance is
TEM
dZ
w
The impedance of the dielectric is
0
1.0376.73 248.4
2.3r
r
TEM
248.4 0.5 mm62.1
2.0 mm
dZ
w
The characteristic impedance is therefore
Lecture 5b Slide 22
Example #1 (3 of 3)
We solve the equation for characteristic impedance for w.
TEMTEM
d d
Z ww Z
To get 50 , wmust be
0
248.4 0.5 mm2.48 mm
50
dw
Z
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Lecture 5b Slide 23
Visualization of TEM Mode
Lecture 5c Slide 24
Summary of TEM Analysis
Field Solution
0
0
, , 0
, ,
, , 0
, ,
, , 0
, , 0
x
j zy
z
j zx
y
z
E x y z
VE x y z e
dE x y z
VH x y z e
d
H x y z
H x y z
Phase Constant
TEM
Cutoff Frequency
c 0f
Characteristic Impedance
TEM
dZ
w
• TEM has no cutoff frequency• TEM is the TM0 mode.
Same as plane wave. No cutoff frequency.Mode supported at DC.
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Lecture 5b Slide 25
TE Analysis
Lecture 5b Slide 26
Recall the Starting Point
The governing equation for TE analysis is
After a solution is obtained, the remaining field components are calculated according to
2 20, 0, 2
0,2 20z z
c z
H Hk H
x y
2 2 2ck k
0,0, 2
0,0, 2
zx
c
zy
c
HjH
k x
HjH
k y
0,0, 2
0,0, 2
0, 0
zx
c
zy
c
z
HjE
k y
HjE
k x
E
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Lecture 5b Slide 27
Simplify Governing Equation
Assuming the waveguide is uniform in the direction of x
The governing equation reduces to
2
20 and 0
x x
20,
2
zH
x
2 20, 0,2 2
0, 0,2 20 0z z
c z c z
H d Hk H k H
y dy
Lecture 5b Slide 28
General Solution
The general solution to the governing equation is
2
0, 20, 0,2
0 sin coszc z z c c
d Hk H H A k y B k y
dy
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Lecture 5b Slide 29
Boundary Conditions (1 of 2)
The electric field must be zero at the plates.
The solution, however, is in terms of the magnetic field.
We must write it in terms of an electric field.
The only component of the electric field tangential to the interface is E0,x.
0,0, 2 2
sin cos
cos sin
zx c c
c c
c cc
Hj jE A k y B k y
k y k y
jA k y B k y
k
Lecture 5b Slide 30
Boundary Conditions (2 of 2)
The first boundary condition is
0,
0,
, 0 0
,0 cos 0 sin 0 0 0
x
xc c
E x
j jE x A B A A
k k
0,
0,
, 0
, sin sin
x
x c cc c
E x d
j jE x d B k d B k d
k k
The second boundary condition is
We cannot choose B = 0 because that would lead to a trivial solution. Instead, it must be the sine term that is zero at y = d.
sin 0 1, 2,3,...c ck d k d n n
The cutoff wave number is then
1, 2,3,...c
nk n
d
Note that n = 0 would force the entire field to be zero so this is not a valid solution.
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Lecture 5b Slide 31
Phase Constant
Recall our definition of the cutoff wave number. Solve this for .
2 2 2 2 2 c ck k k k
We previously derived an expression for kc that arose from the boundary conditions.
22 1, 2,3,...
nk n
d
We see that we have an infinite number of discrete solutions where the order of the mode is n.
22 1, 2,3,...n
nk n
d
Lecture 5b Slide 32
Final Solution
Recall the general solution was
2
0, 20, 0,2
0 sin coszc z z c c
d Hk H H A k y B k y
dy
But now we know that A = 0 and kc = n/d. The final solution becomes
0, , cos , , cos nj zz n z n
n y n yH x y B H x y z B e
d d
The remaining field components are calculated from this result.
2 2
2 2
2 2
cos 0
cos sin
cos sin
n
n n
n
j zzx n
c c
j z j zzy n n
c c c
j zzx n n
c c c
Hj j n yH B e
k x k x d
Hj j n y j n yH B e B e
k y k y d k d
Hj j n y j n yE B e B
k y k y d k d
2 2cos 0
0
n
n
j z
j zzy n
c c
z
e
Hj j n yE B e
k x k x d
E
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Lecture 5b Slide 33
Why No TE0 Mode?
For n = 0, the field components are
0
sin 0 0
cos 0
sin 0 0
0
0
n
n n
n
x
j zy n
c
j z j zz n n
j zx n
c
y
z
H
jH B e
k
H B e B e
jE B e
k
E
E
This is not a physical solution.
Lecture 5b Slide 34
Cutoff Condition
Recall that we calculate or phase constant as
22 2 2 1, 2,3,...n c
nk k k n
d
This becomes imaginary when kc > k. Values of n that cause this condition correspond to modes that are “cutoff.” These are still modes, but they decay very quickly so they are almost never used.
2
2 2
c c
c
cc
nk
dn
fd
knf
d
This is the cutoff frequency for the TEnmode.
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Lecture 5b Slide 35
Characteristic Impedance
The characteristic impedance is defined as
We substitute in our expressions for the field quantities to obtain
0,0,TE
0, 0,
yx
y x
EEZ
H H
0,TE
0,
sin
sin
nx c
yn
c
j n yB
E k d kZ
j n yH Bk d
Lecture 5b Slide 36
Visualization of TE1 Mode
E
H
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Lecture 5b Slide 37
Visualization of TE2 Mode
E
H
Lecture 5b Slide 38
Visualization of TE3 Mode
E
H
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Lecture 5c Slide 39
Summary of TE Analysis
Field Solution
, , sin
, , 0
, , 0
, , 0
, , cos
, , sin
n
n
n
j zx n
c
z
j
y
x
zy n
c
z
j
z n
j n yE x y z B e
k d
E x y z
H x y z
j n yH
E x y z
n yH x y z B e
d
x y z B ek d
Phase Constant2
2n
nk
d
Cutoff Frequency
c,2
n
nf
d
Characteristic Impedance
TE,nn
kZ
• TE0 mode does not exist• TE1 is the lowest order TE mode
Same equation as for TM
Same equation as for TM1,2,3,...n
Lecture 5b Slide 40
TM Analysis
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Lecture 5b Slide 41
Recall the Starting Point
The governing equation for TM analysis is
After a solution is obtained, the remaining field components are calculated according to
2 20, 0, 2
0,2 20z z
c z
E Ek E
x y
2 2 2ck k
0,0, 2
0,0, 2
0, 0
zx
c
zy
c
z
EjH
k y
EjH
k x
H
0,0, 2
0,0, 2
zx
c
zy
c
EjE
k x
EjE
k y
Lecture 5b Slide 42
Simplify Governing Equation
Assuming the waveguide is uniform in the direction of x
The governing equation reduces to
2
20 and 0
x x
20,
2
zE
x
2 20, 0,2 2
0, 0,2 20 0z z
c z c z
E d Ek E k E
y dy
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Lecture 5b Slide 43
General Solution
The general solution to the governing equation is
2
0, 20, 0,2
0 sin coszc z z c c
d Ek E E A k y B k y
dy
Lecture 5b Slide 44
Boundary Conditions
The electric field component E0,z is tangential to the interfaces to the boundary conditions are applied to this directly.
The first boundary condition is
0, ,0 sin 0 cos 0 0 0zE x A B B B
0, , sin 0z cE x d A k d
The second boundary condition is
We cannot choose A = 0 because that would lead to a trivial solution. Instead, it must be sine term that is zero at y = d.
sin 0 0,1, 2,3,...c ck d k d n n
The cutoff wave number is then
0,1, 2,3,...c
nk n
d
Note that n = 0 is allowed in this case because it does not force the field to be zero. It does, however, force the field to be uniform. Thus TM0 is the TEM mode.
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Lecture 5b Slide 45
Phase Constant
Recall our definition of the cutoff wave number. Solve this for .
2 2 2 2 2 c ck k k k
We now have an expression for kc that arose from the boundary conditions.
22 0,1, 2,3,...
nk n
d
We see that we have an infinite number of discrete solutions where the order of the mode is n.
22 ,1, 2,3,.0 ..n
nk n
d
TM0 is the TEM mode
Lecture 5b Slide 46
Final Solution
Recall the general solution was
2
0, 20, 0,2
0 sin coszc z z c c
d Ek E E A k y B k y
dy
But now we know that B = 0 and kc = n/d. The final solution is
0, , sin , , sin nj zz n z n
n y n yE x y A E x y z A e
d d
The remaining field components are calculated from this result.
2 2
2 2
2 2
2
sin cos
sin 0
0
sin 0
n n
n
n
j z j zzx n n
c c c
j zzy n
c c
z
j zzx n
c c
yc
Ej j n y j n yH A e A e
k y k y d k d
Ej j n yH A e
k x k x d
H
Ej j n yE A e
k x k x d
jE
k
2
sin cosn nj z j zzn n
c c
E j n y j n yA e A e
y k y d k d
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Lecture 5b Slide 47
Why Does TM0 Mode Exist?
For n = 0, the field components are
cos 0
0
0
0
cos 0
n n
n
j z j zx n n
c c
y
z
x
j z jy n n
c c
j jH A e A e
k k
H
H
E
j jE A e A e
k k
, , sin 0 0
n
n
z
j zz nE x y z A e
This is a valid solution.
Lecture 5b Slide 48
Cutoff Condition
Recall that we calculate or phase constant as
22 2 2 0,1, 2,3,...n c
nk k k n
d
This becomes imaginary when kc > k. Values of n that cause this condition correspond to modes that are “cutoff.” These are still modes, but they decay very quickly so they are almost never used.
2
2 2
c c
c
cc
nk
dn
fd
knf
d
This is the cutoff frequency for the TMnmode.
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Lecture 5b Slide 49
Characteristic Impedance
The characteristic impedance is defined as
We substitute in our expressions for these field quantities to obtain
0,TM
0,
cos
cos
ny c
xn
c
j n yA
E k dZ
j n yH kAk d
0,0,TM
0, 0,
yx
y x
EEZ
H H
Lecture 5b Slide 50
Visualization of TM0 Mode
TEM
E
H
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Lecture 5b Slide 51
Visualization of TM1 Mode
E
H
Lecture 5b Slide 52
Visualization of TM2 Mode
E
H
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Lecture 5b Slide 53
Example #2 (1 of 2)
Suppose we have the following parallel plate waveguide.
w = 2.48 mm
d = 0.5 mmr 2.3
What is the bandwidth of this waveguide when used as a transmission line?
Lecture 5b Slide 54
Example #2 (2 of 2)
When used as a transmission line, we are interested in using the TEM mode. The bandwidth is the range of frequencies for which the waveguide supports only the TEM mode.
The cutoff frequencies are the same for the TE and TM modes we can essentially check them at the same time.
The second order modes are TE1 and TM1. The bandwidth is simply the cutoff frequency of these modes.
01 299792458 m s
1 197.6 GHz2 2 2 0.5 mm 1.0 2.3
cc
r r
k ncf n
d
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Lecture 5c Slide 55
Summary of TM Analysis
Field Solution
, , 0
, , cos
, , cos
, , 0
, , sin
, , 0
n
n
n
x
j zy n
c
j zx
j zz n
y
z
nc
n yE x y z A
E x y z
j n yE x y z A e
k d
j n yH x y z A e
k d
H x y z
ed
H x y z
• TM0 mode is the TEM mode
Phase Constant2
2n
nk
d
Cutoff Frequency
c,2
n
nf
d
Characteristic Impedance
TM,n
nZk
Same equation as for TE
Same equation as for TE0,1,2,3,...n
Lecture 5b Slide 56
Conclusion
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Lecture 5b Slide 57
Summary of Parallel Plate Waveguide0,1,2,3,...n 1,2,3,...n
Lecture 5b Slide 58
Modes in Parallel Plate Waveguide
n = 0
n = 1
n = 2
n = 3
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Notes
• Supports TEM mode when it has a homogeneous dielectric because it has two conductors.
• Supports TE and TM modes when it has a homogeneous dielectric
• The lowest order mode is TM0 which is the TEM mode.
Lecture 5b Slide 59