Transcript Propagation in dielectrics

ENE 428
Microwave Engineering
Lecture 7 Waveguides
RS
1
Review
• Impedance matching to minimize power reflection from load
• Lumped-element tuners
• Single-stub tuners
• Microstrip lines
• The most popular transmission line
• Knowing the characteristic impedance and the relative dielectric
constant of the material helps determine the strip line
configuration and vice versa.
• Attenuation
• conduction loss
• dielectric loss
• radiation loss
2
A pair of conductors is used to guide TEM
wave
• Microstrip
• Parallel plate
• Two-wire TL
• Coaxial cable
3
The use of waveguide
• Waveguide refers to the structure that does not support TEM mode.
• They are unable to support wave propagation below a certain
frequency, termed the cutoff frequency.
Figures (a) and (b) are metallic waveguide.
The rectangular one is for high-power
microwave applications but is limited in
frequency range and tends to suffer from
dispersion.
The circular waveguide has higher power
handling capability than the rectangular one.
The dielectric waveguide has a smaller loss
than metallic waveguide at high freq. Optical
fiber has wider bandwidth and provides good
signal isolation between adjacent fibers.
4
Rectangular waveguide fundamentals
- A typical cross section is shown. Propagation is in the +z direction. The
conducting walls are brass, copper or aluminum. The inside is
electroplated with silver or gold and smoothly polished to reduce loss.
- The interior dimensions are “a x b”, the longer side is “a”.
- “a” determines the frequency range of the dominant, or lowest order
mode.
- Higher modes have higher attenuation and be difficult to extract from the
guide.
- “b” affects attenuation; smaller “b” has higher attenuation.
- If “b” is increased beyond “a/2”, the next mode will be excited at a lower
frequency, thus decreasing the useful frequency range.
- In practice, “b” is chosen to be about “a/2”.
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Rectangular waveguide fundamentals
- Waveguide can support TE and TM modes.
- The order of the mode refers to the field configuration in the guide and is
given by “m” and “n” integer subscripts, as TEmn and TMmn.
- The “m” subscript corresponds to the number of half-wave variations of
the field in the x direction.
- The “n” subscript is the number of half-wave variations in the y direction.
- “m” and “n” determines the cutoff frequency for a particular mode.
fcmn 
1
2 
2
m n
   
 a  b
2
The relative cutoff
freq for the first 12
modes of waveguide
with “a” = “2b” are
shown.
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7
8
Wave Propagation
(a) A y-polarized TEM plane wave
propagating in the +z direction.
(b) Wavefront view of the
propagating wave
Note: the waves propagate at a
velocity Uu, where u subscript
indicates media unbounded by
guide wall.
In air, Uu = c
9
Now a pair of identical
TEM waves, labeled as U+
and U- are superimposed.
The horizontal lines can be
drawn on the
superimposed waves that
correspond to zero total
field. Along these lines, the
U+ wave is always 180o
out of phase with the Uwave.
Next, replace the horizontal
lines with perfect
conducting walls.
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- “a” is the wall separation and is determined by the angle  and wavelength 
- For a given wave velocity Uu, the frequency is f = Uu/ 
- If we fix the wall separation at “a” and change the frequency, we must then also
change the angle  if we are to maintain a propagating wave.
- The edge of a +Eo wavefront (point A) will line up with edge of a –Eo wavefront
(point B), and two fronts must be /2 apart for the m = 1 mode. For any value of
m, we can write
m 2
sin  
a
or
uu
2a
  sin  
m
f
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The waveguide can support propagation as
long as the wavelength is smaller than a
critical value that occurs at  = 90o, or
2a
2a uu
o
c  sin 90  
m
m fc
where fc is the cutoff frequency for the propagating mode. We can relate the angle 
to the operating frequency (f) and the cutoff frequency (fc) by
fc

sin  

c
f
The time tAC it takes for the wavefront to move from A to C is
t AC
l AC m 2


uu
uu
Meanwhile, a constant phase point moves along the wall from A to D. Calling the phase
velocity, Up:
l AD
m 2

cos 
so
t AD
l AD
m 2


u p u p cos 
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Since the times tAD and tAC must be equal,
we have
uu
up 

cos 
uu
 fc 
1  
 f 
2
The phase velocity can be considerably faster than the velocity of the wave in
unbounded media, tending toward infinity as f approaches fc.
The phase constant associated with the phase velocity is
   u 1  f c f 
2
Where u is the phase constant in unbounded media. The wavelength in the guide
is related to this phase velocity by  = 2

u
 fc 
1  
 f 
2
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The propagation velocity of the superposed
wave is given by the group velocity UG.
uG  uu cos   uu 1  f c f 
2
The group velocity is slower than that of an unguided wave, which is to be expected
since the guided wave propagates in a zig-zag path, bouncing off the waveguide
walls.
Waveguide Impedance
The ratio of the transverse electric field to the transverse magnetic field for a
propagating mode at a particular frequency.
Z
TE
mn

u
 fc 
1  
 f 
2
Z
TM
mn
 u
 fc 
1  
 f 
2
where u is the intrinsic impedance of the propagating media. In air u = o =
120 
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Waveguide impedance of the TE11 and TM11 modes vs frequency for WR90
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Example: determine the TE mode impedance looking into a 20-cm long section of
shorted WR90 waveguide operating at 10 GHz.
Solution: at 10 GHz, only the TE10 mode is supported, so
Z
We can find Zin from
TE
10

120π Ω
 6.5GHz 
1 

 10GHz 
2
Z IN  jZ O tan( l )
 500 Ω
for a shorted line.
2f
2
Now  is found from    u 1   f c f  
1   fc f 
c
9
2 (10 10 Hz )
2

1 6.56GHz 10GHz   158 rad/m
8
3 10 m / s
2
So, l in the Zin equation is
l  (158 rad/m)(0.2 m)  31.6 rad
And Zin is then calculated as
Z IN  j (500 Ω) tan(31.6)  j100Ω
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General wave behaviors along uniform guiding
structures (1)
• The wave characteristics are examined along straight guiding structures with
a uniform cross section such as rectangular waveguides.
We can write E in the instantaneous form as
E ( z , t )  E 0 cos(t   z )
We begin with Helmholtz’s equations:
assume WG is filled in with a charge-free
lossless dielectric
 2 E  u 2 E  0
 2 H  u 2 H  0
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General wave behaviors along uniform guiding
structures (2)
We can write E and H in the phasor forms as
E ( x , y , z )  ( E x a x  E y a y  E z a z )e   z
and
H ( x, y , z )  ( H x a x  H y a y  H z a z )e   z .
Note that E x E y E z H x H y H z are functions of x and y only. The zdependency comes from the term e-z. In other words, for example

E x ( x, y, z )  E x ( x, y )e z aˆ x

E y ( x, y, z )  E y ( x, y )e z aˆ y and so on ….
2

2
2
2
2
2
2
so that  E  ( xy   z ) E  ( xy  2 ) E   xy E   E
z

 2xy E  ( u2   2 ) E  0
and
 2xy H  ( u2   2 ) H  0.
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Use Maxwell’s equations to show E and H in
terms of z components (1)
From
we have
and
  E   j H
Ez
  E y   j H x
y
Ez
 Ex 
  j H y
x
E y Ex

  j H z
x
y
H z
  H y  j Ex
y
H z
 H x 
 j E y
x
H y H x

 j Ez
x
y
  H  j E
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Use Maxwell’s equations to show E and H in
terms of z components (2)
We can express Ex, Ey, Hx, and Hy in terms of Ez and Hz
by substitution so we get for lossless media  = j, and  u   
Ex 
 j H z
j  Ez

u2   2 y u2   2 x
j H z
j  Ez
Ey  2
 2
2
u   x u   2 y
Hx 
j Ez
j  H z

u2   2 y u2   2 x
 j  H z
j Ez
Hy  2
 2
2
u   y u   2 x
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Propagating waves in a uniform
waveguide
• Transverse ElectroMagnetic (TEM) waves, no Ez or Hz
• Transverse Magnetic (TM), non-zero Ez but Hz = 0
• Transverse Electric (TE), non-zero Hz but Ez = 0
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Transverse ElectroMagnetic wave (TEM)
• Since Ez and Hz are 0, TEM wave exists only when
u2   2  0
u     
u p ,TEM 
ZTEM
1

rad / m
m/s
Ex
j



Hy
j



• A single conductor cannot support TEM
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Transverse Magnetic wave (TM)
From
2xy Ez  ( 2  u2 ) Ez  0
We can solve for Ez and then solve for other components from (1)-(4) by setting Hz
= 0, then we have
ZTM
Ex  E y




Hy
Hx
j

Notice that  or j for TM is not equal to that for TEM .
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Eigen values
We define
h 2   2   u2
Solutions for several WG problems will exist only for real
numbers of h or “eigen values” of the boundary value problems, each eigen value
determines the characteristic of
the particular TM mode.
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Cutoff frequency
From
  h2  u2
The cutoff frequency exists when  = 0 or h 2  u2  c2 
or
fc 
h
2 
Hz.
We can write
f 2
  h 1 ( ) .
fc
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a) Propagating mode (1)
2
 f 
 f  1
 c
f  fc
or
and  is imaginary
  2   h2
Then
  j   j u 1  (
h
u
)  j u
2
fc 2
1 ( ) .
f
This is a propagating mode with a phase constant :
  u 1  (
fc 2
)
f
rad / m
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a) Propagating mode (2)
Wavelength in the guide,
g 
2


u
 fc 
1  
 f 
2
m.
where u is the wavelength of a plane wave with a frequency
f in an unbounded dielectric medium (, )
2
up
u 


u f 
f
1
m
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a) Propagating mode (3)
The phase velocity of the propagating wave in the guide is

up  

uu
 f 
1  c 
 f 
2
g
 u

m/s
The wave impedance is then
ZTM
 fc 
  1  
 f 
2

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b) Evanescent mode
2
 f 
 f  1
 c
or
f  fc
  2   h2
Then
    u
fc 2
1  ( )  real
f
Wave diminishes rapidly with distance z.
ZTM is imaginary, purely reactive so there is no power flow
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Transverse Electric wave (TE)
From
2xy H z  ( 2  u2 ) H z  0
• Expanding for z-propagating field gets
where
2 H z 2 H z
2
2


(



)H z  0
u
2
2
x
y
We can solve for Hz and then solve for other components from (1)-(4) by setting Ez = 0, then we
have
H z  H z ( x , y )e   z .
Notice that  or j for TE is not equal to that for TEM .
ZTE
Ex  E y j



Hy
Hx


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TE characteristics
• Cutoff frequency fc,, g, and up are similar to those in TM mode.
• But
ZTE 

 fc 
1  
 f 
2

• Propagating mode f > fc
• Evanescent mode f < fc
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Ex2 Determine wave impedance and guide wavelength (in terms
of their values for the TEM mode) at a frequency equal to twice
the cutoff frequency in a WG for TM and TE modes.
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