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ECE 3065 Homework 6: Diffraction and Waveguides
Solutions
- Below is a sketch of the screen equivalent problem:
30m
15m
200m^ x
200.6m
f
i
O
The actual observation angle is given φ = tan
− 1 (x/ 13 .5). We note that the transmitter is
vertically polarized, corresponding to ‖−incidence in our problem. Thus, we may use the
following diffraction coefficient:
D
‖
(φ, φ i
− exp
−j
π
4
2 π
[
sec
φ − φ i
φ + φ i
)]
The resulting graph is
5 10 15 20 25 30 35 40 45 50
−
−
−
−
−
−
−
−
−
Received Power vs. Distance from Building Edge
x, distance (m)
Received Power (dBm)
Note that the signal performs better as the user steps away from the diffracting edge. However,
there is a local maximum where the increased distance begins to introduce more loss simply
due to extra distance traveled. This is a more graceful degradation, but certainly observable.
- The general solution for the TM n
in the parallel plate waveguide is given by:
Ez (x, y, z) = An sin
nπy
b
exp(−jβz)
E
y
(x, y, z) =
−jβA n
h
cos
nπy
b
exp(−jβz)
H
x
(x, y, z) =
jωAn
h
cos
nπy
b
exp(−jβz)
Now recall the boundary conditions for a perfect electric conductor:
ˆn ·
E = ˜ρS nˆ ×
H =
Js
For the bottom plate, we evaluate these boundary conditions for y = 0 and ˆn = ˆy to produce
˜ρ(z) =
−jβAn
h
exp(−jβz)
Js(z) =
jωAn
h
exp(−jβz) ˆz
For the top plate, we evaluate these boundary conditions for y = b and ˆn = −y to produceˆ
˜ρ(z) =
j(−1)
n βAn
h
exp(−jβz)
Js(z) =
j(−1)
n+ ωAn
h
exp(−jβz) ˆz
Note that currents for TM modes flow forwards and backwards along the waveguide. Inter-
estingly, for even modes the currents on the surface of the PEC flow equal and opposite to
one another, just like a transmission line. Note: it is OK to omit the term exp(−jβz), even
though it is incorrect and lazy – but of course your book did it in one of the solutions in the
back.
- The general solution for the TE n
in the parallel plate waveguide is given by:
Ex(x, y, z) =
jωμB n
h
sin
nπy
b
exp(−jβz)
H
y
(x, y, z) =
jβBn
h
sin
nπy
b
exp(−jβz)
Hz (x, y, z) = Bn cos
nπy
b
exp(−jβz)
For the bottom plate, we evaluate boundary conditions for y = 0 and ˆn = ˆy to produce
˜ρ(z) = 0
Js(z) = Bn exp(−jβz) ˆx
For the top plate, we evaluate boundary conditions for y = b and ˆn = −ˆy to produce
ρ˜(z) = 0
J
s
(z) = (−1)
n+ B n
exp(−jβz) ˆx
Note that currents for TM modes flow laterally in this waveguide. There is no surface charge.
(The EM nerd will, at this point, verify that both solutions in (1) and (2) satisfy the continuity
equation).
Since this is not listed as one of the first three cut-off frequencies, we can be assured that our
answer is correct. The actual cut-off for the TE/TM 11 modes is
(f c
11
μ 0
0
2
2
= 8.1 GHz
which would have been the fourth cut-off frequency.
The field distribution for the TE 10 is sketched below. The fields are uniform from top to bottom
with vertical polarization. Their amplitudes taper sinusoidally across the lateral dimension of
the waveguide.
Waveguide Cross-section
- Circular vs. Rectangular Waveguide:
(a) For a given cut-off frequency f , the perimeter of the circular waveguide is:
r =
f
μ
−→ Perimeter = 2πr =
f
μ
For a square wave with the same cut-off frequency:
a =
f
μ
−→ Perimeter = 4a =
f
μ
Clearly the circular guide is cheaper.
(b) The ratio between the second and the first mode cut-offs for the circular guide is:
Cut-off for TM 01
Cut-off for TE 11
The ratio between the second and the first mode cut-offs for the square waveguide is:
Cut-off for TE/M 11
Cut-off for TE 10
For the same initial cut-off frequency, the square guide should have more single-mode
bandwidth.
