Why electrostatics and magnetostatics are useful
In the last section we proved that the vector wave equation admits of purely
transverse (E and H) solutions that travel at the speed of light in the medium. We found
that these solutions are characterized b
P. 8.6: Given a uniform plane wave
E ( z , t ) x 2 cos(108 t
a)
The p
phasor form
z
3
) y sin(108 t
j z
j z j 2
3
E ( z ) x 2e 3 ye
v
z V
m
3
)
m
Frequency & wave number
108 f / 2 108 / 2 1.59 107 Hz
1/ 3 (m1 )
b)
The phase velocity
u k
Also
u
1
1.
EEE 341
Solution to Assignment #3
P.8-26
Solution
For normal incident
1+=
(1)
| 1.
(2)
1 + |
=3
1 |
(3)
where
when
| | = |, one solution is = 21 and = 12 .
Hence
S=
SdB = 20 log S = 9.54dB
(4)
P.8-38
Given i = c c = 2 , cos t = 0
(a) From (8.207)
(
(b) Fr
EEE 341
Solution to Assignment #4
P.8.37
(a) sin c = 21
Phase matching condition
sin t =
For i > c
1
sin i > 1.
2
cos t = j
From (8.196) and (8.197)
(1)
1
sin2 i 1
2
(2)
i = yEo ej1 (x sin i +z cos i )
E
(3)
i = Eo (
H
x cos i + z sin i )ej1 (x sin i +z
EEE 341
Solution to Assignment #5
P.9-15
Given Zo = 50 (distortion-less), R = 0.5 m
loss tangent
tan c =
=
(7.115)
(1)
G
(9.71)
C
= 0.0018
(2)
(3)
and
= 8000
(4)
R
G
= 0.0018 = 45.24 =
C
L
(5)
Solution
The circuit parameters are:
R
0.5
L= ( ) =
= 0.01105
P.8-add.
A
E0
B
EH
n
Solution. The critical angle,
EV
C sin 1
Et
C
2
1
sin 1
41.810
1
1.5
Total reflection occurs on the slant edge B (from glass to air) because the incident angle i 450 C .
This means the light ray changes direction at the slant edge b
EEE 341 Homework 10
P.11-6. f = 1MHz, length =15 m, r = 2 cm, Center fed copper rod, uniform current.
c 3 108
300
f
10 6
m,
2
2
, d 15 m ,
300 150
d
15
1 and uniform current
300
distribution, Hertz dipole.
2
a) Radiation resistance:
h
Rr 80 1.974 . (11
EEE 341, Homework 11
P.11-10.
Solution.
a)
Using Eq (11-44) with d S
2
Lx Ly
2
Rr 80
20 Lx Ly , or
2
4
M
Rr 0
, where M IS , I current, S=loop area
6 I 2
2
2
b)
Lx Ly
Rl
Rs ,
b
c ) For f 1 ( MH z ),
r
Rr
Rr Rl
Lx Ly 1 ( m),
2 / 300 ( rad / m)
P.10-add
P.10-add. An attenuator can be made using a section of waveguide operating below cutoff, as
shown below. If a = 2.85 cm and the operating frequency is 10 GHz, determine the required
length of the below-cutoff section of waveguide to achieve an at
P.10-add. A hollow circular waveguide of radius a = 5 cm is operating at f = 3 GHz of the lowest mode.
For this mode, find its cutoff frequency and wave impedance. Hint: use Table 10-2 and 10-3 on p. 565.
Fig. P.10-add: geometry of an air
filled circular
EEE 341, Homework 09, Spring 2012
P.10-29
For a straight waveguide of a semicircular cross-section, showing in Fig. 10-26,
a). Write the appropriate expression of Ez0 for TM modes
b). Write the appropriate expression of H z0 for TE modes
c). Explain how t
One more numerical method
Laplaces equation can be written in finite difference form by first
expressing the first derivatives as central differences and then the
second derivative as central differences of the first derivatives.
For instance look at a
The wave equation I:
2. General propagation in one dimension.
Let us now derive the most general TEM wave traveling in the z-direction without
any assumption about its x-y dependence. We will start this time with the E-field for
reasons that will become o
Application of the Method of Moments to the problem
of finding the Impedance of aTransmission Line
Motivation: In electrostatics we first learn that if we had highly
symmetric charge distributions, we could calculate the D field
everywhere around them us
Lecture 5. The Plane Wave - Plane Slab problem in the time domain
The prototypical interaction between waves and materials is the interaction between plane waves
and plane slabs. In this section we analyze this problem in its simplest form by assuming the
Section 2, Lecture 4. The Plane Wave in the Time Domain
This derivation and the next few lectures are inspired by Heaviside's approach in his
Electromagnetic Theory. The idea is to obtain a valid solution to the vector wave equation by
obtaining a space-t
The wave equation I:
1. Simplest propagation in one dimension.
The one-dimensional wave equation has extensive applications in electromagnetic
engineering. The vast majority of them occur in the frequency domain because most of
our engineering instrumenta
Summary of the lecture on the halfspace
All you need to know is the definition of wave impedance and the right
hand rule and you can prove the following.
Assume you have a TEM (plane) wave approaching aboundary along the
z axis. For simplicity I assume Ei
Lecture 6
The Electromagnetic wave equation
in one Dimension
Although we started in the Time Domain
Most EM applications take place in the
frequency domain.
Mostly for historical reasons. Frequency
domain signals are relatively easy to
manipulate and fi
EEE 341 Homework Solutions Week 7
JTA-1. Modify the MATLAB routine for the dielectric slab waveguide and use it to find the solution
for (kyd,y0) for the dominant mode at 10 GHz for 0.030 in thick RO4003C (DK = 3.38) grounded
dielectric slab. (Assume loss