EE383L-Fall 2012
Problem Set 1
Due date: Sep. 11, 2012.
All problems are worth 10 points except those indicated by *, which are worth 20 points.
Problem 1*
Problem 1.1 in the textbook.
Problem 2*
Given F = rk1 / r + zk2 / z , evaluate the surface integral
EE383L Fall 2012
Problem Set 4 - Solutions
Problem 1
Using the vector identity (vg ) g v g v , we have
e jkr
e jkr e jkr
f (, )
f (, )
r f (, )
r
r
e jkr e jkr
e jkr e jkr
f (, )
f (, ) jk
2 f (, )
r
r
r
r
(1)
(2)
In the far-field (kr>1), terms
EE383L-Fall 2012
Problem Set 9
Due date: Tuesday, Dec. 4, 2012.
All problems are worth 10 points except those indicated by *, which are worth 20 points.
Problem 1*
z
A uniform plane wave with Ei ( z , t ) = xEio cos[ (t )] in medium 1 (1 , 1 ) is incident
EE 383L Fall 2012
Problem Set 3 Solutions
Problem 1
Consider the time domain (TD) Poynting Theorem
(HMimp E Jimp )dv (E H)ds
V
S
( | H |2 | E |2 )
d
dv
2
dt V
| E |2 dv
(1)
V
For any two vector quantities A and C represented as phasors, A(t )
Ae j t A
EE383L-Fall 2012
Problem Set 4
Due date: Oct. 4, 2012.
All problems are worth 10 points except those indicated by *, which are worth 20 points.
Problem 1
Show that in the far-field region any function of the form (f (, )e jkr / r ) can be
approximated as
EE383L-Fall 2012
Problem Set 3
Due date: Sep. 27, 2012.
All problems are worth 10 points except those indicated by *, which are worth 20 points.
Problem 1
The (time domain) Poynting theorem is given as follows for a volume V bounded by a
surface S that is
EE383L Fall 2012
Problem Set 2 Solutions
Problem 1
Problem 2
Problem 3
Problem 4
(a)
Consider a small cylindrical volume around the interface boundary as shown below.
Using
J dS 0 over the cylindrical surface and assuming the volume to be small (l 0) , t
NAME: _
PROBLEM 1
Consider a simple conducting medium with constant permittivity , permeability , and
conductivity and let a vector potential A and a scalar potential be defined such that
A(r , t ) = B(r , t ) and (r , t ) A(r , t ) = E (r , t ) . Derive
EE383L Fall 2012
Homework 1 Solutions
Problem 1
Problem 2
The cylinder has three surfaces: the top face, the bottom face, and the side wall. The three integrals can be
evaluated separately:
S
F ds
F ds
top face
2
bottom face
2
2
Fzd d
F ds
top face
F ds
0
EE383L-Fall 2012
Problem Set 2
Due date: Sep. 18, 2012.
All problems are worth 10 points except those indicated by *, which are worth 20 points.
Problem 1
Problem 1.13 in the textbook.
Problem 2*
Problem 1.15 in the textbook.
Problem 3
Problem 1.17 in the