ECE 520 HW#7 Solutions
4.13.
Solution:
(a) Assume that the conductor coincides with the xyplane and the outer surface of the
coating is located at z = t . For the fields, we have
E1 = xE0 e j 0 z + xRE0 e j 0 z
z < t
E 2 = xTE0 e j 2 z xTE0 e j2 z
t < z
HW#10 Solutions
5.8
First, if we assume that
Ez 0
, or
A1 A2 0
, then (5.3.32) can be simplified
to
B1
cos k1 y h
B2
2
1t
k
cos k2 y (b h)
k22t
Comparing the above with (5.3.30):
B1 cos k1 y h B2 cos k2 y (b h)
we find that
k12t k22t
Since
kz
.
is the sa
HW#2 Solutions
1.17.
Solution:
Here we derive
d
dl = dt D dS + I
C
from its differential form and boundary
S
conditions, and the other equations can be derived in a similar way.
The equation in differential form is
H =
D
+J
t
(1)
Consider an open surfac
HW#4 Solutions
3.1)
The current loop is equivalent to a magnetic dipole with Kl = j IS . By using the image
theory, the original system is equivalent to a 4dipole system radiating in free space,
where the four dipoles are:
dipole 1:
dipole 2:
dipole 3:
d
HW#3 Solutions
2.1.
Solution:
Given
A (r ) =
4
V
J (r)
dV
R
and
R = r r
B = A
we can find B(r ) as
B (r ) =
4
V
J (r )
dV
R
The volume integral over the onedimensional current source in question becomes a line
integral over the same integrand:
B (r )
HW#1 Solutions
1.1)
Definition of the divergence:
1
f ds
v 0 v
s
f = lim
RECTANGULAR
In Cartesian space, v can be chosen as a cube of dimensions x, y, and z , such that
v = xyz , centered at (x, y, z). Therefore,
1 6
f ds n
v 0 v
n=1 sn
f = lim
wh
ECE 520
FIRST MIDTERM EXAM
Thursday, September 29,
20ll
7:00 p.m.  9:00 p.m.
Name tfu;*i.cfw_' 1*
Signature
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cfw_*+a Y
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ECE 520
FIRST MIDTERM EXAM
Thursday, October 4, 2OlZ
7=0O p.m. 
Name
9:00 p.m.
//,'rcfw_o* /
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grading. Write your answers clearly in
ECE 520
FIRST MIDTERM BXAM
Thursday, October 3, 2013
7:00 p.m.  9:00 p.m.
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l
ECE 520
SECOND MIDTERM EXAM
Tuesday,
April 12,20ll
7:00 p.m.  9:00 p.m.
Name flq .q:t; an"n f.t*"l,r o "l
J
\D \
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NetID
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NOTE: This is an openbook exam.
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Writ
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ECE 520
SECOND MIDTERM EXAM
Thursday, November 4, 201 0
7:00 p.m.  9:00 p.m.
Name
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/
ECE 520
SECOND MIDTERM EXAM
Thursday, November 3, 2011
7:00 p.m.  9:00 p.m.
Name le  Wui C harl
Signature
.,' dJr'L\*: r.rCt' .' /t,'fC vtu
IJ
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ECE 520
SECOND MIDTERM EXAM
Tuesday, November 13, 2Ol2
7:00 p.m.  9:O0 p.m.
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/1
,>/."( (^/r/

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lLrG ,i.*, !t
:'
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2*
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ECE 520
SECOND MIDTERM EXAM
Wednesday, November 6, 2013
7:00 p.m.  9:00 p.m.
Name Oui,>
Lr] ;
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ij
i
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cfw_
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HW#5 Solutions
3.5)
By the equivalence principle, we can replace the gap between the two halves of the
antenna with a conductor to achieve the shape of the antenna as a cylinder with length L
and (unchanged) radius a as long as we introduce an the appropr
ECE 520 HW#10 Solutions
6.1.
Solution:
A. The analysis can be carried out either through Az and Fz or through Ez and H z . Lets
consider the first approach.
TEz modes:
Starting from the general solution of Fz in cylindrical coordinates:
Fz = AJ m (k ) + B
HW#2 Solutions
1.17)
Here we derive
d
dl = dt D dS + I
C
from its differential form and boundary
S
conditions, and the other equations can be derived in a similar way.
The equation in differential form is
H =
D
+J
t
(1)
Consider an open surface S with c
HW#8 Solutions
4.19) Because the incident wave has a polarization along the passing axis of the first Polaroid,
it will pass through unmodified with amplitude E0. Decomposed into fastaxis and slowaxis components (denoted as u and v, respectively), the wa
HW#3 Solutions
2.1)
Given
A (r ) =
4
V
J (r )
dV
R
R = r r
and
B = A
we can find B(r ) as
B(r ) =
4
V
J (r )
dV
R
The volume integral over the onedimensional current source in question becomes a line
integral over the same integrand:
B(r ) =
J (r )
HW#9 Solutions
5.1)
TM modes:
Az = ( A sin k y y + B cos k y y )e jkz z
1
H=
H = x
E=
E = y
k ykz
A =
ky
1
j
1 Az
A
y z
x
y
x
( A cos k y y B sin k y y )e jkz z
H =
( A cos k y y B sin k y y )e
1 H x
H x
z
y
j z
y
jk z z
+ z
ky2
( A sin k y y
HW#6 Solutions
4.8)
Assume that E = E0 (2 x + jy )e j 0 z , then
H=
1
z E =
0
1
0
E0 (2 y jx )e j 0 z
Therefore, the timeaverage Poynting vector is
S av =
1
5
2
Re(E H*) = z
E0
2
20
From S av = 10 6 W/m 2 we can obtain
E0 = 1.23 102 V/m
and Emax = 2 E0 =
1.1 (a) In rectangular coordinates:
We take v to be the space occupying [ x x / 2, x + x / 2] in the x direction
and [ y y / 2, y + y / 2] in the y direction and [ z z / 2, z + z / 2] in the z
direction. Then v = xyz and
f ds [ f
x
( x + x / 2, y, z ) f
HW#11 Solutions
*
6.1 Q: Analyze TE and TM modes in halfcircular waveguide, determine propagation
constant and modal fields for all possible modes.
TMz mode:
Firstly, we write down general expression of E z :
Ez = E0 ( am J m (k ) + bmYm (k ) ) ( cm cos(
HW#7 Solutions
4.14)
For the perpendicular polarization case, we have the three electric fields
E1 = yE0 e j0 ( x sin i + z cos i ) + yR E0 e j0 ( x sin i z cos i )
E 2 = yAe jd ( x sin d + z cos d ) + yBe j d ( x sin d z cos d )
E 3 = yT E0 e j0 ( x sin
HW Solution 12
0 e jky with constant E0 ,
6.13 Q: timeharmonic plane wave, incident field E inc = zE
incident toward conducting halfcylinder placed on infinitely large
conducting ground. Find field scattered by the halfcylinder.
A:
If the there is no
ECE 520 HW#11 Solutions
6.7.
Solution:
TMPolarization
The electric field of a TMpolarized incident plane wave can be expressed as
Einc = zE0e jkx = zE0
j
n
n =
J n (k )e jn
The scattered field propagates away from the cylinder, and can be expressed as
ECE 520 HW#8 Solutions
4.17.
Solution:
(a) For nonmagnetic media, the Brewster angle will cause total transmission of the
component with parallel polarization, leaving only a reflection of perpendicular
polarization. Thus, the electric field of the reflec
HW #6 Solutions
4.7.
Solution:
(a) The magnetic field is
H= y
E0
(e j z e j z )
Therefore,
Ex
Hy
=
z=d
e j d + e j d
= Zd
e j d e j d
from which we find the reflection coefficient
=
Z d 2 j d
e
Zd +
(b) The SWR for z < d is
j d
j d
e
+ e
E
SWR = max = j
ECE 520 HW#9 Solutions
5.1.
Solution:
TM modes
For the TM modes, we assume that
Az = ( A sin k y y + B cos k y y )e jkz z
from which we obtain the fields:
H=
E=
1
A =
1
j
= y
1 Az
A
y z
x
y
x
H =
k ykz
k
jk z z
y
= x ( A cos k y y B sin k y y )e
1 H x
ECE 520 HW#12 Solutions
7.1.
Solution:
Since the field region is a r b , which doesnt include the origin, both J n (kr ) and
Y (kr ) should be included in the solution for F and A . Since the field region includes
n
r
r
the zaxis, where the field is fini