ECE 350
Homework 1
Due: Tue, Sept 1, 2015, 5 PM
1.
a) Take the divergence of Amperes law, and then make use of r r H = 0 as well as Gausss
law, r D = , to derive the continuity equation
@
+ r J = 0.
@t
b) Which conservation principle does this equation re
ECE 350
Homework 12 Solution
Due: Tue, Nov 15, 2011, 5PM
1. As mentioned, X-band is between 8.2 f 12.4 GHz. Recall, the cuto frequency for the rectangular
waveguide is given as
mc 2
nc 2
fc =
+
,
(1)
2a
2b
where when f > fc the particular mode propagates.
ECE 350
Homework 1 Solution
Due: Tue, Sept 1, 2015, 5PM
1.
a) Since divergence of curl is always zero, we have from Amperes Law
rH=J+
@D
@
) 0 = r J + r D.
@t
@t
Using Gauss law r D = , we obtain
0=rJ+
@
,
@t
which is the continuity equation.
b) This equa
ECE 350
Homework 6 Solutions
Due: Fri, July 11, 2014, 5 PM
1.
a) For r = (0, 0, 10) km, the angles x and y are 90 . Also x = and y = this can be
x
y
r cos y
y
x
x = r cos x and y =
seen gometrically or using the formulas
sin x
sin y , respectively, from
ECE 350
Homework 5 Solutions
Due: Tue, July 8, 2014, 5 PM
1. Graphically, the currents related to each z -polarized short dipole can be represented in the xy-plane
as shown below. Note the observation point is in the far-eld at point (r cos , r sin , 0) =
ECE 350
Homework 1 Solution
Due: Tue, Aug 30, 2011, 5PM
1. Utilize boundary conditions:
n [D+ D ] = s : The normal components of the electric ux density across an interface are
discontinuous
n [B+ B ] = 0: The normal components of the magnetic ux density
ECE 350
Homework 1
Due: Tue, Jan 28, 2014, 5 PM
1.
a) Take the divergence of Amperes law, and then make use of H = 0 as well as Gausss
law, D = , to derive the continuity equation
+ J = 0.
t
b) Which conservation principle does this equation represent mat
ECE 350
1.
Homework 1 Solution
Due: Tue, Jan 27, 2015, 5PM
a) Since divergence of curl is always zero, we have from Ampere's Law
H=J+
Using Gauss' law
, we obtain
D
0=
t
J+
t
D.
D=
0=
J+
,
t
which is the continuity equation.
b) This equation represents
ECE 350
Homework 2 Solution
Due: Tue, Jan 31, 2012, 5PM
1.
a) Using the Stokes theorem for the closed path integral
in terms of the magnetic eld B.
A dl =
C
S
A dl we obtain the following expression
A dS
S
=
C
B dS
b) Circulation of A = xAx over path C1
ECE 350
Homework 1 Solution
Due: Tue, Jan 24, 2012, 5PM
1.
a) Apply the divergence to both sides on the point form of Amperes Law:
H=J+
D
t
D
, the given vector identity states that the divergence
t
of a curl of a vector is zero. Recall from ECE 329 that
ECE 350 Summer 2011
Homework 1 Solutions
Due: Friday, June 17, 2011, 5PM
1. Utilize boundary conditions:
n [D+ D ] = s : The normal components of the electric ux density across an interface are
discontinuous
n [B+ B ] = 0: The normal components of the mag
ECE 350
Homework 3 Solutions
Due: Tue, Sep 15, 2015, 5 PM
1. For a Hertzian radiation source with the current density,
J(r, t) = z I0 4z (x + 100 ) (y
where the oscillation frequency f =
a) The wavelength is
=
c
f
b) The wavenumber is k =
=
2
!
2
3108
310
ECE 350
Homework 9 Solutions
Due: Tue, Mar 31, 2015, 5 pm
1. In a homogeneous collisionless plasma, the plasma frequency is given as
s
N e2
!p =
.
m0
(1)
a) Given the correct numerical values for the electron mass and charge in MKS, the plasma frequency c
ECE 350
Homework 5 Solutions
Due: Tues, Feb 24, 2015, 5 PM
1. (Optional Problem) The superposition of the radiation eld of the two short dipoles given in the
problem is given by
jkr
L
e
E(r) = jo Io k [sin x x + j sin y y ]
.
(1)
2
4r
a) The total electr
ECE 350
Homework 10 Solution
Due: Tue, Nov 1, 2011, 5PM
1.
a) Given (x, t) = ej (tkx) then in
j
2 2
=
,
t
2m x2
the result yields:
j (tkx)
e
t
j
2
ej (tkx)
2m x x
2
jk ej (tkx)
2m x
=
j .j ej (tkx) =
2
2m
=
(jk)2 ej (tkx)
2m
j .j ej (tkx) =
k2 = dispersi
ECE 350
Homework 4 Solutions
Due: Tue, July 1, 2014, 5 PM
1.
a) Refering to the solution of P2 in HW 3, we should generalize the expression of the radiation
eld so that it can be dened for any location in spherical coordinates. Thus, replacing by
x
sin ,
ECE 350
Homework 2
Due: Tue, Jan 31, 2012, 5 PM
1.
a) Given the vector potential A, the magnetic ux density is B = A (by denition). Express
the closed path integral C A dl around a path C in terms of the integral of B over a surface
bounded by path C . Hi
ECE 350
Homework 3
Due: Tue, Sept 18, 2012, 5 PM
1. (continued from HW 2 complete the problem starting with part d)
A Hertzian radiation source with the current density
J(r, t) = z Io z (x
100 ) (y ) (z
100 ) cos(! t)
A
,
m2
!
is embedded in free space an
ECE 350
Homework 6 Solutions
Due: Tue, Mar 3, 2015, 5 PM
1.
a) The Rayleigh distance is given by
Rray =
2L2
=
2 182 1109
= 2160 m.
3 108
The distance to observation point, r must be greater than 216 m for the far-eld approximation
to be valid.
b) A plane
ECE 350
Homework 8
Due: Thu, March 14, 2013, 3 PM
1. A plane-wave with eld phasor
V
m
propagating in air in region 1 (x < 0) is incident on x = 0 plane, which happens to be an air/conductor
interface plane. Assume that the conductivity = 2 in medium 2 (x
ECE 350
Due: Fri, June 27, 2014, 5 PM
Homework 3
1. A Hertzian radiation source with the current density
J (r, t) = xIo x(x 100)(y)(z 100)cos(t)
A
,
m2
is embedded in free space and its oscillation frequency f = 2 = 300 MHz. Assuming that Io x =
1 A.m, de
ECE 350
Homeworks 11&12
Due: Mon, Apr 20, 2015, 5 PM
1. Consider a dielectric-lled parallel-plate waveguide with
lling is
o
and its refractive index is
a=2
cm. The permeability of the dielectric
n = 1.5.
a) Which TEm and TMm modes can
propagate (instead o
ECE 350
Homework 7
Due: Tue, Mar 10, 2015, 5 PM
1. For each uniform plane TEM wave described below in terms of its wave vector
corresponding (i) wavelength
and
,
(ii) wave frequency
k
determine the
(assume free space propagation), (iii) angles
describing
ECE 350
1.
Homework 4 Solutions
(Optional Problem)
Due: Tues, Feb 17, 2015, 5 PM
The four edges of the small squared loop will be constitude by the four Hertzian
dipoles. The dipoles are numbered as seen in the following gure.
y
x #2
#3
y
#1
x
r
#4
a) The
ECE 350
Homework 7
Due: Tue, Oct 11, 2011, 5 PM
1. A TE polarized plane TEM wave with electric phasor (written in mksA units)
Ei = y Eo ej (x+z ) , x < 0,
is incident on the x = 0 plane, which is the interface between medium 1 (x < 0) and medium 2
(x > 0)
ECE 350
Homework 9 Solutions
Due: Thu, Mar 26, 2013
1. In a homogeneous collisionless plasma, the plasma frequency is given as
N e2
.
m0
p =
(1)
a) Given the correct numerical values for the electron mass and charge in MKS, the plasma frequency can be wri
W.C.Chew
ECE 350 Lecture Notes
8. Examples on Using the Smith Chart
(a) Find the voltages at A on the transmission line.
Z0 = 50 W, v = 1.5 108 ms1
20
Vs =
10 sin t
volts
Zs
A
B
ZL (30+j25)
z = l = 1 m
25 MHz
z=0
The voltage source sets up a forward goi
ECE 350
Homework 1
Due: Tue, Sept 5, 2017, 5 PM
1.
a) Take the divergence of Amperes law, and then make use of H = 0 as well as Gausss
law, D = , to derive the continuity equation
+ J = 0.
t
b) Which conservation principle does this equation represent mat
ECE 350 Sum 2011
Homework 8 Solutions
Due: Tue, July 19, 2011, 5 PM
1. Based upon lecture21 Example 1.
a) We have
M
= + kv
E
= kv
if we add two equations together,
M + E = 2
M + E
=
2
= 3109 rad/s.
As for the wavenumber and wavelength, we have
= 10
c
2
ECE 350
Homework 9
Due: Friday, Jul 22, 2011, 5 PM
1. The propagation of ocean waves on the surface of deep water (depth much longer than a wavelength
= 2 ) is governed by the dispersion relation
k
=
gk
where g represents gravitational acceleration and ,
W.C.Chew
ECE 350 Lecture Notes
2. Review of Vector Analysis
A vector A can be written as
A = x^Ax + y^Ay + z^Az :
(1)
Similarly, a vector B can be written as
B = x^Bx + y^By + z^Bz :
(2)
In the above, x^ y^ z^ are unit vectors pointing in the x y z direct
W.C.Chew
ECE 350 Lecture Notes
4. Using Phasor Techniques to Solve Maxwell's Equations
For a time-harmonic (simple harmonic) signal, Maxwell's Equations can
be easily solved using phasor techniques. For example, if we let
H = <eH~ ej!t ]
(1)
E = < E~ j!t]
W.C.Chew
ECE 350 Lecture Notes
11. Lossy Transmission Lines.
When R and G are not zero, we have a lossy transmission line. In this
case,
V (z) = V0(e;z + v e+z )
(1)
where
p
p
= ZY = (j!L + R)(j!C + G) = + j:
The current is derived using the telegrapher'
W.C.Chew
ECE 350 Lecture Notes
27. Radiation Field Approximations
The vector potential due to a source J(r), can be calculated from the
equation
ZZZ
A(r) =
dr 4jJr(;r )r j e j
(1)
0
0
0
jr;r j
;
0
V
where V is the volume occupied by J(r).
J(r)
z
|r r|
r