ECE 329
Homework 3
1. Given that V (x, y, z) = 4x2 y
yz 2 and E =
Due: Fri, June 27, 2014, 5PM
rV , what is r E?
2. Is E = 2 + 4z 2 x a possible electrostatic eld? Discuss.
x
y
3. An important vector identity which is true for any vector eld A(x, y, z) is
ECE 329
Homework 6
Due: Tue, July 8, 2013, 5PM
1. In this problem we will compare the magnitudes of electrical force qE and magnetic force qv B on
a moving test charge q due to elds E and B produced by another charge Q at some distance r.
Q
. Also let B t
ECE 329
Homework 2
Due: Jan 28, 2008, 5PM
1. In this problem we will compare the magnitudes of electrical force qE and magnetic force qv B on a moving test charge q due to elds E and B produced by another charge Q at some distance r. a) Let E be
ECE 329
Homework 3 Solutions
Due: Fri, June 24, 2016, 5PM
1. In electrostatics, we generate a curl-free vector field E(x, y, z) if we take the gradient of a scalar
function V (x, y, z). Therefore, E = (V ) = 0.
2. Maxwell equations for electrostatics stat
ECE 329
Homework 8 Solutions
Due: Tue, July 12, 2016, 5PM
1. Since the closed paths are not varying in time and the magnetic field B is independent of position,
we can rewrite Faradays law as follows
E=
I
C
E dl =
dB
dt
Z
S
dS =
dB
n
Area,
dt
where n
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ECE 329 Fields and Waves I Summer 16
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Exam 1
Fri, July 1, 2016 — 11 AM - 11:50 AM
Name:
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Please clearly PRINT your name in CAPITAL LETTERS and circle your section in the above boxes.
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ECE 329
Homework 8 Solutions
Due: Tue, July 15, 2014, 5PM
1.
a) Solution of the left-hand side of the Faradays law r E =
x
rE=
y
z
@
@x
@
@y
@
@z
0
cos(!t + x)
which must be equal to
@B
@t .
@B
@t
is given by
@
(sin(!t + x) z =
@x
=
0
sin(!t + x),
z
Thus,
ECE 329
Homework 6 Solutions
Due: Wed, July 9, 2013, 5PM
1.
a) Referring to the hint, we expect E, B, and Vs to be orthogonal to one another. Based on this
assumption, we can write
Q
|E| =
4o r2
for the magnitude of the electric eld, and
|B| =
o I|dl|
4r2
ECE 329
Homework 7 Solutions
Due: Fri, July 11, 2014, 5PM
1.
a) The current in the outer loop can be calculated by Ia =
remember from Lecture 10. Thus, we write
Ia =
V
V
= 2a =
R
r2
5
21
4107 0.0012
V
R,
where R
1
G
=
d
A
as you may
= 100 A.
b) Refering
ECE 329
Homework 8
Due: Tue, July 15, 2014, 5PM
1.
a) If
V
,
m
= c, and = o , nd the corresponding H by using Faradays law
E=
!
cos(!t + x) y
rE=
@B
.
@t
Hint: nd a suitable time varying anti-derivative for r E.
b) If
H = sin(!t
= 0,
!
z) x
A
,
m
= 2 c an
ECE 329
Homework 7
Due: Fri, July 11, 2014, 5PM
1. Consider two concentric circular wire loops of radii a = 1 m and b = 0.02 m placed on the x y plane
of the reference coordinate system with their centers at the origin. The medium is free space. The
condu
ECE 329
Homework 4 Solutions
Due: Tue, July 1, 2014, 5PM
1.
a) Using E =
rV , the electric eld E at x =
d
2
is calculated as
@ (x/d)4/3
x
@x
8 x1/3
x
d
3 d4/3 x= 2
8
23 V
x .
3d m
E(x) =
Va
=
=
b) Using Gausss law = r D, the volumetric free-charge density
ECE 329
Homework 5 Solutions
Due: Thu, June 30, 2016, 5PM
1.
S are fresh water, ground conductivity of
a) Examples of materials with conductivity near 103 m
S
S
S
3
4
Earth (10 m ), dry earth (10 m ), and wet earth (102 m
).
b) Examples of materials with
University of Illinois
Spring 2016
ECE 310: Problem Set 13
Due: 5pm, April 29, 2016
1. Complete the following signal flow diagram (butterfly structure) of a 4-pt, radix-2, decimation-intime FFT algorithm. Specify all the connection weights and determine t
ECE 329
Homework 13 Solutions
Due: Thu, July 28 2016, 5 PM
1. Bonus Problem :
The magnetic field of a plane wave propagating in a non-magnetic material ( = o = 4 107 H/m)
is given by
A
H = 25 ez cos(8 106 t 3z ) x
.
3
m
By inspection, we have the
frequen
ECE 329
Homework 12 Solutions
Due: Tue, July 26, 2016, 5 PM
1. Bonus Problem : Solution of this problem will be given within the solutions of HW13.
2.
a) Since
= 2103 49 109 = 89 106 1, we use the approximations for good conductors, i.e.
4
= = f . Theref
ECE 329
Homework 10 Solutions
Due: Tue, July 19, 2016, 5PM
1.
a) The vector wave field E(z, t) is given by
E(z, t) = 4(
t z/c
V
) y .
m
b) The associated wave field H(z, t) is
H(z, t) =
t z/c
4
A
(
)x
.
o
m
c) The Poynting vector is
S=EH=
16 2 t z/c
W
ECE 329
Homework 11 Solutions
Due: Fri, July 22, 2016, 5PM
A
1. The pulse of sheet current Js (t) = x
Jso cos(t) m
will produce magnetic and electric fields as shown
in the following figure.
z
S+ = E+ H+
y
E+
Js(t) = xJso(t)
H+
x
H
E
S = E H
a) The averag
ECE 329
Homework 14 Solutions
Due: Tue, Aug 02, 2016, 5 PM
1. This problem is similar to that given in Example 3 in Lecture 30.
a) The voltage in the first line will be a sum of incident V + and reflected voltages V , i.e. V1 =
V + + V whereas in the seco
ECE 329
Homework 1 Solutions
Due: Fri, June 17, 2016, 5PM
1. Review of vectors :
a) D = A + B = 2
x + 3
y 4
z.
b) A + B 4C = 2
x + 3
y 4
z 4(3
x y + 2
z ) = 10
x + 7
y 12
z.
p
c) |A + B 4C| = (10)2 + 52 + (10)2 = 293 17.12.
d) A + 2B C = 4
x + 5
y 8
z.
p
ECE 329
Homework 9 Solutions
Due: Thu, July 14, 2016, 5PM
1.
a) Referring to the hint, we should first find the divergence of the current density. Thus, we write
J=
2
3
4z +
3x y +
2z(y yo )2 = 3x3 + 2(y yo )2 .
x
y
z
Then, taking the integral of bot
ECE 329
Homework 2 Solutions
Due: Tue, June 21, 2016, 5PM
1.
a) Expressing the equation
H
S
D dS =
R
V
dV in terms of its units, we write
D=
[ mC3 ] [m3 ]
c
= [ 2 ].
2
[m ]
m
This makes sense because electric displacement gives the electric flux density.
ECE 329
Homework 4 Solutions
Due: Tue, June 27, 2016, 5PM
1.
a) In vacuum, the displacement vector is D = o E + P. Thus, the displacement field between the
plates is
C
D = 4o z 2 ,
m
from which we obtain the polarization field as P = 0 mC2 .
b) Since D =
ECE 329
Homework 7 Solutions
Due: Fri, July 8 2016, 5PM
1.
a) The 2-D sketch of the described configuration is shown below.
y(m)
3
1
2
-1
x(m)
1
1
-1
4
b)
i. Case : |x| < 1, and |y| < 1
The magnetic field distribution of the system of current sheets in th
ECE 329
Homework 6 Solutions
Due: Wed, July 6, 2016, 5PM
1.
a) The capacitance of the capacitor will be
C=
A
o (2 102 )2
=
2.23 pF.
d
5 103
b) The conductance of the same capacitir will be
G=
A
103 (2 102 )2
= 8 105 S,
=
d
5 103
whereas corresponding res
ECE 329
Homework 5 Solutions
Due: Thu, July 3, 2014, 5PM
H
1. Applying Gauss law S B dS = 0 and dening the closed surface S composed by the partial cone
surface S1 , the semicircle S2 with radius R, the semicircle S3 with radius 2R, and the trapezoid S4
w
ECE 329
Homework 2 Solutions
Due: Tue, June 24, 2014, 5PM
1.
a) According to Gausss law, the electric ux will be
I
Z
Z
1
1
8
E dS =
dV =
1dV =
Vm,
"0 V
"0 V
0
S
where its unit is volts-times-meter.
b) For (x, y, z) = x2 + y 2 + z 2
I
1
E dS =
"0
S
C
,
m3
Probability with Engineering Applications
ECE 313 Course Notes
Bruce Hajek
Department of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
June 2015
c 2015 by Bruce Hajek
All rights reserved. Permission is hereby given to free
University of Illinois
Fall 2014
ECE 313: Hour Exam I
Monday, March 3, 2014
7:00 p.m. 8:15 p.m.
228 Nat. History (Sec. E/9am and C/10am) & 314 Altgeld (D/11am and F/1pm)
1. (a) The solution is n2 . For each choice that Tom can make (for a total of n), Jer
University of Illinois
Fall 2014
ECE 313: Hour Exam I
Wednesday, October 8, 2014
7:00 p.m. 8:15 p.m.
Section B in 151 Everitt Lab, Sections C & E in 141 Loomis Lab, Section D in 100 MSEB
1. (a) There are n2 possibilities for which two cells fail, and n 1
University of Illinois
Fall 2014
ECE 313: Hour Exam I
Monday, March 3, 2014
7:00 p.m. 8:15 p.m.
228 Nat. History (Sec. E/9am and C/10am) & 314 Altgeld (D/11am and F/1pm)
1. [18 points] The Sweet Dreams Cookie Store sells n types of cookies, kept in n sepa
University of Illinois
Spring 2015
ECE 313: Hour Exam I
Wednesday, March 4, 2015
7:00 p.m. 8:15 p.m.
MSEB 100, ECEB 1002, ECEB 2017
1. [20 points] The two parts of this problem are unrelated.
(a) Suppose that you have five distinct bowls, each can hold up