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ECE 450 Fall 2009
Homework 2
Solutions
Due: Tuesday, Sep 8
th
, 2009
1)
[]
3
ˆˆˆˆ
(,)
(
)
(
)
(
)
(
)co
s
( )
C
t
q
hx
hx
hy
hy
t
m
ρδ
δ
ω
=−
−
+
+
−
−
+
rrrrr
, essentially you have
four point charges in space that exist in the
x

y
plane
only
at points
േ
and
േ
.
ˆ
hx
ˆ
hy
a)
Since we assume
t
ρ
r
is a time harmonic source, i.e., it oscillates at single frequency
we can develop the phasor representation as follows:
{}
{
()
'
()co
s
(
)
'( ) Re
Re
'( )
Re
( )
jt
jt j
tt
e
ee
e
ωα
}
ρω
+
α
=
+
=
=
=
rr
r
r
r
±
(1)
Thus,
r
±
is the phasor or complex quantity. The phase of the sinusoid,
α
,
has been absorbed
into the phasor and must be accounted for. In
t
r
we see that
α
=0, therefore
3
(
)
C
q
h
xh
yh
y
m
=
−−++−−+
r
r
r
±
♠
*To find,
t
t
∂
∂
r
, one can easily take the time derivative of
t
r
and find that,
3
(
,
)
(
)(
)
s
i
n
(
)
tC
t
q
hx
hx
hy
hy
t
tm
δδδδ
∂
==
−
−
−
+
+
−
−
+
∂
r
r
r
r
♠
.
1
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View Full Document In phasor form, one can replace
t
j
ω
∂
∂
→
due to the time harmonic dependence (the time to
frequency domain translation for harmonic sources) and simply multiply
()
ρ
r
±
by
j
and bring
the expression back into the time domain via the expression in (1).
That is,
{}
[]
3
(,)
Re
( )
ˆˆˆˆ
s
i
n
(
)
jt
t
je
t
C
qh
xh
yh
y
t
m
ωρ
ωδ
δ
∂
=
∂
=
−
−−++−−+
r
r
rrrr
±
♠
b)
Use what you’ve determined above, either in the time domain or phasor domain,
respectively that is,
·(,)
o
r ·()
t
tj
t
∂
∇=
−
∇
=
−
∂
r
Jr
r
±
±
. Therefore,
3
·
(
,
)
(
)(
)
s
i
n
(
)
C
t
q
hx
hx
hy
hy
t
m
∇
=
r
r
r
r
♠
c)
Now we know that each of these four ‘point charges” generate an electric field,
associated with the electric field is the scalar potential
V
. So the problem is as simple as
calculating the scalar potential from each of these point charges, and summing the results via
superposition. The only difference now is that the charges are timeharmonic in nature hence the
notion of finding the ‘retarded’ scalar potential.
Start in the phasor domain directly.
'
0
'
0
4'
(
(
)
'
)
4
jk
jk
e
d
x
h
x
h
y
h
y
e
d
π
δδδδ
−−
−
=
−
=
∫
∫
rr
r
Vr
r
r
±
±
ε
ε
(2)
2
What we do know, via
()
ρ
r
±
, the delta functions tells us exactly where the source locations are
located. We have delta functions inside an integral over all space—the only thing to do is to
apply the sifting, sampling property to each charge. Recall in rectangular coordinates,
( ) ( )
ˆˆ
ˆ
ˆ
''
'
'
x
xy
yz
z x
z
−=
++
−
+
+
rr
Therefore, for example, the first point charge located at
ˆ
(
hx
)
δ
−
r
tells us that the charge only
exists at
, then
ˆ
'
hx
=
r
ˆ
ˆ
'
ˆ
x
z h
x
−
=
r
h
x
−
, after applying the sifting property to
(2). The rest follows in the same manner. Therefore,
ˆˆˆˆ
0
,
ˆ
(
ˆ
4
)
jk
hx
jk
hx
jk
hy
jk
hy
qe
e
e
e
V
hx
hx
hy
hy
π
−−
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This note was uploaded on 02/21/2010 for the course ECE 450 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Staff

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