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ME/MSE 485: Homework # 3 Solution
- Let
2
E = 3 xy x + 5 z y + x z z and
2
F = yz x + 3 xy y + y z (15 points)
a. Find divergence of E at (-1,1,0)
Divergence of E or
y x z
E
E E
x y z
E
2
E = 3 xy x + 5 z y + x z z =>
2
x y z
E = xy E = z E = x z
( )
2
2
y x z
x z xy E z E E
y x
x x y y z z
So
2
∇ ⋅ E = 3 y + x , ∇ ⋅ E at x = −1, y = 1, z = 0 is
2
b. Find curl of E at (2,1,-2)
y y z z x x
E E
E E E E
y z x z x y
∇ × = − − − + −
E x y z Or
E
x y z
x y z
E E E
x y z
×
From (a)
2
x y z
E = xy E = z E = x z
( ) ( ) ( )
2 2 2
2
x x x
y y y
z z z
xy xy xy E E E
y x
x x y y z z
E z E z E z
x x y y z z
x z x z x z
E E E
xz x
x x y y z z
∇ × E = x 0 − 5 − y 2 xz − 0 + z 0 − 3 x = − 5 x − 2 xz y − 3 x z
∇ × E at (2,1,-2) = − 5 x − 2(2)( 2)− y − 3(2) z = − 5 x + 8 y − 6 z
c. Find F × E , E × F and E F ⋅
2
E = 3 xy x ˆ + 5 z y ˆ + x z z ˆ =>
2
x y z
E = xy E = z E = x z
2
F = yz x ˆ + 3 xy y ˆ + y z ˆ =>
2
x y z
F = yz F = xy F = y
2 2 2 2
3 2 2 2 3 2 2 2
y z z y x z z x x y y x
F E F E F E F E F E F E
xy x z y z yz x z y xy yz z xy xy
x yz y z x yz xy yz x y
F × E x y z
x y z
x y z
2 2 2 2
2 3 3 2 2 2 2 2
y z z y x z z x x y y x
E F E F E F E F E F E F
z y x z xy xy y x z yz xy xy z yz
y z x yz xy x yz x y yz
E× F x y z
x y z
x y z
2 2
2 2 2
x x y y z z
E F E F E F xy yz z xy x z y
xy z xyz x y z
E F
- Find the current i t ( ) in this R L circuit using Phasor. Note that
di
v L
dt
( ) cos
s o
v t = V ω t +φ
R
L
i t ( )
The voltage source is
( ) cos
s o
v t = V ω t + φ. Applying Kirchhoff’s voltage law (KVL),
we get
s
di t
Ri t L v t
dt
Step 1: Adopt a cosine reference =>
( ) cos
s o
v t = V ω t + φ is already in cosine form
Step 2: Express time-dependent variables as phasors
( )
( ) cos Re Re
j t j t
s o o s
v t V t V e V e
ω φ ω
ω φ
where
( )
j
s o
V V e
φ
( ) Re[ ]
j t
i t Ie
ω
Re[ ]
Re Re
j t
j t
j t
d Ie
di d Ie
Ij e
dt dt dt
ω
ω
ω
ω
Step 3: Recast the equation in phasor form
s
di t
Ri t L v t
dt
j t j t j t
s
R Ie L Ij e V e
ω ω ω
or
s
RI + j ω LI = V
Step 4: Solve equation in phasor domain
02
02
L
L
Z Z
Z Z
c. Find
02
Z that will match this line (
02
Z that makes 50
in
Z = Ω )
From (a), we have
2
02
in
l
Z
Z
Z
02 in l
Z = Z Z
For 50
in
Z = Ω and 100
l
Z = Ω ,
02
Z = 50 × 100 = 70.7Ω
- Two adjacent transmission lines has capacitance and inductance matrix below:
(pF/mm) (nH/mm)
C L
If the length of the line is 50 mm, the input voltage on an active line is 2 Volts and the
signal rise time is 100 picoseconds (20 points)
a. What is the characteristic impedance of each line?
The characteristic impedance
9
11
0 12
11
L L
Z
C C
−
−
×
×
Both lines have the same characteristic impedance
b. Estimated far-end crosstalk amplitude
12
0 0 12
0 0
s in
FE m
r
m dv L V
v d Z c d Z C
Z dt Z T
0 12 12
50 mm, 70.71 , 0.1 pF/mm, 0.3 nH/mm, 2 Volts,
100 psec (rise time)
in
r
d Z C L V
T
FE
v =1.4142 Volts
c. Estimated near-end crosstalk amplitude
12 12
11 11
0.040 Volts
m
NE o in
c m C L
v v V
C L C L
d. Draw crosstalk estimated response with amplitude and time
Phase velocity ( ) ( )
9 12
11 11
9 6
7.0711 10 mm/sec 7.0711 10 m/sec
p
v
L C L C
− −
′ ′ × ×
= × = ×
Time of flight =
6
(50 mm)
7.071 nsec
7.0711 10 m/sec
p
d
v
×
