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Transmission Lines
Lattice (bounce) diagram
This is a space/time diagram which is used to keep track of multiple reflections.
Ideal voltage
source
Voltage at the
receiving end
l
T
U
z

90  30
90  30
z

10  30
10  30
Points to Remember
Transmission Lines
1.
In this chapter we have surveyed several different types of waves
on transmission lines. It is important that these different cases not
be confused. When approaching a transmission-line problem, the
student should begin by asking, “Are the waves in this problem
sinusoidal, or rectangular pulses? Is the line ideal, or does it have
losses?” Then the proper approach to the problem can be taken.
2.
The ideal lossless line supports waves of any shape (sinusoidal or
non-sinusoidal), and transmits them without distortion. The
velocity of these waves is (LC )1 /.2 The ratio of the voltage to current
is Zo  L / C , provided that only one wave is present. Sinusoidal
waves are treated using phasor analysis. (A common error is that
of attempting to analyze non-sinusoidal waves with phasors.
Beware! This makes no sense at all.)
k   LC
UP  1 / LC
Zo  L / C
3.
UP   / 
UG  d / d
When the line contains series resistance and or shunt conductance
it is said to be lossy. Lossy lines no longer exhibit undistorted
propagation; hence a rectangular pulse launched on such a line will
not remain rectangular, instead evolving into irregular, messy
shapes. However, sinusoidal waves, because of their unique
mathematical properties, do continue to be sinusoidal on lossy
lines. The presence of losses changes the velocity of propagation
and causes the wave to be attenuated (become smaller) as it
travels.
Electrostatics



Various charge distributions.
Charge density
Consider a charged wire. Charge on an elemental wire segment l is q so that
the charge per unit length   q / l . The total electric field is the sum of the
field contributions from the individual segments.
E(r)  
i

q
4 r  ri
1
4

i

If we let l  0 we get
E(r) 
3

1
4 

3
(r  ri )
(q / l )l
r  ri
3
(r  ri )
 (r)dl
(r  r)
3
r  r
  q / l
Position of the field point
Position of the source point
Electrostatics
Electrostatic Energy and Potential continued
Example continued:
Segment I contributes nothing to the integral because E is perpendicular to dl
everywhere along it. As a result, the potential is constant everywhere on
Segment I; it is said to be equipotential.
On Segment II, e r  e y , dl  dy e y , and r  y
E 
q
q
e

ey
r
4r 2
4y 2
qdy
qdy
E  dl 
e

e

y
y
4y 2
4y 2
b
V1 V 2
qdy
q 1 1

2
a 4y 4  a  b 
 dy
1
  2   
y
 y
If we move P2 to infinity and set V2=0,
V1 
q
4a
point
where a is the distance between the observation point and the source
In general,
  q
4 r  r
Vr 
qi
i
i
4
dl
i
for line charge
ds for surface charge
dv for volume charge
Electrostatics
Capacitors
Any two conductors carrying equal but opposite charges form a capacitor. The
capacitance of a capacitor depends on its geometry and on the permittivity of
the medium.
C 
The capacitance C of the capacitor
is defined as the ratio of the
magnitude of Q of the charge on
one of the plates to the potential
difference V between them.
Area A
Q
V
 E  dS 
Qenclosed
 E  dS 
E A  A / 
S
d
d is very small compared with
the lateral dimensions of the
capacitor (fringing of E at the
edges of the plates is
neglected)
S
E 

A
C 
d

D   
A parallel-plate capacitor.
Cylindrical
Top face (inside the metal)
Side surface
dS
Bottom face
A
E 0
Gaussian
Surface
E
Close-up view of the upper plate of the parallel-plate capacitor.
5

Electrostatics
Method of Images
A given charge configuration above an infinite grounded perfectly conducting
plane may be replaced by the charge configuration itself, its image, and an
equipotential surface in place of the conducting plane. The method can be used
to determine the fields only in the region where the image charges are not
located.
Original problem
Construction for solving
original problem by the
method of images. Field
lines above the z=0
plane are the same in
both cases.
6
By symmetry, the potential in
this plane is zero, which
satisfies the boundary
conditions of the original
problem.
Magnetostatics
Example. A current element is located at x=2 cm, y=0, and
z=0. The current has a magnitude of 150 mA and flows in
the +y direction. The length of the current element is 1 mm.
Find the contribution of this element to the magnetic field
at x=0, y=3 cm, z=0.
dl  103 ey
r  103 ey r   2 * 102 ex
(r  r )  (2ex  3ey ) * 102
r  r   13 * 102
dl * (r  r )  103 ey * (2ex  3ey ) * 102
 10-5(2ez )
ey  ex   ex
ey  ey  0
dH 
I[dl  (r  r )]
4 r  r 
3
0.15(2 * 10-5 )ez

4( 13 * 102 )3
 5.09 * 10-3 ez A m
Magnetostatics
Magnetic Vector Potential
Some electrostatic field problems can be simplified by
relating the electric potential V to the electric field intensity
E(E  V). Similarly, we can define a potential associated with
the magnetostatic field B :
B  A
where A is the magnetic vector potential.
Just as we defined in electrostatics
dq(r ' )
V(r )  
4 r  r '
we can define
A(r ) 
I(r ' )dl'
L 4 r  r'
(electric scalar potential)
Wb
m
(for line current)
Magnetostatics
Ampere’s Circuital Law
Choosing any surface S bounded by the border line L and
applying Stokes’ theorem to the magnetic field intensity
vector H , we have
S   H   ds   H  dl
L
Substituting the magnetostatic curl equation
we obtain
H  J
I


 J  d s   H  dl
enc
S
L
which is Ampere’s Circuital Law. It states that the circulation
of H around a closed path is equal to the current enclosed by
the path.
Time-Varying Fields
General Forms of Maxwell’s Equations
Differential
1
Integral
 D  dS   
  D  V
V
S
2
Remarks
dv
V
 B  dS  0
B  0
S
3
4
xE  
B
t
xH  J 
D
t
 E  dl  
L
Gauss’ Law

B  dS

t S
D 

H

dl

J


L
S  t   dS
In 1 and 2, S is a closed surface enclosing the volume V
In 2 and 3, L is a closed path that bounds the surface S
Nonexistence of
isolated magnetic
charge
Faraday’s Law
Ampere’s circuital law
Time-Varying Fields
Surface Impedance continued
  1cm  10 4 m
2r  d  0.1mm  100m
Wire (copper) bond
f  10GHz  1010 Hz
 E  5.91  10 7 (  m) 1
z=?
Plated metal connections
   o  4  10 7 H / m
Substrate
1
d

 0.6m   50m
2
 E f
ZS 
(1  j)
 E
(a)
Thickness of current layer = 0.6 μm = δ
 (1  j)x0.026
w
 x 100 μm
 10 m 
l

Z  Z S    Z S 


100

m
 w


4
2 r
 0.86  j0.86Ω
X  Lint
δ = 0.6 μm
Z  R  jX
100 μm
Lint  X / 
(Internal inductance)
d  2r
(b)
(c)
Finding the resistance and internal inductance of a wire bond. The bond, a length of wire connecting two
pads on an IC, is shown in (a). (b) is a cross-sectional view showing skin depth. In (c) we imagine the
conducting layer unfolded into a plane.
Electromagnetic Waves
Z(z)

S av
13
S av