Chapter 1
Transmission Lines
A transmission line guides energy from one point to another in such a way that the energy does
not spread as it propagates. Transmission line examples include
• Coaxial cables (network, television)
• Twisted pair lines (telephone, network)
• Waveguides, optical fibers
• Printed circuit board trace; metalized line on an integrated circuit
• Power lines
• Earth/ionosphere system
The circuit diagram symbol for a transmission line is two wires with junctions marked as small
circles:
Figure 1.1: Circuit diagram symbol for a transmission line.
Circuit theory is an approximation to Maxwell’s equations of electromagnetism. In a circuit, when
the voltage between two nodes on a pair of wires is changed at one location, the voltage between
1
any other pair of nodes on the same wires changes instantaneously. But in reality, the voltage
change cannot propagate faster than the speed of light.
Transmission line theory can be viewed as a correction to circuit theory that is needed when wires
or other conductors in a system are long enough that this propagation delay along the wires or
conductors cannot be neglected.
One important aspect of transmission line theory is that many different types of lines, including
systems that do not appear to be transmission lines at all, such as optical coatings and even
empty space, can be treated using the same basic set of equations. We will study the behavior
of transmission lines for two types of driving sources: transients or pulses, and sinusoidal or time
harmonic excitation.
1.1
Transmission Line Equations
How can we analyze the behavior of currents and voltages on a transmission line? We need a
set of equations that govern the currents and voltages at different locations along the line. One
way to arrive at these equations is to model the transmission line as a sequence of lumped circuit
elements. The inductance and capacitance provide the propagation delay as energy moves along
the transmission line, and the resistance represents losses. One can then take the limit as the length
of the lumped-element section goes to zero, and arrive at a set of partial differential equations for
the current and voltage on the transmission line.
Figure 1.2: Lumped element model for a short section of a transmission line.
In the lumped element model, it is convenient to represent the capacitance, inductance, and resistance as per-unit-length quantities, so that
L = Inductance per unit length (H/m)
(1.1)
C = Capacitance per unit length (F/m)
(1.2)
R = Resistance per unit length (Ω/m)
(1.3)
G = Conductance per unit length (S/m)
(1.4)
It is helpful at this point to understand these quantities in terms of a particular transmission line
example. For a simple pair of parallel wires, L∆z represents energy stored in the magnetic field
around the wires for a section of length ∆z. C∆z represents the capacitance between the two pieces
2
of wire. R∆z represents the series resistance of the wires, and G∆z represents parallel conductance
of the dielectric material around the wires.
Lossless line. If the dielectric material between the conductors is a perfect insulator, then G = 0.
If the conductors making up the transmission line are perfect, then R = 0. In this case, the line
is said to be lossless. Real transmission lines are lossy, but in many cases the loss is small, so it is
very common to approximate a transmission line as lossless.
1.1.1
Telegrapher Equations
Applying Kirchoff’s voltage law (KVL) around the loop leads to
v(z, t) − i(z, t)R∆z − L∆z
∂i(z, t)
− v(z + ∆z, t) = 0
∂t
·
¸
∂i(z, t)
v(z, t) − v(z + ∆z, t) = ∆z i(z, t)R + L
∂t
¸
v(z + ∆z, t) − v(z, t)
∂i(z, t)
−
= Ri(z, t) + L
∆z
∂t
(1.5)
(1.6)
·
(1.7)
When we let ∆z → 0, the left hand side becomes the definition of a derivative, so that
−
∂v(z, t)
∂i(z, t)
= Ri(z, t) + L
∂z
∂t
(1.8)
Using Kirchoff’s current law at the top left node, we obtain
i(z, t) − G∆Zv(z + ∆z, t) − C∆z
∂v(z + ∆z, t)
− i(z + ∆z, t) = 0
∂t
(1.9)
Again taking the limit as ∆z → 0, this equation becomes
−
∂i(z, t)
∂v(z, t)
= Gv(z, t) + C
∂z
∂t
(1.10)
Equations (1.8) and (1.10) are known as the Telegrapher equations.
1.1.2
Wave Equation
The Telegrapher equations are coupled first order partial differential equations. Since it is simpler
to solve a single second order differential equation, we combine these two equations into a single
equation. We first take the derivative of Eq. (1.8) with respect to z, to obtain
−
∂2v
∂i
∂2i
=
R
+
L
∂z 2
∂z
∂t∂z
3
(1.11)
We can substitute ∂t i from Eq. (1.10). We also need ∂tz i, which we can get by differentiating Eq.
(1.10) with respect to t:
−
∂2i
∂v
∂2v
=G
+C 2
∂z∂t
∂t
∂t
(1.12)
Substituting Eqs. (1.10) and (1.12) into Eq. (1.11) leads to
∂i
− ∂z
2
∂ i
− ∂t
∂z
z
}|
·
¸{
∂2v
∂v
∂v
∂2v
− 2 = −R Gv + C
−L G
+C 2
∂z
∂t
∂t
∂t
z·
}|
¸{
(1.13)
∂2v
∂v
∂2v
=
RGv
+
(RC
+
LG)
(1.14)
+
LC
∂z 2
∂t
∂t2
This is a second order partial differential equation, where the only unknown is the voltage v(z, t)
on the transmission line. This is called the wave equation. A similar equation for i(z, t) could be
derived by eliminating the voltage instead, but once we know the voltage on the line, the current
can be found using the transmission line equations.
1.1.3
Wave Solutions
How do we solve the wave equation (1.14)? Most of the time when we use differential equations in
engineering, we look up or remember the general form of the solution and solve for the unknowns
using initial or boundary conditions. For the wave equation, in the lossless case the general solution
consists of two traveling waves of the form
v(z, t) = v + (z − ut) + v − (z + ut)
(Lossless line)
(1.15)
The term v + (z − ut) represents a pulse or wave traveling to the right (+z direction), and v −
represents a pulse traveling to the left (−z direction). The functions v + and v − depend on the
excitation of the transmission line, and the constant u is determined by the coefficients of the wave
equation.
Let’s look at the first part of the general solution where the excitation produces a square pulse as
given by
½
1 |x| < 1
p(x) =
(1.16)
0 otherwise
We will set v + (z − ut) = p(z − ut). At time t = 0, the pulse v + (z) is centered at z = 0. At the
time t = to the pulse becomes
½
1 |z − uto | < 1
+
v (z − uto ) =
(1.17)
0
otherwise
The pulse is now centered at the position z = uto . The pulse has moved in the +z direction. The
resulting velocity is given by
velocity =
∆z
uto
=
=u
∆t
to
(1.18)
The other part of the general solution v − (z + ut) travels at the same velocity in the −z direction.
4
1.1.4
Phase Velocity
To solve for the constant u, we plug either part of the general solution (1.15) into the wave equation.
Using the chain rule for the derivative,
∂2 +
v (z − ut) = v +00 (z − ut)
∂z 2
(1.19)
where the primes denote ordinary differentiation. Follow the same process to get the second derivative with respect to time leads to
∂2 +
v (z − ut) = v +00 (z − ut)(−u)2
∂t2
(1.20)
Substituting these two terms into the wave equation gives
v +00 (z − ut) = LC u2 v +00 (z − ut).
(1.21)
In order for this equality to hold, we must have that u2 LC = 1, so that
1
u= √
LC
(Phase velocity)
(1.22)
This quantity is called the phase velocity of waves on the transmission line.
For some common transmission lines, the phase velocity is
! Ãr
!
Ãs
1
1
2π
ln(b/a)
Coaxial Cable : √
=√
=
µ ln(b/a)
2π²
µ²
LC
Two Wire :
Parallel Plate :
1
√
LC
1
√
LC
=
=
1
√
µ²
1
√
µ²
(1.23)
(1.24)
(1.25)
where µ and ² are parameters of the material separating the conductors. For these transmission
lines, the velocity only depends on the properties of the material around the transmission line,
not the geometry. For transmission lines which consist of a dielectric (insulator) and a pair of
conductors, µ = µo , where µo is the permeability of free space (µo = 4π × 10−7 H/m), and ² = ²r ²o ,
where ²o is the permittivity of free space (²o ' 8.854×10−12 F/m) and ²r is the relative permittivity
of the dielectric. A typical value for the phase velocity is
1
1
c
2
1
)( √ ) = √ ≈ c
u = √ = (√
µ²
µ o ²o
²r
²r
3
where c ' 3 × 108 m/s is the speed of light in a vacuum.
5
(1.26)
1.1.5
Characteristic Impedance
For a lossless line, the first of the two telegrapher’s equations is
−
∂v
∂i
v=L
∂z
∂t
(1.27)
Let us consider just the forward traveling wave,
−
∂ +
∂
v (z − ut) = L i+ (z − ut)
∂z
∂t
(1.28)
Using the chain rule this becomes
−v +0 (z − ut) = L (−u)i+0 (z − ut)
(1.29)
Integrating both side with respect to x = z − ut gives
Z
Z
− v +0 (x) dx = −u L
i+0 (x) dx
(1.30)
−v + = −uLi+ + constant
(1.31)
Since i+ = 0 if v + = 0, the constant is zero, and we have
r
v+
L
L
=uL= √
=
i+
C
LC
(1.32)
The constant on the right-hand side has units of Ohms, and we call this the ”characteristic
impedance” of the line:
r
L
(1.33)
Zo =
C
If we repeat this derivation for the reverse traveling wave, we get
r
L
v−
= −Zo
= −u L = −
−
i
C
(1.34)
The total current can then be related to the total voltage using
v(z, t) = v + (z − ut) + v − (z + ut)
v + (z − ut) v − (z + ut)
i(z, t) =
−
Zo
Zo
(1.35)
(1.36)
This allows us to find the current on a transmission line if we know the forward and reverse voltage
waveforms.
The minus sign in the second equation is important. The characteristic impedance Zo is not a simple
resistance produced by the conductors in the line, because even a transmission line constructed from
perfect conductors has a finite, nonzero characteristic impedance. An impedance is the ratio of total
voltage to total current, whereas the characteristic impedance is the ratio of the forward voltage
waveform to the associated current and the negative of the ratio of the reverse voltage waveform to
the associated current. If Eq. (1.36) had a positive sign instead of a negative sign, then Zo would
be a regular impedance. Can you understand this minus sign physically?
6
1.2
Transients on Transmission Lines
Consider the transmission line circuit in Fig. 1.3. When the switch closes a step voltage appears
Figure 1.3: Transmission line circuit.
across the generator end of the line. What is the value of that voltage? The characteristic impedance
is not the total impedance of the line but rather the relationship between the forward and reverse
traveling voltages and currents. But when the switch first closes there is no reverse wave, because
the forward step has not had time to travel down to the end and back. Therefore, the characteristic
impedance of the line is the total impedance of the line imediately after the switch closes. This
leads to the equivalent circuit shown in Fig. 1.4 when the switch is first closed. From the equivalent
Figure 1.4: Equivalent circuit at t = 0.
circuit we can use a voltage divider relationship to calculate the magnitude of the step voltage,
v1+ (z = 0, t = 0) =
Zo
Vg
Zo + Rg
(1.37)
This initial pulse then starts to travel down the line at speed u. At time t = `/(2u), for example
the step is halfway down the line. At t = `/u, the step arrives at the load.
1.2.1
Reflection Coefficient
What happens when the step hits the load? The pulse will reflect, and the v − term in the wave
equation solution will no longer be zero. We need to find the amplitude of the reflected wave. This
is easy to do using boundary conditions at the load end of the transmission line. The boundary
7
conditions are
v(`, T ) = vL (T )
(1.38)
i(`, T ) = iL (T )
(1.39)
where T = `/u and vL (T ) and iL (T ) are the voltage across and the current through the load
resistor. vL and iL are related by Ohm’s law: vL = iL RL . Putting Ohm’s law together with the
boundary conditions at the load end of the line, we obtain
v(`, T ) = RL i(`, T )
Using Eqs. (1.35) and (1.36), this can be rewritten as
· +
¸
v1
v1−
+
−
v1 + v1 = RL
−
Zo
Zo
(1.40)
(1.41)
where the line voltages and currents are all evaluated at z = ` and t = T . Solving this for v1− gives
v1− =
RL − Zo +
v
RL + Zo 1
(1.42)
We call the constant in this expression the load reflection coefficient:
ΓL =
RL − Zo
RL + Zo
(Load reflection coefficient)
(1.43)
From this we can find the magnitude of the reflected wave:
v1− = ΓL v1+ =
RL − Zo Zo
Vg
RL + Zo Zo + Rg
(1.44)
The voltage waveform on the transmission line at a time that is after the reflection from the load,
and before the reflected pulse arrives at the generator, is shown in Fig. 1.5.
Figure 1.5: Voltage waveform at time t = 3T /2.
What happens when the reflected wave gets back to the generator? Using the same idea as at the
load end, we can show that
v2+ =
Rg − Zo −
v = Γg v1− = Γg ΓL v1+
Rg + Zo 1
Can you derive this expression on your own?
8
(1.45)
Figure 1.6 shows the voltage waveform after the reflected pulse has reflected again from the generator end of the transmission line. The total voltage at a point on the transmission line is the sum
of all the reflected and forward steps (v1+ , v1− , v2+ ) that have occured up until the current time.
Figure 1.6: Voltage waveform at time t = 9T /4.
If we look at the results we have found so far, we can see a pattern:
v1+
v1−
v2+
v2−
v3+
..
.
=
=
=
=
=
Zo
Zo +Rg
ΓL v1+
Vg
Γg ΓL v1+
ΓL Γg ΓL v1+
Γg ΓL Γg ΓL v1+
(Voltage divider at t = 0)
(First reflection at the load)
(Reflection at generator)
(1.46)
We can write out the total voltage at some point on the line at t = ∞ as an infinite series:
v(z, t = ∞) = v1+ + ΓL v1+ + Γg ΓL v1+ + ΓL Γg ΓL v1+ + Γg ΓL Γg ΓL v1+ + · · ·
£
¤
= v1+ (1 + ΓL ) + (1 + ΓL )ΓL Γg + (1 + ΓL )Γ2L Γ2g + · · ·
£
¤
= v1+ (1 + ΓL ) 1 + ΓL Γg + Γ2L Γ2g + · · ·
|
{z
}
Geometric series
1
= v1+ (1 + ΓL )
1 − ΓL Γg
(1.47)
(1.48)
(1.49)
If we plug in the definitions of ΓL and Γg , this reduces to
v(z, t = ∞) =
RL
Vg
Rg + RL
This is the steady state voltage on the transmission line. Does this result make sense?
9
(1.50)
1.2.2
Bounce Diagrams
A convenient tool for understanding transmission line transients is a bounce diagram. One axis of
the diagram is the z coordinate, and the other is time. On it we plot the location of the leading edge
of the reflected pulse as it propagates between the load and generator terminations, and note the
amplitude of each reflection. The bounce diagram for a single transmission line with load reflection
coefficient ΓL and generator reflection coefficient Γg is shown in Fig. 1.7.
Figure 1.7: Bounce diagram for a single transmission line with load reflection coefficient ΓL and
generator reflection coefficient Γg .
The voltage at a point on the transmission line at a time t is the sum of all of the reflections that
have occured up to the given time. So, if we want to plot the voltage at a point on the line as a
function of time, we draw a vertical line on the bounce diagram at the given location. The reflected
pulse passes that point at each time for which the vertical line crosses the line representing the
leading edge of the reflected pulse. At a given time, the voltage on the transmission line is the sum
of all the reflection amplitudes below the current time (Fig. 1.8).
Figure 1.8: Voltage as a function of time at the location z = `/4 obtained from the bounce diagram.
10
We can also find the current using
i=
v+ v−
−
Zo
Zo
(1.51)
If the currents and voltages are evaluated at one of the ends of the transmission line, this can be
rearranged to obtain
i−
−v − /Zo
v−
=
=
−
= −Γ
(1.52)
i+
v + /Zo
v+
where Γ is the voltage reflection coefficient at the load or generator end. So, the current bounce
diagram is the same as the voltage bounce diagram, but with the signs of all the reflection coefficients
reversed and v1+ replaced by i+
1.
1.2.3
Multi-section Lines
Transmission lines can be placed in series or in parallel. By proper application of boundary conditions at each junction, any situation can be handled using the techniques we have developed. Let’s
consider the former case, as shown in Fig. 1.9.
Figure 1.9: Two transmission lines in series.
We first find the initial pulse amplitude
+
v11
= 10
50
= 5V
50 + 50
and compute the reflection coefficients at each junction:
Γg =
Γ12 =
Γ21 =
ΓL =
50 − 50
=0
50 + 50
100 − 50
1
=
100 + 50
3
50 − 100
1
=−
50 + 100
3
300 − 100
1
=
300 + 100
2
−
The amplitude of the first reflection from the junction between the two lines is v11
= 5Γ12 = 5/3 V.
+
−
At the time of the first reflection, the total voltage at the right end of line 1 is v11 + v11
= 5 + 5/3 =
11
20/3 V. Applying the voltage boundary condition, this must also be the voltage at the left end of
+
the second line, so that a pulse with amplitude v21
= 20/3 V is launched down the second line. This
−
+
pulse reflects from the load with amplitude v21 = ΓL v21
= 10/3 V . The load reflection reaches the
+
−
junction between the two lines, and reflects to the right with amplitude v22
= Γ21 v21
= −10/9, while
−
+
= 20/9.
at the same time launching a wave down the first transmission line with amplitude v22
+v21
Because the generator impedance is matched to the characteristic impedance of line 1, there is no
reflection from the generator, but reflections continue between the right and left ends of line 2.
The resulting bounce diagram is shown in Fig. 1.10. Figure 1.11 shows the voltage at z = 15 cm as
a function of time.
Figure 1.10: Bounce diagram for the series transmission lines in Fig. 1.9.
Figure 1.11: Voltage for the series transmission lines in Fig. 1.9 at z = 15 cm as a function of time.
We can also handle a branched transmission line. If we replace the second line in Fig. 1.9 with two
lines in parallel, the system shown in Fig. 1.12 is obtained. The same forward wave with amplitude
+
v11
= 5 V is launched at t = 0. When the pulse arrives at the branch junction, the effective load
impedance seen by the pulse is the parallel combination of the characteristic impedances of lines 2
and 3:
RL = 50k50 = 25 Ω
(1.53)
12
The reflection coefficient is then
Γ1 =
25 − 50
1
=−
25 + 50
3
−
+
The reflection back down line 1 has amplitude v11
= Γ1 v11
= −5/3 V. Pulses with amplitude
−
+
v11 + v11 = 10/3 V are launched down lines 2 and 3. Continuing in this way, we can compute the
values of all the reflections at each junction and complete the analysis of the branched line system.
A bounce diagram can be drawn for the branched lines as shown in Fig. 1.13.
Figure 1.12: Branched transmission lines.
Figure 1.13: Bounce diagram for the branched transmission lines in Fig. 1.12.
13
1.2.4
Reactive Load
What happens if the load is a capacitor? Using the general principles we have already developed,
we can solve this and many other types of new problems.
Figure 1.14: Transmission line with a reactive load.
Figure 1.14 shows a transmission line with a reactive load. At t = 0, the voltage on the capacitor
is assumed to be zero. This remains unchanged until the step launched by the generator arrives at
the load at time T = `/u. At this time, the capacitor begins to charge. The steady state voltage
on the capacitor will be Vg , as shown in Fig. 1.15.
Figure 1.15: Voltage across load capacitor.
We can find the analytical form of the voltage across the capacitor by solving a differential equation.
By the current boundary condition at the load end of the transmission line and the capacitor
voltage-current relationship, we can say that
i+ + i− = C
dvL (t)
dt
Using the relationship between the current and voltage waves on the transmission line, we can
rewrite this as
v+ v−
dvL (t)
−
=C
Zo
Zo
dt
There are two unknowns in this equation, but we can reduce that to one by using the voltage
boundary condition vL = v + + v − to eliminate v − , so that we have
dvL (t)
v + vL − v +
−
=C
Zo
Zo
dt
14
By rearranging this into standard form, we obtain the differential equation that we need to solve:
C
dvL (t) vL
2v +
+
=
dt
Zo
Zo
(1.54)
This can be solved using standard techniques for differential equations. The form of the solution
for t ≥ T is
vL (t) = A + Be−m(t−T )
(1.55)
The constants A, B, and m can be found by substituting this into the differential equation (1.54).
After doing this, the final solution is found to be
vL (t) = 2v + − 2v + e−(t−T )/(Zo C)
(1.56)
The amplitude of the reflected wave at the load end is
v − (t) = v + (1 − 2e−(t−T )/(Zo C) )
(1.57)
This result can be used to find the voltage at, say, the generator end of the transmission line.
The voltage waveform given in Eq. (1.57) arrives at the generator at time t = 2T . Before that
time, the voltage at the generator end is Vg /2. At time t = 2T , the arriving reflected wave has
amplitude −v + = −Vg /2, so the total voltage at the generator changes to zero. As the reflected
wave becomes less negative, the voltage at the generator increases. The steady state voltage is the
sum of the initial forward wave v + = Vg /2 and the limiting value of the reflected wave, which is
v − (∞) = v + = Vg /2, so that the final voltage at the generator is Vg . This is shown in Fig. 1.16.
Figure 1.16: Voltage at the generator end of the transmission line.
If we turn off the source at some time to , then v + goes to zero and the voltage at the generator end
changes to v − = Vg /2. The edge of the step down in the v + wave propagates to the right. When it
arrives at the load, v + is zero all the way along the line, and the capacitor is charged to Vg , so the
reverse wave changes to v − = Vg . At this time, the capacitor begins discharging and v − decays to
zero. The step up in v − propagates to the left until it arrives at the generator at time t = to + T .
The voltage at the generator then changes to Vg and decays to zero (Fig. 1.17).
Figure 1.17: Voltage at the generator end of the transmission line if source is turned off.
15
1.3
Sinusoidal Steady State
Electromagnetics applications can be divided into two broad classes:
• Time-domain: Excitation is not sinusoidal (pulsed, broadband, etc.)
– Ultrawideband communications
– Pulsed radar
– Digital signals
• Time-harmonic: Excitation is sinusoidal
– Narrowband communication schemes - amplitude modulation (AM), frequency modulation (FM), phase shift keying (PSK), etc.
– Continuous wave radar
– Optical communications
Time-harmonic systems are fundamental to applications of electrical engineering. The concept
of sharing a communication channel by using carrier sinusoids with different frequencies together
with receivers tuned to discriminate among the carriers dates back to the earliest days of radio
communications.
1.3.1
Phasor Notation
In analyzing time-harmonic systems, we assume that the signal of interest is narrowband enough
that it can be approximated as a sinusoid. This approximation works very well for many important
applications. Due to capacitance, inductance, and propagation delays in a system, the phase of
a signal depends on where the signal is measured. For this reason, it takes two parameters to
characterize the signal at any point in the system:
v(x, y, z, t) = vo (x, y, z) cos [ωt + φ(x, y, z)]
| {z }
| {z }
Amplitude
Phase
(1.58)
where ω is the time frequency of the signal in radians per meter. It is inconvenient to have the
phase φ(x, y, z) inside the argument of the cosine function. Dealing with time-harmonic signals is
much easier if we express these two degrees of freedom in a more symmetric way, as the real and
imaginary parts of a complex number, which we call a phasor. The definition of a phasor voltage
Ṽ is
n
o
v(x, y, z, t) = Re Ṽ (x, y, z)ejωt
(1.59)
At this point, we will drop the x and y dependence, and assume that for a transmission line the
voltage only depends on time and the position z along the line.
16
How does the complex number Ṽ relate to the real voltage V ? By placing the complex number Ṽ
in polar form, we can express the voltage as
n
o
6
v(z, t) = Re |Ṽ (z)|ej Ṽ (z) ejωt
(1.60)
n
o
6
= |Ṽ (z)|Re ej[ωt+ Ṽ (z)]
(1.61)
h
i
= |Ṽ (z)| cos ωt + 6 Ṽ (z)
(1.62)
By comparing this to Eq. (1.58), we can see that the magnitude of the voltage is equal to the
magnitude of the phasor and the phase shift of the voltage relative to ωt is equal to the phase of
Ṽ . Keep in mind that there is no such thing as a complex voltage. The real and imaginary parts of
the phasor voltage Ṽ simply offer a convenient tool for keeping track of the magnitude and phase
in Eq. (1.58) at different locations in a circuit or system.
Another simplification that results from the use of phasor notation is that time derivatives become
multiplication by jω, through the use of the identity
n
o
∂v(z, t)
= Re jω Ṽ (z)ejωt
∂t
(1.63)
The current-voltage relationship for a capacitor, for example, is
i(t) = C
In the phasor domain, this becomes
dv(t)
dt
I˜ = jωC Ṽ
The inverse relationship is
Ṽ = I˜
1
jωC
| {z }
Impedance
(1.64)
(1.65)
(1.66)
The beauty of this result is that the capacitor current-voltage relationship now has the form of
Ohm’s law, but with an imaginary value in place of resistance. So, we can handle resistors, capacitors, and inductors without having to solve differential equations by using phasor notation.
1.3.2
Time-harmonic Wave Equation
By substituting Eqs. (1.59) and (1.63) into the wave equation (1.14), we obtain the time-harmonic
wave equation,
d2 Ṽ
dz 2
=
£
¤
RG + jω(RC + LG) − ω 2 LC Ṽ
= (R + jωL)(G + jωC)Ṽ
(1.67)
(1.68)
The constant on the right-hand side is the square of the complex propagation constant, with the
symbol γ, so that
γ 2 = (R + jωL)(G + jωC)
(1.69)
17
The solution to the ordinary differential equation (1.68) has the form
Ṽ (z) = Aemz + Be−mz
(1.70)
Substituting this expression into Eq. (1.68) leads to
m2 Ṽ = γ 2 Ṽ
(1.71)
so that the general solution can be expressed as
Ṽ (z) = Aeγz + Be−γz
(1.72)
The constants A and B are determined by the excitation and boundary conditions on the transmission line.
In general, γ is complex, and can be expressed in terms of its real and imaginary parts as
γ = α + jβ
(1.73)
Ṽ (z) = Aeαz ejβz + Be−αz e−jβz
(1.74)
Using this in the general solution leads to
If we use Eq. (1.59) to find the voltage on the transmission line, we obtain
n
o
v(z, t) = Re Aeαz ejβz ejωt + Be−αz e−jβz ejωt
n
o
= Re |A|ejφA eαz ejβz ejωt + |B|ejφB e−αz e−jβz ejωt
= |A|eαz cos [ωt + βz + φA ] + |B|e−αz cos [ωt − βz + φB ]
|
{z
} |
{z
}
Reverse wave
Forward wave
(1.75)
(1.76)
(1.77)
By looking at this expression, we can understand the physical meaning of the real and imaginary
parts of the complex propagation constant γ. The real part α represents attenuation and has units
of Nepers per meter (Np/m). The imaginary part β determines the wavelength of the wave, and
is called the wavenumber with units of radians per meter (rad/m). β is also called the spatial
frequency or phase constant of the wave. The phase velocity of the wave is u = ω/β and the
wavelength is λ = 2π/β (Fig. 1.18).
Figure 1.18: Propagating, attenuating forward wave.
From Eq. (1.77), we can see that the constant A represents the amplitude of the wave moving in the
−z direction, and B represents the wave moving in the +z direction. Because of this, we rename
the constants so that Vo+ = B and Vo− = A, so that (1.72) becomes
Ṽ (z) = Vo+ e−γz + Vo− eγz
18
(1.78)
For a lossless line,
√
γ = jω LC = jβ
(Lossless line)
(1.79)
so that the attenuation constant α is zero, and there is no decay of the amplitude of a wave as it
propagates. The general solution for the voltage simplifies to
Ṽ (z) = Vo+ e−jβz + Vo− ejβz
(Lossless line)
(1.80)
Because many transmission lines can be approximated as lossless, we use this expression in most
analyses instead of (1.72).
1.3.3
Current
To get the current on the line, we use one of the telegrapher’s equations in time-harmonic form:
−
dṼ (z)
˜
= (R + jωL)I(z)
dz
(1.81)
Substituting the general solution (1.72) leads to
¤
d £ + −γz
1
Vo e
+ Vo− eγz
R + jωL dz
£
¤
1
−
−γVo+ e−γz + γVo− eγz
R + jωL
£ + −γz
¤
γ
Vo e
− Vo− eγz
R + jωL
p
¤
(R + jωL)(G + jωC) £ + −γz
Vo e
− Vo− eγz
R + jωL
s
¤
G + jωC £ + −γz
Vo e
− Vo− eγz
R + jωL
|
{z
}
˜
I(z)
= −
=
=
=
=
(1.82)
(1.83)
(1.84)
(1.85)
(1.86)
1
Zo
=
Vo+ −γz Vo− γz
e
−
e
Zo
Z
|{z}
| {z o}
Io+
1.3.4
(1.87)
Io−
Reflection Coefficient
At the load end of a transmission line (Fig. 1.19), we can use boundary conditions to find the ratio
of the forward and reflected waves. For the sinusoidal steady state, it is convenient to shift the
coordinate system so that load end is at z = 0. The goal is to find
ΓL =
Vo−
Vo+
(1.88)
The voltage and current boundary conditions at the load end, together with Ohm’s law for the load
impedance, lead to the following relationship between the total current and voltage at the load end
of the transmission line:
Ṽ (0) = ZL I˜L (0)
(1.89)
19
Figure 1.19: Transmission line and load.
Substituting the general voltage and current solutions leads to
¶
µ +
V−
Vo
− o
Vo+ + Vo− = ZL
Zo
Zo
(1.90)
Now we can solve for the reflection coefficient:
ΓL =
Vo−
ZL − Zo
+ = Z +Z
Vo
o
L
(1.91)
Although this is the same expression as was obtained for the transient case (except that it is a
phasor-domain formula and allows for complex load impedances), the meaning of the reflection coefficient is different. For the sinusoidal steady state, the forward and reverse waves exist everywhere
on the transmission line. If we know Vo+ , for example, we can find Vo− using ΓL , and then we can
use Eq. (1.78) or (1.80) to find the voltage anywhere on the transmission line.
Generalized reflection coefficient. In the lossless case, it is also sometimes useful to define a
generalized reflection coefficient as the ratio of the forward and reverse waves at any point on the
transmission line:
Vo− ejβz
= ΓL ej2βz
(1.92)
Γ(z) = +
Vo e−jβz
20
1.3.5
Standing Waves
When forward and reverse sinusoidal waves are both present on a transmission line, the two propagating waves add to form a standing wave pattern. If we apply a voltmeter to a transmission
line instead of an oscilloscope, at high frequencies the voltmeter cannot respond rapidly enough
to follow the cos ωt time variation, so what we actually measure is the standing wave pattern on
the line. Standing waves are also helpful into gaining insight into transmission line phenomena for
different types of loads.
Using the load reflection coefficient, the phasor voltage on a transmission line can be written as
Ṽ (z) = Vo+ e−jβz + ΓL Vo+ ejβz
(1.93)
In terms of the generalized reflection coefficient in Eq. (1.92), this becomes
Ṽ (z) = Vo+ e−jβz [1 + Γ(z)]
(1.94)
To analyze the standing wave pattern, we look at the magnitude of the phasor:
|Ṽ (z)| = |Vo+ ||1 + Γ(z)|
= |Vo+ ||1 + ΓL ej2βz |
=
|Vo+ ||1
(1.95)
jθL j2βz
+ |ΓL |e e
| {z }
|
ΓL
h
i1/2
= |Vo+ | (1 + |ΓL |ejθL +j2βz )(1 + |ΓL |e−jθL −j2βz )
£
¤1/2
= |Vo+ | 1 + 2|ΓL | cos (2βz + θL ) + |ΓL |2
(1.96)
We can understand this function graphically by going back to Eq. (1.95). If we plot the phasor
voltage in the complex plane, we find that its value is equal to |Vo+ | on the real axis plus a complex
number with magnitude |ΓL ||Vo+ | and phase 2βz, as shown in Fig. 1.20. The magnitude of the
phasor is equal to the distance from the origin to the sum of the two terms. It is easy to see
Figure 1.20: (a) Graphical representation of Eq. (1.95). (b) Corresponding standing wave pattern.
graphically that |Ṽ (z)| is bounded by
|Ṽ (z)|max = |Vo+ |(1 + |ΓL |)
|Ṽ (z)|min =
|Vo+ |(1
− |ΓL |)
(1.97)
(1.98)
The phasor travels around the circle each time z changes by π/β, which is equal to λ/2 or one
half wavelength. Because the magnitude of the phasor voltage is the amplitude of the time-varying
voltage in Eq. (1.58), |Ṽ (z)| is the envelope of the voltage along the transmission line as it oscillates
in time.
21
VSWR. The voltage standing wave ratio (VSWR) is defined to be
S=
|Ṽ (z)|max
1 + |ΓL |
=
1 − |ΓL |
|Ṽ (z)|min
(1.99)
This quantity is useful because it is more easily measured on high frequency transmission lines than
the time-varying voltage itself. For a matched load, ΓL = 0, so that the VSWR is equal to one and
there is no standing wave on the transmission line.
1.3.6
Load Examples
Matched load (ΓL = 0): Only a forward wave exists on the transmission line, and there is no
standing wave.
Figure 1.21: Standing wave pattern for a matched load.
Open circuit (ΓL = 1): In this case, |Ṽ (z)|min is zero and the VSWR is infinite. The standing
wave pattern exhibits nulls spaced one half wavelength apart along the transmission line with a
maximum at the load.
Figure 1.22: Standing wave pattern for an open circuit load.
Short circuit (ΓL = −1): |Ṽ (z)|min is zero as with the open circuit load, and the VSWR is also
infinite. In this case, however, there is a null instead of a maximum at the short circuit load.
Figure 1.23: Standing wave pattern for a short circuit load.
22
1.3.7
Input Impedance
We need one more tool in order to analyze a complete sinusoidal steady state transmission system.
Unlike the transient case, the impedance looking into the generator end depends on the entire
transmission line and the load.
To understand impedance on a transmission line for a time-harmonic excitation, we can define a
line impedance that is the ratio of the phasor voltage to the phasor current at a point on the line:
Zin (z) =
Ṽ (z)
˜
I(z)
(1.100)
Vo+ (1 + ΓL ej2βz )
Vo+ (1 − ΓL ej2βz )/Zo
1 + Γ(z)
= Zo
1 − Γ(z)
=
(1.101)
(1.102)
If the generator is located at z = −`, then the input impedance seen by the source is Zin = Zin (−`).
By substituting Eq. (1.91) for ΓL and applying trigonometric identities to Eq. (1.101), the input
impedance can be placed in an alternate form
Zin = Zin (−`) = Zo
ZL + jZo tan β`
Zo + jZL tan β`
(1.103)
This now allows us to analyze a complete time-harmonic transmission line system (Fig. 1.24). In
Figure 1.24: Transmission line system with sinusoidal excitation.
the steady state, as noted above the impedance seen by the source is Zin as given by Eq. (1.103),
so the voltage on the transmission line at the input port can be found using a voltage divider:
Zin
Zg + Zin
(1.104)
Ṽ (−`)
ejβ` + ΓL e−jβ`
(1.105)
Ṽ (−`) = Ṽg
We can then find Vo+ using Eq. (1.93), so that
Vo+ =
Once Vo+ is known, all currents and voltages anywhere on the transmission line can be determined.
23
Example: Sinusoidal Steady State Transmission Line System
Figure 1.25: Transmission line with source and load.
1. Find ΓL
2. Find Γin = Γ(−`)
3. Find Zin
4. Find Ṽin
5. Find Vo+
With these quantities, we can determine whatever else we may want to know about the system.
How about the power dissipated at the load?
24
Special Cases
Let’s consider a few common loads and line lengths, and for each we will determine the input
impedance looking into the generator end of the line using
Zin (−`) = Zo
e−jβz + ΓL ejβz
e−jβz − ΓL ejβz
(1.106)
or one of the alternate forms of this expression that we derived previously.
Open circuit (ΓL = 1):
ejβ` + e−jβ`
ejβ` − e−jβ`
2 cos (β`)
= Zo
2j sin (β`)
= −jZo cot (β`)
oc
Zin
(−`) = Zo
Short circuit (ΓL = −1):
sc
Zin
(−`) = jZo tan (β`)
(1.107)
(1.108)
Notice that in both the open and short circuit cases, the input impedance is purely imaginary,
corresponding to a reactive load, and the lines appear inductive or capacitative. By changing the
length `, we can make the line look like a capacitor or inductor of any value. This principle is often
used in microwave designs. (Is is possible to realize a reactive impedance corresponding to a very
large inductance or capacitance?)
Half-integer line length (` = nλ/2, n = 1, 2, 3, . . .)
¯
ZL + jZo tan β` ¯¯
Zin (−nλ/2) = Zo
Zo + jZL tan β` ¯tan (βnλ/2)=tan (nπ)=0
= ZL
(1.109)
The load impedance repeats along the line each half wavelength.
Quarter-wave transformer (` = nλ/2 + λ/4, n = 0, 1, 2, . . .) The input impedance looking
into a quarter-wave section (or an integer number of half-wavelengths plus λ/4) as shown in Fig.
1.26 is
¯
ZL + jZo tan β` ¯¯
(1.110)
Zin (nλ/2 + λ/4) = Zo
Zo + jZL tan β` ¯tan (βλ/4)=tan (π/2)→∞
=
Zo2
ZL
(1.111)
In order to see a matched load looking into the quarter-wave line, we can set
p
Z2
Zo1 = o2 ⇒ Zo2 = Zo1 ZL
ZL
This is called quarter-wave matching.
25
(1.112)
Figure 1.26: Quarter-wave transformer.
1.4
Power
There are several ways to quantify power for sinusoidal steady state systems:
1. Instantaneous power: pi (t) = v(t)i(t).
RT
2. Time-average power: pav = T1 0 pi (t) dt, T = 2π/ω.
3. Complex power: P̃ = Ṽ I˜∗
It is easy to show that time-average power is related to complex power by
1
pav = Re{Ṽ I˜∗ }
2
(1.113)
The imaginary part of Ṽ I˜∗ is not associated with dissipated or supplied power, but rather represents
changes in the amount of energy stored in inductive and capacitative elements. We will look at
each of these for the case of a transmission line system.
1.4.1
Instantaneous Power
Energy is carried along a transmission line by both the forward and reverse waves. The instantaneous power arriving at a load is
p+ (t) = v + (t)i+ (t)
½ +
¾
© + jωt ª
Vo jωt
= Re Vo e
Re
e
Zo
¾
½ +
n
o
|Vo | jφ+ jωt
+ jφ+ jωt
e e
= Re |Vo |e e
Re
Zo
|Vo+ |2
=
cos2 (ωt + φ+ )
Zo
(1.114)
What this result means is that each half cycle of the forward wave delivers power to the load. If
we repeat this derivation for the reverse wave, we find that
|Vo− |2
cos2 (ωt + φ− )
Zo
|V + |2
= −|ΓL |2 o cos2 (ωt + φ+ + θL )
Zo
p− (t) = −
26
(1.115)
where ΓL = |ΓL |ejθL . The negative sign means that the reverse wave carries power away from the
load. The net power delivered to the load is equal to the sum of the incident and reflected power:
p(t) = p+ (t) + p− (t)
1.4.2
(1.116)
Time-Average Power
The time-average power associated with the forward wave is
Z
1 T |Vo+ |2
cos2 (ωt + φ+ ) dt
p+
=
av
T 0
Zo
Z
|Vo+ |2 1 T
=
cos2 (ωt + φ+ ) dt
Zo T 0
{z
}
|
1/2
=
|Vo+ |2
(1.117)
2Zo
The time-average power carried away by the reverse wave can be computed in the same way, but
we will use the phasor expression in Eq. (1.113) to illustrate an alternate approach:
1 n − ˜−∗ o
−
pav =
Re Ṽ (0)I (0)
(1.118)
2 ½
µ
¶
¾
∗
1
V−
=
Re Vo− − o
2
Zo
−
2
|V |
= − o
2Zo
|V + |2
= −|ΓL |2 o
(1.119)
2Zo
The net time-average power delivered to the load is
−
pav = p+
av + pav =
|Vo+ |2
(1 − |ΓL |2 )
2Zo
(1.120)
What happens if the load is purely reactive (lossless)?
Another way to arrive at the same result is to compute the power absorbed by the load directly
from the total phasor voltage at the load:
o
1 n
Re Ṽ (0)I˜∗ (0)
pav =
2 (
)
1
Ṽ ∗ (0)
=
Re Ṽ (0) ∗
2
ZL
½ ¾
1
1
|Ṽ (0)|2 Re
=
2
ZL∗
½
¾
ZL
1 +
2
|V (1 + ΓL )| Re
=
2 o
|ZL |2
|Vo+ |2 RL
|1 + ΓL |2
=
2|ZL |2
Although the expression appears different, it gives the same absorbed power as Eq. (1.120).
27
1.5
Smith Chart
The Smith chart provides a graphical way to solve the transmission line equations that we have
derived. Most high frequency engineering is done using computer aided design packages, so we
don’t really need the Smith chart as a calculation tool. But it does provide a powerful way to
communicate the behavior of a transmission line system visually. In fact, software packages and
instruments often present computed or measured values on a Smith chart. Thus, the Smith chart
is mainly a tool for gaining insight into transmission line systems.
The Smith chart is a plot of a reflection coefficient in the complex plane. Superimposed on that is
a curved grid of lines that represent the load impedance corresponding to the reflection coefficient.
A passive load cannot reflect more power than is incident on it, so from Eq. (1.119), we must have
|ΓL | < 1. Thus, for most transmission line systems the reflection coefficient is confined to the unit
circle.
Now, let’s derive equations for the curved grid representing the impedance corresponding to ΓL . If
we solve Eq. (1.91) for the load impedance, we obtain
ZL = Zo
1 + ΓL
1 − ΓL
(1.121)
Because we don’t want to have to have a different Smith chart for every possible value of the
characteristic impedance, we will rearrange this expression and work with normalized impedance,
which we will identify with a lower case symbol:
zL =
1 + ΓL
ZL
=
Zo
1 − ΓL
(Normalized impedance)
(1.122)
Now, we break both zL and ΓL into their real and imaginary parts,
rL + jxL =
=
=
1 + ΓLr + jΓLi
1 − ΓLr − jΓLi
1 + ΓLr + jΓLi 1 − ΓLr + jΓLi
1 − ΓLr − jΓLi 1 − ΓLr + jΓLi
1 − Γ2Lr − Γ2Li
2ΓLi
+j
2
2
(1 − ΓLr ) + ΓLi
(1 − ΓLr )2 + Γ2Li
With some algebra, the real and imaginary parts of this equation can be rearranged into the forms
µ
ΓLr −
¶2
µ
¶2
rL
1
+ Γ2Li =
1 + rL
1 + rL
µ
¶2 µ ¶2
1
1
(ΓLr − 1)2 + ΓLi −
=
xL
xL
(1.123)
(1.124)
Both of these equations represent circles. The first one is centered at (ΓLr , Γli ) = (rL /(1 + rL ), 0)
and has radius 1/(1 + rL ). The second circle is centered at (1, 1/xL ) and has radius 1/xL . Since
the imaginary part of the impedance can be positive or negative, we have to consider two circles
for each value of xL . For a given reflection coefficient in the complex plane, two circles intersect
at that point, one given by (1.123) for a particular value of rL and the other given by (1.124) for
28
a value of xL . On the Smith chart, the rL and xL are labeled, so that the value of zL = rL + jxL
can easily be read from the chart.
Before using the Smith chart to analyze transmission lines, it is helpful to learn our way around
the chart by considering some important landmarks (Fig. 1.27):
1. The unit circle |ΓL | = 1 corresponds to lossless loads (capacitative, inductive, open circuit,
short circuit).
2. The real axis corresponds to purely resistive load impedances. The left side of the real axis
corresponds to RL < Zo , and the right side to RL > Zo .
3. The upper half plane represents inductive loads and the lower half plane represents capacitative loads.
4. The center of the Smith chart (Γ = 0) corresponds to a matched load (zL = 1).
5. The point Γ = −1 corresponds to a short circuit.
6. The point Γ = 1 corresponds to an open circuit or infinite load impedance. Since small
capacitances, large inductances, and large resistances all lead to Γ ' 1, all of the impedance
circles go through that point.
Figure 1.27: Smith chart landmarks.
29
Generalized reflection coefficient. The next key principle for the Smith chart is that we can
also plot the generalized reflection coefficient
Γ(z) =
Vo− ejβz
j2βz
+ −jβz = ΓL e
Vo e
(1.125)
and the corresponding normalized input impedance zin (z) = Zin (z)/Zo as a function of position on
the line. As z moves away from zero, the phase of Γ(z) changes such that Γ(z) on the Smith chart
moves around a circle centered at the origin. The radius of the circle is |ΓL |. Increasing z causes
the phase angle in Eq. (1.125) to increase, which corresponds to counterclockwise rotation. Since
the generator is at z = −`, moving from the load towards the generator corresponds to clockwise
rotation along the circle. If z changes by λ/2, the generalized reflection coefficient moves once
around the circle. If the distance moved along the line is given in wavelengths, so that L = `/λ,
then we rotate 2L times around the circle.
VSWR. The VSWR on a transmission line is
S=
1 + |Γ|
1 − |Γ|
(1.126)
where Γ can be the load reflection coefficient or the generalized reflection coefficient anywhere on
the line. A circle of constant |Γ| on the Smith chart is also a circle of constant VSWR. When the
VSWR circle crosses the positive real axis, the imaginary part of zin is zero and the real part is
equal to some value r > 1, so the magnitude of the generalized reflection coefficient is
|Γ(z)| =
r−1
r+1
(1.127)
If we solve this equation for r and compare the resulting expression to (1.126), we find that S = r.
As we move along a transmission line, the voltage standing wave maxima occur when the generalized
reflection coefficient crosses the positive real axis. The minima occur when it crosses the negative
real axis.
Admittance. When transmission lines or circuit elements are in parallel, it is convenient to
convert from impedances to admittances. There are two ways to work with admittances on a
Smith chart. One is to add another grid for admittances in a different color. The other is to shift
an impedance point to a new point and then reinterpret the impedance circles on the Smith chart
as admittance lines. The load admittance is
yL =
1
1 − ΓL
=
zL
1 + ΓL
By comparing this to
zin (z = −λ/4) =
=
=
30
1 + ΓL e2jβz
1 − ΓL e2jβz
1 + ΓL ejπ
1 − ΓL ejπ
1 − ΓL
1 + ΓL
it can be seen that a λ/4 rotation on the Smith chart transforms an impedance to an admittance.
This corresponds to reflection with respect to the origin. So, if we are given an admittance, we can
plot it on a single color Smith chart using its real and imaginary parts as values for the impedance
circles, and the resulting reflection coefficient is found by reflecting the point about the origin.
1.5.1
Smith Chart Solution Procedure
To solve a lossless transmission line problem with generator and load impedance graphically on a
Smith chart, the following steps are involved:
1. Find the normalized load impedance and plot it on the Smith chart.
2. Rotate the load reflection coefficient clockwise around a circle of constant radius on the Smith
chart by an angle of 2βl or 2`/λ times around the circle.
3. Read the normalized input impedance zin (−`) from the Smith chart, and unnormalize to get
Zin . This can be used in a voltage divider to get Ṽ (−`), from which Vo+ can be found using
equations.
4. Other quantities can be read from the Smith chart as well:
(a) VSWR: Real part of the input impedance when the constant VSWR circle crosses the
positive real axis.
(b) Voltage maxima/minima: The first voltage maximum occurs when the generalized reflection coefficient first crosses the positive real axis. The voltage minima occur when it
crosses the negative real axis. The rotation angles at which the extrema occur can be
read from the Smith chart and converted to distances along the line.
31
Example: Smith Chart Solution
Figure 1.28: Transmission line with source and load.
1. Find the normalized load impedance and plot on Smith chart.
2. Rotate to the generator end.
3. Read off zin and unnormalize to get Zin .
4. Read the VSWR from the Smith chart.
5. Locate the first voltage minimum.
32
1.6
Matching
A mismatched load (ZL 6= Zo ) means that power is reflected and is not delivered to the load. If
this is undesirable, an impedance matching network can be used to make the load appear to be
matched. There are many ways to do this:
• Stub tuning
• Quarter-wave matching
• Lumped element matching
• Matching transformers
• Multistage (broadband) matching networks
• Tapered transmission lines
• Isolators (non-reciprocal devices allowing only one-way power flow)
1.6.1
Shunt Single-Stub Matching
One type of matching network is a short section of transmission line placed in parallel with the main
line at some distance from the load (Fig. 1.29). The stub is typically terminated with an open or
short circuit. Since we are placing two elements in parallel, it is convenient to use admittances for
the design procedure, since the admittance of two parallel elements is the sum of the admittances
of the elements. The basic principle is to place the stub at a location where the input admittance
on the main line is of the form Yo + jB, and then choose the length of the stub so that its input
admittance is −jB. The parallel combination of the stub and main line then has admittance
Yin = Yo , which represents a match.
Figure 1.29: Shunt single-stub matching network.
The goal is to find the length of the stub d2 and the distance of the stub from the load d1 such
that the input impedance looking into the junction of the stub and main line is equal to Zo . The
solution procedure is as follows:
1. Normalize the load impedance.
33
2. Plot yL on the Smith chart. On a single-color Smith chart, this is done by plotting zL and
then rotating by 180◦ to yL .
3. Rotate towards the generator (clockwise) around the constant VSWR circle until the input
impedance reaches the g = 1 circle. At this point, the normalized input admittance of the
line is of the form yd1 = 1 + jb. Read yd1 and d1 in wavelengths from the Smith chart.
4. The input admittance of the sub must be ystub = −jb, so that yin = yd1 + ystub = 1. From
the Smith chart, determine the length of the stub d2 that gives this input admittance. The
generator end of the stub is at the point (0, −1). On a single color Smith chart, we flip this
about the origin to the point (1, 0). We then rotate towards the generator end until the
admittance is −jb, and read off the length of the stub in wavelengths. The lengths can easily
be converted to meters by multiplying by λ.
5. What would d2 be if we want to use an open circuit stub?
There are other solutions for this matching problem. We could continue rotating away from the
load until the main line input impedance hits the g = 1 circle again.
1.6.2
Series Single-Stub Matching
If the stub is in series, then the design procedure changes, so that we rotate from the load end of
the main line to the r = 1 circle, and then add a series stub to cancel the imaginary part of the
impedance.
34
Shunt Stub Matching Example 2
Figure 1.30: Shunt sub matching example.
1. Normalize the load impedance.
2. Plot yL on the Smith chart.
3. Rotate towards the generator (clockwise) around the constant VSWR circle until the input
impedance reaches the g = 1 circle. Read yd1 and d1 in wavelengths from the Smith chart.
4. The input admittance of the sub must be ystub = −jb. For an open circuit stub, we start at
yL,stub = 0 and rotate towards the generator end until the admittance is −jb, and read off
the length of the stub in wavelengths.
5. Find another solution to the problem:
35
1.7
Digital Signaling
One important application of transmission line theory is modeling connections carrying digital
signals between logic elements. Two of the main issues that must be dealt with are terminations
to reduce reflections and cross-talk between closely spaced lines.
1.7.1
Microstrip
One important type of transmission line that is used for both analog and digital systems is the
microstrip, which consists of a conductive strip separated from a ground plane by a dielectric layer
(Fig. 1.31). An example is a printed circuit board trace with a ground plane on the bottom of the
board.
Figure 1.31: (a) Microstrip transmission line. (b) Dielectric and air replaced by an effective medium
with relative permittivity ²0r .
The microstrip produces electric and magnetic fields in both the dielectric and the air above the
dielectric. To a good approximation, the dielectric and air can be replaced by an effective medium
everywhere above the ground plane with relative permittivity
²0r =
1
²r + 1 ²r − 1
p
+
2
2
1 + 10h/w
(1.128)
The inductance per unit length of a microstrip line can be approximated by
¤
 60 £ 8h
w
w
 c ln w + 4h
h ≤1
L'
¡w
¢¤−1 w
 120π £ w
c
h + 1.393 + 0.667 ln h + 1.444
h ≥1
The capacitance per unit length is

²0r

w
 60c ln [ 8h
+ 4h
]
w
C'

¡w
¢¤
 ²0r £ w
120πc h + 1.393 + 0.667 ln h + 1.444
w
h
(1.129)
≤1
(1.130)
w
h
≥1
These formulas allow microstrip transmission lines to be designed for a given characteristic impedance.
36
1.8
Printed Circuit Board (PCB) Termination
Consider a connection between two digital logic elements, as shown in Fig. 1.32. This can be
modeled as a microstrip transmission line. There are several ways to terminate the system to
minimize undesirable reflections on the transmission line.
Figure 1.32: Connection between two digital logic elements.
No termination. The FET at the receiver end appears as a high impedance load, which is
essentially an open circuit. The load reflection coefficient is ΓL ' 1. For a driver impedance of
10 Ω and pulse amplitude 5 V, the initial forward step amplitude is
v1+ =
25
50
5V =
V ' 4.167 V
10 + 50
6
This wave propagates until reaches the receiver, at which time it reflects and a reverse pulse
propagates towards the driver. The source reflection coefficient is
Γs =
10 − 50
2
=−
10 + 50
3
Repeated reflections lead to the voltage signal at the receiver end shown in Fig. 1.33.
Figure 1.33: Voltage as a function of time at the receiver end.
The change in the voltage decreases by ΓL Γs = −2/3 at each bounce. In order for the voltage at
the receiver to settle to within 10%, we must have
(ΓL Γs )N ≤ 0.1
(1.131)
which first occurs when N = 6. The time required for the voltage to settle to this level is
ts = 2N T
37
(1.132)
If the pulse is repeated with a frequency f , we would like to have the voltage settle by at least the
middle of the pulse, so that the next pulse is not disturbed too strongly by reflections still occurring
from the previous pulse. This means that we must have
ts ≤
1
4f
(1.133)
Substituting the definition of phase velocity and using N = 6, for the given example we find that
f < u/(48L).
If the line is very short, then the delay time T is small, and so the settling time ts is also small. In
this case, termination may be unnecessary. A rule of thumb is that if the pulse rise time is greater
than 6T , termination is not needed. For example, a trace 6.6” long on standard FR4 PC board
has a delay of T = 1 ns. The pulse rise or fall time must be greater than 6 ns in order to neglect
termination.
For a longer line or higher pulse repetition rates, reflections may be intolerable. For L = 10 cm and
u = 2 × 108 m/s, we find for the given example that f < 42 MHz. Alternately, the length of the
line must be less than about 0.1 λ. For a digital system, this is a very low operating frequency. To
do better, we need to add some kind of termination to the line to reduce the reflections.
Load termination. Another way to terminate the connection is to add a matching resistor of
value equal to Zo in parallel with the receiver transistor (Fig. 1.34). This eliminates all reflections.
But the power dissipated at the load is
P =
4.1662
V2
'
' 0.35 W
R
50
For a system with many connections, the total dissipated power would be intolerably large. Moreover, this is more power than can be supplied by a typical driver circuit.
Figure 1.34: Matching resistor at the load.
Source termination. Another termination scheme with less power consumption is to add a
resistor at the driver end which increases the effective source impedance to Zo (Fig. 1.35). The
initial forward wave has amplitude v1+ = 2.5 V, and the reflection from the receiver is v1− = 2.5 V.
But the reflection coefficient looking into the driver and source termination is zero, so there is only
one bounce, and the settling time is faster than in the case of no termination. Because the receiver
still appears as a high impedance load, little current flows and the power dissipated is small. Some
potential problems with this approach are that the driver impedance may be different depending
38
Figure 1.35: Matching resistor at the source.
on whether it is sourcing or sinking current; multiple loads do not see a “high” voltage at the same
time; and there is still some power dissipated due to the source resistance.
Figure 1.36: Thevenin load termination.
Thevenin termination. Another approach is to use a pullup resistor at the receiver (Fig. 1.36).
When the driver switches high and a forward wave travels to the load end, there is no reflection,
since the termination has value Zo . But in this case the voltage across the resistor is small and not
very much current flows through it. The current associated with the reflected wave is almost equal
and opposite to the incident current. When the driver goes low, the incident current turns off, so
there is a net current flow towards the driver. This setup has the advantage that the driver does
not have to supply very much current and only has to sink current when the pulse turns off.
Figure 1.37: Diode clamp termination.
Diode clamp termination. If the diode turn-on voltage is vd , and the incident voltage is v1+ ,
when the incident pulse arrives at the load, the upper diode in Fig. 1.37 will turn on. The voltage
at the load will be clamped at vcc + vd , so the amplitude of the reflected wave is only vcc + vd − v1+ .
39
For the given example, with vd = 0.7 V, v1− = 1.53 V, which is smaller than the reflection in the
case of no termination.
When further reflected forward pulses arrive at the diode, as long as the total current through the
diode is positive, the load appears to be a short, since the total voltage at the diode is clamped
and new pulses cannot change the load voltage. At some point, the total current through the diode
drops to zero and the diode shuts off, and the reflected wave must be determined by requiring the
total current (the sum of all forward and reverse currents) through the diode to be zero. After
that, the diode remains off, and the load appears to be an open circuit.
(Time t = 3T ) For the example circuit, the second forward wave has amplitude v2+ = Γs v1− '
−1.022 V. Since the voltage is clamped to 5.7 V, the reflection cannot change the voltage, so v2− =
−v2+ . To check and make sure the diode is still on, we need to look at the current through the
diode, which at the time of the second reflection is (v1+ − v1− + v2+ − v2− )/Zo ' 0.59/Zo . Since this
is positive, the diode is still on.
(Time t = 5T ) At the time of the third reflection from the load, the current would go negative
if we considered the diode to still be on. So, the diode turns off, and the total load current
(v1+ − v1− + v2+ − v2− + v3+ − v3− )/Zo must be zero. Solving for v3− gives a value of −.0926. After this
time, the pulse bounces between the source and open load. In the steady state, no current flows,
and the load voltage approaches 5 V.
Figure 1.38: (a) Bounce diagram and load voltage plot (b) for diode termination.
The bottom diode will never let the voltage at the load become smaller then −vd . The diodes also
protect the other circuit elements from damage due to static discharge.
40
1.9
Cross-talk
Consider two coupled transmission lines. One line (A) is active, and the other line is quiet (Q).
The lines share a common ground. The lines are coupled by mutual capacitance per unit length
Cm (F/m) and mutual inductance per unit length Lm (H/m). The goal is to determine the signal
induced on the quiet line when the active line is driven by a source.
Figure 1.39: Coupled transmission lines. The lines share a common ground.
Capacitative Coupling
Figure 1.40: Capacitative coupling.
As with the distributed capacitance associated with a single transmission line, the distributed
mutual inductance can also be approximated by a lumped element section as shown in in Fig. 1.40.
The current and voltage satisfy
∆im (z, t) = Cm ∆z
∂vA (z, t)
∂t
(1.134)
The current induced on the quiet line splits in half going in each direction, so that
Cf
∆iCr
Q (z, t) = ∆iQ (z, t) =
Cm ∆z ∂vA (z, t)
2
∂t
(1.135)
The ∆ means that this is only the contribution from coupling of one small section of the line, and
we will later add up the contributions all along the line to get the total induced voltage. The
corresponding forward and reverse voltage waveforms are
Cf
Cr
∆vQ
(z, t) = ∆vQ
(z, t) =
41
Zo Cm ∆z ∂vA (z, t)
2
∂t
(1.136)
Figure 1.41: Inductive coupling.
Inductive Coupling
For the mutual inductance in Fig. 1.41, a similar analysis leads to
Lf
Lr
∆vQ
(z, t) = −∆vQ
(z, t) =
Lm ∆z ∂vA (z, t)
2Zo
∂t
(1.137)
Total Coupling
The combined inductive and capacitative forward and reverse voltage contributions from a section
of the line are
µ
¶
Zo Cm
Lm
∂vA (z, t)
f
Cf
Lf
∆vQ = ∆vQ + ∆vQ =
−
(1.138)
∆z
2
2Zo
∂t
|
{z
}
r
∆vQ
=
Cr
∆vQ
+
Lr
∆vQ
Kf
¶
Lm
Zo Cm
∂vA (z, t)
+
=
∆z
2
2Zo
∂t
|
{z
}
µ
(1.139)
Kr
The constants Kf and Kr represent the strength of the coupling between the two lines. Notice that
Kr is always positive, whereas Kf can be positive, negative, or zero.
f
Figure 1.42: Summing up the contributions ∆vQ
(z 0 , t) at each z 0 between 0 and z to get the total
forward wave at z.
Now we want to add up the contributions from the coupling due to each short section of the lines
by integrating. For the forward wave, the voltage at a point z on the quiet line is due to the
forward coupled wave at all points z 0 ≤ z (see Fig. 1.42). The voltage on the active line at z 0 is
vA (z 0 , t) = vA (t − z 0 /u). Using Eq. (1.138), this induces a voltage at z 0 of
f
∆vQ
(z 0 , t) = Kf ∆z 0
42
∂vA (t − z 0 /u)
∂t
(1.140)
This voltage signal takes a time to = (z − z 0 )/u to arrive at the point z, so that
∂vA (t − (z − z 0 )/u − z 0 /u)
∂t
∂v
(t
−
z/u)
A
= Kf ∆z 0
∂t
f
(z, t) = Kf ∆z 0
∆vQ
(1.141)
We now integrate the contributions from all z 0 between the left end of the line and the point z:
X
0 ∂vA (t − z/u)
lim
K
∆z
f
∆z 0 →0
∂t
Z z
∂vA (t − z/u) 0
= Kf
dz
∂t
0
∂vA (t − z/u)
= Kf z
∂t
vQf (z, t) =
(1.142)
Does this result make sense? Let’s look at each term:
Kf : As the coupling capacitance and inductance grow, Kf increases, and the induced voltage also
increases, as expected.
z: The farther we go down the line, the bigger z is, and the bigger the induced voltage. In other
words, one mile of parallel quiet line picks up a lot more voltage than one inch.
vA : Because of the time derivative, the induced voltage doesn’t depend on the DC voltage on the
active line, but on how fast the active signal changes. This makes sense, because it is the
transients that couple energy to the quiet line, not the DC voltage.
r (z 0 , t) at each z 0 between z and ` to get the total
Figure 1.43: Summing up the contributions ∆vQ
reverse wave at z.
Now we will follow a similar procedure for the reverse wave. To find the total induced reverse wave
at a point z on the quiet line, we have to add up the contributions for all z 0 > z. The induced
reverse wave contribution at z 0 is
r
∆vQ
(z 0 , t) = Kr ∆z 0
∂vA (t − z 0 /u)
∂t
(1.143)
As this contribution propagates from z 0 to z, the time shift is to = (z 0 − z)/u, and after that time
shift the signal at z is
∂vA (t − (z 0 − z)/u − z 0 /u)
∂t
∂v
(t
+
z/u
− 2z 0 /u)
A
= Kr ∆z 0
∂t
r
∆vQ
(z, t) = Kr ∆z 0
43
(1.144)
Now we integrate the contributions from z to `:
Z
r
vQ
(z, t) = Kr
z
`
∂vA (t + z/u − 2z 0 /u) 0
dz
∂t
(1.145)
Using the chain rule twice and combining the results,
∂
0
vA (t + z/u − 2z 0 /u) = vA
(t + z/u − 2z 0 /u)
∂t
∂
0
vA (t + z/u − 2z 0 /u) = vA
(t + z/u − 2z 0 /u)(−2/u)
∂z 0
∂
∂
vA (t + z/u − 2z 0 /u) = (−u/2) 0 vA (t + z/u − 2z 0 /u)
∂t
∂z
Now, we can work the integral in Eq. (1.145) using the fundamental theorem of calculus:
Z
r
vQ
(z, t)
`
∂
v (t + z/u − 2z 0 /u) dz 0
0 A
∂z
z
¯`
= (−Kr u/2) vA (t + z/u − 2z 0 /u)¯z
= Kr
(−u/2)
= (−Kr u/2) [vA (t + z/u − 2`/u) − vA (t + z/u − 2z/u)]
= (Kr u/2) [vA (t − z/u) − vA (t + z/u − 2`/u)]
(1.146)
Equations (1.142) and (1.146) allow us to find the induced voltage on the quiet line in terms of the
voltage on the forward line.
44
Example
Consider a pair of coupled transmission lines with length ` = 1, coupling constants Kf = −0.02 ns/m,
Kr = 0.15 ns/m, and phase velocity u = 0.2 m/ns. We want to plot the forward and reverse voltages
at z = 0 and z = ` as functions of time. The active line is driven with the signal shown in Fig. 1.44
Figure 1.44: Voltage signal on active line.
f
Forward wave. At z = 0, we have vQ
(0, t) = 0 because of the factor of z in Eq. (1.142). At
z = `,
∂
f
vQ
(`, t) = (−0.02 ns/m)(1 m) vA (t − 5 ns)
(1.147)
∂t
The derivative of the active voltage is zero except during the rise time of the pulse, when the
derivative is equal to 50 V/ns. The forward voltage on the quiet line at z = ` is then
½
(−.02 ns)(50 V/ns) = −1 V 5 ns ≤ t ≤ 5.1 ns
f
vQ (`, t) =
(1.148)
0
otherwise
as shown in Fig. 1.45.
Figure 1.45: Induced forward wave on quiet line at z = `.
Reverse wave. At z = 0, we have
r
vQ
(0, t) = 0.015 [vA (t) − vA (t − 10 ns)]
(1.149)
This waveform is shown in Fig. 1.46. At z = `, the two terms in the reverse wave cancel, so
r (`, t) = 0.
vQ
Figure 1.46: Induced reverse wave on quiet line at z = 0.
45
1.10
Review
Transmission line equations
– Lumped element equivalent circuit
– Telegrapher equations
– Wave equation
Transients
– Voltage solution - forward and reverse waves
– Phase velocity
– Current solution and characteristic impedance
– Reflection coefficient
– Steady state voltage
– Bounce diagrams
– Multi-section lines
Sinusoidal steady state (time-harmonic)
– Phasor notation
– Time-harmonic wave equation
– Voltage solution - forward and reverse waves
– Complex propagation constant, attenuation constant, wavenumber
– Current solution and characteristic impedance
– Reflection coefficient
– Generalized reflection coefficient
– Standing waves, VSWR
– Input impedance
– Matched load, open circuit, short circuit, half-integer line, quarter-wave line
– Instantaneous power, complex power, time-average power
– Smith chart
∗
∗
∗
∗
∗
Reflection coefficient plane
Impedance circles
Landmarks: unit circle, real axis, upper/lower half planes, center, short, open
Standing waves - VSWR, voltage maxima and minima
Admittance
– Matching - shunt or series single stub
Digital signaling
– Matching terminations - load, source, diode clamp
– Cross-talk - forward and reverse waves on quiet line
46
Fundamentals
Transients
Lossless wave equation:
∂ 2 v(z, t)
∂ 2 v(z, t)
= LC
2
∂z
∂t2
√
Voltage solution: v(z, t) = v + (z − ut) + v − (z + ut), u = 1/ LC
Current solution: i(z, t) = v + (z − ut)/Zo − v − (z + ut)/Zo , Zo =
p
L/C
−
L − Zo
Reflections: Current and voltage boundary conditions at end of line, ΓL = v + = R
RL + Zo
v
Sinusoidal steady state
n
o
Phasor: v(z, t) = Re Ṽ (z)ejωt
Lossless time-harmonic wave equation:
d2 Ṽ (z)
+ β 2 Ṽ (z) = 0, β 2 = ω 2 LC
dz 2
Voltage solution: Ṽ (z) = Vo+ e−jβz + Vo− ejβz
˜ = Vo+ e−jβz /Zo − Vo− ejβz /Zo
Current solution: I(z)
Reflections: Current and voltage boundary conditions at end of line, ΓL =
Generalized reflection coefficient: Γ(z) =
Vo−
L − Zo
=Z
ZL + Zo
Vo+
V0− ejβz
= ΓL ej2βz
V0+ e−jβz
Ṽ (z) = Vo+ e−jβz [1 + Γ(z)]
Standing wave pattern: v(z, t) = |Ṽ (z)| cos [ωt + φ(z)]
Ṽ (z)
1 + Γ(z)
Input impedance: Zin (z) = ˜
= Zo
1 − Γ(z)
I(z)
Zin (−`) = Zo
n
o
Power: pav = 12 Re Ṽ I˜∗
ZL + jZo tan β`
Zo + jZL tan β`
Smith chart: Complex Γ(z) plane together with impedance circles from
Zin (z)
1 + Γ(z)
Zo = 1 − Γ(z)
Shunt stub matching: Find z so that Yin (z) = Yo + jB, add parallel reactance −jB at z so that
the parallel combination is Yo (matched)
Series stub matching: Find z so that Zin (z) = Z0 + jX, add series reactance −jX at z so that the
series combination is Zo (matched)
47
Chapter 2
Electrostatics
2.1
Maxwell’s Equations
Electromagnetic behavior can be described using a set of four fundamental relations known as
Maxwell’s Equations. These equations are not derived, but rather are a mathematical model for
observations. In general, these equations are given by
∇ · D = ρv
∂B
∇×E = −
∂t
∇·B = 0
∂D
∇×H = J +
∂t
(2.1)
(2.2)
(2.3)
(2.4)
where
E:
H:
D:
B:
ρv :
J:
Electric field intensity
Magnetic field intensity
Electric flux density
Magnetic flux density
Electric charge density
Electric current density
V/m
A/m
C/m2
Wb/m2
C/m3
A/m2
There is also an integral form of these equations that can be represented as
Z
I
D · ds =
ρv dV = Q
V
S
Z
I
d
E · d` = −
B · ds
dt A
C
I
B · ds = 0
Z
Z
IS
d
H · d` =
J · ds +
D · ds
dt A
A
C
48
(2.5)
(2.6)
(2.7)
(2.8)
where S is the closed surface bounding the volume V and C is the closed path bounding the area
A. We will actually show how the integral and differential forms of these equations are related a
little later. For now, just take them on faith.
Suppose that the fields do not change in time (static fields). All of the time derivatives go to zero,
and so Maxwell’s equations become
∇ · D = ρv
(2.9)
∇×E = 0
(2.10)
∇·B = 0
(2.11)
∇×H = J
(2.12)
Note that for the case of static fields, the electric and magnetic fields are no longer coupled.
Therefore, we can treat them separately. For electric fields, we are dealing with electrostatics. For
magnetic fields, we are dealing with magnetostatics.
2.2
Charge and Current Distributions
We need to remind ourselves about charges. The volume charge density is the amount of charge
per unit volume. It can vary with position in the volume. The total charge is the integral of the
charge density over the volume V , or
Z
Q=
V
ρv dV
(2.13)
where Q is the charge in coulombs (C). When dealing with things like conductors, the charge may
be distributed on the surface of a material. We therefore are interested in the surface charge density
ρs with units of C/m2 . This is the amount of charge per unit area on the surface. The total charge
Q would be the integral of ρs over the surface. Finally, we can have a line charge density ρ` with
units of C/m which is the amount of charge per unit distance along a line. An example of this
might be a very thin wire. The total charge Q would be the integral of ρ` over the length of the
line segment.
By a similar token, we can discuss current density. J is the volume current density, measured in
units of A/m2 . It represents the amount of current flowing through a unit surface area. The total
current flowing through a surface A is therefore
Z
I=
J · ds
(2.14)
A
We can also define a surface current density J s with units of A/m. This means the current is
confined to the surface, and so J s is the amount of current per unit length, where the length
represents the “cross-section” of the surface.
If we have a volume charge density ρv moving at a velocity u (note that we are including the vector
direction in the velocity), then J = ρv u.
2.3
Electric Fields and Flux
Let’s also quickly review electric fields and flux. Coulomb’s law states that
49
1. an isolated charge q induces an electric field E at every point in space. At an observation
point a distance R from this charge, the electric field is
E = R̂
q
4π²R2
(2.15)
where R̂ is the unit vector pointing from the charge to the observation point. ² is called the
permittivity of the medium.
2. The force on a charge q 0 due to an electric field is
F = q0E
(2.16)
Now, the permittivity is:
² = ²0 ²r
²0 = 8.854 × 10−12 F/m
²r
permittivity of vacuum (free space)
relative permittivity or dielectric constant of material
The electric flux density due to an electric field is
D = ²E
(2.17)
The electric field for a charge distribution is given by
Z
0
1
R−R
E=
dv 0
ρ0v
4π² v0 |R − R0 |3
(2.18)
In cartesian coordinates this becomes
y
RR'
P(x,y,z)
R
R'
x
ρv'
z
R0 = x0 x̂ + y 0 ŷ + z 0 ẑ
(2.19)
R = xx̂ + y ŷ + z ẑ
(2.20)
dv
0
0
0
= dx dy dz
50
0
(2.21)
1
E(x, y, z) =
4π²
Z
v0
ρ0v (x0 , y 0 , z 0 )
(x − x0 )x̂ + (y − y 0 )ŷ + (z − z 0 )
[(x −
x0 )2
+ (y −
y 0 )2
+ (z −
z 0 )2 ]3/2
dx0 dy 0 dz 0
Even for a fairly simple charge distribution ρ0v this equation is very complicated.
2.4
2.4.1
Review of Cylindrical and Spherical Coordinates
Cylindrical Coordinates
z
z
x
φ
r
(r, φ,p
z)
r = x2 + y 2
φ = tan−1 (y/x) = angle from +x axis
Swept Variable
r
φ
z
2.4.2
Differential Length
dr
rdφ
dz
Spherical Coordinates
(R, θ,p
φ)
R = x2 + y 2 + z 2
φ = tan−1 (y/x) = angle from +x axis
θ = cos−1 (z/R) = angle from +z axis
Swept Variable
R
θ
φ
Differential Length
dR
R dθ
R sin θdφ
51
y
(2.22)
z
θ
R
φ
x
2.5
y
Gauss’s Law
Starting with the point form of one of Maxwell’s equations
∇ · D = ρv
(2.23)
we want to convert it into the integral for. We do this by integrating both sides over a particular
volume to get
Z
Z
∇ · Ddv =
ρv dv.
(2.24)
v
v
The divergence theorem is then used to convert the volume integral on the left side to a surface
integral as given by
Z
I
D · ds = ρv dV = Q.
(2.25)
v
S
This equation is Gauss’s Law.
Let’s look at a parallel plate capacitor. We know that we have charge (equal and opposite) on the
two conductors. Gauss’s Law says that the flux equals the charge:
Z
I
ψe =
D · ds =
ρv dV = Q
(2.26)
S
V
z
+V0
0
52
2.5.1
Applying Gauss’s Law
1. Use symmetry to guess direction of D.
2. Express flux as a vector (use unit vectors) with an unknown constant.
3. Pick surface to use for applying Gauss’s Law.
4. Apply Gauss’s Law.
5. Solve for unknown constant.
2.5.2
Point Charge
R
1. Flux must extend radially from point charge.
2. Flux density: D = Do R̂.
3. Surface = sphere of radius R.
4. Gauss’s Law:
Z
I
D · ds =
ρv dV = Q
I
Z π
D · ds =
Do R̂ · R̂R2 sin θdθdφ
S
0
0
Z π
2
= 2πDo R
sin θdθ = 4πDo R2 = Q
S
(2.27)
V
Z 2π
(2.28)
(2.29)
0
5.
Do =
D =
Q
4πR2
Q
R̂
4πR2
(2.30)
(2.31)
So, the flux density falls off as 1/4πR2 = 1/surface area of sphere of radius R.
Note that we often use sin θdθdφ, which is a differential solid angle. A full sphere has 4π steradians
of solid angle.
53
2.5.3
Line Charge
Z
0
L Z 2π
0
D = Do r̂
(2.32)
Do r̂ · r̂rdφdz = 2πrLDo = Q
Q
2πrL
Q
Q 1
r̂ =
r̂
2πrL
L 2πr
Do =
D =
(2.33)
(2.34)
(2.35)
Note that Q/L = ρ` which is the line charge density. So
D=
ρ`
r̂
2πr
(2.36)
z
r
2.5.4
Plane Charge
integration
surface
z
y
x
charge
Suppose we have charge uniformly distributed over a plane. We choose the x-y plane for simplicity.
Since the flux will be uniform in x and y, just look over a small area.
½
Do ẑ
z>0
D=
(2.37)
−Do ẑ
z<0
Our surface for applying Gauss’s law is the cube with the plane of charge cutting it in half.
54
Z
Lx
Z
0
Z
Ly
0
Do ẑ · ẑdxdy − −
Lx
0
Z
0
Ly
Do ẑ · ẑdxdy = Q
(2.38)
2Do Lx Ly = Q
(2.39)
Do =
D =
Q
2Lx Ly
1 Q
ẑ,
2 Lx Ly
(2.40)
z>0
(2.41)
Note that Q/Lx Ly = ρs . Note also that the factor of 2 can be explained by the fact that 1/2 of
the flux goes in the +z direction and 1/2 goes in the −z direction.
2.5.5
Sphere of Charge
DR
ρv
R
2a
3
ρ va 3
R
3R 2
ρv C/m3
a
R
Consider a sphere of radius a filled with charge at a volume charge density of ρv . The flux will be
D = Do R̂.
For R ≤ a:
Z
2π
0
Z
π
Z
2
Do R̂ · R̂R sin θdθdφ =
0
4πR2 Do =
Do =
D =
2π
0
ρv
Z
π
0
Z
R
0
ρv R2 sin θdRdθdφ
R3 4π
3
ρv R
3
ρv
RR̂
3
(2.42)
(2.43)
(2.44)
(2.45)
For R ≥ a:
Z
0
2π
Z
0
π
Z
Do R̂ · R̂R2 sin θdθdφ =
4πR2 Do =
Do =
D =
55
2π
Z
0
0
ρv 3
π
a 4π
3
ρv a3
3R2
ρv a3
R̂
3R2
Z
0
a
ρv R2 sin θdRdθdφ
(2.46)
(2.47)
(2.48)
(2.49)
2.6
Electric Potential
Suppose that we have a positive charge in an electric field. To move the charge against the electric
field requires that we apply a force:
F ext = −qE
(2.50)
The energy expended is:
dW = −qE · d`
(2.51)
The differential energy per unit charge is defined as the differential electric potential:
dV =
dW
= −E · d`
q
(2.52)
with units of volts = J/C.
To find the potential difference, we integrate dV:
Z
Z P2
V21 = V2 − V1 =
dV = −
P1
P2
E · d`
(2.53)
P1
Let’s carefully consider the sign. If we have E = Eo x̂, and we integrate in d` from x1 to x2 , where
x2 > x1 , then V2 < V1 . Therefore, V2 − V1 < 0, meaning we have experienced a voltage drop. The
equation is:
Z
x2
V21 = V2 − V1 = −
x1
Eo dx = −Eo (x2 − x1 )
For static fields, V21 is independent of the path taken. This also implies that
I
E · d` = 0
(2.54)
(2.55)
C
Note that this is one of Maxwell’s equations for statics. In fact, there is a theorem in vector calculus
known as Stokes’ Theorem which states that
Z
I
E · d`
(2.56)
(∇ × E) · ds =
S
C
So, using (2.55) in Stokes’ Theorem results in
∇×E =0
(2.57)
which is the differential form of Faraday’s Law for statics.
2.6.1
Example 1
Let
E = xx̂ + y ŷ = r cos φx̂ + r sin φŷ = r(x̂ cos φ + ŷ sin φ) = rr̂
(2.58)
We want to find the potenial for the path defined by x = y from (0, 0) to (1, 1).
√
Z
VBA = VB − VA = −
0
56
2
¯√2
r2 ¯¯
rdr = − ¯ = −1
2 0
(2.59)
y
B
1
x
A
2.6.2
1
Example 2
Let E = xx̂ = r cos φx̂. We want to find the potenial for the circle of radius r = 1 about the origin
in the x-y plane. Therefore, d` = φ̂rdφ (don’t forget to plug in r = 1 everywhere).
Z
VBA = −
2π
xx̂ · φ̂dφ
0
Z
2π
= −
=
1
2
cos φ(− sin φ)dφ
0
Z 2π
sin 2φdφ = 0
0
y
A
x
B
2.6.3
Example 3
Let E = yx̂ + xŷ = r sin φx̂ + r cos φŷ. We will use the same unit circle as above.
Z
VBA = −
2π
[sin φ(x̂ · φ̂) + cos φ(ŷ · φ̂)]dφ
0
Z
2π
= −
[− sin2 φ + cos2 φ]dφ
0
Z
= −
2π
cos 2φdφ = 0
0
57
2.6.4
Poisson’s and Laplace’s Equations
We have examined the integral relation between electric field and potential. However, if we know
the potential, how do we obtain the electric field? First, we need to recall from vector calculus that
the differential of any scalar can be written as
dT
∂T
∂T
∂T
dx +
dy +
dz
∂x
∂y
∂z
µ
¶
∂T
∂T
∂T
=
x̂ +
ŷ +
ẑ · (dxx̂ + dy ŷ + dz ẑ)
∂x
∂y
∂z
=
= ∇T · d`
(2.60)
(2.61)
(2.62)
Therefore, we can rewrite (2.52) as
dV = −E · d` = ∇V · d`
(2.63)
E = −∇V
(2.64)
and can conclude from this that
Now, using Gauss’s Law in differential form leads to
∇ · D = ∇ · ²E = −²∇ · ∇V = ρv
ρv
∇ · ∇V = −
²
ρv
2
∇ V = −
²
This is known as Poisson’s Equation. If we have a charge-free region (ρv = 0) then
∇2 V = 0
(2.65)
(2.66)
(2.67)
(2.68)
which is known as Laplace’s Equation. Now we have mechanisms for finding the field if we know
the potential Note that in Cartesian coordinates
∇2 V =
∂2V
∂2V
∂2V
+
+
∂x2
∂y 2
∂z 2
(2.69)
We will actually solve these equations in the laboratory.
2.7
Electric Properties of Materials
We classify materials based upon their constitutive parameters
²
µ
σ
electrical permittivity
magnetic permeability
conductivity
F/m
H/m
S/m
We will focus on µ later in our discussion of magnetostatics. For now, let’s focus on the other two.
Conductors and dielectrics are classified by how well they conduct current. Dielectrics are essentially
insulators, meaning that while they may allow small amounts of current flow through them, this
current is quite small. Conductors on the other hand allow free flow of current. Materials that are
in between are called semiconductors. Semiconductors have the interesting property that we can
dope them properly such that we can control current flow and restrict it to given region.
58
2.7.1
Conductors
We are not going into great detail here. But we need to simply recognize that the conductivity relates current flow to the electric field which supplies the force to move the charges. This relationship
is
J = σE
(2.70)
which looks a lot like Ohm’s law (in fact, Ohm’s law is a simplification of this). Given this definition,
we see that a perfect dielectric has σ = 0 so that J = 0. A perfect conductor, on the other hand,
has σ → ∞, which implies that to have finite current density we must have E = 0 since E = J/σ.
For metals σ ∼ 10−7 , so it is common practice to set E = 0. A perfect conductor is an equipotential
medium, meaning that the electric potential is the same everywhere.
2.7.2
Resistance
R
R
− l E · dl
− l E · dl
V
R=
= R
=R
I
S J · ds
S σE · ds
(2.71)
Example: Linear resistor
1. Assume a particular voltage.
2. Calculate the resulting electric field.
3. Calculate the current from the electric field.
4. Divide voltage by current to get resistance.
Z
V
x2
= −
E · dl
(2.72)
x̂Ex · x̂dl
(2.73)
Zx1x2
= −
x1
= Ex l
Z
(2.74)
Z
I=
J · ds =
A
R=
A
σE · ds = σEx A
Ex l
l
V
=
=
I
σEx A
σA
59
(2.75)
(2.76)
How does the resistance apply to integrated circuits? A typical process technology has interconnect
lines that are 0.4µm wide and 0.6µm high. With copper as the metal the resistance is given by
R=
l
= 72 Ω/mm
(5.8107 )(.24µm2 )
(2.77)
Example: Conductance of a Coaxial Cable
What is the conduction between the conductors of a coaxial cable?
This time assume a current flowing between the two conductors I and then calculate the voltage
from this assumed current.
Jv = r̂
I
I
= r̂
A
2πrl
(2.78)
I
2πσrl
(2.79)
Since Jv = σE
Ev = r̂
Calculate the voltage from the electric field
Z a
Z
Vab = −
E · dl = −
b
b
Vab =
G0 =
2.7.3
I
ln
2πσl
a
I r̂ · r̂dr
2πσl r
µ ¶
b
a
G
1
I
2πσ
=
=
=
l
Rl
Vab l
ln(b/a)
(2.80)
(2.81)
(2.82)
Dielectrics
Let’s examine dielectrics a little bit more. We will assume we are dealing with perfect dielectrics
with σ = 0 and therefore J = 0. If we look at the microscopic level, we consider an atom or
molecule with a positively charged nucleus and negatively charged electron cloud. The cloud center
is coincident with the nucleus center, leading to a neutral charge configuration. If an external
electric field E ext is applied to the material, the center of the electron cloud will be displaced from
its equilibrium value. While we still have charge neutrality, we can consider that there will be an
electric field emanating from the positively charged nucleus and ending at the negatively charged
cloud center. This process of creating electric dipoles within the material by applying an electric
field is called polarizing the material.
The induced electric field created by our new dipole is referred to as a polarization field, and it is
weaker and in the opposite direction to E ext . We actually express this using the flux density
D = ²0 E + P
60
(2.83)
+q
- - - +
- - -
- + - - - -
Eext
+
Eext
-
-q
where P is referred to as the Polarization of the material. This quantity is proportional to the
applied field strength, since the separation of the charges is more pronounced for stronger external
fields. We can therefore write
P
= ²0 χe E
(2.84)
D = ²0 E + ²0 χe E = ²0 (1 + χe )E = ²E
(2.85)
(2.86)
² = ²0 (1 + χe )
| {z }
²r
2.8
Capacitance
Capacitance is capacity to store charge.
Q
(2.87)
V
with units of Farads = C/V. V is the potential difference between the conductor with charge +Q
and the conductor with charge −Q.
C=
Steps:
1. Assume a charge on the conductors
2. Find D from Gauss’s Law
3. Find E = D/².
4. Find V from E
5. Find C from C = Q/V
2.8.1
Parallel Plate Capacitor
We assume that the charge will evenly distribute itself over the conducting plates.
61
I
Z
Ly
0
Z
0
D · ds = Q
(2.88)
Do dxdy = Q
(2.89)
Lx
Do =
D =
E =
V
=
C =
2.8.2
Q
Q
=
(2.90)
Lx Ly
A
Q
− ẑ
(2.91)
A
Q
− ẑ
(2.92)
A²
Z d
Q
Qd
−
− ẑ · ẑdz =
(2.93)
A²
A²
0
Q
²A
=
(2.94)
V
d
z
+Q
d
Spherical Shell Capacitor
y
-Q
a +Q
x
ε
b
62
Area = LxLy
-Q
E
y
The charge will evenly distribute itself over the conducting spherical shells.
I
D · ds = Q
Z 2π Z π
Do R2 sin θ dθdφ = Q
0
(2.95)
(2.96)
0
4πR2 Do = Q
Do =
D =
E =
Vab = Va − Vb =
=
C =
Q
4πR2
Q
R̂
4πR2
Q
R̂
2
4π²R
Z a
Q
−
R̂ · R̂dR
2
4π²R
b
¯
µ
¶
Q ¯¯a
Q 1 1
=
−
4π²R ¯b
4π² a b
4π²
Q
= 1 1
Vab
a − b
Question: Why does larger ² increase the capacitance?
63
(2.97)
(2.98)
(2.99)
(2.100)
(2.101)
(2.102)
(2.103)
Chapter 3
Magnetostatics
In electrostatics, we considered the electric field and flux density arising from charges. In magnetostatics, we are concerned with magnetic fields and flux density arising from currents. The basic
equations of interest are Ampere’s Law and Gauss’ Law for magnetostatics:
Point or Differential Form:
∇·B = 0
(3.1)
∇×H = J
(3.2)
Integral Form:
I
IS
B · ds = 0
Z
H · d` =
J · ds
(3.3)
(3.4)
S
C
We will begin by using this integral form of Ampere’s Law to determine magnetic fields much like
we used Gauss’ Law to determine electric fields in the prior section.
3.1
Ampere’s Law
Suppose we are given a static current distribution, which means we have charges that move, but
that motion is constant in time. We can use Ampere’s Law to determine the electric field that
emanates from the current. Let’s try a few examples:
3.1.1
Line Current
Suppose we have a current of I amperes moving through a thin wire. The current flows in the ẑ
direction. Recall that we use the right hand rule which means that we put the thumb of our right
hand in the direction of the current and our fingers will go in the direction of the magnetic field.
The magnetic field will loop around the wire.
39
I
Z
H · d` =
z
Z
I
Ho φ̂ · φ̂rdφ = I
(3.6)
2πHo r = I
(3.7)
Ho =
x
H =
The magnetic flux density is then
B = µH =
3.1.2
(3.5)
S
2π
0
y
C
J · ds = I
C
I
2πr
I
φ̂
2πr
µI
φ̂
2πr
(3.8)
(3.9)
(3.10)
Coaxial Cable
On the conductors, the current density can be written as
(
a
I
ẑ
πa2
−I
ẑ
π(c2 −b2 )
J=
c
inner conductor
outer conductor
(3.11)
b
We can compute the magnetic field in 4 regions:
r < a:
Z
0
Z
2π
Ho φ̂ · φ̂rdφ =
2πrHo =
Ho =
H =
2π
Z
r
I
ẑ · ẑro dro dφ
2
0
0 πa
r2
I
r2
=
I
2π
πa2
2
a2
Ir
2πa2
Ir
φ̂
2πa2
a < r < b: Note that the integral on the left hand side is the same for all regions.
Z 2π Z a
I
I
a2
2πrHo =
ẑ
·
ẑrdrdφ
=
2π
=I
2
πa2
2
0
0 πa
I
Ho =
2πr
I
H =
φ̂
2πr
40
(3.12)
(3.13)
(3.14)
(3.15)
(3.16)
(3.17)
(3.18)
b < r < c:
Z
2π
Z
r
I
ro dro dφ
− b2 )
0
b
µ 2
¶
I
r
b2
I−
2π
−
π(c2 − b2 )
2
2
µ 2
¶
2
r −b
I −I
c2 − b2
· 2
¸
c − b2 − r2 + b2
I
c2 − b2
µ 2
¶
c − r2
I
c2 − b2
¶
µ 2
I
c − r2
φ̂
2πr c2 − b2
2πrHo = I −
=
=
=
=
H =
(3.19)
π(c2
(3.20)
(3.21)
(3.22)
(3.23)
(3.24)
r > c: Note that on the right hand side of (3.23), we simply replace r with the outer limit of
integration c (the left hand side remains the same). Therefore, the right hand side will be zero, so
H=0
(3.25)
This makes sense, since there is no net current enclosed.
3.1.3
Infinite Current Sheet
Consider a surface current in the x-y plane flowing in the x̂ direction. The magnetic field will be
½
−Ho ŷ
z>0
H=
(3.26)
Ho ŷ
z<0
Let’s consider doing the integration over a square path of side length b:
Z
z
0
J s xˆ
Ho dy −
b
Z
0
Ho dy =
0
b
Js dy
2Ho b = Js b
Js
Ho =
2
Js
H = ± ŷ
2
y
0
Z
b
b
(3.27)
(3.28)
(3.29)
(3.30)
Note that we only integrated the surface current density over the line from 0 to b in y. We did not
integrate in z since the surface current density is in units of A/m (ie we only need to integrate in
one dimension to get the total current enclosed).
3.1.4
Solenoid (coil)
In this case, H = Ho ẑ and the total current enclosed by the surface is N I, where N is the number
of turns of wire and I is the current flowing in the wire. Also, we let the contour of the surface
41
outside of the solenoid go off to infinity where the fields are zero. The main contribution to the
line integral then comes from the integration within the solenoid of length L:
Z
N turns, radius a
L
0
z
NI
L
NI
H =
ẑ
L
NI
B = µ
ẑ
L
Ho =
I
L
3.1.5
(3.31)
(3.32)
(3.33)
(3.34)
Toroid
N turns
Z
2π
0
Ho φ̂ · φ̂rdφ = N I
2πrHo = N I
NI
Ho =
2πr
NI
H =
φ̂
2πr
NI
B = µ
φ̂
2πr
b
I
3.2
Ho ẑ · ẑdz = N I
(3.35)
(3.36)
(3.37)
(3.38)
(3.39)
Magnetic Vector Potential
In electrostatics, we had the notion of a potential. This concept is useful, since sometimes it is
more convenient to compute the potential and then compute the electric field using E = −∇V . It
would be convenient to also define a magnetic potential to assist in the computation of magnetic
fields. It turns out that we have not identified a measurable quantity that we could call a magnetic
potential. However, we can define one mathematically to assist us in computation. We call this
quantity the magnetic vector potential and denote it as A.
In constructing an equation for A, we use Gauss’s Law of magnetics that ∇ · B = 0. We also take
advantage of a vector identity that for any vector A,
∇ · (∇ × A) = 0
(3.40)
B =∇×A
(3.41)
Therefore, we will use
so that we are guaranteed to satisfy Gauss’s Law. Since B has units Wb/m2 , A has units of Wb/m.
42
Since B = µH, Ampere’s Law gives
∇ × B = µJ
(3.42)
∇ × (∇ × A) = µJ
(3.43)
Now, we invoke a second vector identity that states
∇2 A = ∇(∇ · A) − ∇ × (∇ × A)
(3.44)
where ∇2 A is the vector Laplacian operation. We can therefore write
∇(∇ · A) − ∇2 A = µJ
(3.45)
Now, when we define B = ∇ × A, we have not uniquely specified the vector A. In order to uniquely
specify a vector, we must specify its curl and it divergence. Therefore, we are still free to choose
the divergence of A any way we like. It is clearly convenient to specify
∇·A=0
(3.46)
∇2 A = −µJ
(3.47)
Our equation therefore becomes
So, we now have a differential equation for A much like we had an equation for V . A solution to
this differential equation (which we will not prove) is
Z
µ
J
A=
dV 0
(3.48)
4π V R0
where R is the distance from the integration point to the point where the field is observed, or
R0 = |R − Ri |. Here, R is the point where A is evaluated. Ri is the point of integration. So, if we
have a current distribution, we can compute A using this integral and then compute B = ∇ × A.
3.3
Magnetic Properties of Materials
We need to briefly discuss the magnetic permeability µ much like we discussed the permittivity
for electrostatics. Using a classical description of matter, all atoms have electrons which orbit the
nucleus. This orbiting charge represents a current loop which creates a magnetic moment. In most
materials, called diamagnetic materials, these atoms are randomly aligned so that there is no net
magnetic effect.
The electrons also spin which creates another contribution to the magnetic moment. In atoms with
even numbers of electrons, there are always two spins that are equal but opposite, resulting in zero
net magnetic moment from spin. For an odd number of electrons, there is a net magnetic effect
from the single unpaired electron.
When a material is exposed to a magnetic field H, we can express the magnetic flux density as
B = µ0 H + µ0 M = µ0 (H + M )
(3.49)
where M is called the magnetization vector of a material. This vector represents the vector sum
of the magnetic dipole moments of the atoms. Physically, the magnetic field is aligning the atomic
43
magnetic dipoles. The degree to which these dipoles can be aligned is represented by the magnitude
of M . Much like we did in electrostatics, we define a magnetic susceptibility χm and write
M
= χm H
(3.50)
B = µ0 (H + χm H) = µ0 (1 + χm ) H
| {z }
(3.51)
µr
B = µH
(3.52)
The units of µ are H/m. For most materials, χm is so small that we can write µr = 1. Ferromagnetic
materials, however, which are suceptible to magnetic alignment, can have high values of µr . For
example, pure iron has µr = 2 × 105 .
1. Diamagnetic χm < 0 (χm ∼ 10−5 , µr ∼ 1)
2. Paramagnetic χm < 0 (χm ∼ 10−5 , µr ∼ 1)
3. Ferromagnetic |χm | À 1
3.4
Inductance
The physical behavior of currents and magnetic fields is interesting. For example, we know that
currents create magnetic fields. However, a static magnetic field (created by a magnet) which cuts
through a loop of wire will not create a current. The key concept is that sources (charges, currents)
create fields.
We will see that time-varying magnetics fields can induce currents. For example, if we have two
separate loops, and drive a current through one (that changes with time), we will observe a current
in the other. This magnetic linking is mutual inductance. Similarly, in a solenoid, the behavior of
the current in one loop is influenced by time-varying current in another loop. This is self inductance,
since the actual current flowing through both loops is the same.
Despite the fact that we need time-varying currents/fields to have this coupling, we can compute
the strength of the coupling using static analysis. The strength of the coupling is the inductance
of the system:
Λ
Inductance: L =
(3.53)
I
where
Λ = magnetic flux linkage = flux linking current
I = current producing flux linkage
3.4.1
Steps for Computing Inductance
1. Assume a current and a form for H
2. Calculate H using Ampere’s Law
44
3. Calculate B = µH
R
4. Calculate Λ = s B · ds
5. Calculate L = Λ/I
3.4.2
Solenoid
We will change the notation so that the length of the solenoid is `, since we will be using the symbol
L to mean inductance.
1. We know the current and H was computed previously in class
2.
H=
NI
ẑ
`
(3.54)
NI
ẑ
`
(3.55)
B · ds
(3.56)
3.
B=µ
4. For a single loop, the flux is:
Z
Λ1 =
2π
0
Z
Z
a
0
2π
Z
a
µ
=
0
= µ2π
0
a2 N I
NI
ẑ · ẑrdrdφ
`
(3.57)
(3.58)
2 `
NI
= µπa2
`
(3.59)
For N loops, the flux linkage is
Λ = N Λ1 = µπa2
5.
L=
N 2I
`
Λ
µπa2 N 2
=
I
`
(3.60)
(3.61)
You may wonder why we use Λ = N Λ1 . The concept is that if flux passes through one loop, it will
influence current in that loop (time-varying fields). If it passes through multiple loops, it will have
an effect on each one. But since the current through each loop is the same, the overall effect on
the solenoid current is additive.
45
3.4.3
Toroid
Based on our prior work, steps 1-3 yield:
NI
φ̂
2πr
B=µ
(3.62)
The integral in step 4 is relatively difficult to perform, so let’s make an approximation. If the radius
a of the core of the toroid is small compared to the radius of the toroid (ie b À a), then
B≈µ
NI
φ̂
2πb
(3.63)
Now, the integration is over the area of the toroid cross-section:
Z Z
NI
φ̂ · φ̂drdz
2πb
NI 2
N Ia2
= µ
πa = µ
2πb
2b
N 2 Ia2
Λ = N Λ1 = µ
2b
Λ1 =
b
a
The inductance is therefore:
L=
3.4.4
µ
Λ
N 2 a2
=µ
I
2b
(3.64)
(3.65)
(3.66)
(3.67)
Coax
We have already computed
µI
φ̂
(3.68)
2πr
The flux linking the two conductors is now the flux passing through the area between the conductors:
B=
Z bZ
Λ =
l
a
=
=
0
`
µI
φ̂ · φ̂dzdr
2πr
µI`
ln r|ba
2π
µ ¶
µI`
b
ln
2π
a
(3.69)
(3.70)
(3.71)
The inductance is:
L =
L
`
=
µ ¶
b
µ`
ln
2π
a
µ ¶
µ
b
ln
2π
a
46
(3.72)
(3.73)
3.4.5
Parallel Flat Conductors
Z
H ≈ Ho ŷ
0
w
Ho ŷ · ŷdy = Ho w = I
I
ŷ
w
µI
ŷ
B =
w
Z dZ
Λ =
H =
w >> d
x
I
d
I
y
0
w
L
`
47
= µ
d
w
0
(3.74)
(3.75)
(3.76)
`
µI
µI
dzdx =
`d
w
w
(3.77)
(3.78)
Chapter 4
Dynamic Fields
We are now ready to look at the behavior of fields that vary in time. To begin, let’s examine a very
interesting law: Faraday’s Law. We will then look more generally at Maxwell’s Equations.
4.1
Faraday’s Law
A time-varying magnetic flux through a loop will cause a current to flow in the loop of wire.
Faraday’s Law describes this effect. To begin, consider a wire loop as shown. A magnetic field
passes through the loop (supplied by an external source) such that the flux density vector B is
normal to the surface defines by the loop. The flux is defined as
Z Z
φ=
B · ds
(4.1)
S
with units of Weber (Wb). When this flux changes in time, a current will flow in the loop. This
means that a voltage has been created across the loop terminals called the electromotive force (emf).
I
Vemf
+
-
Vemf
Charge buildup creates
positive potential
dφ
d
=−
=−
dt
dt
Z Z
B · ds
(4.2)
S
Note that if B does not change in time, Vemf = 0.
The negative sign comes from Lenz’s Law, which states that the induced current will oppose the
change in flux. To ensure the correct polarity of Vemf , we use the right hand rule.
48
Thumb = direction of ds
Fingers = direction of + terminal to − terminal
For the above loop, let ds be out of the page. If φ decreases, Vemf is positive. So, Vemf represents
the voltage available to the load circuit.
I
+
Vemf ~
-
There are two distinct ways to get this emf.
• Transformer emf: A time varying magnetic field linking a stationary loop
• Motional emf: A moving loop with a time varying area (relative to the normal component of
B) in a static magnetic field
4.1.1
Transformer Action
This is what we’ve been talking about. The applied B through a loop changes in time. The induced
potential at the loop terminals is called the transformer emf.
z
y
b
+
- I
x
For example, consider the loop shown with
B = Bo tẑ
Z 2π Z b
b2
φ =
Bo trdrdφ = Bo t 2π = Bo tπb2
2
0
0
dφ
= −Bo πb2
Vemf = −
dt
(4.3)
(4.4)
(4.5)
B is increasing in time, so I is induced as shown to oppose the change. Vemf is therefore negative.
49
Notice also the following. Since Vemf 6= 0 in this system, E 6= 0. If we integrate E around the loop,
we will obtain the voltage Vemf . Using the right hand rule for integration direction gives
I
Z Z
d
Vemf =
E · d` = −
B · ds
(4.6)
dt
C
S
This is Faraday’s Law. Note that in statics, we had a minus sign to maintain the proper potential
reference. However, here the electric field is pushing the charge to make one terminal more positive
than the other. Therefore, we do not have the minus sign.
I
+
Vemf
E
The ideal transformer uses the transformer emf. The time varying voltage creates a time varying
magnetic field in the core that has a high magnetic permeability. The time varying magnetic flux
induced a current in second winding.
4.1.2
Generator Action
This occurs when the loop is mechanically altered while the flux density remains constant.
y
+
Vemf
-
v
B = B0 zˆ
l
x
Z `Z
vt
φ(t) =
0
Vemf
= −
0
Bo dxdy = Bo `vt
d
φ = −Bo `v
dt
(4.7)
(4.8)
Electromagnetic Generator
A loop of length ` and width w is rotating with an angular velocity of ω within a constant magnetic
field given by
B = ẑBo
50
(4.9)
Z
Φ =
B · ds
(4.10)
ẑBo · n̂ds
(4.11)
ZS
=
S
n̂ = sinα ẑ + cosα ŷ
(4.12)
α = ωt + Co
(4.13)
where
Z `Z
w
Φ =
Bo cosαdydz
(4.14)
= Bo w` cos (ωt + Co )
(4.15)
0
4.2
0
Relating Maxwell’s Laws in Point and Integral Form
We need to recall once again the vector derivative operations:
Gradient:
∇φ(x, y, z) =
∂φ
∂φ
∂φ
x̂ +
ŷ +
ẑ
∂x
∂y
∂z
(4.16)
∂Fx ∂Fy
∂Fz
+
+
∂x
∂y
∂z
(4.17)
Divergence:
∇ · F (x, y, z) =
Curl:
µ
∇ × F (x, y, z) = x̂
∂Fy
∂Fz
−
∂y
∂z
¶
µ
+ ŷ
∂Fx ∂Fz
−
∂z
∂x
¶
µ
+ ẑ
∂Fy
∂Fx
−
∂x
∂y
¶
(4.18)
We also have two vector identities:
∇ × ∇f
= 0
(4.19)
∇ · (∇ × F ) = 0
(4.20)
We also need to consider the integral relations:
Stokes Theorem:
Z Z
I
∇ × F · ds =
S
Divergence Theorem:
F · d`
(4.21)
F · ds
(4.22)
C
Z Z Z
I
∇ · F dV =
V
S
51
Let’s use these to derive Maxwell’s equations in point form from the equations in integral form:
Faraday’s Law:
I
Z Z
d
E · d` = −
B · ds
dt
C
S
Z Z
Z Z
d
∇ × E · ds = −
B · ds
dt
S
S
∂
∇×E = − B
∂t
(4.23)
(4.24)
(4.25)
Ampere’s Law:
I
H · d` =
Z Z
C
∇ × H · ds =
S
∇×H =
Z Z
Z Z
d
D · ds +
J · ds
dt
Z ZS
Z ZS
d
D · ds +
J · ds
dt
S
S
∂
D+J
∂t
(4.26)
(4.27)
(4.28)
Gauss’ Laws:
I
Z Z Z
D · ds =
Z Z Z
S
V
ρv dV
(4.29)
ρv dV
(4.30)
Z Z Z
∇ · DdV
=
V
V
∇ · D = ρv
(4.31)
I
B · ds = 0
(4.32)
∇·B = 0
(4.33)
S
4.3
Displacement Current
Consider Ampere’s Law:
I
Z Z
Z Z
Z Z
d
∂
H · d` =
D · ds +
J · ds =
D · ds + Ic
dt
∂t
where Ic represents the conduction current. Notice that the term
Z Z
∂
Id =
D · ds
∂t
has units of Amperes. We call this the displacement current.
52
(4.34)
(4.35)
Consider a parallel plate capacitor as shown. In the wire, E = 0 and so D = 0. Therefore, the
current is
Iw = Ic = conduction current
dvs
= C
= −CVo ω sin ωt
dt
+
vs (t ) ~
-
ε1
Z Z
Icap = Id =
(4.37)
C
y
vs (t ) = V0 cos ωt
In the capacitor, J = 0 so
(4.36)
∂
D · ds
∂t
Since E = (Vo /d) cos ωt ŷ, D = (Vo ²1 /d) cos ωt ŷ.
Z Z
²1 A
Vo ²1
ω sin ωt dxdy = −
Vo ω sin ωt = −CVo ω sin ωt
Icap = −
d
d
|{z}
(4.38)
(4.39)
C
So: Id = Ic so that the displacement current allows continuity of current.
4.4
Continuity of Charge
Consider a volume V containing a charge density ρv and total charge Q. The only way for Q to
change is by charge entering/leaving the surface S bounding V . If I = the net current flowing
across S out of V
Z Z Z
dQ
d
=−
I=−
ρv dV
(4.40)
dt
dt
V
But we can also write
I
d
I=
J · ds = −
dt
S
Z Z Z
V
ρv dV
(4.41)
The last term came from Eq. (4.40). This equation represents the integral form of the continuity
of charge. If we now use the Divergence theorem:
I
Z Z Z
J · ds =
∇ · JdV
(4.42)
S
V
∇·J
∂
= − ρv
∂t
This is the continuity of charge in point or differential form.
53
(4.43)
4.5
Maxwell’s Laws in Time-Harmonic Form
To go to sinusoidal steady state, we assume a time variation of cos ωt. Recall once again that
n
o
v(t) = Re Ṽ ejωt
(4.44)
Now, we haven’t explicitly written it this way yet, but all of the fields are functions of space and
time:
E = E(x, y, z, t) = E(R, t)
n
o
˜
jωt
E(R, t) = Re E(R)e
o
n
∂E(R, t)
˜
jωt
= Re E(R)jωe
∂t
Therefore, Maxwell’s equations in time-harmonic (phasor) form are
I
Z Z
˜
˜ · ds
E · d` = −jω
B
Z Z
Z Z
I
˜
˜
H · d` = jω
D · ds +
J˜ · ds
Z Z Z
I
˜
D · ds =
ρ˜v dV
IS
˜ · ds = 0
B
(4.45)
(4.46)
(4.47)
(4.48)
(4.49)
(4.50)
(4.51)
S
˜ = −jω B
˜
∇×E
˜ + J˜
˜ = jω D
∇×H
˜ = ρ˜
∇·D
v
˜ = 0
∇·B
(4.52)
(4.53)
(4.54)
(4.55)
54
4.6
Boundary Conditions
Our goal is to understand how boundaries in materials impact electric and magnetic fields. Break
the electric field and magnetic field into components that are either tangential or normal to the
boundary as given by
E 1 = E1t t̂ + E1n n̂
(4.56)
E 2 = E2t t̂ + E2n n̂
(4.57)
H 1 = H1t t̂ + H1n n̂
(4.58)
H 2 = H2t t̂ + H2n n̂.
(4.59)
The goal is to find the relationship between the various components.
We will begin with electric fields. Consider a material boundary as shown with a contour that
crosses the boundary. Faraday’s law is
I
Z Z
d
E · d` = −
B · ds
(4.60)
dt
C
S
2
Normal points from 1 to 2
h
n̂
ε 2 , µ2 , σ 2
1
n̂
w
tˆ
ε1 , µ1 , σ 1
To apply this to the contour, let the contour shrink to zero area so that the right hand side of
Faraday’s law goes to zero. If we let w → 0, the left hand side becomes
Z 0
Z 0
Z h/2
Z h/2
En1 n̂ · n̂d` +
En1 n̂ · (−n̂)d` +
En2 n̂ · n̂d` +
En2 n̂ · (−n̂)d` = 0
(4.61)
−h/2
−h/2
0
0
In other words, 0 = 0. This isn’t very useful.
n̂
h
tˆ
If we instead let h → 0, we get
Z
0
Z
w
E1t t̂ · (−t̂)d` +
Z w
0
w
E2t t̂ · t̂d` = 0
(4.62)
(E2t − E1t )d` = 0
(4.63)
E2t − E1t = 0
(4.64)
0
55
So, the tangential component of the electric field is continuous across the boundary. Note that
we can also write
n̂ × (E 2 − E 1 ) = 0
(4.65)
since n̂ × E is the tangential electric field.
Et 2
n̂
w
E t1
tˆ
If we apply the same technique to Ampere’s law, we get the same result WITH ONE TWIST.
I
Z Z
Z Z
d
H · d` =
B · ds +
J · ds
(4.66)
dt
C
S
S
Suppose that J represents a surface current that flows in the
tangent to the surface but points normal to the surface of our
Z w
Z w
H1t t̂ · (−t̂)d` +
H2t t̂ · t̂d` =
0
Z w 0
(H2t − H1t )d` =
0
ŝ = n̂ × t̂ direction. Note that ŝ is
integration contour.
Z w
Js d`
(4.67)
Z0 w
Js d`
(4.68)
0
H2t − H1t = Js
(4.69)
This works out to
n̂ × (H 2 − H 1 ) = J s
(4.70)
So, tangential H can be discontinous across the interface if a surface current exists.
Let’s try Gauss’ law. We use a “pillbox” for the integration.
I
Z Z Z
D · ds =
ρv dV
S
(4.71)
V
h
Dn 2
n̂
Dn1
tˆ
If h → 0, the right hand side will go to zero unless there is a surface charge density. Then,
Z Z
Z Z
Z Z
Dn2 n̂ · n̂ds +
Dn1 n̂ · (−n̂)ds =
ρs ds
(4.72)
A
A
A
Dn2 − Dn1 = ρs
(4.73)
n̂ · (D2 − D1 ) = ρs
(4.74)
56
So normal D is discontinous by the surface charge density. For magnetic fields:
Bn2 − Bn1 = 0
(4.75)
n̂ · (B 2 − B 1 ) = 0
(4.76)
Normal B is continous. Let’s summarize:
Field
Tangential E
Tangential H
Normal D
Normal B
Two Dielectrics
Continuous
Continuous
Continuous
Continuous
Dielectric Conductor
Continous: n̂ × E = 0
Discontinous: n̂ × H 1 = 0, n̂ × H 2 = J s
Discontinous: n̂ · D1 = 0, n̂ · D2 = ρs
Continous: n̂ · B = 0
57