. (7.67) Because of the load, V+ and V can no
longer be arbitrarily chosen but must satisfy
the boundary conditions. These reflection and
transmission coefficients have a number of
interesting properties. If the load impedance is
0 (a short), R = 1 and so

the charge density is q, the total charge in
this slab is Q = qdxA and it feels a net force
F~ = QE~ . Because a current is flowing, charge
is moving relative to this force and so work is
being done. The work associated with the slab
moving from one end o

frequencies can no longer be solved for
analytically, for a coaxial cable a rough
approximation for the TM modes is to ask that
the wavelength be a multiple of radial spacing
c 2 n (ro ri) n = 1, 2, 3, . . . , (7.92) and for
a TE mode that there be an int

radius 1/c located at (1, 1/c) given by the
complex part of the input impedance c. The
intersection of these two circles relates the
input impedance to the reflection coefficient,
conveniently found graphically on a Smith
chart (Figure 7.7). 94 Circuits,

common type. 7.3.1 Governing Equations Start
with the wave form of Maxwells equations
without any sources: 2E~ = 2E~ t2
2H~ = 2H~ t2 . (7.77) We are looking
for waves that travel along the axis of the
waveguide periodically as e itz. The real
part of is t

have been considering conduction in the lowfrequency limit; in this section we will use
Maxwells equations to look at how AC fields
penetrate conductors and are guided by them
at higher frequencies. 7.2.1 Skin Depth
Assume that the conductor is described

than the cut-off wavelength c , is pure real
and so the mode decays exponentially. When
is greater than the cut-off frequency for a
mode, is pure imaginary and the mode
propagates. These modes are labeled TMmn.
Repeating this analysis for the TE wave by

infinitely wide. w h I -I Figure 7.9. Transmission
line for Problem 7.4. (7.5) The most common
coaxial cable, RG58/U, has a dielectric with a
relative permittivity of 2.26, an inner radius of
0.406 mm, and an outer radius of 1.48 mm. (a)
What is the chara

the same. Since the field of a solenoid of
radius r and length l with n turns/meter is H =
nI, the inductance is L = I = N I Z B~ dA~ = nl
I nIr2 = n2 lr2 . (7.24) 7.2 Wires and
Transmission Lines 85 I Figure 7.3. A solenoid;
the dotted line closes the in

Voltage Standing-Wave Ratio (VSWR) VSWR
Vmax Vmin = 1 + |R| 1 |R| (7.69) or |R| =
VSWR 1 VSWR + 1 . (7.70) The VSWR is one
of the most important measurements in an RF
system, used to ensure that impedances are
matched so that all of the power goes in the

related to the capacitance and inductance per
unit length. In a real cable, different
frequencies are damped at 7.2 Wires and
Transmission Lines 91 different rates, changing
the pulse shape as it travels, and if there are
nonlinearities then different fre

Similarly, the current across the load is IL(t) =
I+(t) + I(t) . (7.64) The current across the
termination must equal the current in the
transmission line immediately before the
termination: VL ZL = V+ Z0 V Z0 . (7.65)
Eliminating variables between this a

iEy (7.82) which can be rearranged as Ex =
1 k 2 c Ez x + i Hz y Hx = 1 k 2 c i
Ez x Hz y Ey = 1 k 2 c Ez x + i
Hz y Hy = 1 k 2 c i Ez x + Hz y .
(7.83) If the axial components Ez, Hz are found
from equations (7.80), they completely
determine the transve

arriving at the T. 7 Circuits, Transmission Lines,
and Waveguides Electric and magnetic fields
contain energy, which can propagate. These
are the ingredients needed for
communications; in this chapter we will look
at how electromagnetic energy can be guid

relations among the voltage, electric field,
current, and charge motion. 7.1.2 Kirchhoffs
Laws There are two Kirchhoff Laws that can be
used to analyze the current flow in a circuit:
The sum of currents into and out of a circuit
node must be zero. If mul

relationships will hold as long as the
frequencies are low enough for the size of the
circuit to be much smaller than the
electromagnetic wavelength. Above this there
is a tricky regime in which the entire circuit
acts like a distributed antenna, and then

rewritten suggestively as r = 1 (x 2 + y 2 ) (1
x) 2 + y 2 r(1 x) 2 + x 2 1 + r + y 2 = 1 1 + r x 2
2x r 1 + r + r 1 + r + y 2 = 1 1 + r x 2 2x r 1 + r
+ r1+r2+y2=11+r+ r1+r2r1+r x
r 1 + r 2 + y 2 = 1 (1 + r) 2 . (7.75) In the
complex (x, y) plane, the

vanishes, the only way the transverse
components can be non-zero is for the
denominator k 2 c = 2+k 2 to also vanish.
This means that = ik = i = i/c,
therefore TEM waves travel at the speed of
light in the medium. kc = 0 also reduces
Helmholtz equations t

rectangular slabs that confine modes in both
directions and for circular dielectric
waveguides, although the imposition of these
boundary conditions becomes a more difficult
calculation [Yariv, 1991]. The result for the
circular geometry is that there are

the current flowing through the inductance V
= Ldz I t . (7.52) Current flows from high to
low potential, so an increasing potential drop
across the inductor has the opposite sign from
the decreasing current. Equating these
expressions, V z = LI t . (7.53

load impedance ZL. The load might be a
resistor, or it could be another transmission
line. For a resistor the impedance is associated
with energy dissipated by ohmic heating, and
for a transmission line the impedance is
associated with energy that is tran

t2 1 v 2 2V t2 , (7.56) where v 1 LC .
(7.57) This is a wave equation for the voltage
in the transmission line. It is solved by an
arbitrary distribution traveling with a velocity
v V (z, t) = f(z vt) + g(z + vt) (7.58) = V+ + V .
If we follow a fixed poi

(7.46) 7.2.3 Wave Solutions Now consider the
little differential length of the transmission
line dz shown in Figure 7.5, with parallel
capacitance C dz and series inductance L dz. If
there is an increase in the current flowing
across it I = I z dz (7.47)

varying electric field associated with the
capacitor plates storing or releasing charge
rather than from real charge passing through
the capacitor. If the applied voltage is V = e
it, then the current is I = C dV dt = iCeit .
(7.18) The impedance (complex

twisted pair of conductors surrounded by a
grounded shield. Why the twist? Why the
shield? (7.2) Salt water has a conductivity 4
S/m. What is the skin depth at 104 Hz? (7.3)
Integrate Poyntings vector P~ = E~ H~ to find
the power flowing across a crosssec

positive charge. A charge q such as an electron
in a wire feels a force F~ = qE~ , and so
according to these definitions electrons flow
from low to high potentials (Figure 7.1). The
current ~I, in amperes, at a point in a wire is
equal to the number of co

This defines the sheet resistivity R (R
square). Since L/w is dimensionless, R has
units of resistance without any other length.
R2 V = IR1 +IR2 R1 V = IR2 V = 0 V I R2 V = I 1R1
=I 2R2 R1 V = 0 V I 2 I 1 Figure 7.2. Series and
parallel circuits. Kirchhof

waveguides are usually designed to be
operated with a single mode. 7.3.4 Dielectric
Waveguides and Fiber Optics Fortunately for
telecommunications, waves can be guided by
dielectric rather than conducting waveguides.
The surface resistance that we saw in

the conductors, = Z B~ dA~ = z Z ro ri 0 I
2r dr = z 0I 2 ln ro ri . (7.42) Since the
dielectric is non-magnetic we can take r 1.
Therefore, the inductance per 7.2 Wires and
Transmission Lines 89 length is L = zI = 0 2
ln ro ri H m . (7.43) Similarly, fro

there is no variation in the x direction.
Depending on the relative magnitudes of
and k this can have oscillatory or exponential
solutions. For the solution to be confined, and
reflect the symmetry of the structure, we
require the wave to be exponentiall