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 integer number of
azimuthal cycles c 2 n a + b 2 (7.93) [Ramo
et al., 1994]. In the section on transmission
lines we studied the fu
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 point in the distribution
f(0), z vt = 0 z = vt. The V+ solution travels
to the right, and V to the left. For a sinusoidal
wave V = e
nIr2 = n2 lr2 . (7.24) 7.2 Wires and
Transmission Lines 85 I Figure 7.3. A solenoid;
the dotted line closes the integration path.
Problem 6.3 showed that the energy stored in
a solenoid is LI2/2. If the current flowing
through an inductor is I = e it then
problems can be addressed by carrying light in
a glass fiber instead of RF in a metal box. To
see how a wave can be guided by dielectrics,
consider the slab geometry shown in Figure
7.8. Well look for a mode confined in the y
direction with a periodic z d
Takahashi, 1993]. This corresponds to a loss of
103 over 150 km, making long links possible
without active repeaters. So far weve been
considering step-index fibers that have a
constant dielectric constant in the core. By
varying the core doping as a func
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) there must be a
corresponding decrease in the charge stored in
the capacitance I = Cdz V t . (7.48)
Therefore I z dz = Cdz V t (7.49) o
telecommunications, waves can be guided by
dielectric rather than conducting waveguides.
The surface resistance that we saw in Section
7.2.1 represents a significant drag on a wave
traveling in a waveguide, limiting the distance
over which it is useful. A
input impedance to the reflection coefficient,
conveniently found graphically on a Smith
chart (Figure 7.7). 94 Circuits, Transmission
Lines, and Waveguides r = 0 r = 0.5 r = 1 r = 2 c
= 1 c = 2 c = 0.5 c = -0.5 x y |R| = 1 c = 0 90 135 - c = -1 c = -2 90
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 coulombs of charge
passing that point per second. It is defined to
be in the same direction as the electric field
and hence opposite to th
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 transverse components through
equations (7.83). This set of equations admits
three kinds of solutions: Transverse Electr
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, from Gauss Law
the electric field between the conductors is E =
Q 2r , (7.44) where Q is the charge per unit
length and the field van
Mollenauer et al., 1996]. Using all of these
tricks, fiber links have been demonstrated at
speeds above 1 Tbit/second, approaching the
limit of 1 bit/second per hertz of optical
bandwidth [Ono & Yano, 1998; Cowper, 1998].
7.4 S EL E C T E D R E FE R E N C
(Problem 7.4), and a strip above a ground
plane (called a stripline). Because a
transmission line is operated in a closed circuit
there is no net charge transfer 88 Circuits,
Transmission Lines, and Waveguides I -I z ri ro
e z Figure 7.4. A coaxial cable
relative permittivity of 2.26, an inner radius of
0.406 mm, and an outer radius of 1.48 mm. (a)
What is the characteristic impedance? (b)
What is the transmission velocity? (c) If a
computer has a clock speed of 1 ns, how long
can a length of RG58/U be an
that this can be understood as many different
path lengths reflecting at the corecladding
interface. Because theyre easier to make and
connect to, these are still are used for short
links and for many optical sensors that
measure light coupling into or ou
equal to the skin depth. For example, pure
copper at room temperature has a
conductivity of 5.8 107 S/m and so 7 cm
at 1 Hz, 2 mm at 1 kHz, 70 m at 1 MHz, and 2
m at 1 GHz. Since the skin depth is so small at
even fairly low frequencies, very little
thick
Waveguides 7.3.2 Rectangular Waveguides
Now consider a rectangular waveguide with
width w in the x direction and height h in the y
direction. The transverse equation for a TM
wave is 2 TEz = 2Ez x2 + 2Ez y2 = k 2
cEz . (7.84) Solving this subject to the
b
for a TM mode is 2 TEz = 1 r r r Ez r + 1
r 2 2Ez 2 = k 2 cEz , (7.90) which is solved
by Bessel functions of the first (Jn) and second
(Nn) kind [Gershenfeld, 1999a]: Ez(r, ) =
[AJn(kcr) + BNn(kcr)][C cos(n) + D sin(n)] .
(7.91) The modes TMnl are indexe
magnetic flux = Z B~ dA~ (7.20) linking a
circuit to the current that produces it: L = I .
(7.21) The MKS unit of inductance is the henry,
H. In Figure 7.3 the electric field vanishes
along the dotted line for an ideal solenoid,
therefore the line intergr
surface, and weve ignored the unphysical
possible solution e kz. The total current per
unit width that is produced by this field is
found by integrating the current density over
the depth I = Z 0 J dz = Z 0 E dz = Z 0
E0e kz dz = E0 k . (7.36) 7.2 Wires a
(7.97) The transverse components are found
from equations (7.83), which for Ex is Ex = i
k 2 c Hz y = i k 2 2 Hz y =
i a Aea(yh) (y > h) i b B sin(by)
(|y| < h) i a Aea(yh) (y < h) . (7.98)
Equating these again at the boundaries, A a =
B b sin(bh) . (7
boundary conditions becomes a more difficult
calculation [Yariv, 1991]. The result for the
circular geometry is that there are two modes
with axial H and E components, one called the
HE with H dominant, and an EH mode with E
dominant. Dielectric waveguide
solutions. For the solution to be confined, and
reflect the symmetry of the structure, we
require the wave to be exponentially damped
98 Circuits, Transmission Lines, and
Waveguides x y z h -h e 0 e 1 e 0 Figure 7.8. A
dielectric slab waveguide. outside o
V] = I+ + I , (7.61) where Z = 1 Cv = r L C ().
(7.62) The current is proportional to the
voltage, with the sign difference in the two
terms coming from the difference between the
solutions traveling in the right and left
directions. The constant of propo
applies. A complete solution of Maxwells
equations is then required. Some of these new
modes will prove to be desirable, and some
will not. Waveguides, not surprisingly, guide
electromagnetic waves. Depending on the
geometry they may or may not be able to
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. Kirchhoffs Laws can be used
to simplify the circuits in Figure 7.2. For the
series circuit on the left, V = IR1 + IR2 (7.7) or I
= V R1 +
equal to the square of the current times the
resistance. This appears as heat in the resistor.
7.1.5 Capacitance There will be an electric
field between an electrode that has a charge
of +Q on it and one that has a charge of Q,
and hence a potential diffe
moving relative to this force and so work is
being done. The work associated with the slab
moving from one end of the resistor to the
other is equal to the integral of the force times
the displacement: dW = Z + F~ d~x = Q Z +
E~ d~x = QV = qdxAV (7.14) f
current leads to resistive dissipation, therefore
in making this approximation we are leaving
out the mechanism that damps fields around
conductors. This is very important in resonant
electromagnetic cavities that are designed to
have a high Q (low dampin