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].
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
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 t
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
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 con
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
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 wave
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
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 d
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 equat
Reikd . (7.71) Normalizing this by the
characteristic impedance of the transmission
line, Z(d) Z0 = e ikd + Reikd eikd Reikd 7.2
Wires and Transmission Lines 93 = 1 + Rei2kd 1
Rei2kd r + ic 1 + (x +
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)
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,
(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 W
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 clo
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
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
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 +
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)][
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
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
(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)
Equatin
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 dominan
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
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
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. D
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
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
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
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