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 integration path.
Problem 6.3 showed that the energy stored in
a solenoid is LI2/2. If the current fl
applied electromagnetics. [Hagen, 1996]
Hagen, Jon B. (1996). Radio-Frequency
Electronics: Circuits and Applications. New
York: Cambridge University Press. [Danzer,
1999] Danzer, Paul (ed). (1999). The ARRL
Handbook for Radio Amateurs. 76th edn.
Newington
model used in the last section no longer
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 th
materials this has been reduced below 0.2
dB/km at 1.55 m [Miya et al., 1979;
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
which this is no longer true. There is an
electric field between the inner and outer
conductors, giving rise to a distributed
capacitance between them. Current flowing in
the inner conductor also produces a magnetic
field around it, and hence a distribute
coaxial cable, makes a TEM solution possible.
96 Circuits, Transmission Lines, and
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
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) 90 Circuits,
Transmission Lines, and Waveguides Now take
a time derivative of equation (7
charge. Now taking the curl of the curl of E~ ,
E~ = B~ t 86 Circuits, Transmission
Lines, and Waveguides E~ = t
B~ ( E~ ) | cfw_z 0 2E~ = E~ t
2E~ t2 . (7.28) The first term on the right
hand side is due to real conduction, and the
second term is du
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 crosssectional slice
of a coaxial cable, and relate the answer to
the current and voltage i
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 characteristic impedance? (b)
What is the transmission v
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 there is a reflected
pulse of the same shape but opposite sign. If
the load resistance is 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 transverse components through
equations (7.83). This set of equations admits
three kinds
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 integer number of
azimuthal cycles c 2 n a + b 2 (7.93) [Ramo
et al., 1994]. In the s
the circuit. 7.1.3 Resistance In an isotropic
conductor the current and electric field are
related by J~ = E , ~ (7.3) where is the
materials conductivity. For very large fields
there may be nonlinear deviations from this
linear relationship, and in a com
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 exponentially damped
98 Circuits, Transmission Lines, and
Waveguides x y z h -h e 0 e 1 e 0 Fig
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
intended direction. Z0 ZL d z VL ,IL V+
and n. If we define a characteristic frequency
c associated with each mode by c(m, n) =
kc(m, n) = 1 m w 2 + n h 2 1/2 ,
(7.87) then we can find the propagation
constant 2 = k 2 c k 2 = k 2 c 1 k 2 k 2 c =
k 2 c 1 2 2 c = k 2 c 1 2 2 c .
(7.88) Therefore
current flowing through it times the voltage
drop across it, which by Ohms Law is also
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 th
diameter of 30 mils (a mil is a thousandth of
an inch), what is the inner diameter? (e) For
RG58/U, at what frequency does the
wavelength become comparable to the
diameter? 7.5 Problems 101 (7.6) Consider a
10 Mbit/s ethernet signal traveling in a
RG58/U
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) there must be a
corresponding decrease in the
arising from current flowing through a coil.
The inductance is defined to be the ratio of the
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 e
of the light [Galtarossa et al., 1994]. Weve
also assumed that the medium is linear, but
the intense fields in the small fiber cores can
excite nonlinear effects. Well see more of this
in the Chapter 9, but one of the most
important applications is to the
section as the fundamental mode for
Maxwells equation in a cylindrical geometry. I
C dz L dz V V+DV dz L dz I I+DI -I - - D I I -I
Figure 7.5. Effective circuit model for a
transmission line, and a differential element.
From Stokes Law, the magnetic field
Circular Waveguides For a waveguide with
cylindrical symmetry, the transverse Laplacian
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]:
vt)] (7.60) and integrate over z I = Cv[f(z vt)
g(z + vt)] 1 Z [f(z vt) g(z + vt)] = 1 Z [V+
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 dif
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 frequencies can
travel at different rates causing dispersion: a
sharp pulse will spread o
the electric field is the gradient of a potential
and the value of its line integral is
independent of the path; it can go through
wires or free space as needed and will always
give the same answer. Conversely, if there are
time-varying magnetic fields th
imaginary part defines the surface inductance
Ls . Associated with this current there is
dissipation; in a small volume of crosssectional area A and length along the surface
L the dissipation per volume is I 2 volumeR AL
= 1 ALJ 2A 2 L A = J 2 . (7.38) If
A second relationship comes from the
definitions of a and b a 2 = 2 k 2 0 b 2 = k 2
1 2 a 2 + b 2 = k 2 1 k 2 0 . (7.101) a
and b are restricted to a circle, with a radius
given by the difference of the squares of k 2 =
2 in the media. For a and b to be
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 multiple wires meet at a
point, the sum of all t