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
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). Th
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 sur
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 f
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 prod
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 heig
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, a
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 si
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 ca
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 c
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 sa
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
equ
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 cy
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
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, Transm
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 pow
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
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
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) Cons
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
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
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
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 di
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 th
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 t
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 differen
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. C
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 volu
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 t
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 ze