material per surface area Z 0 |J| 2 2 dz = Z
0 2E2 0 2 e 2z/ dz = E2 0 4 . (7.40)
The amplitude of the field and current are
falling off exponentially with a length scale
equal to the skin depth. For example, pure
copper at room temperature has a
conduct

New York: Wiley. 100 Circuits, Transmission
Lines, and Waveguides A nice introduction to
applied electromagnetics. [Hagen, 1996]
Hagen, Jon B. (1996). Radio-Frequency
Electronics: Circuits and Applications. New
York: Cambridge University Press. [Danzer,
1

with multiple branches because of the 7.4
Selected References 99 periodicity of tan(bh).
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 r

dt = q dx dt AV = JAV = IV = I 2R . (7.15) The
power dissipated in a resistor is equal to the
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 h

moving through the resistor in Figure 7.1. If
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

E~ = = 0 . (7.27) The linear response
coefficients require that there be no free
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 firs

usefulness of the cable. This is why
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.
T

0 ) 1/2 (|y|h) Aea(|y|h) |y| > h B cos (k 2
1 2 ) 1/2y B cos(by) |y| < h . (7.96) Now
the boundary conditions require continuity of
the field at the interfaces, hence A = B cos(bh) .
(7.97) The transverse components are found
from equations (7.83), which

about their long axis; a rectangular pipe is a
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
wavegui

which we have already found to be the
boundary condition for a perfect conductor.
The part of the field that does leak into the
conductor causes a current to flow, and this
current leads to resistive dissipation, therefore
in making this approximation we

outside the solenoid if it is approximated to be
a section of an infinite solenoud, therefore the
integral of the magnetic field across the
surface bounded by the path is equal to the
flux linking the solenoid times the number of
turns of the coil. This l

restricts the reflection coefficient to a circle of
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 grap

the sum of these: VL(t) = V+(t) + V(t) . (7.63)
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

fiber. The first ones were multi-mode fibers
that had core diameters many times the
optical wavelength, resulting in very
dispersive communications. In the next
chapter well see that this can be understood
as many different path lengths reflecting at
the

equivalently when the wavelength is greater
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.

stable shape that can propagate for long
distances without changing. These can be sent
across ocean-scale distances at Gbits/second
without errors [Nakazawa et al., 1993;
Mollenauer et al., 1996]. Using all of these
tricks, fiber links have been demonstra

increases along a path from negative to
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 poin

assuming that the flux linking all the turns is
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 Figur

impedance of RG58/U (to minimize
reflections). If such a cable has an outer
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 Pr

minimum value for this sum defines the
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
match

Capacitances range from picofarads in circuit
components up to many farads in
supercapacitors based on electrochemical
effects [Conway, 1991]. The current across a
capacitor is given by C dV dt = dQ dt = I . (7.17)
A capacitor is a device that stores ener

because the numerator in equations (7.83)
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.

minimum absorption in optical glasses occurs
at infrared wavelengths; by using very pure
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 possi

field at the surface; the real part of this
defines an effective surface resistance Rs , and
the imaginary part defines the surface
inductance Ls . Associated with this current
there is dissipation; in a small volume of crosssectional area A and length al

per unit length C = Q V = 2 ln(ro/ri) F m .
(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

E~ dA~ = t Z B~ dA~ (7.2) will vanish if
there is no time-varying magnetic flux linking
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 la

assuming that they are sections of conductors
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

current, substitute equation (7.58) into
equation (7.50) I z = C[vf (z vt) + vg (z +
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 propo

is R = L A = L TW = T L W R L W .
(7.6) 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

TEM wave cannot be supported. Adding
another conductor, such as the center lead in a
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

current I is therefore V = Z + E~ d~x = Z +
J~ d~x = Z + I A dx = IL A IR . (7.4)
Remember that the integral goes from low to
high potential, but that current flows from
high to low potentials, so J~ d~x = J dx = I
dx/A. This is just Ohms Law, and it def

transverse to the waveguide axis and that are
axial, taken here to be in the ~z direction:
2E~ = 2 TE~ + 2E~ z2 = 2 TE~ + 2E . ~
(7.79) This turns equations (7.78) into
Helmholtz equations for the transverse
dependence of the field 2 TE~ = ( 2 + k 2 )
E~

must be equal since potential is independent
of path, V = I1R1 = I2R2 , (7.10) and the
current in both legs must add up to the total
current I1 + I2 = V Rtotal . (7.11) 7.1 Circuits 83
Therefore V R1 + V R2 = V Rtotal (7.12) or 1
Rtotal = 1 R1 + 1 R2 Rtot

electric field V = Z E~ d ~l . (7.1) As long as
dB/dt ~ = 0 then E~ = 0, which implies that
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 an

for well-behaved functions): 2V z2 = LC 2V
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)