directions and for circular dielectric
waveguides, although the imposition of these
boundary conditions becomes a more difficult
calculation [Yariv, 1991]. The result for the
circular geometry is that
Figure 7.4). Other important transmission line
geometries include parallel wires or strips
(Problem 7.4), and a strip above a ground
plane (called a stripline). Because a
transmission line is operated
x) 2 + y 2 r(1 x) 2 + x 2 1 + r + y 2 = 1 1 + r x 2
2x r 1 + r + r 1 + r + y 2 = 1 1 + r x 2 2x r 1 + r
+ r1+r2+y2=11+r+ r1+r2r1+r x
r 1 + r 2 + y 2 = 1 (1 + r) 2 . (7.75) In the
complex (x, y) plane
2r dr = z 0I 2 ln ro ri . (7.42) Since the
dielectric is non-magnetic we can take r 1.
Therefore, the inductance per 7.2 Wires and
Transmission Lines 89 length is L = zI = 0 2
ln ro ri H m . (7.43) Si
and Waveguides Electric and magnetic fields
contain energy, which can propagate. These
are the ingredients needed for
communications; in this chapter we will look
at how electromagnetic energy can be
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 point in a wire is
equal to the numbe
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 t
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 Figure 7.3. A solenoid;
the dotted line closes t
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 l
Depending on the relative magnitudes of
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
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 associated with the slab
moving from
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 wa
the possible modes as a function of integers m
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 prop
= Ldz I t . (7.52) Current flows from high to
low potential, so an increasing potential drop
across the inductor has the opposite sign from
the decreasing current. Equating these
expressions, V z = LI
waveguide is driven. 7.3 Waveguides 97 7.3.3
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.
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 st
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 +
resistor, or it could be another transmission
line. For a resistor the impedance is associated
with energy dissipated by ohmic heating, and
for a transmission line the impedance is
associated with ene
becomes impractical at higher frequencies to
work with macroscopic objects with
microscopic dimensions. Both of these
problems can be addressed by carrying light in
a glass fiber instead of RF in a me
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 .
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. kc = 0 also reduces
Helm
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 corecladding interface. Becau
grounded shield. Why the twist? Why the
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 acros
capacitor plates storing or releasing charge
rather than from real charge passing through
the capacitor. If the applied voltage is V = e
it, then the current is I = C dV dt = iCeit .
(7.18) The impeda
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 demonstrated at
speeds above 1 Tbit/secon
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 for Ex is Ex = i
k 2 c Hz y =
1 k 2 c Ez x + i Hz y Hx = 1 k 2 c 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
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.
preserving fiber that retains the polarization
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
unit length. In a real cable, different
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 diff