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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 xr1
+ r 2 + y 2 = 1 (1 + r) 2 . (7.75) In the complex
(x, y) plane, the reflection coefficient lies on a
circle of radius 1/(1 + r) with a center at (r/(1 +
r), 0) set by the r
Hagen, Jon B. (1996). RadioFrequency
Electronics: Circuits and Applications. New
York: Cambridge University Press. [Danzer,
1999] Danzer, Paul (ed). (1999). The ARRL
Handbook for Radio Amateurs. 76th edn.
Newington, CT: American Radio Relay League.
Pract
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 difference between the
solutions traveling in the ri
intersection of these two circles relates the
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 = 0.5 c = 0
conductor causes a current to flow, and this
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 cavities t
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]: Ez(r, ) =
[AJn(kcr) + BNn(kcr)][C cos(n)
termination must equal the current in the
transmission line immediately before the
termination: VL ZL = V+ Z0 V Z0 . (7.65)
Eliminating variables between this and
equation (7.63) gives the ratio of the incoming
and reflected voltages, called the reflectio
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/second, approaching the
limit of 1 bit/second per hertz of optical
bandwidth [Ono & Yano, 1998; Cowper, 1
This is why wires carrying high frequency
signals are stranded rather than solid. 7.2.2
Transmission Lines While electromagnetic
fields cannot penetrate far into good
conductors, they can be guided long distances
by them. Distributed objects can have ener
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) (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 right, an
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 velocity? (c) If a
computer has a clock speed
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 their currents must be
equal to zero. This is
line. For a resistor the impedance is associated
with energy dissipated by ohmic heating, and
for a transmission line the impedance is
associated with energy that is transported
away, but in both cases the voltage drop
across the element is equal to the c
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
Helmholtz equations to Laplaces equation,
giving the static field solutions we used when
studying
dielectric is nonmagnetic 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) Similarly, from Gauss Law
the electric field between the conductors is E =
Q 2r , (7.44) where Q is
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Physics 110C Final Exam Practice
Closed Book and Notes, Formula sheet allowed, No electronics
1. (15) Suppose one attempts to send electromagnetic waves down a long copper pipe,
with a rectangular 1 cm x 2 cm cross section, open at both ends. What is the
Analy&cal Mechanics: Lecture 18
Small Oscilla&ons
1
Monday next week, 1pm this room.
6 ques&ons, 5 on material since the midterm
Op&onal Review Session Saturday 4pm5:30pm in Physics 185
(285 is an alternate loca
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 (7.50) 2 I tz
= C 2V t2 (7.54) and a z derivative
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 that has a charge
of +Q on it and one that has
communications. In the next chapter well see
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 optical s
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 cable. (a) What is the physical length
of a bi
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 TM
wave is 2 TEz = 2Ez x2 + 2Ez y2 = k 2
cEz .
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. Series and
parallel circuits. Kirchhoffs Laws can be used
to simplify the circuits in Figure 7.2. For the
series circuit