Chapter 6: Solutions to Problems
Problem 6.1. One dimensional geometry.
We use the solution to Laplace's equation in one dimension given in Eq. (6.8). We apply it directly for the two divisions
shown
Chapter 13: Solutions to Problems
Problem 13.1. Reflection and transmission at air - lossy dielectric interface.
The wave coming from the left propagates without attenuation until it hits the wall. No
Chapter 9: Solutions to Problems
Problem 9.1. Magnetic dipole.
a. m = IS (magnitude) or m = n IS = n a2 I A.m2 , (Eq. (9.13) is the magnetic dipole moment for the circular loop
b. The square loop will
Chapter 18: Solutions to Problems
Problem 18.1. Near and far fields of antennas.
To see if a particular area is in the far zone we calculate the quantity R. If this quantity, which is the electrical l
Chapter 14: Solutions to Problems
Problem 14.1. Application: Co-axial transmission line.
Line parameters are capacitance per unit length, inductance per unit length, intrinsic impedance and phase cons
Chapter 5: Solutions to Problems
Problem 5.1. Solution to Laplace's equation.
In general for the given function we can write:
2
2 f(x)g(y) = g 2 f + f 2 g = 0
2 V = f(x)g(y)
+
x 2
y 2
x 2
y 2
2
2
2
2
Chapter 15: Solutions to Problems
Problem 15.1. Line properties using the Smith chart.
a.
1. Normalize the load impedance: zl = 2.6 + j1.8. Enter this on the Smith chart at the intersection of the res
Chapter 3: Solutions to Problems
Problem 3.1. Electric and gravitation forces.
The gravitational force must equal the electrostatic force for the condition to be satisfied.
a. The gravitational force
Chapter 17: Solutions to Problems
Problem 17.1. Application: TM modes in parallel plate waveguides.
First we calculate the cutoff frequency of the lowest mode. Then the frequency of operation is 20% a
Chapter 12: Solutions to Problems
Problem 12.1. The wave equation.
Solution is by direct manipulation of equations (11.24 through 11.27):
E = B (1)
t
.D= (3)
H = J + D
t
.B=0 (4)
(2)
a. We start by t
Chapter 2: Solutions to Problems
Problem 2.1.
Work is calculated from the integral of F.dl. In general, the work is path dependent but not always, depending on the
vector F.
F = 1 ,
dl = rdr + rd + zd
Chapter 4: Solutions to Problems
Problem 4.1. Charge density required to produce a field.
From the given potential we calculate the electric field intensity using the gradient of potential. Then, the
Chapter 8: Solutions to Problems
Problem 8.1. Magnetic flux density due to filamentary currents.
Consider Figure A. The magnetic flux density at a height h above the lower segment, carrying a current
Chapter 10: Solutions to Problems
Problem 10.1. Motional emf.
R
The induced electric field intensity due to motion of
train (1) points up and therefore the induced emf is
positive on the upper rail, n
Chapter 11: Solutions to Problems
Problem 11.1. Displacement current density.
The displacement current density is evaluated directly from Maxwell's second equation (Ampere's law):
Jd = D = E
t
t
A
m2
Chapter 16: Solutions to Problems
Problem 16.1. Application: Reflectometry: narrow pulses.
The distance between the pulses equals twice the distance to a discontinuity on the line. The type of discont
Chapter 7: Solutions to Problems
Problem 7.1. Current and charge in a battery.
a. dQ/dt = 100 A or: dQ/dt = 100 Coulomb/sec. Total charge supplied in one hour is 1003600 = 360,000 Coulombs.
b. Since Q
David Charles
February 1, 2017
Communication Theory
Reflection 1
When most people think of what is a theory, the first thing that comes to their mind is a
picture of ancient philosopher back in the da
Setting a career in life is the personal goal. One wants to be financially stable, get married
and start a family. This is one of my goals in life, to be a dental hygienist and maybe even further
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