Gerd Keiser,
Optical Fiber Communications
, McGraw-Hill, 4
th
ed., 2011
Problem Solutions for Chapter 2
2.1
E
100cos
2
10
8
t
30
e
x
20 cos
2
10
8
t
50
e
y
40cos
2
10
8
t
210
e
z
2.2
The general form is:
y = (amplitude) cos(
t - kz) = A cos [2
(
t - z/
)]. Therefore
(a) amplitude = 8
m
(b) wavelength: 1/
= 0.8
m
-1
so that
= 1.25
m
(c)
= 2
(2) = 4
(d) At time t = 0 and position z = 4
m we have
y = 8 cos [2
(-0.8
m
-1
)(4
m)]
= 8 cos [2
(-3.2)] = 2.472
2.3
x
1
= a
1
cos (
t -
1
)
and
x
2
= a
2
cos (
t -
2
)
Adding x
1
and x
2
yields
x
1
+ x
2
= a
1
[cos
t
cos
1
+ sin
t
sin
1
]
+ a
2
[cos
t
cos
2
+ sin
t
sin
2
]
= [a
1
cos
1
+ a
2
cos
2
] cos
t + [a
1
sin
1
+ a
2
sin
2
] sin
t
Since the a's and the
's are constants, we can set
a
1
cos
1
+ a
2
cos
2
= A cos
(1)
a
1
sin
1
+ a
2
sin
2
= A sin
(2)
provided that constant values of A and
exist which satisfy these equations. To
verify this, first we square both sides and add:
A
2
(sin
2
+ cos
2
) =
a
1
2
sin
2
1
cos
2
1
1

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