-29-
7.2
The total number of tones, or individual channels is:
W
s
/
f
d
= (400 MHz)/(100 Hz) = 4
×
10
6
.
The minimum number of PN bits =
log
2
(4
×
10
6
)
= 22
where
x
indicates the smallest integer value not less than x. Source: [SKLA01]
7.3
W
s
= 1000
f
d
;
W
d
= 4
f
d
; Using Equation 7.3 ,
G
p
=
W
s
/
W
d
= 250 = 24 dB
7.4
a.
Period of the PN sequence is 2
4
– 1 = 15
b.
MFSK
c.
L
= 2
d.
M
= 2
L
= 4
e.
k
= 3
f.
slow FHSS
g.
2
k
= 8
h.
Time
0
1
2
3
4
5
6
7
8
9
10
11
Input data
0
1
1
1
1
1
1
0
0
0
1
0
Frequency
f
1
f
3
f
3
f
2
f
0
f
2
Time
12
13
14
15
16
17
18
19
Input data
0
1
1
1
1
0
1
0
Frequency
f
1
f
3
f
2
f
2
7.5
a.
Period of the PN sequence is 2
4
– 1 = 15
b.
MFSK
c.
L
= 2
d.
M
= 2
L
= 4
e.
k
= 3
f.
fast FHSS
g.
2
k
= 8
h.
Same as for Problem 7.4
7.6
a.
This is from the example 6.1.
f
1
=
75 kHz
000
f
2
=125 kHz
001
f
3
= 175 kHz
010
This is the end of the preview.
Sign up
to
access the rest of the document.
- Fall '11
- Dr.TimothyCerner
- Frequency, Hertz, 2k, kHz, 2L, 7.3 Ws
-
Click to edit the document details