Principles of Modern Communication Systems (2017) - Agbo & Sadiku - Solutions Manual PDF

ALL 8 CHAPTERS COVERED




SOLUTIONS MANUAL

, 1

CHAPTER 1

P. E. 1.1
3
10 10
10 mW =10 log10 20 dBW
1
3
10 10
10 mW =10 log10 10 3 10 dBm



P. E. 1.2

10 log10 S / N S/N 102 100
20
C
C B log2 (1 S / N B
) log2 (1 S / N )
20
B kHz = 3.0038 kHz = 3003.8 Hz
log2 (101)




Prob.1.1
See text

Prob. 1.2
Noise is anỵ signal which interferes with and corrupts the desired signal. Noise is an
interference and constitutes a major limitation to communication sỵstem.

Prob. 1.3
(a) 10log10 0.036 = - 14.43 dB
(b) 10log 42 = 10.23 dB
(c) 10log 508 = 27.06 dB
(d) 10 log10 (3 103 ) 54.77 dB

Prob. 1.4

14
(a) 140 dB = G 10
10log10G G 10 0.3
0.501
(b)
− 3 dB = 10log10G
(c) 20 dB = 10log10G G 102 100
(d) 42 dB = 10log10G G 104.2 15,848.93

, 2



Prob. 1.5 P P 0.3
(a) -3dBm = 10log 10 0.5
10
1mW  1mW
P 0.5 mW
P 1.2
(b) -12dBW = 10log P 10 0.063 W
10
 1W
P  P
(c) 65 dBm = 10log 106.5 3162277.60
10
1mW  1mW
P 3162.3 W
P
(d) 35 dBW = 10log P 103.5 3162.3 W
10
 1W

Prob. 1.6 4mW
(a) 10log 10log(0.004)= -23.98 dBW
10
1W
4mW
10log 10log(4)= 6.02 dBm
10
 1mW
(b) 10log(0.36)= -4.437 dBW
10log(360)=25.56 dBm
(c) 10log(2) = 3.01 dBW
10log(2000) = 330.1 dBm
(d) 10log(110) = 20.41 dBW
10log(110 103 ) 50.41 dBm

Prob. 1.7 3
10 9
(a) 0 dBm 1mW 10log10 12
10 log10 90 dBrn
 10 
10
1.5 88.5 dBrn
dBm
30 dBrn
60 dBm
(b)

, 3

P P
P dBm = 10log 10PdBm /10
10 
-3 -3
10
P 10 3 PdBm /10
9 PdBm /10

P dBrn = 10log -12
10 10 log 10

1
log 10
10 2




P dBrn= 90+ P dBm
Hence,
dBrn = 90+ dBm
Prob. 1.8

B = 2 MHz, S/N = 102.5 = 316
C 2 log2 (1 316) 16.62 Mbps
L 50, 000 8
t= s 24.067 ms
C 16.62

Prob. 1.9
(a) S=250 W, N = 20 W, B = 3 MHz 250
C B log (1 S / N ) 3 106 log 1 11.26 Mbps
2 2
20
3 2 250
(b) C = 106 log2 1 4.7 Mbps
3 20


Prob. 1.10
25k
25 kbps = Blog2 (1 B 2.78 kHz
500) log2 501

Prob. 1.11
S S
30dB 103 1000
N N
dB
C
C B
log S B 36, 000 3.6118 kHz
1
2
N log 1001 log 1001
2 10

log10 2