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(ELEC214)[2010](s)midterm~116^_10309.pdf
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Question 1 (10)
No unique solution
1.
Radio broadcast (audio signal, air, radio)

2.
TV broadcast (video signal, air, TV)

3.
Mobile telecommunication (human voice, air, mobile phone)

4.
Computer network (computer, cable, computer)

5.
Email (text, cable, computer)

Question 2 (10)
a) No. This is because the spectrum is continuous. 1 11
b) X ( f ) = (24) ( f / 6) + (16) ( f / 2) * ( f / 4)
6 24
Y(f)
x(t) = (24) sinc(6t) + (16) sinc(2t) sinc(4t)
16

c) To= 0.5 fo = 2 12
1
Y ( f ) = X ( f ) ( f . nfo )
To n =.

d) dc value = 16 C 2 0 2 f c) AM

Question 3 (20)
a) AM C 0.06 MHz LSB C 0.15 MHz DSB C 0.3 MHz VSB C 0.7 MHz FM C 90 MHz
b) AM C 0.02 MHz LSB C 0.05 MHz DSB C 0.2 MHz VSB C 0.4 MHz FM C 0.6 MHz

DSB
LSB
f

f
f

C 10k 0 10k C 50k 0 50k C 100k 0 50k 100k
VSB
FM

f

f
C 300k 0 300k C 100k 0 100k
d) AM C 10 kHz LSB C 50 kHz DSB C 100 kHz VSB C 300 kHz FM C 100 kHz
e)
AM Disadvantage: twice the message bandwidth, 100% bandwidth expansion, 50% power efficiency at most Advantage: simple detector for demodulation
DSB Disadvantage: twice the message bandwidth, 100% bandwidth expansion Disadvantage: demodulation is complicated because synchronization is required Advantage: 100% power efficiency
SSB Advantage: same bandwidth as original message, 0% bandwidth expansion, 100% power efficiency Disadvantage: demodulation is complicated because synchronization is required Disadvantage: modulation is complicated because a sharp cut-off filter is required
VSB Advantage/Disadvantage: about 1.25 times the message bandwidth, about 25% bandwidth expansion Advantage: modulation not so complicated because a sharp cut-off filter is not required Advantage: 100% power efficiency Disadvantage: demodulation is complicated because synchronization is required
FM Advantage: better quality Disadvantage: wider bandwidth is needed
Question 4 (20)
a) FM1 : 100 MHz FM2 : 105 MHz
b) FM1 : [109.7 110 110.3] FM2 : [114.7 115 115.3]
c) Bandpass filter

Passband: 89.95 MHz to 95.1 MHz
Stopband: 109.7 MHz to 115.3 MHz
d) Transmission bandwidth = 2 (D + 1) W
FM1 : W = (90.05 C 89.95) M / 10 = 10 kHz FM2 : W = (95.1 C 94.9) M / 10 = 20 kHz
e) Differentiator Envelope detector

11
2RC RC
11 22
Question 5 (20)
a) Peak frequency deviation =fd Am = 100k (10m) = 1 kHz
fd Am 1k
Modulation index = = == 0.05
fm 20k b) Bandwidth = 2 (20k) = 40 kHz c) Modulation index = 0.05 (100) = 5

98.8 99.9 99.92 99.94 99.96 99.98 100 100.02 100.04 100.06 100.08 100.1 100.12 f (MHz)

98.8 99.9 99.92 99.94 99.96 99.98 100 100.02 100.04 100.06 100.08 100.1 100.12 f (MHz)
e) Transmission bandwidth = 6 (20k) = 120 kHz
2 222
f) Power = 2 (8.9m) + 4 [(16.4m) + (2.35m) + (18.25m)] = 2.589 mW g) Immune to noise but a wider bandwidth is needed.