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(ELEC214)[2011](s)midterm~1833^_62460.pdf
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a) Mobile phone, TV, Radio etc b) Simplex : TV Half duplex : Walkie-Talkie c) Application, Presentation, Session, Transport, Network, Data Link, Physical d) Circuit : Telephone Packet : Computer
Question 2 (10)
a) No. The spectrum is not discrete in the frequency domain. .( f / 2) ( f / 2) .
b) X ( f ) = 8* 2 ( f + 5) + 2 ( f . 5) .
.. ..+
22 . 1 . 1
c) Y ( f ) =.j sgn( f ) Z ( f ) where Z ( f ) = X ( f ) ( f /6)* [( f +
.. 6 ..2
Y(f)j / 3

C12 .10 C8
0 8 . j / 3


Question 3 (20)
a) AM C 600 kHz DSB C 400 kHz b) AM C 900 kHz DSB C 1900 kHz c) AM C 400 kHz or 1400 kHz DSB C 1400 kHz or 2400 kHz
d)
400 600 f (kHz)
e) AM


400 600 f (kHz) 400 600 f (kHz) f) Coherent detector g)
AM


.100 100 f (kHz) .100 100 f (kHz)
Full duplex : Cell phone (any three of them)
x(t) = 8 sinc 2 (2t) + 4 cos(10t) 10) +( f .10)]
10
f (Hz)
USB C 100 kHz USB C 3500 kHz USB C 3000 kHz or 4000 kHz
USB

500 600 f (kHz)
USB

.100 100 f (kHz)
h) No. The bandwidth of IF filter would be at least 600 kHz. The carrier frequency of AM or DSB would be changed to another value. Otherwise AM and DSB will fall into the IF stage simultaneously.
fm 0.1 (10k)
a) fd == =10 kHz /VAm 0.1
Peak frequency deviation
b) (50) = Peak frequency deviation = 0.1 (50) (10k) = 50 kHz
fm
c) Power = 22 /2 = 2 W d) 200M 50(1M) = 150 MHz or 250 MHz

199.94 199.95 199.96 199.97 199.98 199.99 200 200.01 200.02 200.03 200.04 200.05 200.06 f (MHz) f) Immune to noise but a wider bandwidth is needed.
g) Center frequency = 200 MHz Minimum bandwidth = 100 kHz h) i)
Differentiator Envelope detector

11
2RC RC
11 22
. m(t) .
a) m(t) = m(t)/2 x (t) = 21 + cos(10000t) = 2 [1 + m (t)]cos(10000t) Modulation index = 1
nc .. .n
2 .
1 2 3 4 5 6 t (s)



a) AM. Envelope detector is the cheapest detector which only consists of a diode, a capacitor and a resistor. b) Maximum number of users = 200k / 11k 18 if SSB is used. c) DSB, SSB and VSB d) Nyquist frequency = 2 (10k) = 20 kHz e) Transmission bandwidth = 5W f) Tmax = 1 / 20k = 0.05 ms max = 0.05 / 5 = 0.01 ms
g)

h) Reduce the pulse width