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Fee 421: Telecommunications And Electroacoustics A Question Paper

Fee 421: Telecommunications And Electroacoustics A

Course:Electrical And Electronics Engineering.

Institution: University Of Nairobi question papers

Exam Year:2008

Page 1 of 3
UNIVERSITY OF NAIROBI
FIRST SEMESTER EXAMINATIONS 2007/2008
FOURTH YEAR EXAMINATIONS FOR THE DEGREE OF BACHELOR OF SCIENCE
IN ELECTRICAL & ELECTRONIC ENGINEERING
FEE 421: TELECOMMUNICATIONS AND ELECTROACOUSTICS A
DATE: 3RD APRIL, 2008 TIME: 2.00 P.M. – 4.00 P.M.
INSTRUCTIONS
1. Attempt any THREE questions
2. All questions carry equal marks
QUESTION ONE: (20 MARKS)
(a) (i) Define the term 'signal' as used in communication.
(ii) State, with reasons, whether or not the signal x(t) = cos 2t + cos5t belongs to each of
the classes: deterministic, complex, periodic, energy
[5 marks]
(b) (i) Obtain the Complex Exponential Fourier Series of the signal
g t f t f t o o ( ) = 1+ cos 2p + sin 4p
(ii) Plot the Amplitude, Phase and Power Spectra of the signal
[5 marks]
(c) (i) A signal q(t) has a Fourier TransformQ( f ) . Obtain the Fourier Transform, G( f ) of
the signal g(t) defined by g(t) q(t) cos(2 f t) o = p in terms ofQ( f ) , where o f is constant.
(ii) Using suitable sketches illustrate the spectral difference between q(t) and g(t) .
[3 marks]
(d) An exponential pulse V (t) i may be defined in the time domain by

<
³
=
-
0, 0
, 0
( )
t
e t
V t
at
i with a being positive and real valued
(i) Determine the total normalized energy of V (t) i
(ii) Determine the Fourier Transform of the exponential pulse.
[3 marks]
(e) The pulse in (d) is applied to a network of impulse response, h(t) (1 exp[ ])u(t) CR
= - - t .
Determine the energy spectral density function of the output, V (t) o .
[4 marks]
Page 2 of 3
QUESTION TWO: (20 MARKS)
(a) (i) Explain how ideal sampling differs from practical sampling.
(ii) Show that any function of time f (t)whose highest frequency is W Hz can be
determined completely by sampled amplitudes spaced at time intervals of at most
2W
1
seconds apart.
[6 marks]
(b) (i) For the periodic signal

= + = ± ±
=
elsewhere
n r r
x n
0,
1, 4 1; 0, 1, 2,...
[ ] 1 determine the DFT
and sketch both the time and frequency domain representations of the signal.
(ii) For the frequency domain signal

= + = + = ± ±
=
elsewhere
k r and k r r
X k
0
1, 16 2 16 14; 0, 1, 2,...
[ ] 2 show that the time domain
representation is given by

=
4
cos
8
1
[ ] 2
n
x n
p
, n=0, 1, 2 …, 15 for the fundamental
interval. Sketch both the frequency and time domain representations of the signal
[8 marks]
(c) (i) Show the FFT butterfly for one decimation-in-time used for determining the DFT of a
periodic signal for which N = 4.
(ii) For the signal

= = ± ±
=
elsewhere
n r r
x n
0,
1, 4 ; 0, 1, 2,...
[ ] 3 , use the FFT algorithm in (c)(i) to
determine the DFT. Sketch both the time and frequency domain representations of
the signal
[6 marks]
QUESTION THREE: (20 MARKS)
(a) Compare the performance of the four linear-modulation schemes; AM, DSBSC, VSB, SSB
in terms of bandwidth, power and transceiver complexity.
[6 marks]
(b) Consider the message signal m(t) = 20cos(2p t) volts and the carrier wave
c(t) = 50cos(100p t) volts.
(i) Determine and sketch the resulting AM wave for 75% modulation, indicating all salient
features.
(ii) Determine the power developed across a load of 100 ohms due to this AM wave.
[6 marks]
(c) The output stage of a certain AM radio transmitter delivers 100 watts of radio frequency
power into the aerial when the depth of modulation is 40%. Calculate the modulation
index necessary to increase the power to 115 watts.
[3 marks]
(d) For coherent detection of an AM signal, show that the output signal-to-noise ratio is given
by the expression,
N W
k A m t
SNR
o
a C
o 2
( ) 2 2 2
= , with the symbols having their usual meanings
[5 marks]
Page 3 of 3
QUESTION FOUR: (20 MARKS)
(a) (i) Give the expression for a general form of an angle modulated wave and indicate how
the expression differs for a phase modulated wave and for a frequency modulated wave.
(ii) Show that the complex envelope of an FM wave for sinusoidal modulation is given by
~s (t) A exp[ j sin(2 f t)] C m = b p where AC is the amplitude of the carrier, b is the
modulation index and fm is the frequency of the modulating signal.
[8 marks]
(b) A carrier of frequency 100MHz is frequency modulated by a sinusoidal wave of amplitude
20V and frequency 100kHz. The frequency sensitivity of the modulator is 10kHz/V.
(i) Determine the approximate bandwidth of the FM wave, using Carson’s rule.
(ii) Determine the bandwidth achieved by transmitting only those side-frequencies with
amplitudes that exceed 1% of the un-modulated carrier amplitude.
(iii)Repeat (i) and (ii) above assuming the amplitude of the modulating wave is doubled.
[8 marks]
(c) A carrier is phase modulated using a sinusoidal signal of frequency m f and amplitude m A
(i) Determine any two (2) values of the modulation indexb , for which the carrier
component of the PM wave is reduced to zero using the table of Bassel function values.
(ii) In an experiment conducted with m f = 1kHz and increasing m A , starting from 0V, it is
found that the carrier component of the PM wave is reduced to zero for the first time
when Am=2V. Determine the phase sensitivity of the modulator.
[6 marks]
TABLE OF BESSEL FUNCTIONS
Jn(b)
b 0 0.2 0.5 1 2 3 4 5 6 8 10 12 15
n
0 1.00 0.990 0.9385 0.7652 0.2239 – 0.2601 – 0.3971 – 0.178 0.1506 0.1717 – 0.2459 0.0477 – 0.01
1 - 0.100 0.2423 0.4401 0.5767 0.3391 – 0.0660 – 0.328 – 0.2767 0.2346 0.0435 – 0.2234 0.21
2 0.005 0.0306 0.1149 0.3528 0.4861 0.3641 0.047 – 0.2429 – 0.1130 0.2546 – 0.0849 0.04
3 - 0.0026 0.0196 0.1289 0.3091 0.4302 0.365 0.1148 – 0.2911 0.0584 0.1951 – 0.19
4 0.0002 0.0025 0.0340 0.1320 0.2811 0.391 0.3576 – 0.1054 – 0.2196 0.1825 – 0.12
5 - 0.0002 0.0070 0.0430 0.1321 0.261 0.3621 0.1858 – 0.2341 – 0.0735 0.13
6 - 0.0012 0.0114 0.0491 0.131 0.2458 0.3376 – 0.0145 – 0.2437 0.21
7 0.0002 0.0025 0.0152 0.053 0.1296 0.3206 0.2167 – 0.1703 0.03
8 - 0.0005 0.0040 0.018 0.0565 0.2235 0.3179 0.0451 – 0.17
9 0.0001 0.0009 - 0.0212 0.1263 0.2919 0.2304 – 0.22
10 - 0.0002 0.0070 0.0608 0.2075 0.3005 – 0.09
11 - 0.0020 0.0256 0.1231 0.2704 0.10
12 0.0005 0.0096 0.0634 0.1953 0.24
13 0.0001 0.0033 0.0290 0.1201 0.28
14 - 0.0010 0.0120 0.0650 0.25
15 - - 0.03 0.18
16 0.01 0.12
NB: J-n(b) = (-1)nJn(b); ½ [J0(b)]2 + [J1(b)]2 + [J2(b)]2 + ….. = ½

Fee 421: Telecommunications And Electroacoustics A Question Paper

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