SOLUTIONS
Module:
Module Code
Time allowed
Rubric
Examiners
Telecoms Systems
EBU5302
Paper
A
2hrs 30 min
Filename
Solutions_1112_EBU5302_A
ANSWER ALL FOUR QUESTIONS
Prof Laurie Cuthbert
Dr Michael Chai
Dr Frank Gao
SOLUTIONS EBU5302 (2012) Paper A
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Question 1
a) A quantiser with maximum range ± 8V is built with an 8-bit A-D converter. The converter is
coded so that there is a quantiser level at ± 8V and also at 0V. Assume the converter output is
coded Pxxxxxxx where P=1 for positive values and 0 for negative.
i)
Calculate the quantisation interval q in mV.
ii)
If a signal with power 100 mW is fed into the quantiser calculate the signal/distortion
ratio (in dB) that would be obtained.
[8 marks]
b) Figure 1 shows the input blocks for the quantisation in a telephone Local Exchange.
i)
State the name of Block A and Block B.
ii)
Briefly explain the purpose of Block A and why it is essential.
iii) State the value of the main parameters for Block A and Block B.
Block A
Block B
Figure 1
[8 marks]
c) Figure 2 shows a very simplified diagram of an ADSL connection.
i)
Explain why the data is separated from the telephony signal and is not put through the
A-law converter.
ii)
Calculate (approximately) the maximum data rate that could be obtained if the data
signal was passed through the A-law converter.
Telephony
ADSL
modem
Telephone line
A-law
A/D
DSLAM
Data
router
Figure 2
[6 marks]

dv(t )
where ∆ is the step
 max
Ts
dt
size, Ts the sampling period, calculate the value of delta if Ts = 10µs and v(t) = 2 sin (2π 103
t)
d) If the maximum rate for a delta modulator is given by
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SOLUTIONS EBU5302 (2012) Paper A
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[3 marks]
Write your answers on the next 3 pages
Answers to Question 1
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a)
4
marks
4
marks
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SOLUTIONS EBU5302 (2012) Paper A
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b)
i) Name of Block A: anti-aliasing low-pass filter
Name of Block B: sampler
1
mk
1
mk
ii) Purpose of Block A
4
marks
iii) Value of main parameter for Block A 3.3kHz
Value of main parameter for Block B 8kHz
c)
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1
mk
1
mk
SOLUTIONS EBU5302 (2012) Paper A
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ii
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SOLUTIONS EBU5302 (2012) Paper A
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
dv(t )
where ∆ is the step
 max
Ts
dt
size, Ts the sampling period, calculate the value of delta if Ts = 10µs
d) If the maximum rate for a delta modulator is given by
3
marks
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SOLUTIONS EBU5302 (2012) Paper A
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Question 2
a) A packet of 1024 bits is to be transmitted over a link between London and Beijing that has a
bit error probability p of 10-6. Calculate the probability of the packet being in error for:
i)
No error correction
ii)
1-bit error correction
iii) 2-bit error correction
[6 marks]
b) The packet in (a) is to be protected by single-bit error correction. Calculate the number of
check bits that need to be added to the 1024 data word in order to give this correction
capability.
[4 marks]
c) ARQ error correction is to be used on this link and there is no FEC so the packet size is 1024.
Assume the one-way propagation time is 25ms, the bit rate on the link is 1Mbit/s and the
length of an ACK packet can be neglected.
i)
Calculate the effective information transfer bit rate if stop and wait transmission is used.
ii)
Calculate the minimum window size in number of packets for a sliding window
protocol to be used.
[5 marks]
d) The sliding window is set up so there is no timeout of the window (with the minimum size)
and the only retransmissions are caused by errors in the packets. Assume that retransmissions
can be modelled by a 1st order approximation. Calculate:
i)
The effective bit rate when there is no error protection within the packet.
ii)
The effective bit rate when there is single-bit FEC within the packet.
[8 marks]
e) Recalculate the effective bit rate of (d)(i) if the retransmissions are modelled as a second
order approximation – i.e. some of the retransmissions have to be retransmitted.
[2 marks]
Write your answers on the next 3 pages
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SOLUTIONS EBU5302 (2012) Paper A
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Answers to Question 2
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a)
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b)
c)
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SOLUTIONS EBU5302 (2012) Paper A
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Still give marks if the neglect 2nd term in denominator or not
ii)
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SOLUTIONS EBU5302 (2012) Paper A
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d)
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SOLUTIONS EBU5302 (2012) Paper A
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e)
2
marks
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SOLUTIONS EBU5302 (2012) Paper A
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Question 3
a) Explain why equaliser is needed at the receiver and how a three-tap zero forcing equaliser
works.
[7 marks]
b) Figure 3 shows an eye diagram of a transmission signal.
Figure 3
i.
With reference to Error! Reference source not found., explain how the signal quality can be
identified from the eye diagram.
ii.
Calculate the peak signal to ISI ratio of this signal (in dB).
iii.
Suggest, with justifications, an initial sampling time, T0 for this signal.
[8 marks]
c) A synchronous transmission system using cable with 28 dB attenuation per km and regenerative
repeaters every 10 km is affected by additive Gaussian noise. A binary bipolar line-code is used
with equally spaced detection thresholds at the sampling points.
i) How much voltage gain (in dB) is required by the repeater to produce an error probability of
-7
10 ?
ii) If the repeater gains cannot be increased because of the interference to other lines, explain
-5
-7
and calculate how can the error probability be reduced from its current value of 10 to 10 ?
(Use the graph of Q(z) against z which located at the end of the exam paper).
[10 marks]
Write your answers on the next 3 pages
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SOLUTIONS EBU5302 (2012) Paper A
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Answers to Question 3
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a)
The frequency response of a channel is not known sufficiently accurately to allow for
the design of a receiver with zero ISI for all time [1 mark]. The input of the receiver is
therefore usually filtered in such a way that the channel-induced distortion is corrected
[1 mark]. This process is called equalisation and usually it is designed to reduce ISI to
a minimum [1 mark]
Zero forcing equaliser uses filter tap weights to clean up the distorted pulses [1mark].
The tap weights could be chosen to force the system impulse response to zero at all but
the sampling time [1 mark]
[1 mark for the diagram and 1 mark for the description]
b)
1 mark for explanation the eye openning – the bigger the eye openning, the
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7 marks
SOLUTIONS EBU5302 (2012) Paper A
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better the quality of the signal
1 mark for identification of time variation
1 mark for identification of amount of distortion
amount of ISI = 1.5 + 0.5 = 2V 1 mark working
SNR = 20log (12/2) = 15.56dB 2 mark answer
T0=0.2s 1 mark for correct estimation
The best initial sampling time, T0 at the widest opening of the eye because of it is in
the highest SNR condition [1 mark]
8 marks
c)
i)
Assume an equal probability of “1”s and “0”s. Therefore probability of an error is:
0.5 x Q(V/σ) + 0.5 x Q(V/σ)
2
where σ is the standard deviation (σ = variance) of the noise. [1 mark]
-5
If Q(V/σ) = 10 , from the graph we find that (V/σ) = 4.25 [1 mark]
Therefore σ = V/4.25. This is the standard deviation of the zero mean Gaussian noise
that is causing the errors.
-7
-5
With the same noise level, to produce an error probability of 10 rather than 10 , we
need the regenerative repeaters to receive higher voltage levels for the symbols.
Assume they are raised from ±V to amplitude ±U at the sampling points.
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SOLUTIONS EBU5302 (2012) Paper A
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-7
Then Q(U/σ) = 10 and from the graph, we find that U/σ = 5.2. [1 mark]
This means that U = 5.2σ = (5.2/4.25)V = 1.22 V. [1 mark]
20log10(1.22) = 1.75dB. [2 marks]
ii)
Therefore we need to arrange that the voltages as received at the receiver are raised by
1.75 dB. If we cannot raise the voltages transmitted by the regenerative repeaters
because of cross-talk, we can only reduce the distance between the repeaters so that
less attenuation occurs between one regenerative repeater and the next. [1 mark]
The attenuation must be reduced from 10x28 = 280 dB to 278.25 dB. [1 mark]
Distance between repeaters must be reduced to 278.25/28 = 9.94 km instead of 10 km.
[2 marks]
10 marks
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SOLUTIONS EBU5302 (2012) Paper A
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SOLUTIONS EBU5302 (2012) Paper A
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Question 4
a) A sine wave is to be used for two different signalling schemes: i) PSK; and ii) QPSK. The
duration of a signal element is 105 s. If the received signal is of the following form,
s(t )  0.005 sin( 2 106 t   ) volts
and if the measured noise power at the receiver is 2.5×10-8 watts, determine Eb/N0 in dB for
each case.
[8 marks]
Answer:
Ts = signal element period; Tb = bit period; A = amplitude = 0.005
i) Ts = Tb = 10-5 sec
P
1
Ts
Ts
A2
2
2
 s t  
0
[2 marks]
A2
8
Eb  P  Tb  P  Ts 
 Ts ; N 0  2.5  10  Ts
2
Eb
A2 2  Ts

 500; Eb N 0 dB  10 log 500  27dB
N 0 2.5  108  Ts
[2 marks]


ii)
Ts
T
; Eb  P  s ; N 0  2.5  108  Ts
[2 marks]
2
2
Eb N0   250; Eb N0 dB  10 log 250 24dB
[2 marks]
Tb 
b) The following table illustrates the operation of an HFSS system for one complete period of
the PN sequence.
Time
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
Input data
0
1
1
1
1
1
1
0
0
0
1
0
0
1
1
1
1
0
1
0
f1
Frequency
PN Sequence
f3
001
f27
f26
f8
110
f10
011
f1
f3
001
To determine:
i)
What is the period of the PN sequence?
ii)
The system makes use of a form of FSK. What form of FSK is it?
iii)
What is the number of bits per symbol?
iv)
What is the number of FSK frequencies?
v)
What is the length of a PN sequence per hop?
vi)
Is this a slow or fast FH system?
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f2
f2
001
SOLUTIONS EBU5302 (2012) Paper A
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vii) What is the total number of possible hops?
viii) Show the variation of the dehopped frequency with time.
[9 marks]
Answer:
i) Period of the PN sequence is 24 – 1 = 15 [1 Mark]
ii) MFSK
[1 Mark]
iii) L = 2
[1 Mark]
L
iv) M = 2 = 4
[1 Mark]
v) k = 3
[1 Mark]
vi) slow FHSS [1 Mark]
vii) 2k = 8
[1 Mark]
viii)
[2 Marks]
Time
0
1
2
3
4
5
Input data
0
1
1
1
1
1
Frequency
f1
f3
f3
Time
Input data
Frequency
12
0
13
1
14
1
f1
15
1
f3
6
1
7
0
8
0
f2
16
1
10
1
f0
17
0
f2
9
0
18
1
11
0
f2
19
0
f2
c) Frequency reuse improves the SNR from co-channel interference but reduces the capacity in
each cell; diversity also improves SNR by up to about 6dB. Using a mathematical
approach, explain how adding diversity to 3-cell cluster can give the same overall SNR from
co-channel interference as a 7-cell cluster, but gives more capacity in each cell.
[8 marks]
Answer:
Diversity gain is about 4-6dB. (1 mark)
Frequency re-use distance Ru  r 3N so if P  1 / r 4 then received power from neighbouring cell
of same frequency is P  P0 /( r 3N ) 4 (2 marks) but in each case there are 6 nearest co-channel
P 3
P /( r 3  3 ) 4
neighbours (2 marks, 1 for each N) P7 / P3  0
(1 mark) so 7    or in dB = 4
P3  7 
P0 /( r 3  7 )
7.3 dB (accept approx -6dB as correct) (1 marks)
If 6dB gain is achieved by adding the diversity the -7dB reduction is more or less cancelled out
(1 mark)
2
2
2
3
3
7
1
1
6
4
5
N=7
N=3
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