Examination Paper
MEng: Radio and Digital Communications 4 Time Allowed: 2 hours
Materials Permitted:
Calculators Permitted: Yes Models Permitted:
Those from the Casio fx-83 and fx-85
Examination Session:
Exam Code: ENGI4507-WE01
Additional Material provided:
series. Visiting Students may use dictionaries: Yes
Instructions to Candidates:
Answer ALL questions.
All relevant workings must be shown.
Page 2 of 5 [ENGI4507/ENGI44G10] Question 1
(ii) (iii)
cos(2𝜋𝜋𝑓𝑓1𝑡𝑡) . cos(2𝜋𝜋𝑓𝑓2𝑡𝑡) = 12 {cos(2𝜋𝜋𝑓𝑓1 + 2𝜋𝜋𝑓𝑓2) 𝑡𝑡 + cos(2𝜋𝜋𝑓𝑓1 − 2𝜋𝜋𝑓𝑓2) 𝑡𝑡}
Unipolar non return to zero (NRZ).
A frequency shift keying (FSK) communication system transmits so(t) to represent binary 1 (mark) and s1(t) to represent binary 0 (space), where
2 2𝑁𝑁 𝑇𝑇0 𝑇𝑇
𝑠𝑠𝑜𝑜(𝑡𝑡) = 𝐴𝐴𝐴𝐴𝐴𝐴𝑠𝑠2𝜋𝜋𝑓𝑓𝑜𝑜𝑡𝑡 0 < 𝑡𝑡 < 𝑇𝑇
𝑠𝑠1(𝑡𝑡) = 𝐴𝐴𝐴𝐴𝐴𝐴𝑠𝑠2𝜋𝜋𝑓𝑓1𝑡𝑡 0 < 𝑡𝑡 < 𝑇𝑇 Assuming that 𝑇𝑇 ≫ 1/𝑓𝑓𝑜𝑜 and 𝑇𝑇 ≫ 1/𝑓𝑓1
Find the energy per bit.
Find an expression for the correlation coefficient, 𝜌𝜌 between the mark and space signals.
Deduce the relationship that gives zero correlation coefficient.
[20%] [10%]
Assume binary coded information is transmitted at 10 kb/s using FSK signal. The received amplitude of each tone is 2x10-2 V. The additive single sided noise power density spectrum is 10×10-9 W/Hz. Find the bit error rate of a coherent detector using the table of the complementary error function for correlation coefficients of (i) 0 and (ii) 0.3 and comment on the result.
Use can be made of the following relationships:
𝑃𝑃 = 1 𝑒𝑒𝑒𝑒𝑓𝑓𝐴𝐴�𝐸𝐸(1−𝜌𝜌) and 𝜌𝜌 = ∫𝑇𝑇 𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚(𝑡𝑡)𝑠𝑠𝑠𝑠𝑠𝑠𝑚𝑚𝑠𝑠𝑠𝑠(𝑡𝑡)𝑑𝑑𝑡𝑡
�∫ 𝑠𝑠(𝑡𝑡)2 𝑑𝑑𝑡𝑡 ∫ 𝑠𝑠(𝑡𝑡)2 𝑑𝑑𝑡𝑡 0 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 0 𝑠𝑠𝑠𝑠𝑚𝑚𝑠𝑠𝑠𝑠
where Pe is the probability of error, E is the energy per bit and 𝜌𝜌 is the correlation coefficient, 𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚(𝑡𝑡),𝑠𝑠𝑠𝑠𝑠𝑠𝑚𝑚𝑠𝑠𝑒𝑒(𝑡𝑡) are the mark and space signals, respectively and T is the bit duration.
A time division multiplexing pulse analogue modulated system transmits eight audio telephony signals with baseband bandwidth equal to 3.4 kHz and two music signals with baseband bandwidth equal to 15 kHz. For an 8-bit analogue to digital converter determine the required bandwidth of transmission for
Code Help
Page 3 of 5 [ENGI4507/ENGI44G10]
(ii) Unipolar return to zero (RZ).
(iii) Manchester code.
Question 2
(a) A mobile user is travelling at a speed of 50 km/hr as shown in Figure Q.2.1.a-b. Assume that the mobile phone requires a signal to noise ratio of 18 dB and that the noise floor of the receiver is -100 dBm. Assume that both base stations transmit 20 dBm and use antennas with 3 dB gain, while the mobile uses an antenna with 0 dB gain and that the foliage attenuation is 10 dB. The base stations have dual frequency bands at 900 MHz and 1800 MHz.
(i) Find the antenna gains in linear scale.
(ii) Find the transmit power in mW for the base station.
(iii) Given the relationship in equation 2.1 for free space path loss, determine the time at which
the mobile phone would need to be handed over from base station 1 BS1 to base station 2 BS2 for 900 MHz and 1800 MHz operating frequencies for Figures Q2.1.a and Q.2.1.b for a 2 dB margin for hand off.
PR=GG c 2 (2.1) P TR4πfd
where PT and PR are the transmit and receive powers respectively, GT and GR are the gains of the transmit and receive antennas respectively, d is the distance from the transmitter and f is the transmission frequency.
(iv) Comment on the success of the handover strategy for both scenarios at the two frequencies.
Figure Q.2.1.a
BS2 5 km BS1
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Page 4 of 5
[ENGI4507/ENGI44G10]
Figure Q.2.1.b
(b) A cellular system has 15 channels to be multiplexed. Each user has a data rate of 10 kbps. Determine the overall bandwidth required for the system using frequency division multiple access.
(c) Discuss the different mechanisms of propagation that connect the transmitter and a receiver.
Page 5 of 5
[ENGI4507/ENGI44G10]
Error function and the complementary error function
erf(𝑥𝑥)= 2 ∫𝑥𝑥𝑒𝑒−𝑢𝑢2 𝑑𝑑𝑑𝑑,erfc(𝑥𝑥)= 2 ∫∞𝑒𝑒−𝑢𝑢2 √𝜋𝜋 0 √𝜋𝜋 𝑥𝑥
Table of Error function and complementary error function
x erf(x) erfc(x) x erf(x)
0.0659921 0.0477149 0.0338949 0.0236516 0.0162095 0.0109095 0.0072096 0.0046777 0.0029795 0.0018628 0.0011432 0.0006885 0.0004070 0.0002360 0.0001343 0.0000750 0.0000411 0.0000221 0.0000116 0.0000060 0.0000031 0.0000015 0.0000007
0.00 0.0000000 0.05 0.0563720 0.10 0.1124629 0.15 0.1679960 0.20 0.2227026 0.25 0.2763264 0.30 0.3286268 0.35 0.3793821 0.40 0.4283924 0.45 0.4754817 0.50 0.5204999 0.55 0.5633234 0.60 0.6038561 0.65 0.6420293 0.70 0.6778012 0.75 0.7111556 0.80 0.7421010 0.85 0.7706681 0.90 0.7969082 0.95 0.8208908 1.00 0.8427008 1.10 0.8802051 1.20 0.9103140
1.0000000 0.9436280 0.8875371 0.8320040 0.7772974 0.7236736 0.6713732 0.6206179 0.5716076 0.5245183 0.4795001 0.4366766 0.3961439 0.3579707 0.3221988 0.2888444 0.2578990 0.2293319 0.2030918 0.1791092 0.1572992 0.1197949 0.0896860
1.30 0.9340079 1.40 0.9522851 1.50 0.9661051 1.60 0.9763484 1.70 0.9837905 1.80 0.9890905 1.90 0.9927904 2.00 0.9953223 2.10 0.9970205 2.20 0.9981372 2.30 0.9988568 2.40 0.9993115 2.50 0.9995930 2.60 0.9997640 2.70 0.9998657 2.80 0.9999250 2.90 0.9999589 3.00 0.9999779 3.10 0.9999884 3.20 0.9999940 3.30 0.9999969 3.40 0.9999985 3.50 0.9999993
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