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Unformatted text preview: City University of Hong Kong Course code & title: EE3008 Principles of Communications Session : SemesterB 2005—2006 Time allowed : Two hours This paper has SEVEN pages (including this cover page) 1. This paper consists of two parts and each part consists of three questions. You
need to answer four questions. Two from part 1 and two from part 2. 2. For those who answer more than four questions, we only count on the ﬁrst two
questions of each part. 3. Start each question on a NEW page. 4. Start each part on a NEW answer book. Materials, aids & instruments permitted to be used during examination: 1. Approved calculator Part 1 Question 1 25 marks (a) Compare and contrast the following AM systems:
(i) DSB—SC, (ii) DSB—LC, and (iii) SSB. [9 marks] (b) A DSBLC AM signal has the following form:
s(t) = 20[1+ km(t)]cos(1067zt) Assume that m(t) = acos(20007a‘) + sin(20007u‘) where a is a constant, and
k = 0.5 .
(i) Show that the modulation index ,u of the system can be expressed as V a2 +1
2 .
Determine the maximum value of a, a ﬂ: such that a simple envelope detector can be used to demodulate the signal.
(ii) With anm , determine the overall transmission power (500hm resistive load
is used) and power efficiency.
(iii) With am, draw the magnitude spectrum of s(t) (you need to specified the frequency and the magnitude values in your graph).
[16 marks] Question 2 25 marks A frequency modulated signal has the following form: s(t)=10cos[2nfct+ ¢(t)] (a) Assume that ¢(t) = 4cos(100007y”t) and fC = 100MHz , (i) Find the peak frequency deviation and determine the transmission
bandwidth of s(t) by using Carson’s rule. (ii) Determine the percentage of the average power within the frequency
range f6 —8kHz to f6 +8kHz. [12 marks] (b) Now let ¢(t) =4cos(100007u‘) +0.4 sin(200007u‘) , determine peak frequency
deviation (any one of them) of s(t). [6 marks] (c) Draw the block diagram of an Armstrong FM modulator. State the major
advantage and limitation for an Armstrong FM modulator. Can it be applied to
the system in (a) without any further frequency processing? [7 marks] A TABLE OF BESSEL FUNCTIONS Bessel Functions of the ﬁrst Kind, 1,, (x) x Jo J. J2 JJ 1,. IS 1.5 J, 13 I... J",
0.0 1.00
.2 .99 .10
.4 .96 .20 .02
.6 .91 .29 .04
.8 .85 .37 .08 .01 _
1.0 .77 .44 .1'1 .02
.2 .67 .50 .16 .03 .01—
.4 .57 .54 .21 .05 .01"
.6 .46 .57 .26 .07 .01
.8 .34 .58 .31 .10 .02
2.0 .22 .58 .35 .13 .03 .01—
.2 .1 1 .56 .40 .16 .05 .01
.4 .00 .52 .43 .20 .06 .02
.6 —.10 .47 .46 .24 .08 .02 ' .01—
.8 —.19 .41 .48 .27 .11 .03 .01—
3.0 — .26 .34 .49 .31 .13 .04 .0]
.2 —.32 .26 .48 .34 .16 .06 .02
.4 — .36 .18 .47 .37 .19 .07 .02 .01"
.6 — .39 .10 .44 .40 .22 .09 .03 .01"
.8 —.40 .01 .41 .42 .25 .11 .04 .01
4.0 — .40 — .07 .36 .43 .28 .13 .05 .02
.2 —.38 —.14 .31 .43 .3] .16 .06 .02 .01“
.4 —.34 —.20 .25 .43 .34 .18 .08 .03 .01—
.6 —.30 —.26 .18 .42 .36 .21 .09 .03 .01
.8 —.24 —.30 .12 .40 .38 .23 .11 .04 .01
5.0 —.18 —.33 .05 .36 .39 .26 .13 .05 .02 .01—
.2 — 11 —.34 .02 .33 .40 .29 .15 .07 .02 .01—
.4  04 — .35 .09 .28 .40 .31 .18 .08 .03 .01“
.6 .03 —.33 .15 .23 .39 .33 .20 .09 .04 .01
.8 .09 —.31 .20 .17 .38 .35 .22 .11 .05 .02 .01“ Question 3 25 marks (a) (b) (C) ((9 Compare and contrast the use of singlewire and twisted pair as transmission
media. You should address any two of the following areas: (i) bandwidth, (ii)
data rate, and (iii) sensitivity to interference. [4 marks] (i) Time dispersion in optical fiber will cause pulse broadening and hence bit rate reduction. Describe the time dispersion due to intramodal dispersion
and intermodal dispersion. (ii) A particular photodetector has a quantum efﬁciency of 0.6 for wavelength 1.2um. Calculate its responsivity. [8 marks]
{ Note: Planck’s constant = 6.6x10_34 Js, electron charge = 1.6x10_19C
Speed of light = 3x108m/s } State the three major multiple access techniques in satellite communications.
Why do the satellites at GEO be normally chosen for direct broadcasting? [6 marks] The required minimum carrier to interference ratio of a TDMA type cellular radio system is 9dB and the path loss exponent is normally within the range of
3.5 to 4.5. (i) What is the minimum cluster size to cover the whole range of path loss
exponent? (ii) If each cell is required to support 64 users simultaneously, what is the total
number of channels required for the whole system in (i)? [7 marks] Part 2 Q4 (25 Marks)
It is desired to set up a central station for simultaneous monitoring of the electrocardiograms
(ECGs) of 20 hospital patients. The data from the room of 20 patients are brought to processing
center over wires and are sampled, quantized, binary coded, and timedivision multiplexed. The
multiplexed data are now transmitted to the monitoring station. The ECG signal bandwidth is 250 Hz. The maximum acceptable error in sample amplitude is 0.15% of the peak signal amplitude.
The system uses the minimum sampling rate. (a) Determine the minimum sampling rate. (4 Marks) (b) Determine the number of bit for each sample. (9 Marks) (c) Determine the bit rate for transmitting signal from each patient. (4 Marks) ((1) Determine the data rate of the multiplexed data. (4 Marks) (6) Determine the minimum cable bandwidth needed to transmit the multiplexed data, assum
ing that NRZ bipolar waveform is used and nulltonull bandwidth is considered. (4 Marks) Q5 (25 Marks)
A QPSK signal canbe represented as 3(t) = A cos(wct + 0m) where each QPSK symbol is modulated into a phase 0,”. Suppose the mapping rule is given by (a) Sketch a generator for this QPSK signal. (8 Marks) (b) Each QPSK symbol contains two bits. Each bit can be detected, separately on the I and
Qchannels, as shown in Fig.1 where ¢1(t) = y/Z/Ts cos wet and ¢2(t) = x/Z/Ts sin wot.
Let (“1 denote the estimated bit from the Ichannel and I3 denote the estimated bit from the
Q—channel. Write down the decision rule for a and (3. Justify your answer. (8 Marks) (c) Usually the two bits in a QPSK symbol are independent. Show that the symbol error prob
ability of the above coherent detection system is given by ' w< am a where E. is the symbol energy and N0 is the noise single sided power spectral density. (9
Marks) a received
signal r(t) Output symbol a b Multiplexer 13 Figure 1: Coherent QPSK receiver for Q5b. Q6 (25 Marks)
Coherent BPSK and BFSK systems. Q6.a) A BPSK signal is given by s(t) = :tA cos(wct) where positive and negative signs corre
spond to bit one and bit zero, respectively. A random phase 00 is introduced during the
transmission, so that the received signal is given by 7'(t) = :tA cos(wct + 00) + n(t) where n(t) is the additive white Gaussian noise with singlesided power spectral density No. i) Suppose the receiver has no information about 00 and still use ¢(t) = Tl!) cos wet as
the local reference signal. Show that the bit error probability of such a receiver is given
by 2E1, COS2 90
P = ——
b Q ( N0
(7 Marks)
ii) Suppose when 00 = 20°, Pb = 1.14 X 10—3. Determine Pb for the case of 00 : 9.63°.
(8 Marks) Q6.b) Suppose a binary coherent FSK system and a binary coherent PSK system operate in the same AWGN environment having bit energy Eb and noise singlesided power spectral
densityNo. i) Sketch their signal points. (5 Marks) ii) Which system, BFSK or BPSK, has a better error performance? Justify your answer
by giving appropriate expressions, (5 Marks) ——Questions end here—— TABLE 1.1 Value d 00) versus x.‘ 0.00 0.01 0.02 0.03 0.04 0.3 0.“ 0.07 0.“ 0.09 . 0.0 .511!) .4960 .4911 .4880 .4840 .4801 .4761 .4721 .4681 .4641
0. 1 .4602 .4562 .4522 .4483 .4443 .4404 .4364 .4375 .4286 .4247
0.2 .4207 .4168 .4129 .40” .4052 .4013 .3974 .3936 .3897 .3859
0.3 .3821 .3783 .3745 .3707 .3669 .3632 .3594 .3557 .35” .3483
0.4 .3446 .3409 .3372 .3336 .3311) .3264 .3228 .3192 .3156 .3121
0.5 .3155 .3050 .3015 .2981 .2946 .2912 .2877 .2843 .2810 .2776
0.6 .2743 .2709 .2676 .2643 .2611 .2578 .2546 .2514 .2483 .2451
0.7 .241) .2389 .358 .327 .2296 .2266 .2236 .2206 .2177 .2148
0.8 .2119 .2090 .2N1 .2033 .2135 .1977 .1949 .1922 .1894 .1867
0.9 .1841 .1814 .1788 .1762 .1736 .1711 .1685 .1660 .1635 .1611
1.0 .1587 .1562 .1539 .1515 .1492 .1469 .1446 .1423 .1401 .1379
1.1 .1357 .1335 .1314 .1292 .1271 .1251 .1230 .1210 .11” .1170
1.2 .1151 .1131 .1112 .1093 .1075 .1056 .1038 .103) .1133 .0985
1.3 .0968 .0951 .0934 .0918 .0901 .1385 .869 .0853 1338 .1313
1.4 .0603 .0793 .0778 .0764 .0749 .0735 .0721 .0713 .1594 .MBI
1.5 .0668 .0655 .1543 .0630 .0618 .06“ .0594 .0582 .0571 .0559
1.6 .0548 .0537 .0526 .0516 .0505 .0495 .0485 .0475 .0465 .0455
1 .7 .0446 .0436 .0427 .0418 .0443 .0401 .0392 .0384 .0375 .0367
1.8 .0359 .0351 .0344 .0336 .0329 .0322 .0314 .0307 .0301 .0294
1.9 .0287 .028 1 .0274 .0268 .0262 .0256 .0250 .0244 .0239 .0233
2.0 .0228 .0222 .0217 .0212 .0207 .0202 .0197 .0192 .0188 .0183 .
2.1 .0179 .0174 .0170 .0166 .0162 .0158 .0154 .0150 .0146 .0143
2.2 .0139 .0136 .0132 .0129 .0125 .0122 .0119 .0116 .0113 .0110
2.3 .0107 .0104 .0102 .00990 .009“ 111939 .009“ 01889 01866 .01842
2.4 .0381) .00793 01776 .00755 01734 “)7 14 .m695 11576 .m7 .MJ9
2.5 411621 111604 £0587 01570 .0059 111539 .00523 .00503 03494 .0040)
2.6 .004“ .(X)453 03440 03427 03415 411402 #1391 01379 #1368 .00357
2.7 £1347 .m336 113326 411317 “1307 .m298 .m289 M1280 .0027: 01264
2.8 411256 .111248 117240 111233 .WZZ6 #1219 .M12 .00205 .00199 .00193
2.9 .00137 .0015: .0017: “1169 01164 01159 .00154 .00149 01144 .00139 TABLE 1.2 Values of 00:) for large x. 10 10 10
x log x Q(x) x 103 x Q(x) 1: log 1: Q(x) 3.00 4.77 1.35503 4.00 6.02 3.17505 5.00 6.99 2.87507
3.05 4.84 1.14503 4.05 6.07 2.56505 5.05 7.03 2.21507
3.10 4.91 9.68504 4.10 6.13 207505 5.10 7.08 1.70507
3.15 4.98 8.16504 4.15 6.18 1.66505 5.15 7.12 1.30501.
3.20 5.05 6.87504 4.20 6.23 1.33505 5.20 7.16 9.96508
3.25 5.12 5.77504 4.25 6.28 1.07505 5.25 7.20 7.61508
3.30 5.19 4.83504 4.30 6.33 8.54506 5.30 7.24 5.79508
3.35 5.25 4.04504 4.35 6.38 6.81506 5.35 7.28 4.40508
3.40 5.31 3.37504 4.40 6.43 5.41506 5.40 7.32 3.33508
3.45 5.38 2.80504 4.45 6.48 4.29506 5.45 7.36 2.52508
3.50 5.44 2.33504 4.50 6.53 3.40506 5.50 7.40 1.90508
3.55 5.50 1.93504 4.55 6.58 2.68506 5.55 7.44 1.43508
3.60 5.56 1.59504 4.60 6.63 2.11506 5.60 7.48 1.07508
3.65 5.62 1.31504 4.65 6.67 1.66506 5.65 7.52 8.03509
3.70 5.68 1.08504 4.70 6.72 1.30506 5.70 7.56 6.00509
3.75 5.74 8.84505 4.75 6.77 1.02506 5.75 7.60 4.47509
3.80 5.80 7.23505 4.80 6.81 7.93507 5.80 7.63 3.32509
3.85 5.85 5.91505 4.85 6.86 6.17507 5.85 7.67 2.46509
3.90 5.91 4.81505 4.90 6.90 4.79507 5.90 7.71 1.82509
3.95 5.97 3.91505 4.95 6.95 3.71507 5.95 7.75 1.34509 Figure 2: Qfunction: Q(x) as a function of x. ...
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