eee206_2001_sols - Solutions to EE206 / EEE274...

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Unformatted text preview: Solutions to EE206 / EEE274 COMMUNICATION SYSTEMS May 2001 Set by B Chambers 1. (a) DSB takes up too much bandwidth, but need to transmit sidebands down to d.c. as these contain mean picture brightness information. Cannot just transmit one sideband because of impossible filtering requirement. Hence transmit one full sideband plus a vestige of the other. (b) (1) Total power = Pc(1 + m2/2) = 1.5 x 103 when m = 1 Hence PC = 1000 watts (ii) m = 0.6 Power in one sideband = m2 Pc/4 = 0.36 x 1 x 103 / 4 = 90 watts If PRC is power in the reduced carrier, 10 log10(PC / PRC) = 20, hence PRC = 10 watts Total power radiated is thus 90 + 10 = 100 watts (c) Most of the transmitted power is being used to convey information. The residual carrier is enough to provide a reference for demodulation. Use of only one sideband conserves spectrum. (d) m1 = 0.5, m2 = 035, Hence effective m = 5/(052 + 0.352) = 0.61 Total radiated power = 1 x 103 (1 + 0612/2) = 1186 watts 2. (c) When control voltage = 0 V, the oscillation frequency is determined by L = 15 11H and C = 100 pF. 1 27m LC Control voltage has the form v = 5 cos(21t.2000t), i.e. Vpeak = 5 volts and f = 2 kHz Hence fc = = 4.109363 MHz Hence peak capacitance change = 5 X 2 = 10 pF Maximum oscillator frequency = 1 / 2m/(15 x 10-6 x 90 x 10-12) = 4.331649 MHz Minimum oscillator frequency = l / 21t\/(15 x 10'6 x 110 x 10‘12) = 3.918124 MHz Frequency deviation = 0.5 x (4.331649 - 3.918124) = 2067625 Hz = ff Hence approximate bandwidth = 6fm + 2.13ff= 6 x 2 + 2.13 x 206.763 = 452.405 kHz This is increased by a factor of 10 in the frequency multiplier Hence final bandwidth = 4.524 MHz 3.(a) Main advantages are: flexibility (can deal with analogue or digital signals), multiplexing, signal regeneration, error detection and correction. (b) Main processes to be stressed are the signal conditioning, sampling, quantisation, coding. (c) Telephone channel bandwidth = 4000 Hz, hence sample at 8000 / sec. (i) N = 12. Bandwidth = 12 x 8000 x 8 = 768,000 = 768 kHz, assuming NRZ pulse representation. (ii) Each channel requires 8 x 8000 = 64000 samples / sec Time for 1 sample = 1 / 64000 = 1.5625 x 10‘5 sec One sample from each channel fits into one frame. Hence number of channels in frame = 125 x 10-6/ 1.5625 x10'5 = 8 (iii) Other 2 time slots are used for system purposes, e.g. synchronisation, routing, telemetry 4.(a)(i) C = information carrying capacity of the channel in bits / sec B = bandwidth in Hz S = signal power in watts N = noise power in watts (ii) Hartley Shannon gives the theoretical upper limit on C. Practical systems perform less well. (b) (i) Minimum bandwidth = 2 x sampling frequency = 2 x 44.1 = 88.2 kHz (ii) Number of quantisation levels q = 216. Hence S/N = q2 = 232 C = 88.2 x 103 x log2(1 + 232) = 2.8224Mbits /sec (iii) S/N = 10 log10(232) = 96.33 dB (c) Oversampling eases the design of the low-pass filter since the skirts can be made less steep and hence fewer elements are required (hence cheaper to manufacture) (d) Because of interlacing and field flyback, each picture appears to have 575 lines. Assume each pixel is square, then number of pixels/sec = number on 1 line / time for 1 line = 4/3 x 575 / ((64-1205) x 10'6 ) = 1.476 x 107 bits / sec Let there be 10 resolvable brightness levels. Then information / pixel = logz (10) = 3.32 bits Information = 3.32 x 1.476 x 107 bits / sec Received S/N = 30 dB = 1000 7 Hence from Hartley Shannon, minimum bandwidth required = = 4.92 MHz log2 (1 + 1000) To get an estimate of the maximum bandwidth required, let the picture be made up of a checkerboard of alternating black and white pixels. Video signal is then a square wave of period 2 pixels z sine wave of period 2 pixels. Bandwidth to transmit this is approximately ———1———7 = 7.38 MHz 0.5x1.476x10 ...
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This note was uploaded on 09/25/2008 for the course EEE 206 taught by Professor Langley during the Spring '08 term at University of Sheffield.

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eee206_2001_sols - Solutions to EE206 / EEE274...

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