Question

KS31603 SULIT/CONFIDENTIAL QUESTION 4 (20 marks) (a) A message signal has a peak voltage of 1V and its average power is 125 mW. Design a PCM system (find minimum value of R of the R bit A/D) with uniform quantization to achieve at least a signal to quantization noise ratio of 36 dB. What is the signal to quantization noise ratio obtained in your design? [6 marks] (b) If the signal is sampled at 9kHz in 4(a), what is the bit rate for your design? What is the minimum bandwidth required? (Note that minimum bandwidth is achieved by pulse shaping with a "brick wall" filter. Also the signal is a baseband signal.) [2 marks] (c) It is required to transmit baseband digital data at 2 Mbit/s. The power spectral density of white noise No = 10. Assume that the amplitude of the baseband pulse 4 = 1. Determine the symbol rate, R, the bit energy E, bandwidth for a "brick wall" filter and bit error rate for the following baseband modulation schemes: (1) QPSK and, (11)16PSK [6X2-12 marks]

Answer 1

Question 4 IV 125 mw Am (peal Value) (Pi) peak power = Minimum required signal to quantization Noise Ratio - 36 dB 125 mW Signal power si si 10 10 2 3 6 iog Ni - ) Ni 125 50.081398 mW 11 103.6 Nia 12 12 A (1 + Stef size) (12 Ni)!/2 30.6188. : 0.6138 = 2" : 2 x1 2'Am 2n 0.6138

(12 ni) 1/2 < 2 Am 2n 0.0194 2 0.0194 -> an ? 2103.032 n> 6.6877 Noise Bower n=7 2 A? 2 Am 12 12 2 2 12x 2 an 12x214 -0.01017 mw. 125 Weise bow 0.00017 = 40,89 dB S'NR =1010 Bit Rate 10 10g (signal bowo rolog SNR of New design nfs = 94 HZ Bit Rate : 9X103x4 = 63 kbps Rb = 2 mbps To A al no=107 fs / 7 symbol Rate Rb 2 mbps IX A²Tb Eb: = 0.25x10m 2 2x106x2

for probibility of Error 1) QPSK Omin 2 Gs sin (T/m) sin (7/4) 2 2 Eb 2NE Pe=0 . Domin & Eb 2010 2 No (. ED ] 0.25 X 10.6 2 X 10 -7 2 No (V1125) 16 - psk Omina 2 4 Eb sin (7/6) 4x Eb x00195 = 0.780 VED ce : كم Dmin Q ( . 0.609x ED NO 2010 0.60gx 10-6 - Q 1o- QC 2.46