Assignment 4 – EKT358 1. Consider a compact disc that uses pulse-code modulation to record audio signals whose bandwidth 15 kHz. Specifications of the modulator include the following: Quantization : uniform with 512 levels Encoding: Binary Determine: (a) The Nyquist rate (b) Minimum permissible bit rate Nyquist rate, fs= 2B = 2(15 kHz) = 30 kHz R = nfs n = log 512/ log 2 =9 R = nfs= 9 (30kHz) = 270 kbits/s 2. A television signal (video and audio) has a bandwidth of 4.2 MHz. This signal is sampled, quantized and binary coded to obtain a PCM signal. (a) Determine the sampling rate if the signal is to be sampled at a rate 20% above the Nyquist rate. (b) If the samples are quantized into 1024 levels, determine the number of binary pulses required to encode each sample. (c) Determine the binary pulses rate(bits per second) of the binary-coded signal, and the minimum bandwidth required to transmit this signal Nyquist rate, fs= 2B = 2(4.2 MHz) = 84 MHz 20% above Nyquist rate, fs = 120/100 x 84 MHz = 10.08 MHz. 2n=1024 n = log21024 = 10 R = nfs= 10(10.08MHz) = 100.8Mbits/s Bmin = R/2 = 100.8 MHz/ 2 = 50.4 MHz 3. Determine the minimum number of bits required for PCM codes with the following dynamic ranges and determine the coding efficiencies: a) 24 dB b) 48 dB c) 72 dB DR(dB) 20 log( 2 n 1) 24 / 20 log( 2 n 1) 2 n 15.85 1 log 16.85 4.07 log 2 n 4.07, n 5 4.07 efficiency x100% 81.5% 5 n DR(dB) 20 log( 2 n 1) 48 / 20 log( 2 n 1) 2 n 251.19 1 log 252.19 7.98 log 2 n 7.98, n 8 7.98 efficiency x100% 99.73% 8 n DR(dB ) 20 log( 2 n 1) 72 / 20 log( 2 n 1) 2 n 3981.07 1 log 3982.07 11.96 log 2 n 11.96, n 12 11.96 efficiency x100% 99.66% 12 n 4. For a PCM system with the following parameters: Maximum analog input frequency = 4 kHz Maximum decoded voltage at the receiver= ±2.55 V Minimum dynamic range=46 dB Determine (a) (b) (c) (d) (e) Minimum sample rate Minimum number of bits used in the PCM code Resolution Quantization error Coding efficiency (a) Nyquist rate, fs= 2B = 2(4 kHz) = 8 kHz (b) DR(dB) 20 log( 2 n 1) 46 / 20 log( 2 n 1) 2 n 199.53 1 log 200.53 n 7.64 log 2 n 7.64, n 8 8 bits must be used for the magnitude but one additional sign bit is required to indicate +ve and –ve voltage, thus the minimum number of bits=9 (c) resolution (d) Qe Vmax V 2.55 nmax 8 0.01V DR 2 1 2 1 resolution 0.01 0.005V 2 2 (e) Coding efficiency = minimum number of bits(including sign bit) x 100% Actual number of bits(including sign bit) = 8.64/9 x 100% =96% 5. Briefly explain the process of digital companding. With digital companding, the analog signal is first sampled and converted to a linear PCM code and then the linear code is digitally compressed. In the receiver, the compressed PCM code is expanded and then decoded (i.e., converted back to analog). 6. Determine the voltage of the input signals if the SQR = 40 dB and q =0.2 V. ⁄ [ ] ⁄ 7. A companding system with µ = 100 used to compand from 0V to 10 V sinusoid signal. Draw the characteristic of the typical system. Vout Vin Vout 0 0 1 5.2 2 6.57 Vmax ln(1 Vin Vmax ) ln(1 ) 3 7.44 4 8.05 5 8.52 6 8.91 7 9.24 8 9.52 12 10 8 Vin 6 Vout 4 2 0 1 2 3 4 5 6 7 8 9 10 11 9 9.77 10 10 8. Suppose analog signal has the bandwidth of 15 kHz (uniform). Find the minimum first null bandwidth (square wave pulses) at which it would be possible to transmit a PCM signal while maintaining average SNR of at least 58 dB. Assume Vpeak = 5V.
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