For a full scale sinusoidal modulating signal with peak value A, show that, output signal to quantization noise ratio in binary PCM system is given by,Ī compact disc recording system samples each of the two stereo signals with a 16 bit A/D converter at 44.1 Kb/sec. (iii) The signal to quantization noise ratio in dB may be calculated as under:ĮXAMPLE 4.28. ![]() The bit transmission rate or signaling rate is the number of bits transmitted by the PCM system per second.Īs the signal is sampled at Nyquist rate, f s = 2f m. Now, let us calculate the bit transmission rate (r). Substituting above value of D in equation (i), we obtain (i) Normalized power for quantization noise (N q) is given by ![]() It is given that uniform quantization is used.Īlso, = Mean square value of message signal is 2 Volt 2 (iv) Derive the expressions used in (i) and (iii). (iii) The signal to quantization noise ratio in dB. (i) The normalized power for quantization noise. The mean squared value of message signal is 2 Volt-squared. A low pass signal of 3 kHz bandwidth and amplitude over – 5 Volts to + 5 Volts range is sampled at Nyquist rate and converted to 8-bit PCM using uniform quantization. Hence, n = 7 = number of bits required per sample.ĮXAMPLE 4.27. Where L = 2 n, n is the number of bianry digits. Assume uniform quantization Calculate the number of bits required per sample. The input varies from – 3.8 V to 3.8 V and has the average power of 30 mW, the required signal to quantization noise power ratio is 20 dB. Bandwidth of the input to pulse code modulator is restricted to 4 kHz. If a value larger than the minimum will be choosen, then granular noise will occur.ĮXAMPLE 4.26. Hence, larger step size out of two will be the required step size. Where D is step size and f s is sampling rate.ĭ 1max (minimum value of step size) = EQUITATION What would be the disadvantages of choosing a value of larger than the minimum? Find the minimum value of step size which will avoid slope overload distortion. The input signal is 5 cos (2r 1000t) + 2 cos (2 2000 t) V, with t in sec. The pulse rate in a DM system is 56,000 per sec. (ii) Assuming that the cutoff frequency of the low-pass filter is f m, we haveĮXAMPLE 4.25. The maximum allowable amplitude of the input sinusoid is (ii) Determine the posfiltered output signal-to-quantizing-noise ratio for the signal of part (i) (i) Determine the maximum amplitude of a 1-kHz input siinusoid for which the delta modulator does not show slope overload. A DM system is designed to operate at 3 times the Nyquist rate for a signal with a 3 kHz bandwidth. Then, assuming that the quantizing noise power P q is uniformly distributed over the frequency band up to f s, the output quantizing noise power within the bandwidth f M isĬombining equation (ii) and (iii), we see that the maximum output signal-to-quantizing-noise ratio isĮXAMPLE 4.24. Let the bandwidth of a postreconstruction low-pass filter at the output end of the reciever be f M ≥ f m and f M << f s. We know that the mean quantizing error, or the quantizing noise power, = D 2/3. Thus, the maximum permissible value of the output signal power equals Solution: For no-slope-overload, we must have Where f s = 1/T s is the sampling rate and f M is the cutoff frequency of a low-pass filter at the output end of the recover. Prove that the maximum output signal-to-quantizing-noise ration in a DM system under the assumption of no slope overload is given by Thus, if A >D/(), slope overload distortion will occur. To avoid the slope overload, we require that ![]() Where f s = 1/T s is the sampling frequency. Show that the slope overload distortion will occur if Consider a sinusoidal signal m(t) = A cos applied to a delta modulator with step size D. The minimum bandwidth required for this case will beį pcm = f s = (8000) = 28000 Hz = 28 kHz Ans.ĮXAMPLE 4.22. ![]() The minimum number of bits per sample is 7. Thus, the minimum number of quantizing levels needed is 102. (ii) The minimum required system bandwidth will be The minimum number of bits per sample is 5. Thus, the minimum number of uniform quatizing levels required is 26. (iii) Repeat parts (i) and (ii) when a -law compander is used with = 255. (ii) Calculate the minimum system bandwidth required. (i) What is the minimum number of uniform quantizing levels needed, and what is the minimum number of bits per sample needed? The required output signal-to-quantizing-noise ratio is 30 dB. A PCM signal is generated with a sampling rate of 8000 samples/s. Consider an audio signal with spectral components limited to the frequency band of 300 to 3300 Hz.
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