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Digital Communication (GTU) 512 Pulse Code Modulation ...
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Chap 5
1. 1. Digital Communication (GTU) 51 Pulse Code Modulation Chapter 5 : Pulse Code ModulationSection 5.9:Ex. 5.9.4 : A PCM system uses a uniform quantizer followed by a 7 bit encoder. The system bit rate is 50Mbits/sec. Calculate the maximum bandwidth of the message signal for which this system operates satisfactorily..Page No. 526.Soln. : It has been given that : Bit rate r = 50 Mbits/sec and N = 7 We know that bit rate r = Nfs ∴ fs = = = 7.14 MHz ∴ Maximum signal bandwidth BW = fs /2 = ∴ BW = 3.57 MHz ...Ans.Ex. 5.9.5 : Thebandwidth of a video signal is 4.5 MHz. This signal is to be transmitted using PCM with the number ofquantization levels Q = 1024. The sampling rate should be 20% higher than the Nyquist rate. Calculate thesystem bit rate. .Page No. 526.Soln. : Bandwidth W = 4.5 MHz ∴ As per Nyquist rate fs = 2W = 9 MHzButfs should be 20% higher than Nyquist rate ∴ fs = 1.2 × 9 MHz = 10.8 MHz ...(1) We know that, Q = N 2 , ∴1024 = 2N ∴ N = 10 ...(2) ∴ System bit rate r = N fs = 10 × 10.8 MHzEx. 5.9.7 : Plot the characteristics of aµlaw compressor. .Page No. 530Soln. :To plot the characteristics of µlaw compressor : The expression for thenormalized output of a µlaw compressor is given by, Z (x) = ± ..... x ≤ 1 Let µ = 255.
2. 2. Digital Communication (GTU) 52 Pulse Code Modulation Note that instead of |x| / xmax we have writtenonly |x| and restricted the values of x only upto 1.This has the same effect as that of normalizing. ∴ Z (x) = ± ...(1) Substitute x = 0, 0.2, 0.4, 0.6, 0.8 and 1 in Equation (1) to get the corresponding values of Z asshown in thefollowing table. |x| 0 0.2 0.4 0.6 0.8 1 Z 0 ± ± ± ± ±1 0.71 0.84 0.9 0.96 Therefore the compressorcharacteristics is as shown in Fig. P. 5.9.7. Fig. P. 5.9.7 : µlaw compressor characteristicsSection 5.15 :Ex.5.15.7 : If a voice frequency signal is sampled at the rate of 32,000 samples/sec and characterized by peak valueof 2 Volts, determine the value of step size to avoid slope overload. What is quantization noise power Nq andcorresponding SNR ? Assume bandwidth of signal as 4 kHz. .Page No. 560.Soln. :Given : fs = 32,000samples/sec. Peak value of the signal A = 2V. Bandwidth B = 4 kHz.1. Step size δ to avoid slope overload : Toavoid slope overload the following condition should be satisfied. A ≤ = Substituting the values we get,
3. 3. Digital Communication (GTU) 53 Pulse Code Modulation 2 ≤ ∴ δ ≥ ∴ δ ≥ 1.57 Volt ...Ans.2. Quantizationnoise power (Nq) : The quantization noise power for a delta modulator is given by, Nq = = ∴ Nq = 0.822 W...Ans.3. Signal to noise ratio : SNR = = = 19.45Ex. 5.15.8 : A compact disc (CD) records audio signals digitallyby PCM. Assume audio signals bandwidth to be 15 kHz. If signals are sampled at a rate 20% above Nyquistrate for practical reasons and the samples are quantised into 65,536 levels, determine bits/sec required to encodethe signal and minimum bandwidth required to transmit encoded signal. .Page No. 560Soln. : W = 15 kHz, fs =1.2 × 2 W = 2.4 × 15 kHz = 36 kHz, Q = 65,536.1. Signaling rate (r) : We know that Q = 2N ∴ N = log2 Q
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∴ N = = 16 ...(1) Signaling rate r = N fs = 16 × 36 kHz = 576 kbits/sec. ...Ans. Thus the signaling rate r is 576kbits/sec.2. Minimum bandwidth : BT = Signaling rate = kbits/sec ∴ Minimum bandwidth BT = 288 kHz...Ans.Ex. 5.15.9 : In a single integration DM scheme, the voice signal is sampled at a rate of 64 kHz. Themaximum signal amplitude is 1 Volt. 1. Determine the minimum value of step size to avoid slope overload. 2.Determine granular noise power N0, if the voice signal bandwidth is 3.5 kHz. 3. Assuming signal to be sinusoidal,calculate signal power So and signal to noise ratio (SNR). 4. Assuming that the voice signal amplitude is uniformlydistributed in the range, (– 1, 1), determine So and SNR. .Page No. 560Soln. :Given : fs = 64 kHz A = 1Volt1. Minimum step size to avoid slope overload : A ≤ ,
4. 4. Digital Communication (GTU) 54 Pulse Code Modulation ∴ δmin = = = 0.3436 Volt...Ans.2. Granular noisepower : Nq = × = × = 2.15 × 10– 3 W ...Ans.3. Signal power So and SNRo : As the signal is sinusoidal, thenormalized output signal power So = [A/]2 = A/2 = 1/2 Watt. ...Ans. ∴ SNRo = = = 232.3 or 23.66 dB.4.Signal power for uniformly distributed signal : The signal PDF for a uniformly distributed signal is as shown in Fig.P. 5.15.9. Fig. P. 5.15.9 ∴ Mean square value of the signal, E[S2] = S2 ⋅ fs (S) dS = S2 dS = = 1/3 AssumingR = 1. Normalized signal power So = Mean square value = 1/3 W ...Ans. Signal to noise ratio = SNR = =155.03 or 21.9 dB ...Ans.Ex. 5.15.10 : The information in an analog signal voltage waveform is to be transmittedover a PCM system with an accuracy of ± 0.1 % full scale accuracy. The analog voltage waveform has abandwidth of 100 Hz and an amplitude range of – 10 to + 10 Volts. (a) Determine the minimum sampling raterequired. (b) Determine the number of bits in each PCM word. (c) Determine the minimum bit rate required in thePCM system. (d) Determine the minimum absolute channel bandwidth required for the transmission of the PCMsignal. .Page No. 561Soln. : It has been given that,1. Accuracy of ± 0.1 % of full scale is expected.2. W = 100Hz and amplitude range is – 10 to + 10 V(a) Sampling rate fs : By sampling theorem the minimum sampling rate isfs (min) = 2W = 200 Hz ...Ans.(b) Number of bits per word (N) : As accuracy is expected to be ± 0.1 % of fullscale, the maximum quantization error should be± 0.1 % of full scale.
5. 5. Digital Communication (GTU) 55 Pulse Code Modulation ∴ ∈max = ± 0.1 % of full scale = ± 0.001 [10 –(– 10)] = ± 0.001 × 20 ∴ ∈max = ± 0.02 Volts ...(1) We know that the maximum value of the quantizationerror is ∈max = ± S/2 ...(2) ∴ ± S/2 = ± 0.02 ∴ S = 0.04 Volt ...(3) But S = where VH = 10 V and VL = –10 V ∴ Q = = = 500 ...(4) But Q = 2N ∴ N log10 2 = log10 500 ∴ N = 8.96 ≈ 9 ...Ans.(c) System bit rate :System bit rate (r) = N fs = 9 × 200 = 1800 bits/sec ...Ans.
6. 6. Digital Communication (GTU) 56 Pulse Code Modulation(d) Transmission channel bandwidth (BT) : BT ≥ Nfs ∴ BT ≥ 900 Hz ...Ans.Ex. 5.15.11 : The information in an analog waveform with a maximum frequency fm = 3kHz is to be transmitted over an M level PCM system, where the number of pulse levels is M = 16. Thequantization distortion is specified not to exceed 1% of peak to peak analog signal. 1. What is the maximumnumber of bits per sample that should be used in this PCM system ? 2. What is the minimum sampling rate andwhat is the resulting bit transmission rate ? .Page No. 561Soln. :Given : fm = 3 kHz, Number of quantizationlevels M = 16.1. Number of bits/sample (N) : We know that, number of quantization levels M = 2N. ∴ 2N = 16∴ N = 4 ...Ans.2. Minimum sampling rate : By sampling theorem : fs (min) = 2 fm = 2 × 3 kHz ∴ fs (min) = 6kHz ...Ans.3. Bit transmission rate : r = N fs = 4 × 6 kHz ∴ r = 24 kbits/sec. ...Ans. ∴ Number of quantizationlevels, Q = 2 = 2 = 128N 7 ...Ans.Ex. 5.15.12 : A TV signal with a bandwidth of 4.2 MHz is transmitted usingbinary PCM. The number of quantization levels is 512. Calculate : 1. Code word length 2. Transmissionbandwidth 3. Final bit rate 4. Output signal to quantization noise ratio. .Page No. 561Soln. :Given : fm = 4.2MHz and Q = 512.1. Code word length (N) : Q = 2N ∴ N = ∴ N = 9 bits/word ...Ans.2. Transmissionbandwidth :
7. 7. Digital Communication (GTU) 57 Pulse Code Modulation BT = N fs = N (2 fm) ∴ BT = 9 × 4.2 MHz =37.8 MHz ...Ans.3. Final bit rate (r) : r = N fs = 9 × 2 × fm = 18 × 4.2 MHz ∴ r = 75.6 Mb/s ...Ans.4. Signalto quantization noise ratio : Since the TV signal is not a sinusoidal signal, let us use the general expression of signaltoquantization noise ratio. = 4.8 + 6 N dB = 4.8 + (6 × 9) ∴ = 58.8 dB ...Ans. This is the maximum signal tonoise ratio that we are expected to get from this system.Ex. 5.15.13 : The output signal to noise ratio (SNR) of a
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10 bit PCM was found to be 30 dB. The desired SNR is 42 dB. It was decided to increase the SNR to thedesired value by increasing the number of quantization levels. Find fractional increase in the transmissionbandwidth required for this SNR. .Page No. 561Soln. :Given : SNR = 30 dB, N = 10 Desired value of SNR =42 dB.1. To find “N” for SNR = 42 dB With increase in N by 1 bit the value of SNR increases by 6 dB.Therefore to increase the valueof SNR by 12 dB it is necessary to increase N by 2. ∴ N = 10 + 2 = 12 ...Ans.2.Fractional increase in BW : BW of PCM system = N fs ∴ BW with N = 10 is given by, BW10 = × 10 fs = 5 fsand BW with N = 12 is given by, BW12= × 12 fs = 6 fs ∴ Change in BW = ∆ BW = 6 fs – 5 fs = fs ∴Fractional change in BW = = × 100 % = 20 %...Ans.Ex. 5.15.14 : A telephone signal with cutoff frequency of 4kHz is digitised into 8 bit PCM, sampled at Nyquist rate. Calculate the baseband transmission bandwidth andquantization S/N ratio. .Page No. 561Soln. :1. The Nyquist rate = 2 × 4 kHz = 8 kHz.
8. 8. Digital Communication (GTU) 58 Pulse Code Modulation2. Transmission bandwidth, BT = ⋅ N fs = × 8 × 8kHz = 32 kHz ...Ans.3. SNRq = (1.8 + 6N) dB = 1.8 + (6 × 8) = 49.8 dB ...Ans.Ex. 5.15.15 : In a singleintegration DM scheme the voice signal is sampled at a rate of 64 kHz. The maximum signal amplitude is 2 Volts.Voice signal bandwidth is 3.5 kHz. Determine the minimum value of step size to avoid slope overload andgranular noise power. .Page No. 561Soln. :Given : fs = 64 kHz, Amax = 2V, fm = 3.5 kHz1. Minimum stepsize to avoid slope overload : We know that Amax = ∴ δmin = = = 0.6872 Volt ...Ans.2. Granular noise power: Nq = × = × = 8.6 × 10– 3 W ...Ans.Ex. 5.15.16 : A television signal (video and audio) has a bandwidth of 4.5MHz. This signal is sampled, quantized and binary coded to obtain a PCM signal. 1. Determine the sampling rateif the signal is to be sampled at a rate 20% above the Nyquist rate. 2. If the samples are quantized into 1024levels, determine the number of binary pulses required to encode each sample. 3. Determine the binary pulse rate(bits per second) of binary coded signal and the minimum bandwidth required to transmit the signal. If abovelinear PCM system is converted to companded PCM, will the output bit rate changed ? Justify. .Page No. 562
9. 9. Digital Communication (GTU) 59 Pulse Code ModulationSoln. :Given : W = 4.5 MHz1. Sampling rate = 1.2× Nyquist rate = 1.2 × 2 W = 1.2 × 2 × 4.5 × 106 = 10.8 MHz ...Ans.2. Given that number of quantizationlevels, Q = 1024 But Q = 2N ∴ Number of binary pulse per word, N = log2 Q ∴ N = log2 1024 ∴ N = 10...Ans.3. Binary pulse rate (bit rate) = N fs = 10 × 10.8 MHz = 108 Mbps ...Ans. Bandwidth = bit rate = 54MHz ...Ans.Ex. 5.15.17 : A compact disc (CD) records audio signals digitally by using PCM. Assume the audiosignal bandwidth to be 15 kHz. 1. What is Nyquist rate ? 2. If the Nyquist samples are quantized into L =65,536 levels and then binary coded,determine the number of binary digits required to encode a sample. 3.Determine the number of binary digits per second (bit/s) required to encode the audio signal. 4. For practicalreasons, the signals are sampled at a rate well above Nyquist rate at 44,100 samples per second. If L = 65,536,determine number of bits per second required to encode the signal and transmission bandwidth of encoded signal..Page No. 562.Soln. :Given : W = 15 kHz1. Nyquist rate = 2 W = 2 × 15 kHz = 30 kHz. ...Ans.2. Number ofquantization levels Q = 65,536 we know that Q = 2N ∴ 2N = 65,536 ∴ N = 16 bits. ...Ans.∴ Number of bitsto encode each sample is N = 16.3. Number of bits per second = Number of samples/sec. × Number ofbits/sample = 15 × 103 × 16 = 240 k bits/sec. ...Ans.4. Practical sampling rate fs = 44.1 kHz ∴ Number of bitsper second = 44.1 × 16 × 103 = 705.6k bits/sec. ∴ Transmission bandwidth B = = = 352.8 kHz. ...Ans.
10. 10. Digital Communication (GTU) 510 Pulse Code ModulationEx. 5.15.18 : A Delta Modulator systemoperates at 3 times Nyquist rate for signal with 3.3 kHz bandwidth. The quantisation step is 250 mV. Determinethe maximum amplitude of a 1 kHz input sinusoid for which the DM does not show slope overload. .Page No. 562.Soln. :Given : W = 3.3 kHz, fm = 1 kHz, δ = 250 mV, fs = 3 × 2W = 19.8 kHz.1. Let the maximumamplitude of 1 kHz input sinusoid be “A”. The condition to avoid the slope overload is A ≤ ∴ Maximum value ofA = = Substituting the values we get, Amax = ∴ Amax = 0.787Ex. 5.15.19 : An audio signal with highestfrequency component 3300 Hz is pulse code modulated with a sampling rate of 8000 samples/sec. The requiredsignaltoquantisation noise ratio is 40 dB. 1. What is the minimum number of uniform quantising levels needed ?2. What is the minimum number of bits per sample needed ? 3. Calculate the minimum number of bits per sampleneeded ? .Page No. 562.Soln. :Given : fs = 8000 samples/sec. , fm = 3300 Hz. = 40 dB1. To calculate Q and
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N : Assuming the input signal to be non sinusoidal we write S/N = (4.8 + 6 N) dB ∴ 40 = 4.8 + 6N ∴ N = 5.86≈ 6 ...Ans. Thus the number of bits per word is 6. Q = 2N = 26 = 64 ...Ans. Thus the number of quantizationlevel is 64.2. The signaling rate (r) : r = N fs = 6 × 8 × 103 = 48 kbits/sec.3. Transmission bandwidth : MinimumBW, BT = r = 24 kHz.Ex. 5.15.20 : Delta modulator (DM) gives output pulses + p (t) of – p (t). The output ofDM is + p (t) when instantaneous sample is larger than previous sample value and is – p(t) when instantaneoussample is smaller than previous sample value (last sample). The p(t) has 2 microsecond duration and 628 mVamplitude and repeats every 10 microsecond. Plot
11. 11. Digital Communication (GTU) 511 Pulse Code Modulation the input and output of DM on graph paper onebelow other to same scale if the input to DM is 1 Volt sine wave of frequency 10 kHz for one cycle of inputwave. What is the maximum frequency with 1 Volt amplitude that can be used in this system without slopeoverload distortion. .Page No. 562.Soln.Part I : To plot the DM signal The D. M. Signal is as shown in Fig. P.5.15.20. Fig. P. 5.15.20 : DM signalPart II : Maximum frequency :Given : A = 1V, fm = 10 kHz, fs = 1/Ts =1/10 µ s = 100 kHz. δ = 265 mV. The condition for avoiding the slope overload is A > ∴ 1 = ∴ ωn = 25.6 ×103 rad /sec.
12. 12. Digital Communication (GTU) 512 Pulse Code Modulation ∴ fmax = 4074.36 Hz ...Ans. This is themaximum input frequency without introducing any slope overload distortion.
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