Short Questions and Answers on Waveform Coding Techniques
Q.1. How is the ‘information’ transmitted in a PCM system?
Ans. In PCM, the ‘information’ is transmitted in the form of ‘code words’. The amplitude, width and position of the pulses at the PCM transmitter output are always constant.
Q.2. What is quantization?
Ans. It is the process of approximation. The samples are approximated or rounded off to the nearest quantization level.
Q.3. What is the relation between number of quantization levels and the number of bits per word?
Ans. Q = 2N
Q.4. What is quantization error?What is the maximum value?
Ans. It is the difference between quantized signal and original signal. Its maximum value is ± Δ/2.
Q.5. How can we reduce the quantization error?
Ans. By reducing the step size Δ. This is possible by increasing the number of quantization levels Q.
Q.6. What are the advantages of PCM?
Ans. (i) High noise immunity
(ii) Repeaters can be used
(iii) Storage and processing of the signal is possible
Q.7. What is the disadvantage of PCM?
Ans. Very large bandwidth requirement.
Q.8. What is the signaling rate in terms of N and fs?
Ans. Signaling rate = N fs
Q.9. Why is companding used?
Ans. In order to improve the signal to quantization noise ratio of weak signals.
Q.10. How bandwidth required for the DM signal is less than that of PCM?
Ans. In PCM, we have to transmit N bits per quantized sample but in DM, we have to transmit only 1 bit per sample. This reduces the bandwidth.
Q.11. What information do you transmit in DM system?
Ans. In delta modulation (DM), the information about the result of comparison between present and previous samples is transmitted.
Q.12. What is the cause of slope overload error in DM?
Ans. When the slope of analog input signal is much higher than the slope of approximated staircase signal, the slope overload error is observed.
Q.13. How can we reduce the slope overload error?
Ans. By increasing the sampling frequency fs or by increasing the step size Δ.
Q.14. What is the signaling rate of a DM system?
Ans. As only one bit is transmitted per sample, the signaling rate is equal to sampling rate fs.
Q.15. What is granular noise?
Ans. Granular noise is similar to the quantization noise in PCM.
Q.16. How can we reduce granular noise?
Ans. By reducing the step size Δ.
Q.17. Can we use DM practically as an alternative to PCM?
Ans. No, we can not use because in order to eliminate the slope overload, practically, the sampling rate fs is required to be so high that the bandwidth requirement of DM exceeds that of PCM.
Q.18. How is slope overload reduced in ADM?
Ans. In ADM, the step size is increased or reduced progressively to track the analog signal faithfully. This reduces the possibility of slope overload.
Q.19. Is it possible to use the baseband transmission in PCM?
Ans. Yes, it is possible.
Q.20. What are the advantages of digital representation of a signal?
Ans. The advantages of the digital representation are immunity to noise, regeneration of the coded signal is possible, communication can be kept secret or private due to the use of codes.
Q.21. What are the disadvantages of digital representation of a signal?
Ans. Some of the disadvantages are: increased transmission bandwidth and increased system complexity.
Q.22. Why is PCM not used for radio broadcasting?
Ans. Due to higher BW requirement and complex circuitry.
Q.23. The granular noise in ADM will be ………… than that in DM.
Ans. Higher
Q.24. In DM, we actually transmit ……………….
Ans. The encoded error signal.
Q.25. The predictive coding theory tells that ……………..
Ans. It is possible to reconstruct the original message signal by accumulating a series of error signals.
Q.26. For increasing the signal to quantization noise by 6 dB, we have to increase the number of bits per PCM word by …………..
Ans. 1
Q.27. Name different source coding techniques used in digital communication systems.
Ans. Various source coding techniques are pulse code modulation (PCM), delta modulation (DM), differential pulse code modulation (DPCM), adaptive delta modulation (ADM).
Q.28. Are the signal and noise separable in PCM?
Ans. Yes, the signal and noise are separable in PCM.
Q.29. Is it possible to use the repeaters (regenerators) in a PCM system?
Ans. Yes, it is possible to use repeaters which are actually regenerators. They can separate noise from the PCM signal and regenerate the original PCM signal.
Q.30. What is the advantage of using repeaters?
Ans. The advantage of using repeaters is reduced effect of noise.
Q.31. What is the relation between number of quantization levels and the number of digits per word?
Ans. The relation is Q = 2N, where, Q is the number of quantization levels and N is the number of digits per word.
Q.32. What is the relation between the quality of PCM signal and the number of digits per word?
Ans. The quality of signal will improve with increase in number of digits per word N.
Q.33. Why is quantization necessary?
Ans. If we do not use the quantization, then, we will have to convert each and every sampeld value into a separate digital word. This will need a large number of bits per word (N). This will increase the signaling rate (bits/sec), and hence the bandwidth requirement of the system.
Q.34. What is the maximum value of signal to quantization noise of a PCM system?
Ans. The maximum value of signal to quantization noise with a sinusoidal input signal is SNRQ(max) = (1.8 + 6N) dB.
Q.35. How is uniform quantization is different from non-uniform quantization?
Ans. The quantization is called as uniforms quantization if the step size remains constants throughout the input range. However, if the step size varies depending on the size of input, then, the quantization is known as non-uniform quantization.
Q.36. Where is the non-uniform quantization used?
Ans. Non-uniform quantization is used to quantize the speech and music signals, as they have low values of crest factor.
Q.37. What is the bandwidth of a PCM system?
Ans. The minimum bandwidth of a PCM system is half the signalling rate i.e., it is 1/2 N fs .
Q.38. For a PCM system having 128 quantization levels and sampling frequency of 8 kHz, what is the number of bits per word, signaling rate and minimum bandwidth?
Ans. Q = 128,\ N = 7 bits/word, signaling rate is 56 kbits/sec, and bandwidth BT = 28 kHz.
Q.39. What is the relation between power contents and crest factor of a signal?
Ans. The relation between normalized signal power and its crest factor is given by, P = 1/CF 2
Q.40. State the types of nonuniform quantizers.
Ans. There are two types namely-the μ-law and A-law companding.
Q.41. What is the difference between the m-law and A-law compressor characteristics?
Ans. The μ-law compressor characteristics is linear characteristics for smaller amplitudes and it is logarithmic for the large values of the input signal. The A-law compressor characteristics is made up of piecewise lienar segments for the low level inputs and piece-wise logarithmic segments for the high level inputs.
Q.42. What is the main disadvantage of a PCM system?
Ans. The main disadvantage of a PCM system is its high signaling rate and therefore large bandwidth requirement. This is because each quantized sample is converted into an N bit digital word.
Q.43. How is this disadvantage overcome using the linear delta modulation?
Ans. This disadvantage is overcome using the linear delta modulation because in delta modulation system, only one bit is transmitted per clock cycle. Instead of transmitting the actual sampled value (as in case of PCM), the information about the comparison of present value and previous value of the signal is transmitted. The signaling rate of delta modulation is thus only ‘fs’ instead of Nfs in case of PCM.
Q.44. Practically is the signaling rate of DM system less than that of the PCM system.
Ans. No, practically, the signaling rate of DM system is higher than that of a PCM system.
Q.45. What is the condition for avoiding the slope overload error?
Ans. The condition for avoiding the slope overload error is that the slope of the DM approximated signal x(t), should be greater than or equal to the continuous time signal x(t). Mathematically this condition is stated as:
A ≤ δ fs /2π fm
Q.46. What is the difference between DM and ADM?
Ans. The step size of the ADM approximated signal is variable. This is the difference between DM and ADM systems.
Q.47. State the advantages of ADM over DM.
Ans. Reduction in slope overload error, improvement in signal to quantization noise ratio, wide dynamic range, better utilization of bandwidth as compared to DM are some of the important advantages of ADM.
Q.48. What is the operating principle of differential PCM system?
Ans. In the DPCM system, the error between a signal and its approximate version is encoded and transmitted, instead of transmitting the original signal.
Q.49. Why is the signaling rate of DPCM is less than that of a PCM?
Ans. As the number of bits required for encoding an error signal is always less than the number of bits needed to encode the actual input signal, the bit rate (signaling rate) of DPCM is always less than that of a PCM signal.
Q.50. What is the role of a predictor in DPCM system?
Ans. A predictor will predict the value of the next sample by taking into account all the previous sample values. This will reduce the error to a very small value which needs a small number of bits for its encoding. This will reduce the signaling rate of DPCM to a great extent.
Q.51. What is the spectrum occupied by the speech signal?
Ans. Typically, the speech signal occupies the range from 200 Hz to 3200 Hz.