Waveform Coding
Raghudathesh G P
Asst Professor
DIGITAL COMMUNICATION (VTU) - 10EC61 UNIT – 2:
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WAVEFORM CODING TECHNIQUES: PAM, TDM. Waveform Coding Techniques, PCM, Quantization noise and SNR, robust quantization. 6 Hours TEXT BOOK: 1. Digital communications, Simon Haykin, John Wiley India Pvt. Ltd, 2008.
Special Thanks To:
India
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REFERENCE BOOKS: 1. Digital and Analog communication systems, Simon Haykin, John Wildy India Lts, 2008 2. An introduction to Analog and Digital Communication, K. Sam Shanmugam, John Wiley Pvt. Ltd, 2008. 3. Digital communications - Bernard Sklar: Pearson education 2007
1. Faculty (Chronological): Arunkumar (STJIT), Raviteja B (GMIT). 2. Students: Shubham S Dhivagnya (6th sem GMIT)
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PREPARED BY:
RAGHUDATHESH G P
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Asst Prof ECE Dept, GMIT Davangere 577004 Cell: +917411459249 Mail:
[email protected]
Quotes:
A picture is a poem without words. Every time you paint a portrait you lose a friend. The days you work are the best days. When you give yourself, you receive more than you give. Art is the objectification of feeling. Some people drink from the fountain of knowledge, others just gargle.
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Page No - 1
Waveform Coding
Raghudathesh G P
Asst Professor
WAVEFORM CODING TECHNIQUES Introduction: Analog waveforms or signals are sampled into pulses. These analog pulses are also called carrier pulses. When amplitude of these pulses varies according to amplitude of analog waveform, it becomes pulse amplitude modulation (PAM).
The pulses of PAM can be converted to digital form. Thus the analog waveform is converted to sequence of binary digital format. It is also called pulse code modulation. When the analog pulse is converted to digital form, it is quantized to nearest digital level. This introduces quantization noise in the signal.
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The digital waveform coding techniques are always selected on the basis of quantization noise and bandwidth of transmission.
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Pulse Amplitude Modulation (PAM):
Definition: In pulse-amplitude modulation (PAM), the amplitude of a carrier consisting of a periodic train of rectangular pulses is varied in proportion to sample values of a message signal. In this type modulation, the pulse duration is held constant.
The carrier is a train of rectangular pulses. By making the amplitude of each rectangular pulse the same as the value of the message signal at the leading edge of the pulse, PAM so defined is exactly the same as flat-top sampling. Thus PAM Wave s(t), is defined as
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Here
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g(nTs) = Sample values of the message (modulating) signal g(t). Ts = the sampling period. V(t) = rectangular pulse.
Transmission bandwidth requirement of PAM: According to the definition given before in terms of rectangular pulses, we would require a very wide band of frequencies to transmit PAM.
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Page No - 2
Waveform Coding
Raghudathesh G P
Asst Professor
Disadvantages of PAM:
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1. As we have seen just now, the bandwidth needed for transmission of PAM signal is very large compared to its maximum frequency content. 2. The amplitude of PAM pulses varies according to modulating signal Therefore interference of noise is maximum for the PAM signal and this noise cannot be removed very easily. 3. Since amplitude of PAM signal varies, this also varies the peak power required by the transmitter with modulating signal.
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Other Forms of Pulse Modulation:
Figure below shows various pulse modulation methods.
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There are two more types of pulse modulation other than PAM: 1. Pulse Duration Modulation (PDM): In this technique the width of the pulse changes according to amplitude of the modulating signal at sampling instant. 2. Pulse position Modulation(PPM): In this technique the position of the pulse changes according to amplitude of the modulating signal at sampling instant. Pulse position modulation (PPM) and pulse duration modulation (PDM or PWM) both modulate the time parameter of the pulses. PPM has fixed width pulses where as width of PDM pulses varies. Both the methods are of constant amplitude.
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Page No - 3
Waveform Coding
Raghudathesh G P
Asst Professor
Comparison between Various Pulse Modulation Methods: Sl No.
PAM
PWM
PPM
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Bandwidth of transmission channel depends on rise time of the pulse The instantaneous power of the transmitter varies.
Noise interference is high.
Noise interference is minimum. Simple to implement. Similar to frequency modulation.
System is complex. Similar to amplitude modulation
Bandwidth of transmission channel depends on rising time of the pulse. The instantaneous power of the transmitter remains constant. Noise interference is minimum. Simple to implement. Similar to phase modulation.
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The bandwidth of the transmission channel depends on width of the pulse. The instantaneous power of the transmitter varies.
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Amplitude of the pulse is Width of the pulse is The relative position of pulse proportional to amplitude of proportional to amplitude of is proportional to amplitude modulating signal. modulating signal. of modulating signal.
TIME-DIVISION MULTIPLEXING (PAM System):
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An important feature of pulse-amplitude modulation is a conservation of time. Which means, for a given message signal, transmission of the associated PAM wave engages the communication channel for only a fraction of the sampling interval on a periodic basis. Due to above reason, some of the time interval between adjacent pulses of the PAM wave is cleared for use by other independent message signals on a time-shared basis. By so doing, we obtain a time-division multiplex system (TDM), which enables the joint utilization of a common channel by a plurality of independent message signals without mutual interference.
The concept of TDM is illustrated by the block diagram shown in Figure below.
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Raghudathesh G P
Asst Professor
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Waveform Coding
Each input message signal is first restricted in bandwidth by a low-pass pre-alias filter to remove the frequencies that are nonessential to an adequate signal representation.
The pre-alias filter outputs are then applied to a commutator, which is usually implemented using electronic switching circuitry.
The function of the commutator is two-fold: 1. To take a narrow sample of each of the N input messages at a rate fs that is slightly higher than 2W, where W is the cutoff frequency of the pre-alias filter. 2. To sequentially interleave these N samples inside a sampling interval T s = 1/fs. This latter function is the essence of the time-division multiplexing operation.
Following the commutation process, the multiplexed signal is applied to a pulseamplitude modulator, the purpose of which is to transform-the multiplexed signal into a form suitable for transmission over the communication channel.
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Suppose that the N message signals to be multiplexed have similar spectral properties. Then the sampling rate for each message signal is determined accordance with the sampling theorem.
Let Ts = sampling period determined for each message signal. Tx = time spacing between adjacent samples in the time-multiplexed signal. Thus,
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Thus, use of time-division multiplexing introduces a bandwidth expansion factor N, because the scheme must squeeze N samples derived from N independent message signals into a time slot equal to one sampling interval.
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Page No - 5
Waveform Coding
Raghudathesh G P
Asst Professor
At the receiving end of the system, the received signal is applied to a pulse amplitude demodulator, which performs the reverse operation of the pulse amplitude modulator.
The short pulses produced at the pulse demodulator output are distributed to the appropriate low-pass reconstruction filters by means of a decommutator, which operates in synchronism with the commutator in the transmitter.
The synchronization is essential for a satisfactory operation of the TDM system, and provisions have to be made for it.
The figure below illustrating TDM for 2 message signals.
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Salient Features of TDM:
Full available channel bandwidth can be utilized for each channel. Intermodulation distortion is absent. TDM circuitry is not very complex. The problem of crosstalk is not severe.
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1. 2. 3. 4.
Drawbacks of TDM: 1. Synchronization is essential for proper operation. 2. Due to slow narrowband fading, all the TDM channels may get wiped out.
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Page No - 6
Waveform Coding
Raghudathesh G P
Asst Professor
Formulas related to TDM: 1. Spacing between two samples 2. Number of pulses per second
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3. Number of pulses per second is called as Signaling rate r Also thus 4. Signaling rate = 2 × Transmission Bandwidth. 5. Speed of the commutator in revolution per second(rps) = 2W Here W = minimum bandwidth of the message signal 6. Speed of commutator (samples/sec) = Total number of segments × Speed of commutator(rps) 7. Minimum transmission bandwidth = ½ [sum of nyquist rate] 8. Angle of separation b/w each segment(pole)= 360/Total number of segments(n).
Problems on TDM
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Solution
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1. A Signal m1(t) is band limited to 3.6khz and three signals m2(t), m3(t) and m4(t) are band limited to 1.2 kHz each. These signals are to be transmitted by means of TDM. Sketch set up a scheme for realizing this multiplexing requirement with each signal sampled at its nyquist rate. Determine the speed of commutator in samples per second.
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Message Signal
m1(t) m2(t) m3(t) m4(t)
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Bandwidth
Nyquist rate fs = 2W
Number of segments N
Angle of separation of corresponding segments = 3600/N
3.6 kHz 1.2 kHz 1.2 kHz 1.2 kHz
7.2 kHz 2.4 kHz 2.4 kHz 2.4 kHz
3 1 1 1
1200 3600 3600 3600
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Page No - 7
Waveform Coding
Raghudathesh G P
m1(t)
m1(t)
2.4 kHz
PAM
Channel
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m4(t)
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m2(t) m1(t)
Commutator Speed in rps = 2 × W = 2 × 1.2 kHz = 2400 rps.
Thus if commutator is rotated at 2400 revolution/second then in each revolution, we get one sample each for m2(t), m3(t) and m4(t) and three sample for m1(t).
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Minimum transmission bandwidth Minimum transmission bandwidth = ½ [sum of nyquist rate] = ½ [7.2 kHz + 2.4 kHz + 2.4 kHz + 2.4 kHz] = 7.2 kHz.
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Speed of commutator in samples/second Speed of commutator (samples/sec) = Total number of segments × Speed of commutator (rps) = 6 × 2400 = 14400 samples/sec
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2. Eight message signals are sampled and time multiplexed using PAM. The time multiplexed signal is passed through a LPF before transmission. Six of the input signals have a bandwidth of 4 kHz and the other two are band-limited to 12 kHz. (a) What is the minimum overall sampling rate if all the messages are sampled at the same rate? (b) Design an asynchronous TDM for this application. (c) Compare the transmission bandwidth requirements of parts (a) and (b). Solution:
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Waveform Coding
Raghudathesh G P
Asst Professor
(a) If all the messages are sampled at the same rate, then f, = 2 x 12,000 = 24,000 samples/second per message. Hence, overall sampling rate = 8 x 24, 000 = 192, 000 samples/second. (b) Let g1(t), g2(t) be the 12 kHz message signals and g3(t), g4(t), ….,g8(t) be the 4 kHz message signals. Also let the switch make 8000 rotations/second. It samples 4 kHz messages once per rotation and the 12 kHz messages three times per rotation. g1(t)
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g3(t) g8(t)
g2(t) g4(t)
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Commutator
g2(t)
PAM
g1(t) g6(t) g5(t)
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g1(t) g2(t)
Channel
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g7(t)
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(c) Synchronous sampling rate = 192,000 samples/second. BW = ½[Synchronous sampling rate] = ½[192,000] = 96 kHz Asynchronous sampling rate = 8000 x 12 = 96,000 samples/second. BW = ½[Synchronous sampling rate] = ½[96,000] = 48 kHz.
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3. A signal x1(t) is bandlimited to 3 kHz. There are three more signals x2(t), x3(t) and x4(t) which are bandlimited to 1 kHz each. These signals are to be transmitted by a TDM system. (i) Design a TDM scheme where each signal is sampled at its Nyquist rate. (ii) What must be the speed of the commutator? (iii) Calculate the minimum transmission bandwidth of the channel. Solution: (i) Table below shows different message signals with corresponding Nyquist rates. Message Signal
Bandwidth
Nyquist rate fs = 2W
Number of segments N
Angle of separation of corresponding segments = 3600/N
x1(t) x2(t) x3(t) x4(t)
3 kHz 1 kHz 1 kHz 1 kHz
6 kHz 2 kHz 2 kHz 2 kHz
3 1 1 1
1200 3600 3600 3600
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Page No - 9
Waveform Coding
Raghudathesh G P
Asst Professor
If the sampling commutator rotates at the rate of 2000 rotations per second then the signals x2(t), x3(t) and x4(t) will be sampled at their Nyquist rate. But, we have to sample x1(t) also at its Nyquist rate which is three times higher than that of the other three. In order to achieve this, we should sample x1(t) three times in one rotation of the commutator. Therefore, the commutator must have atleast 6 poles connected to the signals as shown in figure
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x3(t) x1(t)
2000 rpm
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x1(t)
PAM
x4(t) x1(t)
x2(t)
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Channel
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(ii) The speed of rotation of the commutator is 2000 rotations/sec. (iii) Number of samples produced per second is calculated as under: X1(t) produces 3 x 2000 = 6000 samples/sec. x2(t), x3(t) and x4(t) produce 2000 samples/sec. each. Therefore, number of samples per second = 6000 + (3 x 2000) = 12000 samples/sec. Signaling rate = 12000 samples/sec. (iv) The minimum channel bandwidth will be BW = ½[Synchronous sampling rate] = ½[12000] = 6000 Hz.
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4. Six message signals each of bandwidth 5 kHz are time division multiplexed and transmitted. Determine the signaling rate and the minimum channel bandwidth of the PAM/TDM channel. Solution: The number of channels N = 6 Bandwidth of each channel, fm = 5 kHz Minimum sampling rate = 2 x 5 kHz = 10 kHz Signaling rate = Number of bits per second = 6 x 10 kHz = 60 K bits/sec. Minimum, channel bandwidth to avoid cross talk in PAM/TDM is, BW Nfm = 6 x 5 kHz = 30 kHz.
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Page No - 10
Waveform Coding
Raghudathesh G P
Asst Professor
Now, time taken by the commutator for 1 rotation
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Number of pulses produced in 1 rotation = 24 + 1 = 25
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5. Twenty-four voice signals are sampled uniformly and then time division multiplexed. The sampling operation uses flat top samples with 1 µs duration. The multiplexing operation includes provision for synchronization by adding an extra pulse of appropriate amplitude and 1 µs duration. The highest frequency component of each voice signal is 3.4 kHz. (i) Assuming a sampling rate of 8 kHz, calculate the spacing between successive pulses of the multiplexed signal. (ii) Repeat (i) assuming the use of Nyquist rate sampling. Solution: (i) Given that Sampling rate = 8 kHz = 8000 samples/sec. There are 24 voice signals + 1 synchronizing pulse. Pulse width of each voice channel and synchronizing pulse is 1 µs.
Therefore, the leading edges of the pulses are at
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below
distance as shown in figure
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Hence, spacing between successive pulses = 5 - 1 = 4 µs (ii) Nyquist rate of sampling = 2 x 3.4 kHz = 6.8 kHz. This means that 6800 samples are produced per second. One rotation of commutator takes 1/6800 = 147 µs time. Therefore, 147 µ sec corresponds to 25 pulses. Therefore, 1 pulse corresponds to 5.88 t sec. As the pulse width of each pulse is 1 µ sec, the spacing between adjacent pulses will be 4.88 µsec and if we assume τ= 0 then the spacing between the adjacent pulses will be 5.88 µsec.
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Page No - 11
Waveform Coding
Raghudathesh G P
Asst Professor
Pulse Code Modulation (PCM) Introduction: Pulse-code modulation (PCM) provides one method to generate a coded version of the signal.
Broadly speaking PCM is essentially analog-to-digital conversion where the analog samples are represented by digital words in a serial bit stream.
The use of digital representation of analog signals (e.g., voice, video) offers us the following advantages: 1. Ruggedness to transmission noise and interference. 2. Relatively inexpensive digital circuitry may be used extensively in the system. 3. Digital information can be encrypted for security, coded against errors and compressed to reduce storage and transmission costs. 4. The possibility of a uniform format for different kinds of pulse signals. 5. In long-distance digital telephone systems, a clean PCM waveform can be generated using regenerative repeaters kept at regular locations all along the communication route. 6. PCM signals derived from all types of analog sources (audio, video, etc.) may be interleaved with data signals (e.g., from digital computers) and transmitted over a common channel. This technique is called time-division multiplexing. 7. The noise performance of a digital communication system can be superior to that of an analog communication system.
These advantages, however, are attained at the cost of increased transmission bandwidth requirement and increased system complexity.
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Pulse Code Modulation (PCM): Pulse-code modulation systems are complex in that the message signal is subjected to a large number of operations.
The essential operations in the transmitter of a PCM system are sampling, quantizing, and encoding, as shown in the figure below.
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Page No - 12
Waveform Coding
Raghudathesh G P
Asst Professor
The sampling, quantizing, and encoding operations are, usually performed in the same circuit, which is called an analog-to-digital converter.
Regeneration of impaired signals occurs at intermediate points along the transmission path (channel) as indicated in the Figure below.
At the receiver, the essential operations consist of one last stage of regeneration followed by decoding, then demodulation of the train of quantized samples, as in the Figure below.
The operations of decoding and reconstruction are usually performed in the same circuit, called a digital-to-analog converter.
When time-division multiplexing is used, it becomes necessary to synchronize the receiver to the transmitter for the overall system to operate satisfactorily.
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Note:
Pulse-code modulation is not modulation in the conventional sense.
The term "modulation" usually refers to the variation of some characteristic of a carrier wave in accordance with an information-bearing signal.
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The only part of pulse-code modulation that conforms to this definition sampling.
The subsequent use of quantization, which is basic to pulse-code modulation, introduces a signal distortion that has no counterpart in conventional modulation.
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Page No - 13
Waveform Coding
Raghudathesh G P
Asst Professor
Basic signal-processing operations involved in PCM: 1. Sampling: The incoming message wave is sampled with a train of narrow rectangular pulses so as to closely approximate the instantaneous sampling process.
To ensure perfect reconstruction of the message at the receiver, the sampling rate must be greater than twice the highest frequency component W of the message wave (in accordance with the sampling theorem).
Low-pass pre-alias filter is used at the front end of the sampler in order exclude frequencies greater than W before sampling.
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2. Quantizing:
An analog signal, such as voice, has a continuous range of amplitudes and therefore its samples cover a continuous amplitude range.
It is not necessary in fact to transmit the exact amplitudes of the samples. Any human sense (the ear or the eye), as ultimate receiver, can detect only finite intensity differences.
The existence of a finite number of discrete amplitude levels is a basic condition of PCM.
Definition: The conversion of an analog (continuous) sample of the signal into a digital (discrete) form is called the quantizing process. Graphically, the quantizing process means that a straight line representing the relation between the input and the output of a linear analog system is replaced by a transfer characteristic that is staircase-like in appearance. Figure below depicts one such characteristic. The quantizing process has a two-fold effect: 1. the peak-to-peak range of input sample values is subdivided into a finite set of decision levels or decision thresholds that are aligned with the "risers" of the staircase, and 2. The output is assigned a discrete value selected from a finite set of representation levels or reconstruction values that are aligned with the "treads" of the staircase.
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For a uniform quantizer, whose characterized as in Figure below, the separation between the decision thresholds and the separation between the representation levels of the quantizer have a common value called the step size.
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Waveform Coding
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Symmetric quantizer of the midtread type:
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According to the staircase-like transfer characteristic of Figure above, the decision thresholds of the quantizer are located at
,
,
, . . . , and the
Symmetric quantizer of the midriser type:
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representation levels are located at 0, , , . . . , where is the step size. A uniform quantizer characterized in this way is referred to as a symmetric quantizer of the midtread type, because the origin lies in the middle of a tread of the staircase.
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Page No - 15
Waveform Coding
Raghudathesh G P
Asst Professor
Figure above shows another staircase-like transfer characteristic, in which the decision thresholds of the quantizer are located at 0, , , . . . , and the representation levels are located at
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, . . . , where
is again
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the step size. A uniform quantizer having this second characteristic is referred to as a symmetric quantizer of the midriser type, because in this case the origin lies in the middle of a riser of the staircase. A quantizer of the midtread or midriser type, as defined, is memoryless in the quantizer output is determined only by the value of a corresponding sample, independently of earlier (or later) analog samples applied to the input.
The memoryless quantizer is the simplest and most often used quantizer. The transfer characteristics of Symmetric quantizer of the midtread type and Symmetric quantizer of the midriser type, we have included a parameter labeled the overload level, the absolute value of which is one half of peak-to-peak range of input sample values. The number of intervals into which the peak-to-peak excursion is divided, or equivalently the number of representation levels, is equal to twice the absolute value of the overload level divided by the step size.
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Thus, for an analog input sample that lies anywhere inside an interval of either transfer characteristic, the quantizer produces a discrete output equal to the midvalue of the pair of decision thresh in question.
In so doing, however, a quantization error is introduced, the value of which equals the difference between the output and input values of the quantizer.
Both figures show the variations of the quantization error with the input for the two uniform quantizer types. In both cases, we see that the maximum instantaneous value of this error is half of one step size, total range of variation is from minus half a step to plus half a step.
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3. Encoding:
An encoder translates quantized samples into digital codewords. A particular arrangement of symbols used in a code to represent a single value of the discrete set is called a code-word or character.
There can be different kinds of code like:
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Page No - 16
Waveform Coding
Raghudathesh G P
Asst Professor
Binary code: Each symbol may be either of two distinct values or kinds, such as the presence or absence of a pulse. The two symbols of a binary code customarily denoted as 0 and 1. Ternary code: Each symbol may be one of three distinct values or kinds. So on for other codes. Generally binary codes are used due to maximum advantage over the effects of noise in a transmission medium is obtained by using a binary code, because a binary symbol withstands a relatively high level of noise and is easy to regenerate.
In a binary code, each code-word consists of n bits. Then, such a code, we may represent a total of 2n distinct numbers.
Ex., a sample quantized into one of 24 = 16 levels may be represented by a 4-bit -word. There are several formats (waveforms) for the representation of binary sequences produced by analog-to-digital conversion (or by other sources). Figure below depicts two such formats.
In Figure below, binary symbol 1 is represented by a pulse of constant amplitude for the duration of one bit, and symbol 0 is represented by switching off the pulse for the same duration. This format is called nonreturn-to-zero unipolar signal, or on-off signal.
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In Figure below, symbols 1 and 0 are represented by pulses of positive and negative amplitude, respectively with each pulse occupying one complete bit duration. This second format is called a nonreturn-to-zero polar signal.
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4. Regeneration:
The most important feature of PCM systems lies in the ability to control the effects of distortion and noise produced by transmitting a PCM wave through a channel.
Shape of the pulse is affected by two mechanism: 1. Unwanted electrical noise or other disturbances. 2. Nonideal Transfer function of the transmission medium.
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Waveform Coding
Raghudathesh G P
Asst Professor
Figure below shows the pulse distortion over a distance.
This capability is accomplished by reconstructing the PCM wave by means of a chain of regenerative repeaters located at sufficiently close spacing along the transmission route.
Figure below show the block diagram of regenerative repeater.
Three basic functions are performed by a regenerative repeater are: 1. Equalization: The equalizer shapes the received pulses so as to compensate for the effects of amplitude and phase distortions produced by imperfections in the transmission characteristics of the channel. 2. Timing: The timing circuit provides a periodic pulse train, derived from the received pulses, for sampling the equalized pulses at the instants of time where the signal-to-noise ratio is a maximum. 3. Decision making: The decision device is enabled when, at the sampling time determined by the timing circuit, the amplitude of the equalized pulse plus noise exceeds a predetermined voltage level.
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Scenario: In a PCM system with on-off signaling, the repeater makes a decision in each bit interval as to whether or not a pulse is present.
Operation: If the decision is "yes," a clean new pulse is transmitted to the next repeater. If, on the other hand, the decision is "no," a clean base line is transmitted. In this way, the
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Waveform Coding
Asst Professor
accumulation of distortion and noise in a repeater span is completely removed, provided that the disturbance is not too large to cause an error in the decision-making process. Ideally, except for delay, the regenerated signal is exactly the same as the signal originally transmitted. The regenerated signal departs from the original signal for 2 main reasons: 1. The presence of channel noise and interference causes the repeater to make wrong decisions occasionally, thereby introducing bit errors into the regenerated signal; 2. If the spacing between received pulses deviates from its assigned value, a jitter is introduced into the regenerated pulse position, thereby causing distortion.
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The first operation in the receiver is to regenerate (i.e., reshape and clean up) received pulses. These clean pulses are then regrouped into code-words decoded (i.e., mapped back) into a quantized PAM signal. The decoding process involves generating a pulse the amplitude of which is the linear sum of the pulses in the code-word, with each pulse weighted by its place-value (2°, 21 , 22, 23, . . .) in the code.
6. Reconstruction:
Final operation in the receiver is to recover the analog signal. This is done by passing the decoder output through a low-pass reconstruction filter whose cutoff frequency is equal to the message bandwidth W. Assuming that the transmission path is error-free, the recovered signal includes no noise with the exception of the initial distortion introduced by the quantization process.
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5. Decoding:
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7. Multiplexing and Synchronization:
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Page No - 19
Raghudathesh G P
Asst Professor
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Waveform Coding
Figure above illustrates the concept of time-division multiplexing for three bit PCM signals.
At the receiver, decommutator is required to sort out various signals.
If the decommutator in the receiver not synchronized to that in the transmitter, then the messages can be interchanged. This could be most annoying in a telephone conversation; if we were talking to one person and the response was from another person.
In data communication system, loss of such synchronization could prove fatal. It can be seen from Figure that there are 3 X 3 = 9 information bits in each frame and a 10th bit is added for frame synchronization.
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The framing signal is a fixed pattern of l’s and 0’s in every 10th position and such an alternating pattern of 1’s and 0’s will he rarely formed in any other positions for two or three consecutive frames. Therefore, frame synchronization is fairly easy to achieve. As the number of independent message sources is increased, the time Intervalallotted to each message source has to be reduced, since the duration of the frame Ts = 1/fs fixed.
Thus, the width of pulses becomes too short. The process of generating and transmitting very narrow pulses are very difficult.
Hence, in practice, it is necessary to restrict the number of independent message sources that can be included within frame.
Some of the applications of PCM are: 1. Telephone system 2. Digital audio recording
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Waveform Coding
Raghudathesh G P
Asst Professor
3. CD laser disks 4. Voice mail 5. Digital Video etc.
Advantages of PCM (Short):
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Advantages of PCM (Detailed):
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Robustness to noise and interference Efficient regeneration Efficient SNR and bandwidth trade-off Uniform format Secure
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1. 2. 3. 4. 5.
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1. Low Noise Susceptibility: The PCM signal is a digital waveform. Digital waveforms are less susceptible to interference and noise than analog signals. This is because a digital waveform does not have to reproduce the exact data being transmitted. A transmitted pulse that is close enough to the expected value of a binary one can be reliably reproduced into a binary one. This low noise susceptibility allows PCM signals to transmit farther than analog signals without signal degradation, information loss, and distortion. 2. Repeatability: A PCM signal can be received by a repeater device that decodes the data and retransmits it. This allows PCM signals to be sent very long distances without data corruption. Repeaters must be placed close enough to the signal source so that extreme noise does not corrupt the signal. Noise does not accumulate even after many passes through multiple repeaters. This is because the signal is completely regenerated by each repeater, making it noise-free at the start of each repeated transmission. 3. Storage: A PCM waveform may be saved for later recreation or playback. Since PCM data is digital in origin, it can be stored using a computer or similar device. An example of a consumer device that stores PCM data is the Digital Versatile Disc (DVD) technology. The audio portion of a DVD movie is encoded using PCM with a sampling rate as high as 192 thousand samples per second. This PCM stream can be piped directly to an amplifier using a digital audio cable, where it is then decoded into an audible signal. 4. Encoded Signal: A PCM signal can be modulated in such a way that only a specific decoder can make sense of the underlying data. This is useful when the data being sent requires a level of security. The transmitter and receiver each have circuitry that is analogous to a dictionary. This circuit maps the binary pulse-codes to their definitions.
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When a pulse-code is received, the receiver looks up the meaning in the dictionary. Anyone who intercepted the PCM signal would be left with meaningless binary data.
Disadvantages of PCM:
P
1. Complex Circuitry 2. Requires large bandwidth 3. Synchronization is required between transmitter and receiver
Quantization:
Definition1: The process of transforming sampled amplitude values of a message signal into a discrete amplitude value is referred to as Quantization.
Definition2: In digital signal processing, quantization is the process of approximating a continuous range of values (or a very large set of possible discrete values) by a relativelysmall set of discrete symbols or integer values.
The quantization Process has a two-fold effect: 1. The peak-to-peak range of the input sample values is subdivided into a finite set of decision levels or decision thresholds that are aligned with the risers of the staircase, and 2. The output is assigned a discrete value selected from a finite set of representation levels that are aligned with the treads of the staircase.
A common use of quantization is in the conversion of a discrete signal (a sampled continuous signal) into a digital signal by quantizing. Both the steps (sampling and quantizing) are performed in analog-to-digital converters with the quantization level specified in bits.
H
TH ES
DA
GH U
G
A specific example would be compact disc (CD) audio which is sampled at 44,100 Hz and quantized with 16 bits (2 bytes) which can be one of 65,536 (i.e. 216) possible values per sample.
A quantizer is memory less in that the quantizer output is determined only by the value of a corresponding input sample, independently of earlier analog samples applied to the input.
RA
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Asst Professor
H
G
P
Waveform Coding
It is classified into 2 types: 1. Uniform quantization: Here step size remains constant through the input range. 2. Non-uniform quantization: Here step size varies according to the input signal values and mostly the relation (variation) is logarithmic.
GH U
DA
TH ES
Quantization Process can be Classified on the basis of step size as:
Types of Uniform quantization based on based on I/P - O/P Characteristics: Mid-Rise type Quantizer: It has even number of Quantization levels. In the stair case like graph, the origin lies in the middle of the rise portion.
RA
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Mid-Tread type Quantizer: It has odd number of Quantization levels. In the stair case like graph, the origin lies the middle of the tread portion
GH U
DA
TH ES
H
G
P
Waveform Coding
Working Principle of Quantizer: Consider a uniform quantizer of midrise type. Figure below shows the transfer characteristics of a uniform quantizer of midrise type.
RA
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Raghudathesh G P
Asst Professor
TH ES
H
G
P
Waveform Coding
In figure let us assume that the input to the quantizer x (nTs) varies from - 4Δ to + 4Δ. Thus, the peak to peak value of x (nT s) will be between - 4Δ to + 4Δ. Here ‘Δ’ is the step size.
Thus, input x (nTs) can take any value between - 4Δ to + 4Δ.
Now, the fixed digital levels are available at
DA
. These levels are
available at quantizer because of its characteristics. Hence, according to figure above
GH U
If x (nTs) = 4Δ then xq (nT s) =
.
and if x (nT s) = - 4Δ then xq (nTs) =
.
Thus, it may he observed from figure (b) that maximum quantization error would be ± .
Thus quantization error may be expressed as
RA
-------- (1)
Scenario: Let either or
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, the quantizer will assign any one of the nearest binary levels i.e., .
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Say if is assigned then quantization error will be,
If
If
, then , then
.
G
P
Also from figure (a) we can see that,
Thus, from above situation we see that maximum quantization error will be
TH ES
H
Hence Maximum quantization error will be,
.
-------- (2)
Quantization Noise/Error in PCM:
Here we derive an expression for quantization noise (i.e., error) in a PCM system for linear quantization or uniform quantization.
Because of quantization, inherent errors introduced in the signal. This error is called quantization error. The quantization error is given as
GH U
DA
Assuming that the input x(nT s) to a linear or uniform quantizer has continuous amplitude in the range -xmax to +xmax.
RA
------- (1)
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Asst Professor
TH ES
H
G
P
Waveform Coding
From figure above, it may be observed that the total excursion of input x(nT s) is mapped into 'q' levels on vertical axis.
This means that when input is 4Δ, output is 7Δ/2 and when input is - 4 Δ, output is -7Δ/2. Thus, +xmax represents 7Δ/2 and - xmax represents -7Δ/2.
Therefore, the total amplitude range is given by,
------- (3)
Again, now if signal x(t) is normalized to minimum and maximum values equal to 1, then we have, xmax = 1 and - xmax = - 1. Therefore, step side would be,
RA
----- (2)
Now, if this total amplitude range is divided into 'q' levels of quantizer, then the step size ' Δ ' will be,
GH U
DA
(For normalized signal)
-------- (4)
Now, if step size 'Δ' is considered as sufficiently small, then it may be assumed that the quantization error will be an uniformly distributed random variable. We know that the maximum quantization error is given as,
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------- (5)
Hence, over the interval
quantization error may be assumed as an uniformly
TH ES
H
G
P
distributed random variable.
Figure (a) above shows a uniformly distributed random variable 'X' over an interval (a, b).
PDF of uniformly distributed random variable 'X' is given as
DA
Thus, with the help of above equation, the probability density function (PDF) for quantization error ` ' may be defined as
------- (7)
RA
GH U
-------- (6)
Also, from figure (b), it may be observed that quantization error `ε' has zero average value. In other words, the mean 'mε ' of the quantization error is zero.
Now, the signal to quantization noise ratio of the quantizer is defined as, -------- (8)
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If type of signal at input i.e., x(t) is known, then it is possible to calculate signal power. The noise power is expressed as, ------- (9)
Here, As, here noise is defined by random varible 'ε', and PDF fε(ε) therefore, its mean square value is given as,
G
P
= the mean square value of noise voltage.
--------- (11)
Using equation (7), above equation may be written as,
GH U
DA
The mean square value of a random variable 'ε' is expressed as,
TH ES
H
--------- (10)
Thus,
RA
------- (12)
If load resistance, R = 1 Ω, then the noise power is normalized Thus,
----------- (13)
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Hence, above is the value for Normalized noise power or Quantization noise power Or Quantization error (in terms of power) for linear quantization.
Signal to Quantization Noise Ratio for Linear Quantization: In PCM system for linear quantization the signal to quantization noise ratio is given as,
But, In PCM the normalized noise power is Δ2/12. Therefore,
------- (3)
Relation between the number of bits v and quantization levels is given as, -------- (4)
Now, substituting the value of q from equation (2) in equation (4), we get
RA
-------- (2)
Now, the expression for the step size is given as,
GH U
--------- (1)
Assuming that input x(nTs) to a linear quantizer has continuous amplitude in the range xmax to +xmax. Therefore, total amplitude range is given as
DA
TH ES
H
G
P
--------- (5)
Now substituting the above equation in equation (1) we get,
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Let normalized signal power be denoted as ‘P’ thus,
P
---------- (6)
This is the required relation for signal to quantization noise ratio for linear quantization in a PCM system.
This expression shows that signal to noise power ratio of quantizer increases exponentially with increasing bits per sample.
Now, if we assume that inpur x(t) is normalized, i.e., xmax = 1. Also, if the destination signal power 'P' is normalized, i.e., P ≤ 1. Then, signal to quantization noise ratio will be,
H
TH ES
As, xmax = 1 and P≤1, the signal to noise ratio given by equation (8) is said to be normalized. Expressing the signal to noise ratio in decibels is given as below,
RA
GH U
-------- (7)
DA
G
-------- (8)
Thus, signal to quantization noise ratio for normalized values of power P and amplitude of input x(t) is given as above.
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Waveform Coding
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Expression for Signal to Quantization Noise Ratio for PCM that Employs Linear Quantization assuming the PCM system has a Sinusoidal Signal input: Assuming the Sinusoidal Signal input is a voltage signal having a peak voltage of A m. Power of the sinusoidal voltage signal is given as,
Here,
P
G
, thus ,
------ (2)
Signal to quantization noise ratio is given as,
In the above expression
------ (3)
and xmax =Am, thus,
RA
GH U
TH ES
Consider a case where R =1 and power P is normalized, then normalized power is given as,
DA
H
---------- (1)
Representing the above signal in dB we get,
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Waveform Coding
Raghudathesh G P
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Signal to Quantization Noise Ratio for Mid Tread Type:
P
--------- (3)
Let Quantizer input x represents the sampled value of random variable X with zero mean and variance .
The Quantizer is assumed to be uniform, symmetric and mid tread type.
Let xmax denotes absolute value of the overload level of the Quantizer and Δ represent Step size.
Then number of Quantization level L is given by
TH ES
H
-------- (2)
Equating equation (1) and (2) we get,
-------- (3)
RA
GH U
---------- (1)
For a binary code with a code-word of n bits we can have upto 2n representation level. As number of representation level for midtread quantizer is odd, thus L is given as below,
DA
G
The ratio is called the loading factor. To avoid significant overload distortion, the amplitude of the Quantizer input x extend from to , which corresponds to loading factor of 4. Thus with we can write equation (3) as ------- (4)
Now expression for signal to quantization noise ratio is given as,
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------ (5)
For larger value of n (typically n>6), we may approximate the result as
Hence expressing SNR in db
G
P
------ (6)
------ (7)
This formula states that each bit in codeword of a PCM system contributes 6db to the signal to noise ratio.
The equation (7) gives a good description of the noise performance of a PCM system provided that the following conditions are satisfied. 1. The Quantization error is uniformly distributed. 2. The system operates with an average signal power above the error threshold so that the effect of channel noise is made negligible and performance is there by limited essentially by Quantization noise alone. 3. The Quantization is fine enough (say n>6) to prevent signal correlated patterns in the Quantization error waveform. 4. The Quantizer is aligned with input for a loading factor of 4.
In a PCM system, Bandwidth B = nW or [n=B/W] substituting the value of ‘n’ we get,
--------- (8)
RA
GH U
DA
TH ES
H
Quantization Levels, Signalling Rate and Transmission Bandwidth In a PCM System:
Let us assume that the quantizer use ‘v’ number of binary digits to represent each level. Then, the number of levels that may be represented by ‘v’ digits is given as, ------ (1)
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Raghudathesh G P
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Here, q= total number of digital levels of a q-level quantizer.
E.x., if v = 4 bits, the total number of levels will be, q = 2 4 = 16 levels.
Each sample is converted to 'v' binary bits. i.e., Number of bits per sample is v. We know that, Number of samples per second are represented as fs. Therefore, Number of bits per second are expressed as
P
----- (2)
H
Here, the number of bits per second is known as signaling rate of PCM and is denoted by `r' and is given by,
TH ES
G
----- (3)
------ (4)
Here, fs ≥ 2 fm.
Also, bandwidth needed for PCM transmission is half of the signaling rate therefore, Transmission Bandwidth in PCM,
GH U
DA
As, r = v x fs, thus,
-------- (6)
As fs ≥ 2 fm, Thus,
RA
------ (5)
------ (7)
This is the required expression for bandwidth of a PCM system.
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Waveform Coding
Raghudathesh G P
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Robust Quantization: Necessity of Nonuniform Quantization in a PCM System: In case of uniform quantization, the quantizer has a linear characteristics. The step size also remains same throughout the range of quantizer.
Thus, over the complete range of inputs, the maximum quantization error also remains same.
As the quantization error is given as,
G
P
---- (1)
Since, step size 'Δ' is expressed as,
Say let x(t) is normalized, its maximum value i.e, xmax= 1. Therefore, step size 'Δ' is,
DA
---- (2)
E.x.: Let consider a PCM system in which we take v = 4 bits. Then number of levels q will be,
GH U
TH ES
H
Thus, from equation (2) the step size Δ will be,
RA
quantization error is given from equation (1) as,
Thus, we see that the quantization error is 1/16th part of the full voltage range.
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Waveform Coding
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Say if we assume that full range voltage is 16 volts. Then maximum quantization error will be 1 volt.
But, for the low signal amplitudes like 2 volts, 3 volts etc., the maximum quantization error of 1 volt which is quite high i.e., about 30 to 50%. This means that for signal amplitudes which are close to 15 volts, 16 volts etc., the maximum quantization error (which is same throughout the range) of 1 volt can be considered to be small. In fact, this problem arises because of uniform quantization. Therefore non-uniform quantization should be used in such cases.
P
As speech and music signals are characterized by large crest factor. This means that for such signals the ratio of peak to rms value is quite high. Which is given as below,
If we normalize the signal power i.e., if P = 1, then above equation becomes,
RA
------ (2)
Expressing in decibles, the last expression becomes
GH U
----- (1)
The signal to noise ratio is given by,
DA
TH ES
H
G
Necessity of Nonuniform Quantization For Speech Signal:
----- (3)
Here, power P is defined as,
= mean square value of signal voltage = x2(t)
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Hence, normalized power will be, ------ (4)
From equation (1), crest factor is given as,
P
----- (5)
G
When we normalize the signal x(t), then xmax = 1 Substituting this value of xmax in equation (5), we get ----- (6)
TH ES
H
For a large crest factor of voice (i.e., speech) and music signals, P should be very very less than one in above equation. i.e.,P << 1 for large crest factor
Therefore, actual signal to noise ratio would be significantly less than the value which is given by equation (3) since in this equation P = 1. Again, consider equation (2).
---- (7)
GH U
DA
This equation illustrates that the signal to noise ratio for large crest factor signal (P << 1) would be very very less than that of the calculated theoretical value. The theoretical value is obtained for normalized power (P = 1) by equation (3).
RA
Nonuniform Quantization:
Definition: If the quantizer characteristics is nonlinear and the step size is not constant instead if it is variable, dependent on the amplitude of input signal then the quantization is known as nonuniform quantization.
In non-uniform quantization, the step size is reduced with the reduction in signal level. For weak signals (P < < 1), the step size is small, therefore the quantization noise reduces, to improve the signal to quantization noise ratio for weak signals. The step size
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is thus varied according to the signal level to keep the signal to noise ratio adequately high. This is nonuniform quantization.
The non-uniform quantization is practically achieved through a process called companding.
COMPANDING (i.e., COMPANDED PCM): compading is bascically nonuniform quantization. It is required to be implemented to improve the signal to quantization noise ratio of weak signals.
The quantization noise is given by
From above equation we see that in the uniform quantization, once the step size is fixed, the quantization noise power remains constant. However, the signal power is not constant. It is proportional to the square of signal amplitude.
Hence signal power will be small for weak signals, but quantization noise power is constant. Therefore, the signal to quantization noise for the weak signals is very poor. This will affect the quality of signal.
The remedy is to use companding. Companding is a term derived from two words i.e., compression and expansion as under:
GH U
DA
TH ES
H
G
P
Practically, it is difficult to implement the non-uniform quantization as it is not known in advance about the changes in the signal level.
Due to above reason, a particular method is used. The weak signals are amplified and strong signals are attenuated before applying them to a uniform quantizer. This process is called as compression and the block that provides it is called as a compressor.
RA
At the receiver exactly opposite is followed which is called expansion. The circuit used for providing expansion is called as an expander.
The compression of signal at the transmitter and expansion at the receiver is combined to be called Compressor companding. The process of companding has been shown output in the form of a block diagram in figure the fiure below
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Raghudathesh G P
Asst Professor
P
Waveform Coding
Advantages of Non – Uniform Quantization :
TH ES
H
G
1. Higher average signal to quantization noise power ratio than the uniform quantizer when the signal pdf is non uniform which is the case in many practical situation. 2. RMS value of the quantizer noise power of a non – uniform quantizer is substantially proportional to the sampled value and hence the effect of the quantizer noise is reduced.
Types of compressor characteristics:
There are two type of compressor characteristics based on logarithmic compression laws they are: 1. µ - law Companding and 2. A – law Companding.
μ-law:
In this companding, the compressor characteristics is defined by equation below,
GH U
DA
The normalized form of compressor characteristics is shown in the figure below,
RA
---- (1)
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The μ-law is used for PCM telephone systems in the USA, Canada and Japan. A practical value for μ is 255.
A-law: In A-law companding the compressor characteristics is defined by equation below,
P
H
The normalized form of A-law compressor characteristics is shown in the figure below,
GH U
DA
TH ES
G
---- (2)
The A-law is used for PCM telephone systems in Europe.
A practical value for A is 100.
RA
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Formulas 1) Number of Quantization levels
GH U
6) Bit duration
DA
5) Transmission bandwidth
TH ES
4) Signalling rate or bit transmission rate
H
3) Sampling rate Where ‘W’ is the highest frequency of message signal
G
P
2) Number of bits
7) Bit rate
RA
8) Sampling frequency
9) Message bandwidth
10) Maximum signal to quantization ratio
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Note:
P
Where
H
G
11)
TH ES
12) Quantisation noise power or quantisation error or normalised noise power
Note: 1) ∂ or
GH U
DA
2)
Here V is RMS value
And R = 1ohm for normalised power
RA
13) Step size
14) Normalised signal to quantisation noise ratio
15) Signal to quantisation noise ratio for sinusoidal signal
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16)
P
17) Bit duration ‘ ’=
when compression parameter ‘μ’ is given by
TH ES
H
18)
G
and
DA
19) Maximum quantisation error for an uniform quantiser is given by
GH U
20) RMS quantisation error
RA
21) Noise power or mean square value of quantisation error
22)
23) Nyquist rate = 2W 24)
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for non-sinusoidal signal (eg: telephone signal)
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Problems:
P
1. The output signal to noise ratio of a 10 bit PCM was found to be 30 dB. The desired SNR is 42 dB. It was decided to increase the SNR to the desired value increasing the number of quantization levels. Find the fractional increase in transmission bandwidth required for this increase in SNR. Solution: (i) To obtain number of bits for 42 dB . Signal to noise ratio of PCM is given as,
G
(S/N) = (4.8 + 6v) dB
H
Above equation shows that signal to noise ratio increases by 6 dB with ev bit. It is given that
TH ES
for 10 bits
The desired signal to noise ratio is 42 dB. Hence rise in We know that
ratio is 42-30 = 12 dB N
ratio increases by 6 dB for 1 bit. Hence 2 bits are required t ) increase signal
DA
to noise ratio by 12 dB. Hence,
v = 10 + 2 = 12 bits are required.
GH U
(ii) To obtain fractional increase in bandwidth. Bandwidth in PCM is given as, = v
RA
(10 bits) =
≫ Fractional increase in
(12 bits) = =
×
=5
×
=6
100% = 20 %
2. A telephone signal with cut-off frequency of 4 kHz is digitized into 8 bit PCM, sampled at Nyquist rate. Calculate baseband transmission bandwidth and quantization S/ N ratio.
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Asst Professor
Solution : Given data is, W = 4 kHz v = 8 bits transmission bandwidth is given as, = vW = 4k x 8 = 32 kHz
G
P
telephone signal is non sinusoidal signal. Its signal to quantization noise ratio is given in equation as, 4.8 + 6v
H
= 4.8 + 6x 8 = 52.8 dB.
Given
= 40 dB
TH ES
3. A telephone signal bandlimited to 4 kHz is to be transmitted by PCM.The signal to quantization noise is to be atleast 40 dB. Find the number of levels into which signal has to be encoded. Also find the bandwidth of transmission. Solution:
DA
W = 4 kHz n number of levels (q) quantization noise ratio is given as,
GH U
4.8 + 6 v
RA
40 = 4.8 + 6 v v = 5.866 or v = 6 bits. number of levels will be, q= = = 64 levels ii) to obtain transmission bandwidth ( Transmission bandwidth is given as,
= 24kHz
4. Show that for μ = A, the μ -law and theA-law have the same companding gain. Solution: The μ-law is
where u and v are the input and output of the compander respectively.
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Waveform Coding
Raghudathesh G P
Asst Professor
P
Companding gain,
H
GH U
DA
the A –law is
≫
TH ES
hence,
G
For large values of μ,we can write
we find that the two companding gains are equal, A=μ.
RA
5. A Television signal having a bandwidth of 4.2 MHz is transmitted using binary PCM system. Given that the number of quantization levels is 512. Determine: (i) Code word length (ii) Transmission bandwidth (iii) Final bit rate (iv) Output signal to quantization noise ratio. Solution: Given that the bandwidth is 4.2 MHz. This means that highest frequency component will have frequency of 4.2 MHz i.e., 4.2 MHz Also, given that Quantization levels, q = 512
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G
P
(1) We know that the number of bits and quantization levels are related in binary PCM as under:
TH ES
H
Simplifying, we get, v = 9 bits Hence, the code word length is 9 bits.
(ii) We know that the transmission channel bandwidth is given as, 9 x 4.2 x
Hz ≥37.8 MHz
DA
(iii) The final bit rate is equal to signaling rate. We know that the signaling rate is given as,
GH U
Here, sampling frequency is given as Thus,
RA
or Substituting this value of r = 9 x 8.4 x
in equation (i) for signaling rate, we get
bits/sec = 75.6 x
bits/sec
The transmission bandwidth may also be obtained as,
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Or
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which is same as the value obtained earlier.
(iv) The output signal to noise ratio is expressed as
H
G
P
But Therefore,
GH U
DA
TH ES
6. The bandwidth of an input signal to the PCM is restricted to 4 kHz. The input signal varies in amplitude from - 3.8 V to + 3.8 V and has the average power of 30 mW. The required signal to noise ratio is given as 20 dB. The PCM modulator produces binary output. Assuming uniform quantization, i. Find the number of bits required per sample. ii. Outputs of 30 such PCM coders are time multiplexed. What would be the minimum required transmission bandwidth for this multiplexed signal? Solution: The given value of signal to noise ratio is 20 dB. This means that.
Hence,
We know that the signal to quantization noise ratio is given as,
RA
(i)
Here, we are given = 3.8 V P = 30 mW and
= 100
Therefore,
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Solving, we get
G
P
(ii) The maximum frequency is given as We know that the transmission bandwidth is expressed as,
TH ES
H
Since there are 30 PCM coders which are time multiplexed, the transmission bandwidth must be,
We also know that the signaling rate is two times the transmission bandwidth, i.e. Signaling rate,
r = 840 x 2 bits/sec = 1680 bits/sec. Ans.
RA
GH U
DA
7. The information in an analog signal voltage waveform is to be transmitted over a PCM system with an accuracy of + 0.1% (full scale). The analog voltage waveform has a bandwidth of 100 Hz and an amplitude range of - 10 to + 10 volts. (i) Find the minimum sampling rate required. (ii) Find the number of bits in each PCM word. (iii) Find minimum bit rate required in the PCM signal. (iv) Find the minimum absolute channel bandwidth required for the transmission of the PCM signal. Solution: Here an accuracy is given as ± 0.1%. This means that the quantization error must be ± 0.1% or the maximum quantization error must be ± 0.1%. Thus, = ± 0.1% = ± 0.001 We know that the maximum quantization error for an uniform quantizer is expressed as,
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Waveform Coding
Raghudathesh G P
Asst Professor
Or
P
Therefore, step size ∆ We know that the step size, number of quantization levels and maximum value of the signal are related as
TH ES
H
Substituting, values of A and xmax in equation (i), we get
G
Here, given
We know that minimum 10,000 levels should be used to quantize the signal. If binary PCM is used, then number of bits for each samples may be calculated as under, i.e.
GH U
ii.
DA
Hence, the number of levels are 10,000. i. The maximum frequency in the signal is given as 100 Hz, i.e., By sampling theorem minimum sampling frequency should be,
RA
Here
Thus,
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10,000 = 2
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Waveform Coding
(i)
Raghudathesh G P
Asst Professor
The bit rate or signaling rate is expressed as,
P
iv)
. The transmission bandwidth for PCM is expressed as, m
DA
TH ES
H
G
8. Twenty four voice signals are sampled uniformly and then have to be time division multiplexed. The highest frequency component for each voice signal is equal to 3.4 kHz. Now (i) If the signals are pulse amplitude modulated using Nyquist rate sampling, what would be the minimum channel bandwidth required. (ii) If the signals are pulse code modulated with an 8 bit encoder, what would be the sampling rate? The bit rate of system is given as 1.5 x bits/sec. Solution: (i) As a matter of fact, if N channels are time division multiplexed, then minimum transmission bandwidth is expressed as,
GH U
Here, is the maximum frequency in the signals. Given , = 3.4KHz Therefere BW = 24 x 3.4 kHz = 81.6 kHz (iii) The signaling rate of the system is given as, r = 1.5 x
bits/sec
RA
Since there are 24 channels, the bit rate of an individual channel is,
Further, since each sample is encoded using 8 bits, the samples per second will be,
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Waveform Coding
Raghudathesh G P
Asst Professor
Note that the samples per seconds is nothing but sampling frequency Thus, we have ,
= 7812.5 Hz or samples per second .
P
Solving, we get,
9. A PCM system uses a uniform quantizer followed by a 7-bit binary encoder. The bit rate
TH ES
H
G
of the system is equal to 50 x bits/sec. (i) What is the maximum message signal bandwidth for which the system operates satisfactorily? (ii) Calculate the output signal to quantization noise ratio when a full load sinusoidal modulating wave of frequency 1 MHz is applied to the input. Solution: (i) Let us assume that the message bandwidth be Hz. Therefore sampling frequency should be,
DA
The number of bits given as v = 7 bits We know that the signaling rate is given as,
GH U
Or
Substituting value for r, we get 50 x
RA
Thus, the maximum message bandwidth is 3.57 MHz. (ii) The modulating wave is sinusoidal. For such signal, the signal to quantization noise ratio is expressed as,
Substituting value of v, we get
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Waveform Coding
Raghudathesh G P
Asst Professor
GH U
DA
TH ES
H
G
P
10. The information in an analog waveform with maximum frequency = 3 kHz is to be transmitted over an M-level PCM system where the number of quantization levels is M = 16. The quantization distortion is specified not to exeed 1% of peak to peak analog signal. (i) What would be the maximum number of bits per sample that should be used in this PCM system? (ii) What is the minimum sampling rate and what is the resulting bit transmission rate? Solution: (i) Since the number of quantization levels given here are M = 16, q = M = 16 We know that the bits and levels in binary PCM are related as, q= Here, v = number of bits in a codeword Thus, 16 = Or v = 4bits (ii)again since By sampling theorem, we know that
RA
Hence, the minimum sampling rate is 6 kHz Also, bit transmission rate or signaling rate is given as,
11. A signal having bandwidth equal to 3.5 kHz is sampled, quantized and coded by a PCM system. The coded signal is then transmitted over a transmission channel of supporting a transmission rate of 50 k bits/sec. Determine the maximum signal to noise ratio that can be obtained by this system. The input signal has peak to peak value of 4 volts and rms value of 0.2 V. Solution:
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Waveform Coding
Raghudathesh G P
Asst Professor
The maximum frequency of the signal is given as 3.5 kHz, i.e.,
P
Therefore sampling frequency will be
7x Simplifying, we get
7x
Hz in above equation, we get
DA
8 bits
bits/sec and
TH ES
Substituting values of r = 50 x
H
G
We know that the signaling rate is given by
GH U
The rms value of the signal is 0.2 V. Therefore the normalized signal power will be, Normalized signal power
RA
i.e Further, the maximum signal to noise ratio is given by,
Substituting the values of P = 0.04, v = 8 and
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= 2 in above equation, we have
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Waveform Coding
Raghudathesh G P
Asst Professor
G
P
12. A signal x(t) is uniformly distributed in the range ± • Evaluate maximum signal to noise ratio for this signal. Solution: Given that the signal is uniformly distributed in the range ± xmax, therefore we can write its PDF (using the Standard Uniform Distribution) as under:,
GH U
DA
TH ES
H
The mean square value of random variable X is given as ,
RA
Therefore ,mean square value of
The signal power is
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Waveform Coding
Raghudathesh G P
Asst Professor
Normalised signal power
from (i), we get
G
P
Substituting the value of
TH ES
H
We know that the relation between step size, maximum amplitude of signal and number of levels is given as Step size
DA
Therefore, normalized signal power,
GH U
We also know that Normalized noise power =
RA
Therefore, signal to noise power ratio
Since q =
above equation will be,
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Waveform Coding
Raghudathesh G P
Asst Professor
or
13. Given an audio signal consisting of the sinusoidal term given as
G
(ii)
Determine the signal to quantization noise ratio when this is quantized using 10 bit PCM. How many bits of quantization are needed to achieve a signal to quantization noise ratio of atleast 40 dB?
H
(i)
P
This is required expression for maximum value of signal to noise ratio.
TH ES
Solution: Here,
GH U
DA
This is sinusoidal signal applied to the quantizer. (i) Let us assume that peak value of cosine wave defined by x(t) covers the complete range of quantizer. i.e., covers complete range It is expressed as
RA
Since here 10 bit PCM is used i.e; V=10v Thus,
ii)
For sinusoidal signal, again, let us use the same relation i.e ,
To get signal to noise ratio of at least 40 dB we can write above equation as,
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Waveform Coding
Raghudathesh G P
Asst Professor
Solving this we get
Hence, at least 7 bits are required to get signal to noise ratio of 40 dB. Ans.
DA
TH ES
H
G
P
14. A 7 bit PCM system employing uniform quantization has an overall signaling rate of 56 k bits per second. Calculate the signal to quantization noise that would result when its input is a sine wave with peak amplitude equal to 5 Volt. Find the dynamic range for the sine wave inputs in order that the signal to quantization noise ratio may be less than 30 dBs. What is the theoretical maximum frequency that this system can handle? Solution: The number of bits in the PCM system are v = 7 bits Assume that 5 V peak to peak voltage utilizes complete range of quantizer. Then, we can find the signal to quantization noise ratio as,
RA
GH U
We know that the signaling rate is given as, r=v Substituting r = 56 x 103 bits/second and v = 7 bits in above equation, we obtain 56 x =7• Simplifying, we get Sampling frequency, fs = 8 x 103 Hz Further, using sampling theorem we have,
Thus, maximum frequency that can be handled is given as
15. Given a sine wave of frequency fu, and amplitude Am applied to a delta modulator having step size 4. Show that the slope overload distortion will occur if
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Waveform Coding
Raghudathesh G P
Asst Professor
here Ts is the sampling period. Solution:
P
Let us consider that the sine wave is represented as,
TH ES
H
G
It may be noted that the slope of x(t) will be maximum when derivative of x(t) with respect to `t' will be maximum. The maximum slope of delta modulator may be given as,
RA
GH U
DA
We know that, slope overload distortion will take place if slope of sine wave is greater than slope of delta modulator i.e.,
Hence proved.
15. Determine the output signal to noise ratio of a linear delta modulation system for a 2 kHz sinusoidal input signal sampled at 64 kHz. Slope overload distortion is not present and the post reconstruction filter has a bandwidth of 4 kHz. Solution : We know that
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Waveform Coding
= 2 kHz and
Asst Professor
= 4kHz
G
P
Here, = 64 kHz, Therefore,
Raghudathesh G P
TH ES
H
16. For the same sinusoidal input of example 4.15, calculate the signal to quantization noise ratio of a PCM system which has the same data rate of 64 kbits/s. The sampling frequency is 8 kHz and the number of bits per sample is N = 8. Comment on the result. Solution : The signal to noise ratio of a PCM system is given by, (SNR)q = (1.8 + 6 N) dB = 1.8 + (6 x 8) = 49.8 dB
DA
COMMENTS: The SNR of a DM system is 27.94 dB which is too poor as compared to 49.8 dB of an 8 bit PCM system. Thus, for all the simplicity of DM, it cannot perform as well as an 8 bit PCM.
RA
GH U
17. A binary channel with bit rate r= 36000 bits per second (b/s) is available for PCM voice transmission .Evaluate the appropriate values of the sampling rate , the quantizing level q, and the number of binary digits v. Assume fin= 3.2 kHz. Solution: Here, we require that
Therefore, we have, v 5, and also, and
18. An analog signal is sampled at the Nyquist rate and quantized into q levels. Find the time duration τ of 1 bit of the binary-encoded signal. Solution:
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Waveform Coding
Raghudathesh G P
Asst Professor
Let v be number of bits per sample. Then we have
is not integer value i.e,
is the Nyquist interval.
H
where
G
P
where indicates the next higher integer to be taken if binary pulses must be transmitted per second.
TH ES
19. The output signal-to-quantizing-noise ratio (SNR)0 in a PCM system is defined as the ratio of average signal power to average quantizing noise power. For a full-scale sinusoidal modulating signal with amplitude A, prove that
GH U
DA
Where q is the number of quantizing levels. Solution: Since, here peak-to-peak excursion of the quatizer input is 2A. Therefore, the quantizer step size will be
RA
Then, the average quantizing noise power is
The output signal-to-quantizing-noise ratio of a PCM system for a full scale test tone is, therefore,
Expressing this in decibels, we have
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Waveform Coding
Raghudathesh G P
Asst Professor
GH U
Thus we have ,
TH ES
Therefore ,
DA
Now,since
H
G
P
20. In a binary PCM system, the output signal-to-quantizing-noise ratio is to be held to a minimum value of 40 dB. Determine the number of required levels, and find the corresponding output signal-to-quantizing-noise ratio. Solution: In a binary PCM system, q = , where v is the number of binary digits. Then, we have
RA
and the number of binary digits v is
Then, the number of levels required is q = quantizing noise ratio will be
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= 128, and corresponding output signal-to-
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Waveform Coding
Raghudathesh G P
Asst Professor
NOTE: Equation W indicates that each bit in the code word of a binary PCM system contributes 6 dB to the output signal.to-quantizing noise ratio. In fact, this is called the 6 dB rule.
DA
TH ES
H
G
P
21. Consider an audio signal with spectral components limited to the frequency band of 300 to 3300 Hz. A PCM signal is generated with a sampling rate of 8000 samples/s. The required output signal-to-quantizing-noise ratio is 30 dB. (i) What is the minimum number of uniform quantizing levels needed, and what is the minimum number of bits per sample needed? (ii) Calculate the minimum system bandwidth required. (iii) Repeat parts (i) and (ii) when a g-law compander is used with g = 255. Solution: (i) Here, we have
GH U
Thus, the minimum number of uniform quatizing levels required is 26.
RA
The minimum number of bits per sample is 5. ii) The minimum required system bandwidth will be
We have
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Waveform Coding
Raghudathesh G P
Asst Professor
G
P
Thus, the minimum number of quantizing levels needed is 102. Also,
TH ES
H
The minimum number of bits per sample is 7. The minimum bandwidth required for this case will be
= 1/Ts is the sampling frequency.
RA
GH U
where Solution: We have so that
DA
22. Consider a sinusoidal signal m(t) = A cos wmt applied to a delta modulator step size A. Show that the slope overload distortion will occur if
To avoid the slope overload, we require that
Thus, if A > ∆/
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slope overload distortion will occur.
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Waveform Coding
Raghudathesh G P
Asst Professor
H
G
P
23. A DM system is designed to operate at 3 times the Nyquist rate for a signal with a 3 kHz bandwidth. The quantizing step size is 250 mV. (i) Determine the maximum amplitude of a 1-kHz input siinusoid for which the delta modulator does not show slope overload. (ii) Determine the posfiltered output signal-to-quantizing-noise ratio for the signal of part (i) Solution: We have
DA
TH ES
The maximum allowable amplitude of the input sinusoid is
GH U
ii) Assuming that the cutoff frequency of the low-pass filter is I'm, we have
RA
24. The pulse rate in a DM system is 56,000 per sec. The input signal is
Find the minimum value of step size which will avoid slope overload distortion. What would be the disadvantages of choosing a value of larger than the minimum? Solution: Input signal,
To avoid slope overloading, we have
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Waveform Coding
is step size and
Asst Professor
is sampling rate.
TH ES
H
G
P
where
Raghudathesh G P
DA
Hence, larger step size out of two will be the required step size. i.e., = 0.56 V. If a value larger than the minimum will be choosen, then granular noise will occur.
RA
GH U
25. Bandwidth of the input to pulse code modulator is restricted to 4 kHz. 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. The modulator produces binary output. Assume uniform quantization Calculate the number of bits required per sample. Solution: Given that
Quantizer step size, where L = , n is the number of bianry digits
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Waveform Coding
Raghudathesh G P
Asst Professor
TH ES
H
G
P
then, average quantizing power is,
Hence, n = 7 = number of bits required per sample.
RA
GH U
DA
26. 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. The mean squared value of message signal is 2 Volt-squared. Determine the following : (i) The normalized power for quantization noise. (ii) The bit transmission rate. (iii) The signal to quantization noise ratio in dB. (iv) Derive the expressions used in (i) and (iii). Solution : Given that
It is given that uniform quantization is used. Also,
Normalized power for quantization noise (NI) is given by
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Waveform Coding
Raghudathesh G P
Asst Professor
P
Therefore,
H
G
Substituting above value of A in equation (i), we obtain
TH ES
Now, let us calculate the bit transmission rate (r). The bit transmission rate or signaling rate is the number of bits transmitted by the PCM system per second.
we have
DA
Therefore, As the signal is sampled at Nyquist rate,
RA
GH U
(iii) The signal to quantization noise ratio in dB may be calculated as under : Mean square value of signal The normalized signal power P
Therefore,
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Waveform Coding
Raghudathesh G P
Asst Professor
P
27. 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,
TH ES
H
G
where M= Number of quantization levels. A compact disc recording system samples each of the two-stereo signals with a 16 bit A/D converter at 44.1 Kb/sec. (i) Determine output S/N ratio for a full scale sinusoid. (ii) The bit stream of digitized data is augmented by addition of error correcting bits, clock extraction bits etc. and these additional bits represent 100% overhead. Determine output bit rate of CD system. (iii) The CD can record an hour's worth of music. Determine number of bits recorded on CD. Solution :
DA
There are two stereo channels. .
Now, let us evaluate the output bit rate of the CD system. The bit rate for each of two stereo channels = Therefore, the bit rate of two channels
RA
ii)
GH U
(i) Output signal to noise ratio for full scale sinusoid is given by
Including the additional 100% overhead, the output bit rate will be
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Waveform Coding
iii)
Raghudathesh G P
Asst Professor
Next, we calculate the number of recorded on CD. The CD can record an hour's worth of music. Therefore, the number of bits recorded on CD = bit rate x Number of seconds/hour = 2.822 x x 3600 = 10.16 x bits or 10.16 gigabits Ans.
H
G
P
28. Determine the output SNR in a DM system for 1 kHz sinusoid, sampled at 32 kHz without slope overload and followed by a 4 kHz post construction filter. Derive the formula used. Solution : Given that,
DA
Therefore,
TH ES
It is given that there is no slope overload. The output signal to noise ratio in a DM system is expressed as
RA
GH U
29. The bandwidth of TV video plus audio signal is 4.5 MHz. If this signal is converted into PCM bit stream with 1024 quantization levels, determine number of bits/sec of the resulting signal. Assume that the signal is sampled at the rate 20% above Nyquist rate. Solution : Given that
Let us calculate the number of bits/sec
But, we do not know the value of v. We know that
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Waveform Coding
Raghudathesh G P
Therefore, bits/sec
Asst Professor
.
TH ES
H
G
P
30. If a voice frequency signal is sampled at the rate of 32,000 samples/sec and characterized by peak value of 2 Volts, determine the value of step size to avoid slope overload. What is quantization noise power Ng and corresponding SNR ? Assume bandwidth of signal as 4 kHz. Solution : Given that . Bandwidth . Peak value of the signal i) Step size A to avoid slope overload can be calculated as under : To avoid slope overload the following condition must be satisfied :
GH U
DA
Substituting the values, we obtain
RA
ii)Next, we find the quantization noise power (
).
The quantization noise power for a delta modulator is given by
iii)We know that the signal to noise ratio is given by
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Waveform Coding
Raghudathesh G P
Asst Professor
H
G
P
31. A compact disc (CD) records audio signals digitally by PCM. Assume audio signal's bandwidth to be 15 kHz. If signals are sampled at a rate 20% above Nyquist rate for practical reasons and the samples are quantized into 65,536 levels. Determine bits/sec required to encode the signal and minimum bandwidth required to transmit encoded signal. Solution : Given that
Now, signaling rate,
DA
TH ES
Signaling rate (r) can be calculated as under : We know that
is 576 Kbits/sec.
GH U
Hence, the signaling rate
RA
(ii) Minimum bandwidth can be calculated as under :
Therefore, minimum bandwidth,
.
32. In a single integration DM scheme, the voice signal is sampled at a rate of 64 kHz. The maximum signal amplitude is 1 Volt. (i) Determine the minimum value of step size to avoid slope overload. (ii) Determine granular noise power No, if the voice signal bandwidth is 3.5 kHz.
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Waveform Coding
Raghudathesh G P
Asst Professor
(iii)
Assuming signal to be sinusoidal, calculate signal power So and signal to noise ratio (SNR). (iv) Assuming that the voice signal amplitude is uniformly distributed in the range (- 1, 1), determine So and SNR. Solution : Given that
ii)
TH ES
H
G
P
(i) Minimum step size to avoid slope overload is given by
Granular noise power is expressed as
DA
Solving, we get
GH U
Question Bank:
RA
1. What are the advantages of digital representation of analog signals? December 2011 (04 M) 2. What is the need for non-uniform quantization? Explain the µ-law companding. December 2011 (09 M), December 2013 (08 M) 3. Derive an expression for the SNR of a PCM system. December 2011 (07 M) 4. Derive an expression for output SNR of the quantizer and show that in decibles if a sinusoidal signal is quantized. December 2013 (08 M) 5. For a binary PCM signal, determine ‘L’ if the compression parameter and the minimum . Determine the with this value of L. December 2013 (04 M), June 2012 (06 M) 6. With a neat block diagram and waveform, explain TDM. June 2012 (06 M)
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Waveform Coding
Raghudathesh G P
Asst Professor
7. Show that the signal to quantization noise power ratio of a uniform quantizer is . June 2012 (08 M) 8. Write a note on robust quantization. December 2012 (04 M) 9. Determine the probability of symbols error for binary encoded PCM wave and is given by
. December 2012 (10 M)
RA
GH U
DA
TH ES
H
G
P
10. Derive the expression for signal to quantization noise ratio (SNR) and show that for uniform quantization, each bit in the codeword of a PCM contributes 6 dB to SNR. June 2013 (08 M) 11. Six independent message sources of bandwidths w, w, 2w, 2w, 3w and 3w hertz are to be transmitted on TDM. Set up a scheme to accomplish this requirement, with each message signal sampled at its Nyquist rate. June 2013 (05 M) 12. The signal m(t) = 6 sin (2πt) Volts, is transmitted using 4-bit binary PCM system. The quantizer is of midriser type with a step size of 1 Volt. The sampling frequency is 4 Hz with samples taken at t = ±1/8, ±3/8, ±5/8, … sec. Sketch the PCM wave for one complete cycle of the input. June 2013 (07 M) 13. Explain regenerative repeater in a PCM system with a block diagram. June 2014 (05 M) 14. The bandwidth of a signal is 3.4 kHz. If the signal is converted to PCM bit stream with 1024 levels, determine the number of bits per second generated by the PCM system. Assume that the signal is sampled at the rate of 20% above the Nyquist rete. June 2014 (06 M) 15. Derive an expression for the output SNR of s uniform quantizer in terms of step size of the quantizer. Hence show that for mid-tread type uniform quantizer the SNR is (SNR)output = 6n – 7.2 dB, whrer ‘n’ is the number of bits per sample. Assume a loading factor of 4. June 2014 (12 M)
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