# Chapter 4 Digital Transmission PCM Sampling Quantization

## Chapter 4 Digital Transmission PCM Sampling Quantization

Booth, Mary, Morning Host;Midday Host has reference to this Academic Journal, PHwiki organized this Journal Chapter 4 Digital Transmission Copyright © The McGraw-Hill Companies, Inc. Permission required as long as reproduction or display. 4-2 ANALOG-TO-DIGITAL CONVERSION A digital signal is superior to an analog signal because it is more robust to noise in addition to can easily be recovered, corrected in addition to amplified. For this reason, the tendency today is to change an analog signal to digital data. In this section we describe two techniques, pulse code modulation in addition to delta modulation. Pulse Code Modulation (PCM) Delta Modulation (DM) Topics discussed in this section: PCM PCM consists of three steps to digitize an analog signal: Sampling Quantization Binary encoding Be as long as e we sample, we have to filter the signal to limit the maximum frequency of the signal as it affects the sampling rate. Filtering should ensure that we do not distort the signal, ie remove high frequency components that affect the signal shape.

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Figure 4.21 Components of PCM encoder Sampling Analog signal is sampled every TS secs. Ts is referred to as the sampling interval. fs = 1/Ts is called the sampling rate or sampling frequency. There are 3 sampling methods: Ideal – an impulse at each sampling instant Natural – a pulse of short width with varying amplitude Flattop – sample in addition to hold, like natural but with single amplitude value The process is referred to as pulse amplitude modulation PAM in addition to the outcome is a signal with analog (non integer) values Figure 4.22 Three different sampling methods as long as PCM

According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency contained in the signal. Figure 4.23 Nyquist sampling rate as long as low-pass in addition to b in addition to pass signals For an intuitive example of the Nyquist theorem, let us sample a simple sine wave at three sampling rates: fs = 4f (2 times the Nyquist rate), fs = 2f (Nyquist rate), in addition to fs = f (one-half the Nyquist rate). Figure 4.24 shows the sampling in addition to the subsequent recovery of the signal. It can be seen that sampling at the Nyquist rate can create a good approximation of the original sine wave (part a). Oversampling in part b can also create the same approximation, but it is redundant in addition to unnecessary. Sampling below the Nyquist rate (part c) does not produce a signal that looks like the original sine wave. Example 4.6

Figure 4.24 Recovery of a sampled sine wave as long as different sampling rates Consider the revolution of a h in addition to of a clock. The second h in addition to of a clock has a period of 60 s. According to the Nyquist theorem, we need to sample the h in addition to every 30 s (Ts = T or fs = 2f ). In Figure 4.25a, the sample points, in order, are 12, 6, 12, 6, 12, in addition to 6. The receiver of the samples cannot tell if the clock is moving as long as ward or backward. In part b, we sample at double the Nyquist rate (every 15 s). The sample points are 12, 3, 6, 9, in addition to 12. The clock is moving as long as ward. In part c, we sample below the Nyquist rate (Ts = T or fs = f ). The sample points are 12, 9, 6, 3, in addition to 12. Although the clock is moving as long as ward, the receiver thinks that the clock is moving backward. Example 4.7 Figure 4.25 Sampling of a clock with only one h in addition to

An example related to Example 4.7 is the seemingly backward rotation of the wheels of a as long as ward-moving car in a movie. This can be explained by under-sampling. A movie is filmed at 24 frames per second. If a wheel is rotating more than 12 times per second, the under-sampling creates the impression of a backward rotation. Example 4.8 Telephone companies digitize voice by assuming a maximum frequency of 4000 Hz. The sampling rate there as long as e is 8000 samples per second. Example 4.9 A complex low-pass signal has a b in addition to width of 200 kHz. What is the minimum sampling rate as long as this signal Solution The b in addition to width of a low-pass signal is between 0 in addition to f, where f is the maximum frequency in the signal. There as long as e, we can sample this signal at 2 times the highest frequency (200 kHz). The sampling rate is there as long as e 400,000 samples per second. Example 4.10

A complex b in addition to pass signal has a b in addition to width of 200 kHz. What is the minimum sampling rate as long as this signal Solution We cannot find the minimum sampling rate in this case because we do not know where the b in addition to width starts or ends. We do not know the maximum frequency in the signal. Example 4.11 Quantization Sampling results in a series of pulses of varying amplitude values ranging between two limits: a min in addition to a max. The amplitude values are infinite between the two limits. We need to map the infinite amplitude values onto a finite set of known values. This is achieved by dividing the distance between min in addition to max into L zones, each of height = (max – min)/L Quantization Levels The midpoint of each zone is assigned a value from 0 to L-1 (resulting in L values) Each sample falling in a zone is then approximated to the value of the midpoint.

Quantization Zones Assume we have a voltage signal with amplitutes Vmin=-20V in addition to Vmax=+20V. We want to use L=8 quantization levels. Zone width = (20 – -20)/8 = 5 The 8 zones are: -20 to -15, -15 to -10, -10 to -5, -5 to 0, 0 to +5, +5 to +10, +10 to +15, +15 to +20 The midpoints are: -17.5, -12.5, -7.5, -2.5, 2.5, 7.5, 12.5, 17.5 Assigning Codes to Zones Each zone is then assigned a binary code. The number of bits required to encode the zones, or the number of bits per sample as it is commonly referred to, is obtained as follows: nb = log2 L Given our example, nb = 3 The 8 zone (or level) codes are there as long as e: 000, 001, 010, 011, 100, 101, 110, in addition to 111 Assigning codes to zones: 000 will refer to zone -20 to -15 001 to zone -15 to -10, etc. Figure 4.26 Quantization in addition to encoding of a sampled signal

We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample Solution The human voice normally contains frequencies from 0 to 4000 Hz. So the sampling rate in addition to bit rate are calculated as follows: Example 4.14 PCM Decoder To recover an analog signal from a digitized signal we follow the following steps: We use a hold circuit that holds the amplitude value of a pulse till the next pulse arrives. We pass this signal through a low pass filter with a cutoff frequency that is equal to the highest frequency in the pre-sampled signal. The higher the value of L, the less distorted a signal is recovered. Figure 4.27 Components of a PCM decoder

We have a low-pass analog signal of 4 kHz. If we send the analog signal, we need a channel with a minimum b in addition to width of 4 kHz. If we digitize the signal in addition to send 8 bits per sample, we need a channel with a minimum b in addition to width of 8 × 4 kHz = 32 kHz. Example 4.15 Delta Modulation This scheme sends only the difference between pulses, if the pulse at time tn+1 is higher in amplitude value than the pulse at time tn, then a single bit, say a 1, is used to indicate the positive value. If the pulse is lower in value, resulting in a negative value, a 0 is used. This scheme works well as long as small changes in signal values between samples. If changes in amplitude are large, this will result in large errors. Figure 4.28 The process of delta modulation

Isochronous In isochronous transmission we cannot have uneven gaps between frames. Transmission of bits is fixed with equal gaps.

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