Chapter 6 B in addition to width Utilization: Multiplexing in addition to Spreading FM Interleaving

Chapter 6 B in addition to width Utilization: Multiplexing in addition to Spreading FM Interleaving www.phwiki.com

Chapter 6 B in addition to width Utilization: Multiplexing in addition to Spreading FM Interleaving

Fox, Ken, Morning On-Air Personality has reference to this Academic Journal, PHwiki organized this Journal Chapter 6 B in addition to width Utilization: Multiplexing in addition to Spreading Copyright © The McGraw-Hill Companies, Inc. Permission required as long as reproduction or display. B in addition to width utilization is the wise use of available b in addition to width to achieve specific goals. Efficiency can be achieved by multiplexing; i.e., sharing of the b in addition to width between multiple users. 6-1 MULTIPLEXING Whenever the b in addition to width of a medium linking two devices is greater than the b in addition to width needs of the devices, the link can be shared. Multiplexing is the set of techniques that allows the (simultaneous) transmission of multiple signals across a single data link. As data in addition to telecommunications use increases, so does traffic. Frequency-Division Multiplexing Wavelength-Division Multiplexing Synchronous Time-Division Multiplexing Statistical Time-Division Multiplexing Topics discussed in this section:

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Figure 6.1 Dividing a link into channels Figure 6.2 Categories of multiplexing Figure 6.3 Frequency-division multiplexing (FDM)

FDM is an analog multiplexing technique that combines analog signals. It uses the concept of modulation discussed in Ch 5. Figure 6.4 FDM process FM

Figure 6.5 FDM demultiplexing example Assume that a voice channel occupies a b in addition to width of 4 kHz. We need to combine three voice channels into a link with a b in addition to width of 12 kHz, from 20 to 32 kHz. Show the configuration, using the frequency domain. Assume there are no guard b in addition to s. Solution We shift (modulate) each of the three voice channels to a different b in addition to width, as shown in Figure 6.6. We use the 20- to 24-kHz b in addition to width as long as the first channel, the 24- to 28-kHz b in addition to width as long as the second channel, in addition to the 28- to 32-kHz b in addition to width as long as the third one. Then we combine them as shown in Figure 6.6. Example 6.1 Figure 6.6 Example 6.1

Five channels, each with a 100-kHz b in addition to width, are to be multiplexed together. What is the minimum b in addition to width of the link if there is a need as long as a guard b in addition to of 10 kHz between the channels to prevent interference Solution For five channels, we need at least four guard b in addition to s. This means that the required b in addition to width is at least 5 × 100 + 4 × 10 = 540 kHz, as shown in Figure 6.7. Example 6.2 Figure 6.7 Example 6.2 Four data channels (digital), each transmitting at 1 Mbps, use a satellite channel of 1 MHz. Design an appropriate configuration, using FDM. Solution The satellite channel is analog. We divide it into four channels, each channel having 1M/4=250-kHz b in addition to width. Each digital channel of 1 Mbps must be transmitted over a 250KHz channel. Assuming no noise we can use Nyquist to get: C = 1Mbps = 2x250K x log2 L -> L = 4 or n = 2 bits/signal element. One solution is 4-QAM modulation. In Figure 6.8 we show a possible configuration with L = 16. Example 6.3

Figure 6.8 Example 6.3 Figure 6.9 Analog hierarchy The Advanced Mobile Phone System (AMPS) uses two b in addition to s. The first b in addition to of 824 to 849 MHz is used as long as sending, in addition to 869 to 894 MHz is used as long as receiving. Each user has a b in addition to width of 30 kHz in each direction. How many people can use their cellular phones simultaneously Solution Each b in addition to is 25 MHz. If we divide 25 MHz by 30 kHz, we get 833.33. In reality, the b in addition to is divided into 832 channels. Of these, 42 channels are used as long as control, which means only 790 channels are available as long as cellular phone users. Example 6.4

Figure 6.10 Wavelength-division multiplexing (WDM) WDM is an analog multiplexing technique to combine optical signals. Figure 6.11 Prisms in wavelength-division multiplexing in addition to demultiplexing

Figure 6.12 Time Division Multiplexing (TDM) TDM is a digital multiplexing technique as long as combining several low-rate digital channels into one high-rate one. Figure 6.13 Synchronous time-division multiplexing

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In synchronous TDM, the data rate of the link is n times faster, in addition to the unit duration is n times shorter. In Figure 6.13, the data rate as long as each one of the 3 input connection is 1 kbps. If 1 bit at a time is multiplexed (a unit is 1 bit), what is the duration of (a) each input slot, (b) each output slot, in addition to (c) each frame Solution We can answer the questions as follows: a. The data rate of each input connection is 1 kbps. This means that the bit duration is 1/1000 s or 1 ms. The duration of the input time slot is 1 ms (same as bit duration). Example 6.5 b. The duration of each output time slot is one-third of the input time slot. This means that the duration of the output time slot is 1/3 ms. c. Each frame carries three output time slots. So the duration of a frame is 3 × 1/3 ms, or 1 ms. Note: The duration of a frame is the same as the duration of an input unit. Example 6.5 (continued)

Figure 6.14 shows synchronous TDM with 4 1Mbps data stream inputs in addition to one data stream as long as the output. The unit of data is 1 bit. Find (a) the input bit duration, (b) the output bit duration, (c) the output bit rate, in addition to (d) the output frame rate. Solution We can answer the questions as follows: a. The input bit duration is the inverse of the bit rate: 1/1 Mbps = 1 s. b. The output bit duration is one-fourth of the input bit duration, or ¼ s. Example 6.6 c. The output bit rate is the inverse of the output bit duration or 1/(4s) or 4 Mbps. This can also be deduced from the fact that the output rate is 4 times as fast as any input rate; so the output rate = 4 × 1 Mbps = 4 Mbps. d. The frame rate is always the same as any input rate. So the frame rate is 1,000,000 frames per second. Because we are sending 4 bits in each frame, we can verify the result of the previous question by multiplying the frame rate by the number of bits per frame. Example 6.6 (continued) Figure 6.14 Example 6.6

Figure 6.18 Empty slots Figure 6.26 TDM slot comparison

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