Chapter 5 Multiplexing PDF

Chapter 5 Multiplexing 6.1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.  Note Bandwidth utilization is the wise use of available bandwidth to achieve spe...

Chapter 5 Multiplexing 6.1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.  Note Bandwidth utilization is the wise use of available bandwidth to achieve specific goals. Efficiency can be achieved by multiplexing; privacy and anti-jamming can be achieved by spreading. 6.2  MULTIPLEXING Whenever the bandwidth of a medium linking two devices is greater than the bandwidth 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 and telecommunications use increases, so does traffic. Topics discussed in this section: Frequency-Division Multiplexing Wavelength-Division Multiplexing Synchronous Time-Division Multiplexing Statistical Time-Division Multiplexing 6.3  Figure 6.1 Dividing a link into channels ◼ In a multiplexed system, n devices share the capacity of one link ◼ The sender direct the data stream to a multiplexer (MUX) and the stream is separated back into its components at the receiver by demultiplexer (DEMUX) ◼ The word path refers to the physical link and the word channel refers to a portion of path 6.4  Figure 6.1 Dividing a link into channels 6.5  Figure 6.2 Categories of multiplexing 6.6  Figure 6.3 Frequency-division multiplexing ◼ Signal is generated by each sending device modulate different frequencies ◼ Modulated signal are then combined into a single composite signal that can be transported by the link ◼ Channel are separated by unused bandwidth (guard band) to prevent signal from overlapping 6.7  Note FDM is an analog multiplexing technique that combines analog signals. 6.8  Figure 6.4 FDM process 6.9  Figure 6.5 FDM demultiplexing example 6.10  Example 6.1 Assume that a voice channel occupies a bandwidth of 4 kHz. We need to combine three voice channels into a link with a bandwidth of 12 kHz, from 20 to 32 kHz. Show the configuration, using the frequency domain. Assume there are no guard bands. Solution We shift (modulate) each of the three voice channels to a different bandwidth, as shown in Figure 6.6. We use the 20- to 24-kHz bandwidth for the first channel, the 24- to 28-kHz bandwidth for the second channel, and the 28- to 32-kHz bandwidth for the third one. Then we combine them as shown in Figure 6.6. 6.11  Figure 6.6 Example 6.1 6.12  Example 6.2 Five channels, each with a 100-kHz bandwidth, are to be multiplexed together. What is the minimum bandwidth of the link if there is a need for a guard band of 10 kHz between the channels to prevent interference? Solution For five channels, we need at least four guard bands. This means that the required bandwidth is at least 5 × 100 + 4 × 10 = 540 kHz, as shown in Figure 6.7. 6.13  Figure 6.7 Example 6.2 6.14  Applications of FDM Note Television broadcasting The coaxial cable has used a bandwidth of approximately 500Hz Each TV channel has its own 6 MHz bandwidth AM and FM radio broadcasting AM has a band from 530 to 1700 kHz. Each stations needs 10 kHz of bandwidth. FM has a band from 88 to 108 MHz. Each stations needs 200 kHz of bandwidth. 6.15  Note Today, a new method being developed to implement FDM over fiber optic called wavelength division multiplexing (WDM) WDM is an analog multiplexing technique to combine optical signals. Application of WDM is the SONET network; which multiple optical fiber lines are multiplexed and demultiplexed 6.16  Figure 6.10 Wavelength-division multiplexing 6.17  Figure 6.11 Prisms in wavelength-division multiplexing and demultiplexing 6.18  Note TDM is a digital multiplexing technique for combining several low-rate channels into one high-rate one. Can be applied when the data rate capacity of the transmission medium is greater than the data rate required by sending and receiving devices 6.19  Figure 6.12 TDM 6.20  Note In synchronous TDM, the data rate of the link is n times faster, and the unit duration is n times shorter. 6.21  Figure 6.13 Synchronous time-division multiplexing In Figure 6.13, the data rate for each 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, and (c) each frame? 6.22  Example 6.5 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). 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. The duration of a frame is the same as the duration of an input unit. 6.23  Figure 6.14 Example 6.6 Figure 6.14 shows synchronous TDM with a data stream for each input and one data stream for 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, and (d) the output frame rate. 6.24  Example 6.6 Solution We can answer the questions as follows: a. The input bit duration is the inverse of the bit rate: 1/1 Mbps = 1 μs. b. The output bit duration is one-fourth of the input bit duration, or ¼ μs. c. The output bit rate is the inverse of the output bit duration or 1/(4μs) 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. 6.25  Figure 6.15 Interleaving 6.26  Example 6.8 Four channels are multiplexed using TDM. If each channel sends 100 bytes /s and we multiplex 1 byte per channel, show the frame traveling on the link, the size of the frame, the duration of a frame, the frame rate, and the bit rate for the link. Solution The multiplexer is shown in Figure 6.16. Each frame carries 1 byte from each channel; the size of each frame, therefore, is 4 bytes, or 32 bits. Because each channel is sending 100 bytes/s and a frame carries 1 byte from each channel, the frame rate must be 100 frames per second. The bit rate is 100 × 32, or 3200 bps. 6.27  Figure 6.16 Example 6.8 6.28  Example 6.9 A multiplexer combines four 100-kbps channels using a time slot of 2 bits. Show the output with four arbitrary inputs. What is the frame rate? What is the frame duration? What is the bit rate? What is the bit duration? Solution Figure 6.17 shows the output for four arbitrary inputs. The link carries 50,000 frames per second. The frame duration is therefore 1/50,000 s or 20 μs. The frame rate is 50,000 frames per second, and each frame carries 8 bits; the bit rate is 50,000 × 8 = 400,000 bits or 400 kbps. The bit duration is 1/400,000 s, or 2.5 μs. 6.29  Figure 6.17 Example 6.9 6.30  Figure 6.18 Empty slots 6.31  Data rate management in TDM 1 problem with TDM is how to handle a disparity in the input data rates. 3 strategies / technique:- Multilevel multiplexing Multiple-slot allocation Pulse stuffing 6.32  Figure 6.19 Multilevel multiplexing 6.33  Figure 6.20 Multiple-slot multiplexing 6.34  Figure 6.21 Pulse stuffing 6.35  Figure 6.22 Framing bits ◼ One or more synchronization bit are added to the beginning of each frame ◼ Synchronization bit called framing ◼ This synchronization bit consists of 1 bits per frame, alternating between 1 and 0 6.36  Example 6.10 We have four sources, each creating 250 characters per second. If the interleaved unit is a character and 1 synchronizing bit is added to each frame, find (a) the data rate of each source, (b) the duration of each character in each source, (c) the frame rate, (d) the duration of each frame, (e) the number of bits in each frame, and (f) the data rate of the link. Solution We can answer the questions as follows: a. The data rate of each source is 250 × 8 = 2000 bps = 2 kbps. 6.37  Example 6.10 (continued) b. Each source sends 250 characters per second; therefore, the duration of a character is 1/250 s, or 4 ms. c. Each frame has one character from each source, which means the link needs to send 250 frames per second to keep the transmission rate of each source. d. The duration of each frame is 1/250 s, or 4 ms. Note that the duration of each frame is the same as the duration of each character coming from each source. e. Each frame carries 4 characters and 1 extra synchronizing bit. This means that each frame is 4 × 8 + 1 = 33 bits. 6.38  Applications of Synchronous TDM Note Telephone services Digital signal (DS) service or digital hierarchy T-Lines & E-Lines (europeans) 2nd generation of cellular telephone network Bandwidth of 30 kHz bands (6 users/band) 6.39  Note In statistical TDM, slots are dynamically allocated to improve bandwidth efficiency. The number of slots in each frame is less than the number of input lines. 6.40  Figure 6.26 TDM slot comparison 6.41  Figure 6.13 Statistical time-division multiplexing Slot size: - the ratio of the data size to address size must be reasonable to make transmission efficient. - usually block of data is many bytes while the address is just a few bytes. No synchronization bit Bandwidth: - the capacity of the link is normally less than the sum of the capacities of each channel. - Capacity of the link based on the statistics of the load for each channel. → during peak times, some slots need to wait. 6.42  Applications of Statistical TDM Note Telephone services Digital signal (DS) service or digital hierarchy T-Lines & E-Lines (Europeans) Digital TV broadcasting Data transmitting via LAN or WAN network 6.43  SUMMARY Discussed on: ✓ Concept of multiplexing. ✓ Multiplexing techniques: ✓ FDM – its process, minimum bandwidth, applications. ✓ WDM – its concepts and differences between WDM & FDM. ✓ Synchronous TDM – configuration and calculations, empty slots, data rate management, applications. ✓ Statistical TDM – data rate management, applications. 2.44  The Any Question End 1.45 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.