ITT300 - CHAPTER 5 MULTIPLEXING.pdf

ITT300 INTRODUCTION TO DATA COMMUNICATION AND NETWORKING CHAPTER 5 MULTIPLEXING ADAPTED FROM: Behrouz A. Forouzan Multiplexing Techniques that allows the simultaneous transmission of multiple signals across a single data link If the bandwidth o...

ITT300 INTRODUCTION TO DATA COMMUNICATION AND NETWORKING CHAPTER 5 MULTIPLEXING ADAPTED FROM: Behrouz A. Forouzan Multiplexing Techniques that allows the simultaneous transmission of multiple signals across a single data link If the bandwidth of a link is greater than the bandwidth needs of the devices connected to it, the bandwidth is wasted. Multiplexing n lines share the bandwidth of one link Multiplexer (MUX) – combines lines into a single stream (many-to-one) Demultiplexer (DEMUX) – separates the stream back into its component transmission (one-to- many)and directs them to their corresponding lines 3 Multiplexing The word link refers to the physical path. The word channel refers to the portion of a link that carries a transmission between a given pair of lines. One link can have many n channels. There are three basic multiplexing techniques: 4 Frequency Division Multiplexing FDM – an analog technique used when the bandwidth of a link (in hertz) is greater than the combined bandwidths of the signals to be transmitted. In FDM, signals generated by each sending device modulate different carrier frequencies. These modulated signals are then combined into a single composite signal that can be transported by the link. 5 FDM Carrier frequencies are separated by sufficient bandwidth to accommodate the modulated signal Channels can be separated by strips of unused bandwidth – guard bands- to prevent signals from overlapping 6  Multiplexing process Each source generates a signal of a similar frequency range Inside the multiplexer, these similar signals modulates different carrier frequencies (f1, f2, f3) The resulting modulated signals are then combined into a single composite signal that is sent out over a media link that has enough bandwidth to accommodate it. 7 Demultiplexing process The demultiplexer uses a series of filters to decompose the multiplexed signal into its constituent component signals The individual signals are then passed to a demodulator that separates them from their carriers and passes them to the output lines 8  Example 1.1 Assume that a voice channel occupies a bandwidth of 4kHz. We need to combine three voice channels into a link with a bandwidth of 12 kHz, from 20 to 32kHz. Show the configuration, using the frequency domain. Assume there are no guard bands. 9 Example 1.2 Five channels, each with a 100kHz bandwidth, are to be multiplexed together. What is the minimum bandwidth of the link if there is a need for a guard band of 10kHz 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 x 100 + 4 x 10 = 540kHz 10 Example 1.3 Noraini & Co, consist in total 4 worker. If all of the worker have their own voice channel with a bandwidth of 15-kHz each. Calculated the minimum bandwidth required if the multiplexing concept is applied in the company. Assume that the guard band used is 3-kHz. Solution For four channels, we need at least three guard bands. This means that the required bandwidth is at least :(4 ×15)+(3 × 3) = 69 kHz OTHER APPLICATION OF FDM A familiar application of FDM is cable television and AM and FM radio broadcasting. The coaxial cable used has a bandwidth of approximately 500 MHz. Individually television channel requires about 6 MHz (theoretically 83 channels provided by coaxial). [83 channel x 6 Mhz = 498 Mhz] A DEMUX at our television allow us to select which of those channels we wish to receive. Wavelength-Division Multiplexing (WDM) Used the high-data-rate capability of fiber-optic cable The fiber-optic data rate is higher than the data rate of metallic transmission cable Using a fiber-optic cable for one single line wastes the available bandwidth WDM is conceptually the same as FDM, except that the multiplexing and demultiplexing involve optical signals transmitted through fiber-optic channels The difference is that the frequencies are very high 13  Figure 3.6 Wavelength-division multiplexing Prisms in wavelength-division multiplexing and demultiplexing WDM Very narrow bands of light from different sources are combined to make a wider band of light At the receiver, the signals are separated by the demultiplexer The combining and splitting of light sources are easily handled by a prism A prism bends a beam of light based on the angle of incidence and the frequency 15  Synchronous Time-Division Multiplexing (TDM) Statistical ▪ TDM is a digital process that allows several connections to share the high bandwidth of a link. ▪ TDM is a digital multiplexing technique for combining several low-rate channels into one high-rate one. ▪ Instead of sharing a portion of the bandwidth as in FDM, the time is shared 16 Synchronous TDM In synchronous TDM, the data flow of each input connection is divided into units, where each input occupies one input time slot. A unit can be 1 bit, one character, or one block of data. Each input unit becomes one output unit and occupies one output time slot. Sends signal equally At the same time Not most efficient 17 Synchronous TDM Time slots are grouped into frames. A frame consists of one complete cycle of time slots, with one slot dedicated to each sending device. If the duration of the input unit is T, the duration of each slot is T/n and the duration of each frame is T. The data rate of the output link must be n times the data rate of a connection to guarantee the flow of data. 18  Example 2.1 The duration of a frame is the same as the duration of an input unit. The data rate for each input connection is 1kbps. According to the figure below, 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? Solution: 1. The data rate of each input connection is 1 kbps. This means that the bit duration is 1/1000 s or 1 ms OR 0.001 s 2. The duration of each output time slot is one-third of the input time slot (data rate out output link is 3x faster). This means that the duration of the output time slot is 1/3 ms. 3. Each frame carries three output time slots. So the duration of a frame is 3 × 1/3 ms, or 1 ms. 19 Example 2.2 Four channels are multiplexed using TDM. If each channel sends 100 bytes /s and we multiplex 1 byte per channel, a. show the frame traveling on the link, b. the size of the frame, c. the duration of a frame, the frame rate, d. the bit rate for the link. 20 Solution Figure below shows the multiplexer. 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, frame rate must be 100 frames per second. The bit rate is 100 × 32, or 3200 bps.  TDM Interleaving Process TDM can be visualized as two fast-rotating switches, one on each side of a link. The switches are synchronized and rotate at the same speed, but in opposite directions. On the multiplexing side, as the switch opens in front of a connection, that connection has the opportunity to send a unit. This process is called interleaving. On the demultiplexing side, as the switch opens in front of a connection, that connection has the opportunity to receive a unit. 22 Empty Slots Synchronous TDM is not as efficient as it could be. If a source does not have data to send, the corresponding slot in the output frame is empty. One problem with TDM is how to handle a disparity in the input data rates. Previously, we assumed that the data rates of all input lines were the same. However, if data rates are not the same, three strategies can be used. 1.Multilevel multiplexing 2.Multiple-slot allocation 3.Pulse stuffing 23  Data Rate Management Multilevel Multiplexing A technique used when the data rate of an input line is a multiple of others For example, we have two inputs of 20kbps and three inputs of 40kbps The first two input lines can be multiplexed together to provide a data rate equal to the last three A second level of multiplexing can create an output of 160kbps 24  Multiple-Slot Allocation Allot more than one slot in a frame to a single input line For example, an input line that has a data rate that is multiple of another input The input line with a 50-kbpsdata rate can be given two slots in the output 25 Pulse Stuffing/ Bit Padding/ Bit Stuffing Sometimes the bit rates of sources are not multiple integers of each other One solution is to make the highest input data rate the dominant data rate and then add dummy bits to the input lines with lower rates 26 Frame Synchronizing In TDM, synchronization between multiplexer and the demultiplexer is very important. If they are not synchronized, a bit belonging to one channel may be received by the wrong channel One or more synchronization bits are usually added to the beginning of each frame (called framing bits). 27 Frame Synchronizing Framing bits allow the demultiplexer to synchronize with the incoming stream so that it can separate the time slots accurately In most cases, this synchronization information consists of 1 bit per frame, alternating between 0 and 1. 28 Example 2.3 We have four sources, each creating 250 8-bit 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, (f) the data rate of the link.  We have four sources, each creating 250 characters per second. If the interleaved Solution unit is a character and 1 synchronizing bit is added to each frame. 1 byte = 8 bits We can answer the questions as follows: a. The data rate of each source is 250 × 8 = 2000 bps = 2 kbps. 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.  Example 2.4 Two channels, one with a bit rate of 100 kbps and another with a bit rate of 200 kbps, are to be multiplexed. a) How this can be achieved? Channel A: Bit rate = 100 kbps = 1 slot per frame b) What is the frame rate? Channel B: Bit rate = 200 kbps = 2 slot per frame c) What is the bit rate of the link? Channel A will contribute 1 bit and Channel B will contribute 2 bits. Solution: We can allocate one slot to the first channel and two slots to the second channel. Each frame carries 3 bits. The frame rate is 100,000 frames per second because it carries 1 bit from the first channel. The bit rate is 100,000 frames/s × 3 bits per frame, or 300 kbps. 31 Statistical Time-Division Multiplexing In synchronous TDM, each input has a reserved slot in the output frame. This can be inefficient if some input lines have no data to send. In statistical time-division multiplexing, slots are dynamically allocated to improve bandwidth efficiency. Only when an input line has a slot's worth of data to send is it given a slot in the output frame. In statistical multiplexing, the number of slots in each frame is less than the number of input lines. The multiplexer checks each input line in round-robin fashion; it allocates a slot for an input line if the line has data to send; otherwise, it skips the line and checks the next line. 32 Statistical Time-Division Multiplexing In statistical TDM, a slot needs to carry data as well as the address of the destination. The addressing in its simplest form can be n bits to define different output lines with n = log2 N. For example, for eight different output lines, we need a 3-bit address. 33 TDM comparison Slot comparison 34 TDM Comparison Addressing comparison In synchronous TDM, there is no need for addressing; synchronization and pre-assigned relationships between the inputs and outputs serve as an address In statistical TDM, there is no fixed relationship between the inputs and outputs because there are no reserved slots. We need to include the address of the receiver inside each slot. 35 TDM Comparison Slot Size comparison ▪ Since a slot carries both data and an address in statistical TDM, the ratio of the data size to address size must be reasonable to make transmission efficient. ▪ For example, it would be inefficient to send 1 bit per slot as data when the address is 3 bits. ▪ In statistical TDM, a block of data is usually many bytes while the address is just a few bytes. 36 TDM Comparison Synchronization Bit comparison No Synchronization Bit in statistical TDM. The frames in statistical TDM need not be synchronized, so we do not need synchronization bits. Bandwidth comparison In statistical TDM, the capacity of the link is normally less than the sum of the capacities of each channel. 37