Multiplexing and Data Rate Assignment

Multiplexing and Data Rate Assignment Words: 2133

Multiplexing CHAPTER 4 Outline * Frequency Division Multiplexing(FDM) * Synchronous Time Division Multiplexing * Statistical Time Division Multiplexing * Asymmetric Digital Subscriber Line(ADSL) Multiplexing * Set of techniques that allows the simultaneous transmission of multiple signals across a single link * allows several transmission sources to share a larger transmission capacity Link = physical path Channel = portion of a link that carries a transmission between a given pairs of lines 2 CATEGORY OF MULTIPLEXING WDM FDM TDM ADSL Frequency Division Multiplexing * FDM – numerous signals are combined for ransmission on a single communications line or channel. Each signal is assigned a different frequency (subchannel) within the main channel. * Useful bandwidth of medium exceeds required bandwidth of channel * e. g. broadcast radio and cable television * Channel allocated even if no data Frequency Division Multiplexing Diagram * Each signal is modulated to a different carrier frequency * Carrier frequencies separated by guard bands (unused bandwidth) – to prevent interference so signals do not overlap. 3 FDM System FDM is an analog multiplexing technique that combines signals. FDM process

FDM Demultiplexing Example 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 without the use of guard bands. Solution Shift (modulate) each of the three voice channels to a different bandwidth, as shown in next figure Example 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

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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 = 540 KHz, as shown in next Figure. Example 3 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 a 250-KHz bandwidth. Each digital channel of 1 Mbps is modulated such that each 4 bits are modulated to 1 Hz. One solution is 16- QAM modulation. Figure 6. 8 shows one possible configuration. Analog Carrier Systems AT&T (USA) * Hierarchy of FDM schemes * Group 1. 12 voice channels (4kHz each) = 48kHz 2. Range 60kHz to 108kHz * Supergroup 1. 60 channel 2. FDM of 5 group signals on carriers between 420kHz and 612 kHz * Mastergroup 1. 10 supergroups Analog Hierarchy To maximize the efficiency infrastructure, multiplexed signals from lower bandwidth lines onto higher-bandwidth signals FDM of Three Voiceband Signals FDM Applications 1. Common used radio broadcasting – AM and FM * AM (530 – 1700KHz) – shared with all radio stations * FM uses a wider band (88 – 108MHz) – each station needs more bandwidth, 200KHz 2.

Television Broadcasting * Each TV channel has own bandwidth of 6 Mhz 3. 1st Generation of Cellular telephones * Voice signal 3KHz (300 – 3300Hz) channels * Bt = 10 x Bm , therefore each channel has 30KHz channels * each user has been allocated two 30KHz channel, therefore 60KHz. Example 4 The Advanced Mobile Phone System (AMPS) uses two bands. The first band, 824 to 849 MHz, is used for sending; and 869 to 894 MHz is used for receiving. Each user has a bandwidth of 30 KHz in each direction. The 3- KHz voice is modulated using FM, creating 30 KHz of modulated signal. How many people can use their ellular phones simultaneously? Solution Each band is 25 MHz. If we divide 25 MHz into 30 KHz, we get 833. 33. In reality, the band is divided into 832 channels. Wavelength Division Multiplexing * Multiple beams of light at different frequency * Carried by optical fiber * A form of FDM (same concept) * Each colour of light (wavelength) carries separate data channel * 1997 Bell Labs * 100 beams * Each at 10 Gbps * Giving 1 terabit per second (Tbps) * Commercial systems of 160 channels of 10 Gbps now available * Lab systems (Alcatel) 256 channels at 39. 8 Gbps each * 10. 1 Tbps * Over 100km WDM Operation Same general architecture as other FDM * Number of sources generating laser beams at different frequencies * Multiplexer consolidates sources for transmission over single fiber * Optical amplifiers amplify all wavelengths * Typically tens of km apart * Demux separates channels at the destination * Mostly 1550nm wavelength range * Was 200MHz per channel * Now 50GHz Dense Wavelength Division Multiplexing * DWDM * No official or standard definition * Implies more channels more closely spaced that WDM * 200GHz or less TDM digital process that allows several connections to share the high bandwidth of a link ach connection occupies a portion of time in the link TDM is a digital multiplexing technique to combine data. TDM : Time Slots and Frames In a TDM, the data rate of the link is n times faster, and the unit duration is n times shorter. Time Division Multiplexing Example 5 Four 1-Kbps connections are multiplexed together. A unit is 1 bit. Find : (1) the duration of 1 bit before multiplexing, (2) the transmission rate of the link, (3) the duration of a time slot, and (4) the duration of a frame? Solution We can answer the questions as follows: 1. The duration of 1 bit is 1/1 Kbps, or 0. 001 s (1 ms). 2.

The rate of the link is 4 Kbps. 3. The duration of each time slot 1/4 ms or 250 ms. 4. The duration of a frame 1 ms. Interleaving • switches are synchronized and rotate at the same speed but opposite direction •process of sending a unit data onto the path when the connection on the multiplexing and de-multiplexing is open Example 6 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

Example 7 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 7 shows the output for four arbitrary inputs. Synchronous Time Division Multiplexing * Possible when data rate of medium exceeds data rate of digital signal to be transmitted * Multiple digital signals interleaved of each signal in time during transmission * Interleaving can be at bit level of blocks * Time slots preassigned to sources and fixed Time slots allocated even if no data * Time slots do not have to be evenly distributed amongst sources Synchronous TDM System TDM Link Control * No headers and trailers * Data link control protocols not needed * Flow control * Data rate of multiplexed line is fixed * If one channel receiver can not receive data, the * others must carry on * The corresponding source must be quenched * This leaves empty slots * Error control Errors are detected and handled by individual channel systems Data Link Control on TDM Framing •Time slot length = transmitter buffer length •Each frame contains a cycle of time slot Framing bits follow a pattern to ensure incoming stream synchronized with demux to separate time slots accurately * No flag or SYNC characters bracketing TDM frames to manage the overall TDM link * Must provide synchronizing mechanism * Added digit framing * One control bit added to each TDM frame * Looks like another channel – “control channel” * Identifiable bit pattern used on control channel * e. g. alternating bit pattern 01010101…unlikely to be sustained on a data channel * Can compare incoming bit patterns on each * channel with sync pattern * If pattern not match, successive bit position re search until persist over multiple frame * When establish, receiver continue monitoring the framing bit channel * If the pattern break, the receiver must again enter a framing search mode Example 8 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 (1) the data rate of each source, (2) the duration of each character in each source, (3) the frame rate, (4) the duration of each frame, (5) the number of bits in each frame, and (6) the data rate of the link. Solution Answer as follows: 1.

The data rate of each source is 250 x 8 = 2000 bps = 2 Kbps. 2. The duration of a character is 1/250 s = 4 ms. 3. The link needs to send 250 frames per second. 4. The duration of each frame is 1/250 = 4 ms. 5. Each frame is 4 x 8 + 1 = 33 bits. 6. The data rate of the link is 250 x 33 = 8250 bps. Example 9 Two channels, one with a bit rate of 100 Kbps and another with a bit rate of 200 Kbps, are to be multiplexed. How this can be achieved? What is the frame rate? What is the frame duration? What is the bit rate of the link? 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 frame duration is 1/100,000 = 10 ms. The bit rate is 100,000 frames/s x 3 bits/frame = 300 Kbps. Pulse Stuffing * Problem – Synchronizing various data sources * Each source has separate clock, variation among clock cause loss synchronization * Data rates from different sources not related by simple rational number * Solution – Pulse Stuffing as effective remedy * Outgoing data rate (excluding framing bits) higher than sum of incoming rates * Stuff extra dummy bits or pulses into each ncoming signal until it matches local clock * Stuffed pulses inserted at fixed locations in frame and removed at demultiplexer TDM of Analog and Digital Sources Digital Carrier Systems * Long distance carrier system * Designed to transmit voice signal over high capacity transmission lonks usuch as optical fiber, coax and microwave * AT&T developed hierarchy of TDM structures of various capacities * USA/Canada/Japan use one system * ITU-T use a similar (but different) system * US system based on DS-1 format which Multiplexes 24 channels * Each frame has 8 bits per channel plus one raming bit, 24 x 8 +1 =193 bits per frame Digital Carrier Systems (2) * For voice each channel contains one word of digitized data (using PCM at 8000 samples per sec) * Data rate 8000×193 = 1. 544Mbps * Five out of six frames have 8 bit PCM samples * Sixth frame is 7 bit PCM word plus signaling bit * Signaling bits form stream for each channel containing control and routing info * Same format for digital data * 23 channels of data * 7 bits per frame plus indicator bit for data or systems control * 24th channel is sync DS Hierarchy TDM Carrier Standard North American and International TDM Carrier Standards

Statistical TDM * In Synchronous TDM many slots are wasted * Statistical TDM allocates time slots dynamically based on demand * Multiplexer scans input lines and collects data until frame full * Data rate on line lower than aggregate rates of input lines * Improve efficiency is to allow multiple data resources to be packed in one single frame Statistical TDM Frame Formats Performance * Output data rate less than aggregate input data rates cause by average amount of input < capacity of multiplexed line * Difficulty: May cause problems during peak periods when the input exceed capacity * Solution Buffer inputs to hold temporary excess input * Keep buffer size to minimum to reduce delay Cable Modem Outline * To support data transfer to and from a cable modem, two channels from cable TV provider dedicated to data transfer * One in each direction * Each channel shared by number of subscribers * Scheme needed to allocate capacity * Statistical TDM Cable Modem Operation Downstream * Cable scheduler delivers data in small packets * If more than one subscriber active, each gets fraction of downstream capacity • May get 500kbps to 1. 5Mbps * Also used to allocate upstream time slots to subscribers

Upstream * User requests timeslots on shared upstream channel • Dedicated slots for this * Headend scheduler sends back assignment of future time slots to subscriber Cable Modem Scheme Asymmetrical Digital Subscriber Line ADSL Link between subscriber and network * Local loop Uses currently installed twisted pair cable * Can carry broader spectrum * 1 MHz or more ADSL Design Asymmetric -Greater capacity downstream than upstream -Expected for video on demand and related services – high speed access -Users require higher capacity for downstream than upstream Frequency division multiplexing (FDM) element of ADSL strategy • Reserve lowest 25kHz for voice – Plain old telephone service (POTS) – Voice carried only 0 – 4KHz band – Additional bandwidth for prevent crosstalk between voice and data channel • Use echo cancellation or FDM to give two bands – smaller upstream band and larger downstream band • Use FDM within upstream and downstream bands – Single bit stream multiple parallel bit streams – each portion carried separate frequency band Range up to 5. 5km -Depending of diameter cables and quality ADSL Channel Configuration Discrete Multitone * DMT * Use multiple carrier signals at different requencies * Sending some bits on each channel * Available transmission band =4kHz subchannels * Send test signal and use subchannels with better signal to noise ratio * ASL/DMT design employ 256 downstream subchannels at 4kHz (60kbps) * Possible to transmit at rate15. 36MHz * Impairments bring this down to 1. 5Mbps to 9Mbps DTM Bits Per Channel Allocation •Each channel can carry data rate from 0-60 Kbps •Shows increasing attenuation and decreasing signalto- noise ratio at higher frequencies DMT Transmitter xDSL * High data rate DSL * Single line DSL * Very high data rate DSL

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