COMMUNICATION SYSTEM EEEB453 Chapter 7 (Part II) MULTIPLEXING & DEMULTIPLEXING

1. COMMUNICATION SYSTEM EEEB453Chapter 7 (Part II)MULTIPLEXING & DEMULTIPLEXING

2. Time Division Multiplexing • Definition – TDM is the time interleaving of samples from several sources so that the info from these sources can be transmitted serially over a single communication channel. • In brief, TDM is a digital multiplexing technique for combining several low-rate channels into one high-rate one. • Can be used for analog & digital information signal. Figure gives a conceptual view of TDM. Note that the same link is used as in FDM; here the link is sectioned by time rather than frequency

3. Synchronous vs Asynchronous TDM • TDM can be implemented in two ways, synchronous & asynchronous TDM. Synchronous TDM • In synchronous TDM, the data rate of the link is n times faster, and the unit duration is n times shorter.

4. Synchronous TDM • Both transmitter and receiver must be synchronized. • Also called fixed cycle operation i.e data carry in a repetitive frames • Each frame consists of a set of time slots and each source is assigned one or more time slots per frame. • The channel is divided into time slots and each user is allocated a slot – whether it is empty or not.

5. Synchronous TDM • TDM can be visualized as two fast rotating switches, one on the mux side and the other on the demux side. • The switches are synchronized and rotate at the same speed, but in opposite directions.

6. Asynchronous TDM (Statistical TDM) • Time slots are allocated on demand instead of a fixed cycle • Allows unused slots to be allocated to active users • Many more devices can be connected and provides more efficient use of the available BW. • Time slots are not preassigned to particular data sources. • Rather, user data are buffered and transmitted as rapidly as possible using available time slots

8. Synchronous vs Asynchronous TDM • Figure 8.12 shows four data sources produced in four time periods (t0, t1, t2, t3, t4). • In case of sync TDM, during each period, data are collected from all 4 sources. Resulted sources C and D produce no data. Thus 2 of 4 time slots transmitted by mux are empty. • For statistical TDM, the mux does not send empty slots if there are data to send. During the first period, only slot A and B are sent. Address info is required to assure proper delivery. Thus, more overhead per slot is needed because each slot carries an address as well as data.

9. TDM PCM System

10. TDM PCM System • Figure 3.35 illustrates the TDM concept applied to 3 analog sources multiplexed over a PCM system. • An electronic switch is used for the commutation (sampler) • Pulse width for TDM PAM is Ts/3 = 1/3fs • Pulse width for TDM PCM is Ts/3n where n is the number of bits used in PCM word and fs = 1/Tsi.e the freq of rotation • fs should satisfies the Nyquist rate for the analog source with the largest BW. • Larger BW sources may be connected to several switch position on the sampler. • At the Rx, the sampler has to be synchronized with the incoming waveform. • LPF are used to reconstruct the analog signals from the PAM signals

11. TDM of Digital Signal (Identical bit rate) • Multiplexing can be done on a bit-by-bit basis (bit or digit interleaving) or on a word-by-word basis ( byte or word interleaving)

12. TDM of Digital Signal (Different bit rate) • When the bit rates of incoming channels are not identical, the high-bit-rate channel is allocated proportionately more slots –(Figure c and d) • Figure show 4-channel multiplexing consisting of 3 channels (B, C and D) of identical bit rate R and 1 channel (A) with a bit rate of 3R

13. It is possible for a synchronous TDM device to handle sources of different data rates. For example, the slowest input device could be assigned one slot per cycle, while faster devices are assigned multiple slots per cycle. The most difficult problem in the design of a synchronous TDM system is the synchronization of various data sources. 3 types of handling different data rates: Multilevel multiplexing Multiple-slot multiplexing Pulse stuffing Handling Different Data Rates

14. Multilevel multiplexing Handling Different Data Rates

15. Multiple Slot Multiplexing Handling Different Data Rates

16. Pulse Stuffing With pulse stuffing, the outgoing data rate of the mux, is higher than the sum of the maximum instantaneous incoming rates. The extra capacity is used by stuffing extra dummy bits or pulses into each incoming signal until its rate is raised to that of a locally generally clock signal. The stuffed pulses (bits) are inserted at fixed locations in the mux frame format so that they may be identified and removed at the demux Handling Different Data Rates

17. Digital Carrier Signal (North American)

18. Digital Carrier Signal (North American) • A DS-0 service is a single digital channel of 64Kbps • DS-1 is a 1.544Mbps service – 1.544Mbps is 24 times 64kbps plus 8kbps of overhead. • It can be used to mux 24 DS-0 channels or to carry any other combination desired by the user that can fit within its 1.544Mbps capacity. • DS-2 is a 6.312Mbps service – 6.312Mbps is 96 times 64kbps plus 168kbps of overhead. • It can be used as a single service for 6.132Mbps tx or to mux 4 DS-1 channels, 96 DS-0 channels or to carry any other combination of these service type. • Etc..

19. Digital Carrier Signal (CCITT Standard)

20. Example Design a TDM that will accommodate 11 sources with this specification: • Source 1: Analog, 2 kHz bandwidth • Source 2: Analog, 4 kHz bandwidth • Source 3: Analog, 2 kHz bandwidth • Source 4-11: Digital, 7200 bps synchronous Suppose the analog sources are converted to digital using 4-bit PCM words.

21. Solution

22. Exercises: • In a certain telemetry system, there are four analog signals u(t), v(t), w(t) and x(t). The bandwidth of u(t) is 3.6kHz but those of the remaining signals are 1.4kHz each. These signals are to be sampled at rates no less that respective Nyquist rates and are to be analog TDM multiplexed. This can be achieved by multiplexing the PAM samples of the four signals and then binary coding the multiplexed samples. • Design a suitable multiplexing scheme for this purpose • What is the commutator frequency in rotation per second? • What is the total bit rate? • Repeat question 1 if there are four analog signals u(t), v(t), w(t) and x(t) with bandwidth of 1200Hz, 700Hz, 300Hz and 200Hz, respectively.

23. Exercises: • A signal u(t) is band-limited to 3.6kHz and the three other signals v(t), w(t) and x(t) are band-limited to 1.2kHz each. These signals are samples at their Nyquist rate and binary coded using 512 levels (L=512). • Determine the individual Nyquist sampling rate • Find the individual bit rate required for each of the signal • Design a suitable multiplexing scheme for this purpose • What is the commutator frequency in rotation per second? • What is the total bit rate?

24. Four signals m(t), n(t), p(t), and q(t) have bandwidths of 600Hz, 1200 Hz, 600Hz and 1500Hz respectively. These signals are sampled at their Nyquist rates, quantised and then binary coded using 8 bits before multiplexing. Design a bit-by-bit multiplexing scheme for these signals using two commutators. Label your diagram fully. What are the commutator frequencies in rotations per second? What is the total bit rate? FE Nov 06

25. A coaxial transmission line passes frequencies in the range 0 – 50 MHz. Determine the maximum number of speech channels, each of bandwidth limited to 4 kHz, that can be transmitted over the transmission line via : a) A combination of FDM/TDM is used, and each frequency channel can allocate 8 users. Assume the guard bands of 1 kHz exist between adjacent channels. b) A time division multiplexed PCM system. Assume the following: sampling frequency equals 1.2 times Nyquist frequency; 6 bit PCM words; each binary pulse occupies 50% of its assigned time slot. FE Sem 1 10/11