Lecture 3

Applications of TDM ( T & E Lines ) &

Statistical TDM

Digital Signal Service

• Telephone companies implement TDM through a hierarchy of signals, called Digital Signal (DS) service or digital hierarchy.

• Figure given next shows data rates supported by each level.

6.3

Figure: Digital hierarchy

T Lines• DS-0 & DS-1 are the names of the services.• Telecom companies use T lines (T-1 to T-4).• Capacity of these lines precisely matches with DS-1 to DS-4

services.• Today T-1 & T-3 are commercially available.

T Lines for Analog transmission

• T lines are digital lines• They can be used for analog service such as

voice calling• For this purpose analog signal is first sampled

& then time division multiplexed.

6.6

Figure: T-1 line

6.7

Figure: T-1 line frame structure

E Lines

• E Lines are European version of T lines.• T lines & E lines are conceptually same

• STDM, or statistical time division multiplexing, is one method for transmitting several types of data simultaneously across a single transmission cable or line (such as a T1 or T3 line). STDM is often used for managing data being transmitted via a local area network (LAN) or a wide area network (WAN). In these situations, the data is often simultaneously transmitted from any number of input devices attached to the network, including computers, printers, or fax machines.

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.

Statistical Time Division Multiplexing

Statistical Time Division Multiplexing

• Statistical TDM is a more flexible method of TDM. With static TDM the length of time allocated is not fixed for each device but time is given to devices that have data to transmit.

• STDM can also be used in telephone switchboard settings to manage the simultaneous calls going to or coming from multiple, internal telephone lines

Example• a busy laser printer shared by many users might need to

receive or transmit data 80-90% of the time at a much higher transmission rate than a seldom-used, data-entry computer attached to the same T-1 line. With TDM, even though the printer's transmission needs are greater, both devices would still be allocated the same duration of time to transmit or receive data.

• In comparison to TDM, the STDM method analyzes statistics related to the typical workload of each input device (printer, fax, computer) and determines for proper functioning how much time each device should be allocated for data transmission on the cable or line.

• In the above example, STDM would allocate more time to the group printer, based on its past and current transmission needs and less time to the data-entry computer. Many believe the STDM method is a more efficient use of total bandwidth available than the TDM method.

• The main statistics used in STDM are: each input device's peak data rates (in kbps, or kilobytes per second), and each device's duty factors (which is the percentage of time the device typically spends either transmitting or receiving data).

6.14

Figure: TDM slot comparison

Asynchronous TDMFigure

WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998

Frames and AddressesFigure

a. Only three computer/lines out of 5 are sending data

Frames and AddressesFigure -continued

b. Only four lines sending data

Frames and AddressesFigure -continued

c. All five lines sending data

Animation of STDM

• http://www.mhhe.com/engcs/compsci/forouzan/dcn/graphics/animations/08_17.swf

addressing• Figure also shows a major difference between slots in synchronous TDM

and statistical TDM. An output slot in synchronous TDM is totally occupied by data; in statistical TDM, a slot needs to carry data as well as the address of the destination. In synchronous TDM, there is no need for addressing; synchronization and preassigned relationships between the inputs and outputs serve as an address. We know, for example, that input 1 always goes to input 2. If the multiplexer and the demultiplexer are synchronized, this is guaranteed. In statistical multiplexing, there is no fixed relationship between the inputs and outputs because there are no preassigned or reserved slots. We need to include the address of the receiver inside each slot to show where it is to be delivered. The addressing in its simplest form can be n bits to define N different output lines with n = log2N. For example, for eight different output lines, we need a 3-bit address.

Slot size

• 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. This would mean an overhead of 300 percent. In statistical TDM, a block of data is usually many bytes while the address is just a few bytes.

No Synchronization Bit• There is another difference between synchronous and statistical TDM,

but this time it is at the frame level. The frames in statistical TDM need not be synchronized, so we do not need synchronization bits.

Bandwidth• In statistical TDM, the capacity of the link is normally less than the sum

of the capacities of each channel. The designers of statistical TDM define the capacity of the link based on the statistics of the load for each channel. If on average only x percent of the input slots are filled, the capacity of the link reflects this. Of course, during peak times, some slots need to wait.

Comparison of data multiplexer techniques

Sr.No.

Parameter FDM Synchronous TDM

Statistical TDM

1 Line utilization efficiency

Poor Good Very good

2 Flexibility Poor Good Very good

3 Channel capacity Poor Good Excellent

4 Error control Not possible

Not possible Possible

5 Multidrop capacity Very good Difficult to achieve

Possible

6 Transmission delay

Does not exist

Low Random

7 Cost High Low Moderate

Animation of TDM Switching

• http://www.mhhe.com/engcs/compsci/forouzan/dcn/graphics/animations/14_10.swf

• http://www.mhhe.com/engcs/compsci/forouzan/dcn/graphics/animations/14_11.swf

Q.38 A host is connected to 16 asynchronous terminals through a pair of statistical time division multiplexers utilizing the bit map protocol. The sixteen asynchronous terminal ports operate at 1200 bps. The line port has a bit rate of 9600 bps. The data link control protocol is HDLC.

(i) Calculate the maximum line utilization efficiency and throughput.(ii) Will there be any queues in the stat MUXIf the average character rate at all the ports in cps?If the host sends full screen display of average 1200 characters to each terminal?How much time will the stat MUX take to clear the queues? Ans. (i) Since HDLC frame transmitted on the line contains seven overhead bytes, the line

utilization efficiency is given as – E = 16/(7+16) = 0.696 Throughput T = E x 9600 = 16/(7+16)x9600 = 6678 bps

(ii) (a)Aggregate average input = 16 x 10= 160 cps = 160 x 8 = 1280 bps

Since the throughput is 6678 bps, it is very unlikely there will be queues at the terminal ports.

(b) With start and stop bits, the minimum size of a character is 10 bits. Thus, at

1200 bps, the host will take 10 seconds to transfer 1200 characters of one screen of a terminal. The stat MUX will get 1200 x 16 = 19200 characters in 10 seconds from the host. The throughput is

6678 bps = 6678/8 = 834.75 cpsThe stat MUX will transmit 834.75 x 10 characters in 10 seconds. Thus, the

queue at the end of 10 seconds = 19200 – 8347.5 = 10852.5 characters

• (iii) The stat MUX will take 10852.5/834.75 = 13 seconds

• to clear the queue. These are the additional seconds.

Q39. A simple system is designed to multiplex 20 users. Each user has a 20 kHz bandwidth. Sampling is at twice the Nyquist rate and the digitization is to 8 bits.(i) What bit rate does each user signal require?(ii) What clock rate does the multiplexer require?(iii) The resolution is increased to 16 bits, what multiplexer clock rate is needed. Ans. (i) Since, the system is designed to multiplex 20 users with each user having 20 kHz bandwidth. Thus, overall bandwidth requirement = 20 x 20 = 400 kHz. Further, the sampling is at twice the Nyquist rate and the digitization is to 8 bits. Therefore the overall resultant bit rate for 20 users

= 4 x 400 x 8 = 12800 kbps. The bit rate required by each user signal = (12800/20) kbps = 640 kbps

(ii) The desired clock rate = 1/f = 1/(1/640 kbps) = 640 kHz

(iii) If the resolution is increased to 16 bits then

each user signal require the bit rate of (16 x 20 x 20 x 4/20) = 25600/20 = 1280 kbps Thus, the desired clock rate = 1/(1/1280) = 1280

kHz