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TEKNIK PENYAMBUNGAN IT-041246 P2-7 | Evolusi dan Jenis-Jenis Sistem Switching.

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Presentation on theme: "TEKNIK PENYAMBUNGAN IT-041246 P2-7 | Evolusi dan Jenis-Jenis Sistem Switching."— Presentation transcript:

1 TEKNIK PENYAMBUNGAN IT-041246 P2-7 | Evolusi dan Jenis-Jenis Sistem Switching

2 Switching  (definition) “The establishment on demand, of an individual connection from a desired inlet to a desired outlet within a set of inlets and outlets for as long as is required for the transfer of information.”

3 Intro to Switching  A switch routes a call based on a number system – e.g13023696923 – access code – area code – exchange code – subscriber code

4 Intro to Switching Local End Office Tandem Switch Trunk Group Transit End Office Transit Concentrator Transit Local (line-to-line) switching Transit (tandem) Call distribution

5 Switching System Switching Matrix Subscriber Lines TrunksTrunks Signaling Control

6 Essential Switch Functions Interconnection Control Alerting Attending Information receiving Information sending Busy Testing Supervising

7 Functions of Switching Systems  Signaling – monitor line activity – send incoming information to control function – send control signals to outgoing lines  Control – process signaling and set-up/knock down connections  Switching – make connections between input and output lines

8 Basic Switch Requirements  A switch must be able to connect any incoming call to one of a multitude of outgoing calls  A switch must have the ability to hold and terminate calls  A switch has to prevent new calls from intruding into circuits already in use

9 General Switch Requirements  Speed of call setup should be kept short relative to the call holding time  Grade of service should be high –.99 overall –.95 busy-hour  HIGH availability!

10 Switching Methods  Connectivity – Full: any input to any output  Blocking – Blocking: Possibility exists that call setup may fail due to insufficient switching resources – Non-blocking: If any input I j and output O j are free, they can be connected

11 Time Division Switching  Time mapping of inputs and outputs I1I1 I1I1 I2I2 I2I2... InIn InIn O1O1 O1O1 O2O2 O2O2 O 36...

12 Space Division Switching  Spatial mapping of inputs and outputs – Used primarily in analog switching systems Space I1I1 InIn............ O1O1 OmOm

13 Space-Time-Space Switching I1I1 I1I1 I2I2 I2I2... InIn InIn O1O1 O1O1 O2O2 O2O2 O 36...

14 Single Stage Switches  Crosspoint switches – Complex - many crosspoints ( i × j ) – Poor utilization of crosspoints – Not fault tolerant

15 Multiple Stage Switches  Input connected to output via two or more smaller switches  Crosspoints shared by several possible connections (potential for blocking)  Possible to provide multiple paths between any 2 ports

16 Three Stage Switch Matrix (Multiple Stage Switch example)

17 Reducing Cross-point Complexity  Increase number of stages  Allow some blocking  Switch in more than one dimension

18 Number of Cross-points Non-blocking Switch

19 19 The basic idea of crossbar switching is to provide a matrix of n×m sets of contacts with only n+m activators or less to select one of the n×m sets of contacts. This type of switching is also known as coordinate switching as the switching contacts are arranged in a xy- plane. Principal of Crossbar Switching

20 20 A set of horizontal and vertical wires (shown by solid lines) A set of horizontal and vertical contact points connected to these wires. The contact point form pairs, each pair consisting of a bank of three or four horizontal and a corresponding bank of vertical contact points. A contact point pair acts as a crosspoint switch. The contact points are mechanically mounted and electrically insulated on a set of horizontal and vertical bars shown as dotted lines. The bars, in turns are connected to a set of electromagnets. Components of Crossbar switch

21 21 The crosspoint switches remains separated or open when not in use. When an electromagnetic, say in the horizontal direction, is energized, the bar attached to it slightly rotates in such a way that the contact points attached to the bar move closer to its facing contact points make do not actually make any contact. Now if an electromagnetic in the vertical direction is energized, the corresponding bar rotates causing the contact points at the intersection of the two bars to close. This happens because the contact points move towards each other. Working of Crossbar Switch

22 22 Cont. As an example, if electromagnets M2 and M3 / are energized, a contact is established at the crosspoint 6 such that the subscriber B is connected the subscriber C.

23 23 Energizing sequence for latching the crosspoints Let us consider a 6×6 crossbar schematic shown below.

24 24 Let us consider the establishment of the following connections in sequence: A to C and B to E. First the horizontal bar A is energized. Then the vertical bar B is energized The crosspoint AC is latched and the conversation between A and C can now proceed. Suppose we now energize the horizontal bar of B to establish the connection B-E, the crosspoint BC may latch and B will be brought into the circuit of A-C. This is prevented by an energizing sequence for latching the crosspoints. A crosspoint latches only if the horizontal bar is energized first and then the vertical bar. Cont.

25 25 Cont. In order to establish the connection B-E, the vertical bar E need to be energized after the horizontal bar is energized. In this case the crosspoint AE may latch as the horizontal bar A has already been energized for establishing the the connection A-C. This should also be avoided and is done by de- energizing the horizontal bar A after the crosspoint is latched and making a suitable arrangement such that the latch is maintained even though the energisation in the horizontal direction is withdrawn. The crosspoint remains latched as long as the vertical bar E remains energized.

26 26 Cont. The complete procedure for establishing a connection in a crossbar switch: 1.energize horizontal bar 2.energize vertical bar 3.de-energize horizontal bar

27 27 Design parameters In a non-blocking crossbar configurations, there are N 2 switching elements for N subscribers. When all the subscribers are engaged, only N/2 switches are actually used for connections.

28 28 Crossbar switch configurations Different switch points are used to establish a connection between two given subscribers depending upon who initiate the call. For example when the subscriber C wishes to call subscriber B, crosspoint CB is energized. On the other hand when B initiates the call to contact C, the switch BC is used. By designing a suitable control mechanism, only one switch may be used to establish a connection between two subscribers, irrespective of which of them initiates the call. The crosspoints in the diagonal connect the inlets and the outlets of the same subscriber. Hence they can also be eliminated.

29 29 Cont.

30 30 The crosspoints in the diagonal connect the inlets and the outlets of the same subscriber. Hence they can also be eliminated. Cont.

31 31 Class work Calculate the number of switches in a diagonal crosspoint matrix if the number of Subscribers is N.

32 32 Blocking Crossbar switch The diagonal crosspoint matrix is a nonblocking configuration. The number of crosspoint switches can be reduced significantly by designing blocking configurations. The number of vertical bars is less than the number of subscribers. The vertical bars determines the number of simultaneous calls that can be out through the switch.

33 33 Cont. Let a connection be required to be established between the subscriber A and B. The sequence to be followed in establishing the A-B circuit may be summarized as: Energize horizontal barA Energize free verticalbarP De-energize horizontal barA Energize horizontal bar B Energize vertical barP’ De-energize horizontal barB

34 34 Cont. Alternative Energizing sequence: Energize horizontalA and B Energize verticalP De-energize horizontalA and B The number of switches required is 2NK, where N is the number of subscribers and K is the number of vertical bars that are used to establish the connections.

35 35 Definition: space-division switching Space division switching was originally developed for the analog environment and has been carried over into the digital realm. A space division switching is one in which the signal paths are physically separate from one another (divided in space).

36 36 Limitations of crossbar switch The basic building block of the switch is crossbar switch in which a metallic cross-point or semiconductor gate is enabled or disabled by a control unit for the establishment of a physical path. But the crossbar switch has a no. of limitations: The no. of cross-points grows with the square of the no. of attached stations. This is costly for a large switch. The loss of a cross-point prevents connection between the two devices whose lines intersect at that cross-point. Cross-points are inefficiently utilized. Only a small fraction of cross-points are engaged even when all devices are active.

37 37 Cont’d.  To overcome these limitations of crossbar switch, multiple stage switches are employed. Although a multistage network requires a more complex control scheme, it has several advantages over a single stage switch.

38 38 Single stage vs Multistage networks

39 39 Single stage vs Multistage networks

40 40 Two stage representation of N X N network Theorem: For any single stage network there exists an equivalent multistage network. So, N X N single stage network with capacity k can be realized by a two stage network of N X K and K X N stages.

41 41 Cont.  Any of the N inlets can be connected to any of the K outputs of 1 st stage.  Similarly, Any of the K inputs can be connected to any of the N outputs of 2 nd stage.  So, there are K alternative paths and 2NK switching elements. Any of the N inlets can be connected to any of the N outlets.

42 42 Cont.  Each stage has NK switching elements  Assume only a fraction of the subscribers to be active on an average  K can be equal to N/16  So, no. of switching elements, S = 2NK = N 2 /8--- (1) Example: N = 1024, K = 64 S = 131,027 So, for large N, the switching matrix NxK may still be difficult to realize practically.

43 43 Two stage network with multiple switching matrices  M inlets are divided into r blocks of p inlets. M = pr  N inlets are divided into s blocks of q outlets. N = qs

44 44 Cont.  For full connectivity there must be at least one outlet from each block in the 1 st stage terminating as inlet on every block of the 2 nd stage.  So, block sizes are p x s and r x q respectively  So, S = psr + qrs --- (2)  Putting values for M, N  S = Ms + Nr --- (3)  The number simultaneous calls in the network, switching capacity, SC = rs --- (4)

45 45 Cont.  For rs connections to be simultaneously active, the s active inputs in one block of the 1 st stage must be uniformly distributed across all the s blocks in the 2 nd stage at the rate of one per block.  Blocking may occur in two conditions: I.Calls are uniformly distributed (there are rs calls in progress and (rs + 1)th calls arrives) II.Calls are not uniformly distributed, there is a call in progress from I-th block from the first stage to the J- th block in the 2 nd stage and another call originates in the I-th block destined to J-th block.

46 46 The blocking probability Let α be the probability the a given inlet is active. Now, probability that an outlet at the I-th block is active is, β = (pα)/s The probability that another inlet becomes active and seeks an outlet other than the one which already active is given by (p - 1)α/(s - 1) Now, Probability that an ready active outlet is sought P B = ((pα)/s)[1 – (p-1)α/(s-1)] Substituting, p = M/r, we have P B = ((Mα)/rs)[1 – ((M/r)-1)α/(s-1)] --- (5)

47 47 Discussion.  If s and r decrease then S can be minimized  But if we decrease s and r we are increasing blocking probability!  So, we have to choose values for s and r as small as possible but giving sufficient links to provide a reasonable grade of service.

48 48 Cont.  If N > M, network is expanding traffic  If M > N, concentrating the traffic  If N = M, matrix size is uniform i.e. r=s, p=q So, S = 2Nr --- (6) SC = r 2 --- (7)

49 49 Cont.  For square switching matrices as in standard ICs.  p=r=s=q = √N Thus the network has √N blocks each in the 1 st and 2 nd stages and each block is a square matrix of √N X √N inlets and outlets. So, S= N √N + N √N = 2N √N --- (8) SC = √N X √N = N --- (9)

50 50 Cont.  In the two stage network discussed so far, there is only one link between a block in the 1 st stage and a block in the 2 nd stage.  What will happen if this particular link failure?  Rise of severe blocking in the network!!  How can we improve this performance?

51 51 Cont.  Increase number of links between the blocks of the stages.  Consider k links beings introduced between every 1 st and 2 nd stage pair.  Design parameters for M = N are as follows: p = q = √N, s = r = k√N S = 2Nk √N --- (10) SC = N --- (11)

52 52 Cont.  In order to make the network non-blocking, must have K = √N  Now, S = 2N 2 --- (12)  And, SC = N --- (13)  So, a two-stage non-blocking network requires twice the number of switching elements as the single stage non-blocking network.

53 53 Three-stage network  Switching matrices  Stage 1: p x s  Stage 2: r x r  Stage 3: s x p

54 54 Cont.  What is the improvement here?  It has s alternatives from stage 1 to stage 3.  S= rps + sr 2 + spr = 2Ns + sr 2 =s(2N+r 2 ) --- (14)  If we use square matrices in stage 1 and 3, then p = s = (N/r) S=2N 2 /r + Nr --- (15)

55 55 Cont.  There is an optimum value for r which would minimize value of S. (equation 15)  To obtain this optimum value differentiate this equation and set to zero.  dS/dr = -2N 2 /r 2 + N = 0 => r = √(2N)  S min =2N √ (2N) and p = N/r = √(N/2) --- (16)  Optimum ratio of the number of blocks to the number of inputs per block is r/p= √(2N)/ √(N/2) = 2 --- (17)

56 56 Lee’s graph for 3 stage network β = probability of a link being busy β’ = probability of a link is free Now, β = 1 - β’ If there are s parallel links, the blocking probability is the probability that all the links are busy: P B = β s, Q B = 1 – P B = 1 - β s

57 57 Cont. When a series of s links are needed to complete a connection, the blocking probability is easily determined as one minus the probability that they are available: P B = 1 – (β’ ) s = 1 – (1 – β ) s For a three stage network, there are two links in series for every path and there are s parallel paths. Therefore, P B = [1 – (β’ ) 2 ] s = [1 – (1 – β ) 2 ] s --- (18)

58 58 Cont.  If α is the probability that an inlet at first stage is busy, then β = pα/s = α/k --- (19) Now, substitute the value of β in equation (18), we get the blocking probability for three- stage switch as: P B = [1 – (1 - α/k) 2 ] s --- (20) k represents either space expansion or concentration.

59 59 Analysis Look at equation (20), the term α/k is the factor to decide the value for blocking probability.  If α/k is small probability is low.  If α is large then k must be large, i.e. if inlets are well loaded, we need an expanding first stage.  On the other hand, if α is small k may be small, i.e. if inputs are lightly loaded, the first stage may be a concentrating one.

60 60 Three stage non-blocking

61 61 Cont. Multistage non-blocking and fully available networks are known as Clos networks. 3 stage switching network can be made non-blocking. How??? By providing adequate number of blocks in 2 nd stage, i.e. increase value of s.

62 62 Cont.  Worst situation can occur when the situations arise: 1.(p-1) inlets in a block I in 1 st stage are busy 2.(p-1) inlets in a block O in 3 rd stage are busy 3.The (p-1) 2 nd stage blocks, on which (p-1) outlets from block I are terminated, are different from the (p-1) 2 nd stage blocks from which the links are established to the block O. 4.The free inlet of block I needs to be terminated on the free outlet of block O.

63 63 Definition: time-division switching Switching of time-division multiplexed (TDM) channels by shifting bits between time slots in a TDM frame. Time-division multiplexing (TDM) is a type of digital or (rarely) analog multiplexing in which two or more signals or bit streams are transferred apparently simultaneously as sub- channels in one communication channel, but physically are taking turns on the channel. The time domain is divided into several recurrent timeslots of fixed length, one for each sub- channel.

64 64 Statistical Time-division Multiplexing (STDM)  STDM is an advanced version of TDM in which both the address of the terminal and the data itself are transmitted together for better routing. Using STDM allows bandwidth to be split over 1 line.  If there is one 10MBit line coming into a building, STDM can be used to provide 178 terminals with a dedicated 56k connection (178 * 56k = 9.96Mb).  A more common use however is to only grant the bandwidth when that much is needed.  STDM does not reserve a time slot for each terminal, rather it assigns a slot when the terminal is requiring data to be sent or received.

65 65 Preliminaries…  8 kHz sampling rate -> a sample occurs in every 125 μ sec.  In this 125 μ s sampling interval about 120 μ s are unused !!!  How can we utilize this efficiently? Establish a dynamic control mechanism, whereby a switching element can be shared by a number of simultaneously active speech circuit.

66 66 Basic Time division space switching  N X N time division space switch  Each inlet/outlet is a single speech circuit corresponding to a subscriber line.  Speech is carried as either PAM or PCM samples.

67 67 Two stage Equivalent  Matrices: (1 st and 2 nd stage) N X 1 and 1 X N

68 68 Cont.  If PAM samples are switched on then Analog time division switch  If PCM binary samples are switched on then Digital time division switch  Interconnection is through a bus  Number of simultaneous conversations SC = 125/t s  t s is time in μ sec to setup a connection and transfer sample value.

69 69 Cont.  Selection of inlet/outlet is controlled dynamically.  Simplest manner is to select in a cyclic manner  Cyclic control is organized using a modulo-N counter and a k-to-2 k decoder  Ceil(log 2 N) = k

70 70 Cont.  All the inlets/outlets are scanned within 125 μ sec, the switching capacity, SC, of the network is the same as number of inlets or outlets in the system.  Switching is non-blocking  Lacks full availability -> Not possible to connect any inlet to any outlet.  How to obtain it full availability? Make one of the controls (input or output), memory based Figure at next slide

71 71 Input driven TDSS  A control memory on output side.  modulo-N counter also acts as memory address register (MAR)  Control memory has N words corresponding to N inlets and has a width of ceil(log 2 N) bits.

72 72 Cont.  For an active inlet i, the corresponding outlet address is is contained in the i-th location of the control memory.  Address is decoded by MDR (memory data register)  Then proper outlet is enabled.  Then the sample value is transferred.  Thus any inlet i can be connected to any outlet k ensuring full availability.

73 73 Finally, Definition  Since a single switching element, the bus, is being time shared by N connections, all of which can be active simultaneously, and a physical connection is established between the inlet and outlet for the duration of the sample transfer, the switching technique is known as time division space switching

74 74 Output controlled TDSS  Each location of the control memory is rigidly associated with a given outlet.

75 75 Analysis  For both input and out controlled configurations, the number of inlets or outlets N is equal to the switching capacity SC.  N = SC = 125/(t i + t m + t d + t t ) t i = time to increment modulo-N counter t m = time to read control memory t d = time to decode address and select inlet or outlet t t = time to transfer the sample value from inlet to outlet

76 76 Cont. Number of switching elements: on the input side = N on the output side = N total= 2N Switching capacity, SC= N Traffic handling capacity, TC = SC/(theoretically max load) = 1 Cost of the switching network = cost of the switching element + cost of the control memory = 2N + N = 3N Cost capacity index, CCI = SC/(cost per subscriber line) = N/ (3N/N) = N/3

77 77 Discussion  Use of cyclic control in input or output-controlled switches restricts the number of subscribers on the system rather than the switching capacity.  That’s why cyclic control demands all the lines to be scanned irrespective of whether they are active or not.  Practically, number of active subscriber is around 20% of the total.

78 78 Generalized TDSS  Control memory for controlling both inlets and outlets.  Permits a larger number of subscribers than the switching capacity of the network.  Each word in control memory has two addresses: inlet & outlet  Control memory width is 2*ceil(log 2 N).

79 79 Operation procedure of the switch  modulo-SC counter is updated at the clock rate  Control memory words are read one after another  Inlet address is used to connect the corresponding inlet to the bus and so also the outlet address to connect the outlet  The sample is then transferred from inlet to outlet  Next, clock updates the counter and the cycle of operation is repeated.

80 80 Cont.  Recall the switching capacity,  SC = 125/t s t s =(t i + t m + t d + t t ) If time to read memory i.e. t m of t s is dominating factor in the equation then it means that control memory is busy through out the sampling interval of 125 μ s

81 81 Cont.  Time slot is 125/M μ s, duration of 1 sample.  In 1 time slot N samples are switched.  The output is cyclically scanned.  1-to-M relationship between the outlets and control memory, i.e. M location in control memory (CM) corresponding to each outlet.  CM has MN words.  Number of trunks can be supported, N = 125/(M*t s ) Where, t s is the switching time as earlier. Cost of the switch, C = Number of switch + Number of memory words = 2N + MN

82 82 Exercise  Calculate number of trunks that can be supported on a time multiplexed space switch, given that 32 channels are multiplexed, control memory access time is 100 ns, bus switching and transfer time is 100 ns per transfer. Solution: Here, M = 32, t s =100 + 100 = 200ns N = 125/(M*t s ) = 125/(32 * 200 * 10 -3 ) = 20 (approx.)

83 83 Time Multiplexed Time Switch  Unlike time multiplexed space switches, it permits time slot interchange (TSI) of sample values.  Such an operation necessarily implies a delay between the reception and the transmission of a sample.  Illustration in next graph.

84 84  M channels are multiplexed on each trunk  The switch is organized in sequential write/ random read fashion  Time slot duration, t TS = 125/M  Time slot clock runs at the time slot rate  Time slot counter is incremented by 1 at the end of each time slot.

85 85 Cont.  Contents of counter provides location addresses for the data memory and the control memory.  Data memory and control memory access take place simultaneously at the starting of the time slot.  Contents of the control memory are used as the address of the data memory and the data read out to the output trunk

86 86 Cont.  Even if there is no time slot interchange, a sample is delayed by a minimum of one time slot in action.  Depending on the output time slot, delay range is t TS to Mt TS microsec  In the example given in figure 1 st location in CM contains value 1 which implies that the contents of input time slot 1 is switched to output time slot 1. Delay for this sample is t TS microsec

87 87  Location 2 contains 7. Therefore, input time slot 7 is switched to output time slot 2. Delay for this sample: (M – (7-2) + 1)t TS μ s =(M – 4)t TS μ s  Location 3 contains 4. Therefore, input time slot 4 is switched to output time slot 3. Delay for this sample: (M – (4-3) + 1)t TS μ s =Mt TS μ s = 125 μ s There are two sequential memory access per time slot. So, time constraint may be stated as t TS = 2t m, 125 = 2Mt m

88 88 Cost estimation  No switching elements!  Cost is equal to the number of memory locations.  There are M locations each in the control and data memory  So, total cost is given by C = 2M units

89 89 Exercise  Calculate maximum access time that can be permitted for the data and control memories in TSI switch with a single input and single output trunk multiplexing 2500 channels. Estimate the cost of the switch and compare the result with a single stage space division switch. Solution: t m =(125*10 3 )/(2500*2) = 25 ns C = 2 * 2500 = 5000 units This switch is non-blocking and full available. An equivalent single stage space division switch uses a matrix of 2500 X 2500. So, cost is 6.25 million units Cost advantage of time switch = (6.25*10 6 )/5000 = 1250

90 90 Combination Switches

91 91 What we study so far?  Time division space switch  do not provide full availability as they are not capable of time slot interchange  Time slot interchange switch  not capable of switching sample values across trunks without the help of some space switching matrices.  So, here comes the idea of combining both techniques and getting their advantages in a single platform. This is combination switch.

92 92 Combination switch

93 93 Contd.


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