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Chapter 4 · 8 hours

Digital Switching

IOE past exam questions

Past questions and answers

44 questions set from this chapter, 3 of them more than once. Most asked first.

  • Asked 4 times
  • 2070 Asar · 10 marks
  • 2069 Bhadra (old course) · 12 marks
  • 2065 Magh (old course) · 10 marks
  • 2065 Baisakh (old course) · 4+12 marks

With a block diagram explain the working principle of a Digital Telephone Exchange.

Answer

A digital telephone exchange is a stored-program-controlled (SPC) exchange in which speech is switched in digital (PCM) form, using time and space switching of 64 kbps channels, under the control of computer processors. Examples: Alcatel E10, Siemens EWSD, Ericsson AXE, and the exchanges installed by Nepal Telecom.

Block diagram

 analogue      +-------------+  +------+
 subscribers --| Subscriber  |--|      |
 (2-wire)      | line units  |  | conc-|   +-----------+
 (BORSCHT)     |  + codec    |  |entr. |===|           |
               +-------------+  +------+   |  Digital  |
 digital trunks (E1)--[trunk interface]====| switching |
 analogue trunks--[ATI + codec]===========| network   |
                                           |  (TST)    |
 tones/announcements --[tone generator]====|           |
 signalling --[CAS/CCS (SS7) units]=======|           |
                                           +-----+-----+
                                                 |
                 +-------------------------------+---+
                 |  Control subsystem (processors,   |
                 |  memory, stored programs)         |
                 +----+-------------------------+----+
                      |                         |
              [O&M terminal, billing]   [network sync clock]

Main blocks

  1. Subscriber line interface (line card). Each analogue line needs the BORSCHT functions:
    • Battery feed (−48 V DC for the telephone),
    • Overvoltage protection (lightning, power-line contact),
    • Ringing (75 V, 25 Hz AC applied to the line),
    • Supervision (detecting off-hook, on-hook, dial pulses),
    • Coding (codec: A-law PCM at 64 kbps, with filters),
    • Hybrid (2-wire to 4-wire conversion),
    • Testing (access for line tests).
  2. Concentrator / remote line unit. Since only a fraction of subscribers are busy at once, many lines (e.g. 128–1000) are concentrated onto fewer PCM channels (E1 links) to the switching network. Remote units can be placed near subscriber clusters.
  3. Trunk interfaces. Digital trunks (E1, 2.048 Mbps, HDB3) are connected directly after frame alignment and clock adaptation; analogue trunks need codecs and signalling conversion.
  4. Digital switching network. Usually a TST (time–space–time) or multi-stage network. Time switches (time slot interchangers) move a sample from one time slot to another using speech and control memories; space switches connect a time slot on one PCM highway to the same slot on another. This connects any 64 kbps channel to any other. The network is duplicated for reliability.
  5. Signalling equipment. Receivers for dial pulses and DTMF, CAS units (E1 TS16) and common channel signalling (SS7) terminals for inter-exchange signalling.
  6. Service circuits. Tone generators (dial, busy, ring-back), announcement machines, conference bridges.
  7. Control subsystem. Duplicated processors with stored programs carry out call processing, routing translation, charging and resource management. Control may be centralised or distributed.
  8. Operation and maintenance (O&M) and billing. Terminals for subscriber data changes, traffic measurement, fault alarms, diagnostics and call detail records.
  9. Synchronisation. A clock locked to the national network clock keeps all PCM links in step to avoid slips.

Working principle (a local call)

  1. Off-hook detection: subscriber A lifts the handset; loop current flows; the line card's supervision scans and reports to the processor.
  2. Dial tone: the processor checks A's class of service, assigns a free path and a digit receiver, and connects dial tone.
  3. Digit reception: dialled digits (pulses or DTMF) are received and stored; dial tone is removed after the first digit.
  4. Analysis and routing: the processor translates the number. For a local number it checks B's line; for a distant number it selects an outgoing trunk and sends signalling (e.g. SS7 IAM).
  5. Path set-up: a free time slot path through the TST network is found and the speech and control memories are written.
  6. Ringing: ringing current is sent to B and ring-back tone to A.
  7. Answer: when B lifts the handset, ringing is tripped, the speech path is through-connected and charging starts.
  8. Conversation: speech from each phone is coded to 64 kbps, switched digitally in its time slots and decoded at the other line card.
  9. Clear-down: on on-hook, the processor releases the path, stops charging and records the call details.

Advantages

  • No noise build-up; uniform digital quality and easy integration with digital transmission (no codec at each trunk).
  • Small size, low power, fast set-up, high reliability (duplicated control).
  • Many subscriber services (call waiting, forwarding, conference, abbreviated dialling), easy changes by software, centralised O&M and accurate billing.
  • Supports ISDN, data and common channel signalling.
  • Asked 4 times
  • 2079 Bhadra · 6 marks
  • 2076 Asoj · 4 marks
  • 2073 Chaitra · 8 marks
  • 2069 Chaitra · 6 marks

Explain the various modes of operation of dual processor configuration used in a centralized digital exchange.

Answer

In centralised SPC, a single processor controls the whole exchange, so its failure would stop all calls. To give high availability (an exchange may be down only about 2 hours in 40 years), the central processor is duplicated. The two processors work in one of three modes.

1. Standby mode

 +-------+         +-------+
 | P1    |         | P2    |
 |active |         |standby|
 +---+---+         +---+---+
     +----[shared memory / disk]----+
           |
     exchange equipment
  • One processor (P1) is active and handles all calls; the other (P2) is idle in standby.
  • Both have access to a common secondary memory; the active one periodically copies the exchange state (call data, tables) to it.
  • When P1 fails, P2 takes over, loads the latest state and continues.
  • Cold standby: P2 starts from the last saved data, so calls being set up are lost. Hot standby: P2 is kept updated and takes over almost at once.
  • Simple, but takeover takes time and transient calls may be lost.

2. Synchronous duplex (match) mode

 +-------+   comparator   +-------+
 | P1    |<------C------->| P2    |
 |  M1   |                |  M2   |
 +---+---+                +---+---+
     +-------- exchange --------+
  • Both processors run the same program on the same data, in step, each with its own memory (M1, M2).
  • A comparator checks their results continuously. One processor's outputs actually control the exchange.
  • If the results disagree, a fault is suspected; both run diagnostics. The faulty processor is isolated and the good one continues alone, with almost no loss of calls.
  • Gives fast fault detection and no takeover delay, but both processors do the same work, so no extra capacity, and a transient fault may be hard to locate.

3. Load-sharing mode

 +-------+   exclusion    +-------+
 | P1    |<-----ED------->| P2    |
 | calls |   (shared      | calls |
 | 1,3,5 |    data)       | 2,4,6 |
 +---+---+                +---+---+
     +-------- exchange --------+
  • Each processor handles part of the traffic; an incoming call is given randomly or in turn to one of them.
  • Both share the exchange resources and data; an exclusion device (ED) stops them from seizing the same resource at the same time.
  • Each can handle the full load. If one fails, the other carries all the traffic (with possible congestion at peak).
  • Gives higher capacity in normal operation and good reliability, but software is more complex.

Comparison

ModeNormal capacityTakeoverComplexity
StandbyOne processorSlowest, calls may be lostSimple
Synchronous duplexOne processorImmediateComparator needed
Load sharingUp to two processorsQuick, other takes all loadExclusion logic, complex software
  • Asked 2 times
  • 2079 Bhadra · 3+4 marks
  • 2073 Shrawan · 2+6 marks

What is the advantage of multi-stage switching system over single stage switching system? Explain TST Switching with neat diagram and its blocking probability.

Answer

Advantages of multi-stage over single-stage switching

  • Far fewer crosspoints: a single-stage N×NN \times N switch needs N2N^2 crosspoints (e.g. 10610^6 for 1000 lines); a three-stage switch needs only a small fraction, so cost and size fall.
  • Better crosspoint utilisation: in a single stage only N/2N/2 of N2N^2 crosspoints can be busy at once; in multi-stage each crosspoint is shared by many connections.
  • Alternative paths: a call can go through any of several middle-stage arrays, so failure of one crosspoint does not isolate a subscriber (better reliability).
  • Modular growth: capacity can be expanded by adding arrays.
  • Concentration and expansion can be built into the stages to suit traffic.
  • Cost: some blocking may occur, and path search (control) is more complex.

TST switching

A Time–Space–Time (TST) switch is a three-stage digital switching network: an incoming time stage, a space stage in the middle, and an outgoing time stage. It is the most common structure in large digital exchanges.

 incoming     T stage      S stage       T stage   outgoing
 highways   (time slot   (crosspoint   (time slot  highways
            interchange)   matrix)    interchange)
 PCM 1 --->[ TSI ]--->+-------------+--->[ TSI ]---> PCM 1
 PCM 2 --->[ TSI ]--->|  N x N      |--->[ TSI ]---> PCM 2
  ...                 |  space sw.  |
 PCM N --->[ TSI ]--->+-------------+--->[ TSI ]---> PCM N
 slot i  -> slot j (internal) -> slot j -> slot m

Operation (connect slot ii of PCM 1 to slot mm of PCM N):

  1. First time stage: the time slot interchanger (TSI) on PCM 1 writes incoming samples into its speech memory in order and reads the sample of slot ii out in a free internal slot jj (chosen by the control).
  2. Space stage: in internal slot jj the crosspoint connecting row 1 to column N is closed (set by the space-switch control memory), passing the sample to the outgoing side. The same crosspoint is reused by other connections in other slots.
  3. Second time stage: the TSI on output highway N stores the sample and reads it out in the required outgoing slot mm.
  4. The reverse direction uses a related internal slot (often j+c/2j + c/2) so both directions are set together.

Because the space stage only needs to find one free internal slot common to both links, the many internal time slots act like many middle-stage arrays, giving very low blocking at low cost. The time stages can also expand (l>cl > c internal slots) to reduce blocking further.

Blocking probability of TST (Lee graph)

           slot 1
       +--- ... ---+
  A ---+--- slot j-+--- B     l parallel paths
       +--- ... ---+          (internal slots)
           slot l

Let pp = occupancy of an incoming channel, cc = channels per highway, ll = internal slots, time expansion α=l/c\alpha = l/c. Each internal link is busy with probability p′=p/αp' = p/\alpha. A path is blocked if either of its two links is busy: 1−(1−p′)21-(1-p')^2. The call is blocked only if all ll paths are blocked:

p′=pα,α=lcPB(TST)=[1−(1−p′)2]l\begin{aligned} p' &= \frac{p}{\alpha},\quad \alpha = \frac{l}{c} \\ P_B(\text{TST}) &= \left[1 - (1-p')^2\right]^{l} \end{aligned}

Example: p=0.2p = 0.2, l=c=32l = c = 32: PB=(1−0.82)32=0.3632≈6.3×10−15P_B = (1 - 0.8^2)^{32} = 0.36^{32} \approx 6.3 \times 10^{-15}, practically non-blocking.

  • 2081 Chaitra · 2+3+3 marks

Define what the Clos network is. Derive the expression for blocking probability of both STS and TST switches, and compare them.

Answer

Clos network

A Clos network is a three-stage (multi-stage) switching network, proposed by Charles Clos (1953), in which NN inputs are divided into N/nN/n first-stage arrays of size n×kn \times k, connected to kk middle arrays of size Nn×Nn\frac{N}{n} \times \frac{N}{n}, which connect to N/nN/n third-stage arrays of size k×nk \times n. Every array of one stage has exactly one link to every array of the next stage.

Nx=Nn(nk)⏟stage 1+k(Nn)2⏟stage 2+Nn(kn)⏟stage 3=2Nk+k(Nn)2\begin{aligned} N_x &= \underbrace{\tfrac{N}{n}(nk)}_{\text{stage 1}} + \underbrace{k\left(\tfrac{N}{n}\right)^2}_{\text{stage 2}} + \underbrace{\tfrac{N}{n}(kn)}_{\text{stage 3}} \\ &= 2Nk + k\left(\frac{N}{n}\right)^2 \end{aligned}

Clos showed that the network is strictly non-blocking when k=2n−1k = 2n - 1 (worst case: n−1n-1 other inputs of the calling array and n−1n-1 outputs of the called array use different middle arrays, so one more is needed). With the best choice n≈N/2n \approx \sqrt{N/2}, the minimum is Nx≈4N(2N−1)N_x \approx 4N(\sqrt{2N} - 1), much less than N2N^2.

Blocking probability of STS

 STS:     S (N x k) -> T (k TSI modules) -> S (k x N)
 Lee graph: k parallel paths, each = 2 links in series

In STS, a call from input highway A to output highway B in given slots can go through any of the kk time-switch modules. Each path uses a link from the first space stage to a module and a link from the module to the second space stage. With link occupancy p′=p/βp' = p/\beta (β=k/N\beta = k/N space expansion):

  • Probability a link is free =1−p′= 1 - p'; both links of a path free =(1−p′)2= (1-p')^2.
  • Path blocked =1−(1−p′)2= 1 - (1-p')^2; all kk paths blocked:
p′=pβ,β=kNPB(STS)=[1−(1−p′)2]k\begin{aligned} p' &= \frac{p}{\beta},\quad \beta = \frac{k}{N} \\ P_B(\text{STS}) &= \left[1 - (1-p')^2\right]^{k} \end{aligned}

Blocking probability of TST

 TST:     T (c -> l slots) -> S (N x N) -> T (l -> c)
 Lee graph: l parallel paths (internal time slots)

In TST, the path from A to B passes through the space stage in any of the ll internal time slots. Each path uses one internal slot on A's highway and the same slot on B's highway. With time expansion α=l/c\alpha = l/c, p′=p/αp' = p/\alpha, and by the same argument:

p′=pα,α=lcPB(TST)=[1−(1−p′)2]l\begin{aligned} p' &= \frac{p}{\alpha},\quad \alpha = \frac{l}{c} \\ P_B(\text{TST}) &= \left[1 - (1-p')^2\right]^{l} \end{aligned}

Comparison

Since the number of paths in TST equals the number of internal time slots ll (32 to several hundred), while in STS it equals the number of time-switch modules kk (usually small), TST blocking is far lower for the same hardware. Example with p′=0.2p' = 0.2:

STS, k=8: (1−0.82)8=0.368≈2.8×10−4TST, l=32: (1−0.82)32=0.3632≈6.3×10−15\begin{aligned} \text{STS},\ k = 8 &: \ (1-0.8^2)^{8} = 0.36^{8} \approx 2.8 \times 10^{-4} \\ \text{TST},\ l = 32 &: \ (1-0.8^2)^{32} = 0.36^{32} \approx 6.3 \times 10^{-15} \end{aligned}
PointSTSTST
Number of alternative pathskk (time modules)ll (internal slots)
BlockingHigherMuch lower
Expansion byMore space arrays/modulesMore internal slots (cheap memory)
Cost for large exchangeHigherLower (time stages cheap)
ControlSimpler for small sizesMore memory, path search over slots
UseSmall exchangesMost large digital exchanges
  • 2080 Chaitra · 2+6 marks

What are advantages of multi-stage switching system over single stage switching system? Compare TST and STS switch in digital telephone exchange system with necessary diagram and its blocking probabilities.

Answer

Advantages of multi-stage switching

  • Far fewer crosspoints: a single-stage N×NN \times N switch needs N2N^2 crosspoints (e.g. 10610^6 for 1000 lines); a three-stage switch needs only a small fraction, so cost and size fall.
  • Better crosspoint utilisation: in a single stage only N/2N/2 of N2N^2 crosspoints can be busy at once; in multi-stage each crosspoint is shared by many connections.
  • Alternative paths: a call can go through any of several middle-stage arrays, so failure of one crosspoint does not isolate a subscriber (better reliability).
  • Modular growth: capacity can be expanded by adding arrays.
  • Concentration and expansion can be built into the stages to suit traffic.
  • Cost: some blocking may occur, and path search (control) is more complex.

TST switch

 incoming     T stage      S stage       T stage   outgoing
 highways   (time slot   (crosspoint   (time slot  highways
            interchange)   matrix)    interchange)
 PCM 1 --->[ TSI ]--->+-------------+--->[ TSI ]---> PCM 1
 PCM 2 --->[ TSI ]--->|  N x N      |--->[ TSI ]---> PCM 2
  ...                 |  space sw.  |
 PCM N --->[ TSI ]--->+-------------+--->[ TSI ]---> PCM N
 slot i  -> slot j (internal) -> slot j -> slot m

An incoming time slot ii is moved by the first TSI to a free internal slot jj, passed through the space matrix in slot jj, and moved by the outgoing TSI to the wanted slot mm. Blocking happens only if no internal slot is free on both the input and output highways. With ll internal slots and p′=p c/lp' = p\,c/l:

p′=pα,α=lcPB(TST)=[1−(1−p′)2]l\begin{aligned} p' &= \frac{p}{\alpha},\quad \alpha = \frac{l}{c} \\ P_B(\text{TST}) &= \left[1 - (1-p')^2\right]^{l} \end{aligned}

STS switch

 incoming    S stage      T stage      S stage   outgoing
 PCM 1 --->+--------+--->[ TSI 1 ]--->+--------+---> PCM 1
 PCM 2 --->| N x k  |--->[ TSI 2 ]--->| k x N  |---> PCM 2
  ...      | space  |     ...         | space  |
 PCM N --->+--------+--->[ TSI k ]--->+--------+---> PCM N

The first space stage connects the incoming highway (in its own slot) to one of kk time switches; the TSI moves the sample to the outgoing slot; the second space stage connects it to the outgoing highway. Blocking happens if none of the kk modules has both links free. With p′=p/βp' = p/\beta:

p′=pβ,β=kNPB(STS)=[1−(1−p′)2]k\begin{aligned} p' &= \frac{p}{\beta},\quad \beta = \frac{k}{N} \\ P_B(\text{STS}) &= \left[1 - (1-p')^2\right]^{k} \end{aligned}

Comparison of TST and STS

PointTSTSTS
StagesTime – Space – TimeSpace – Time – Space
Alternative pathsll internal time slotskk time modules
BlockingVery low (ll large)Higher (kk small)
Main hardwareMemories (cheap) + one space matrixTwo space matrices + TSIs
Cost for large NLowerHigher
ExpansionAdd internal slots or TSIsAdd space arrays and TSIs
ControlMore memory, simple space controlSimpler for small sizes
Typical useLarge digital exchanges (e.g. AXE, E10)Small/medium exchanges, some early systems

Example (p′=0.2p' = 0.2): STS with k=8k = 8 gives PB=0.368≈2.8×10−4P_B = 0.36^8 \approx 2.8 \times 10^{-4}, while TST with l=32l = 32 gives 0.3632≈6.3×10−150.36^{32} \approx 6.3 \times 10^{-15}. Hence TST is preferred for large exchanges.

  • 2079 Chaitra · 3+5 marks

What are the advantages of multi-stage switching over single-stage switching? Calculate and draw, how many cross points are found in three stages switching system, whereas 3 stages array of 4 input lines and 5 second stages array.

Answer

Advantages of multi-stage switching over single-stage

  • Far fewer crosspoints: a single-stage N×NN \times N switch needs N2N^2 crosspoints (e.g. 10610^6 for 1000 lines); a three-stage switch needs only a small fraction, so cost and size fall.
  • Better crosspoint utilisation: in a single stage only N/2N/2 of N2N^2 crosspoints can be busy at once; in multi-stage each crosspoint is shared by many connections.
  • Alternative paths: a call can go through any of several middle-stage arrays, so failure of one crosspoint does not isolate a subscriber (better reliability).
  • Modular growth: capacity can be expanded by adding arrays.
  • Concentration and expansion can be built into the stages to suit traffic.
  • Cost: some blocking may occur, and path search (control) is more complex.

Crosspoint calculation

Reading of the data (assumed): the first stage has 3 arrays of 4 input lines each, so N=3×4=12N = 3 \times 4 = 12 lines and n=4n = 4, and there are k=5k = 5 second-stage (middle) arrays. The network is symmetric (third stage = mirror of first).

Array sizes:

  • First stage: N/n=3N/n = 3 arrays of n×k=4×5n \times k = 4 \times 5.
  • Second stage: k=5k = 5 arrays of Nn×Nn=3×3\frac{N}{n} \times \frac{N}{n} = 3 \times 3.
  • Third stage: 3 arrays of k×n=5×4k \times n = 5 \times 4.
  stage 1 (3 x [4x5])  stage 2 (5 x [3x3])  stage 3
 4 in->[4x5]-+-----> [3x3] -----+->[5x4]-> 4 out
             |-----> [3x3] -----|
 4 in->[4x5]-+-----> [3x3] -----+->[5x4]-> 4 out
             |-----> [3x3] -----|
 4 in->[4x5]-+-----> [3x3] -----+->[5x4]-> 4 out
 each 4x5 array has 1 link to each of the 5 middle
 arrays; each middle array has 1 link to each 5x4
Nx=2Nk+k(Nn)2Stage 1=3×(4×5)=60Stage 2=5×(3×3)=45Stage 3=3×(5×4)=60Nx=60+45+60=165\begin{aligned} N_x &= 2Nk + k\left(\frac{N}{n}\right)^2 \\ \text{Stage 1} &= 3 \times (4 \times 5) = 60 \\ \text{Stage 2} &= 5 \times (3 \times 3) = 45 \\ \text{Stage 3} &= 3 \times (5 \times 4) = 60 \\ N_x &= 60 + 45 + 60 = 165 \end{aligned}

Answer: 165 crosspoints (60 + 45 + 60).

Remarks:

  • A single-stage 12×1212 \times 12 matrix needs 122=14412^2 = 144 crosspoints; for such a small NN the three-stage network gives no saving. The saving appears for large NN, e.g. N=128N = 128, n=8n = 8, k=15k = 15 needs 7680 crosspoints against 16 384.
  • With k=5<2n−1=7k = 5 < 2n - 1 = 7, this network is blocking; a strictly non-blocking Clos version needs k=7k = 7, i.e. 2(12)(7)+7(3)2=2312(12)(7) + 7(3)^2 = 231 crosspoints.
  • 2081 Bhadra · 4+4 marks

What are the roles of time-switches and space-switches in digital exchanges? How do combinations of these switches (e.g., STTS, TSST) enhance the capabilities of telecommunications networks?

Answer

Role of time switches

A time switch (time slot interchanger, TSI) moves PCM samples from one time slot to another on the same highway.

 incoming slots -> [speech memory] -> outgoing slots
   write in order     ^ read address
                [control memory]
  • Incoming samples are written into a speech memory in slot order (sequential write) and read out in the order set by a control memory (random read), or the reverse.
  • It connects channel ii to channel jj on the same highway with a delay of up to one frame (125 µs).
  • Built from RAM, so it is cheap, and it is non-blocking for one highway; the size is limited by memory speed (cc slots × 2 accesses within 125 µs).

Role of space switches

A space switch is a crosspoint matrix (electronic gates) that connects an input highway to an output highway, keeping the same time slot.

  • Each crosspoint is opened or closed every time slot according to its control memory, so it is shared by many calls (time-shared space switching).
  • It enlarges the switch to many highways but cannot change time slots; used alone it blocks when two calls need the same output in the same slot.

How combinations enhance capability

Real exchanges have hundreds of highways, so time and space switches are combined in multi-stage networks:

  • TS / ST (two stage): simple, but blocking is high.
  • TST (time–space–time): incoming TSI moves the call to any free internal slot, the space stage crosses highways, the outgoing TSI moves it to the wanted slot. The number of alternative paths equals the number of internal slots, so blocking is very low and memory is cheap. Standard in large exchanges.
  • STS (space–time–space): space stages spread calls over kk time modules; suits smaller systems.
  • TSST / TSSST: two or more space stages in the middle allow very large numbers of highways (large transit exchanges) without huge single matrices.
  • STTS: two time stages in the middle give more slot-changing freedom and paths between groups of highways, useful in modular/distributed designs.

Benefits of these combinations:

  1. Scalability: tens of thousands of lines and trunks switched in one network by adding modules.
  2. Low blocking: many alternative paths (internal slots × middle arrays); with time expansion near non-blocking.
  3. Lower cost and size: fewer crosspoints; most switching done in cheap RAM.
  4. Reliability: alternative paths and duplicated planes let traffic avoid faults.
  5. Flexibility: supports any-to-any connection of 64 kbps channels, broadcast/conference connections and n×64n \times 64 kbps data paths, which digital networks and ISDN need.
  • 2081 Bhadra · 2+6 marks

What do you know about cross point and switching array in switching structure? Explain three stage switching and calculate how many cross point used in three stage switching.

Answer

Crosspoint and switching array

A crosspoint is the basic switching element at the intersection of an input line and an output line; when it is closed (a relay contact, reed or electronic gate) it connects that input to that output. A switching array (matrix) is a rectangular arrangement of crosspoints with NN inputs and MM outputs (N×MN \times M crosspoints); any input can be connected to any free output. A single-stage N×NN \times N array needs N2N^2 crosspoints (or N(N−1)/2N(N-1)/2 if folded), which grows too quickly for large exchanges.

        out1 out2 out3
 in1 ----x----x----x---
 in2 ----x----x----x---    x = crosspoint
 in3 ----x----x----x---

Three-stage switching

The NN inputs are divided into groups of nn.

  first stage      middle stage     third stage
 (N/n arrays,     (k arrays,       (N/n arrays,
   n x k)        N/n x N/n)          k x n)
 n -->[n x k]--+->[N/n x N/n]--+-->[k x n]--> n
 n -->[n x k]--+->[N/n x N/n]--+-->[k x n]--> n
   ...         +->   ...       +-->   ...
 n -->[n x k]--+->[N/n x N/n]--+-->[k x n]--> n
  (every first-stage array has one link to
   every middle array, and so on)
  • Stage 1: N/nN/n arrays, each n×kn \times k.
  • Stage 2: kk arrays, each Nn×Nn\frac{N}{n} \times \frac{N}{n}.
  • Stage 3: N/nN/n arrays, each k×nk \times n.

Each input reaches any output through any one of the kk middle arrays, so there are kk alternative paths. Total crosspoints:

Nx=Nn(nk)⏟stage 1+k(Nn)2⏟stage 2+Nn(kn)⏟stage 3=2Nk+k(Nn)2\begin{aligned} N_x &= \underbrace{\tfrac{N}{n}(nk)}_{\text{stage 1}} + \underbrace{k\left(\tfrac{N}{n}\right)^2}_{\text{stage 2}} + \underbrace{\tfrac{N}{n}(kn)}_{\text{stage 3}} \\ &= 2Nk + k\left(\frac{N}{n}\right)^2 \end{aligned}

If k=2n−1k = 2n - 1 the network is strictly non-blocking (Clos); if k<2n−1k < 2n-1 some blocking occurs but the cost is lower.

Example calculation

Take N=128N = 128 lines, n=8n = 8 (so N/n=16N/n = 16), non-blocking k=2n−1=15k = 2n - 1 = 15:

Stage 1=16×(8×15)=1920Stage 2=15×(16×16)=3840Stage 3=16×(15×8)=1920Nx=2(128)(15)+15(16)2=7680\begin{aligned} \text{Stage 1} &= 16 \times (8 \times 15) = 1920 \\ \text{Stage 2} &= 15 \times (16 \times 16) = 3840 \\ \text{Stage 3} &= 16 \times (15 \times 8) = 1920 \\ N_x &= 2(128)(15) + 15(16)^2 = 7680 \end{aligned}

A single-stage 128×128128 \times 128 switch would need 1282=16 384128^2 = 16\,384 crosspoints. The three-stage switch saves about 53 % while still being non-blocking. Here n=8=N/2n = 8 = \sqrt{N/2} is the optimum, and the result equals 4N(2N−1)=4(128)(15)=76804N(\sqrt{2N} - 1) = 4(128)(15) = 7680.

Answer: 7680 crosspoints for a 128-line non-blocking three-stage switch (vs 16 384 single-stage).

  • 2081 Baisakh · 4+4 marks

What are the advantages and issues of PCM switching? Briefly elaborate the concept of Store Program Control (SPC).

Answer

PCM (digital) switching

In PCM switching, speech is switched in its digital form: 64 kbps PCM samples are moved between time slots and highways by time and space switches, without converting back to analogue.

Advantages

  • Direct integration with digital PCM transmission (E1/T1): no codec or 2W/4W conversion at every exchange, so no noise or loss build-up.
  • Time switches made of cheap RAM; very few crosspoints; small size and low power.
  • Four-wire switching throughout, so no hybrid echo problems inside the network.
  • Easy to switch data and n×64n \times 64 kbps channels; supports ISDN.
  • Fast connection, high reliability, easy software control and maintenance.

Issues (problems)

  • Synchronisation: all exchanges must be clock-locked; clock differences cause slips (lost or repeated frames).
  • Delay: each time switch adds up to 125 µs; many in tandem add delay and need echo control.
  • Codec per line: analogue subscriber lines need BORSCHT functions and a codec on each line card, which is costly.
  • Jitter and wander on incoming links need elastic buffers.
  • Control complexity: path search over slots and highways needs powerful processors and software.
  • Interworking with analogue exchanges and signalling conversion.

Stored Program Control (SPC)

SPC is the control of a telephone exchange by a digital computer (processor) that runs a stored program, instead of hard-wired relays or electromechanical logic.

  [processor] <--> [program + data memory]
       |
  [scanners / distributors / signal units]
       |
  switching network  <-->  lines and trunks
  • Scanners read the state of lines and trunks (off-hook, digits); the processor runs call-processing programs, consults the data memory (subscriber class, routing tables, charging) and orders the switching network through distributors/markers.
  • Centralised SPC: one main processor (duplicated in standby, synchronous-duplex or load-sharing mode). Distributed SPC: many processors, each controlling part of the exchange or a set of functions.

Benefits: new services (call forwarding, waiting, conference, abbreviated dialling) by software change; easy changes of subscriber data and routing; automatic fault diagnosis and traffic statistics; centralised maintenance; accurate billing; faster call set-up; smaller size.

  • 2081 Baisakh · 8 marks

Describe the working principle of TSI or Time switch in random read, sequential write mode.

Answer

A time switch or Time Slot Interchanger (TSI) moves the speech sample (PCM byte) in one time slot of an incoming TDM frame into a different time slot of the outgoing frame. In sequential write, random read mode (also called output-associated control), samples are written into memory in order and read out in the order set by a control memory.

Main parts

  • Speech (data) memory, SM: cc locations, one per time slot, each 8 bits wide (one PCM sample). For a 32-channel PCM frame, c=32c = 32.
  • Control (connection) memory, CM: cc locations, each holding an SM address (log⁡2c\log_2 c bits, 5 bits for 32 slots).
  • Time-slot counter: a modulo-cc counter driven by the channel clock. It gives the current slot number 0,1,…,c−10, 1, \dots, c-1.
  • Write/read control logic and the processor interface that loads the CM when a call is set up.
 Input TDM                              Output TDM
 frame                                  frame
 ----->+---------------------+--------->
       |   Speech memory     |
       |   (c x 8 bits)      |
       +---------------------+
   write addr ^          ^ read addr
 (sequential) |          | (random)
       +-------------+  +------------+
       | Time-slot   |->| Control    |
       | counter     |  | memory     |
       +-------------+  +------------+
                             ^
                     processor (call set-up)

Working principle

  1. Write phase (sequential): in time slot ii of the incoming frame, the sample is written into SM location ii. The write address comes straight from the time-slot counter, so SM fills in order 0, 1, 2, ... every frame.
  2. Read phase (random): in output time slot jj, the counter addresses CM location jj. CM(jj) contains the SM address of the input slot that must go out in slot jj. That SM location is read and placed in output slot jj.
  3. When a call is set up, the processor writes the input slot number into the CM location of the wanted output slot. When the call is released, the entry is cleared.
  4. Each slot needs one write and one read, so in every slot time the memory is accessed twice.

Example

Connect input TS3 to output TS7 (32-slot frame):

  • In TS3 the sample is written into SM(3).
  • The processor has stored CM(7) = 3.
  • In TS7 the logic reads CM(7) = 3, then reads SM(3) and sends it in output slot 7.
  • The sample is delayed by 7−3=47 - 3 = 4 slots. If the output slot comes before the input slot, the sample goes out in the next frame (delay up to one frame, 125 µs).

Timing limit

Frame time is 125 µs, so slot time is 125/c125/c µs. Since there are two accesses per slot, the memory cycle time must satisfy

taccess≤125 μs2ct_{access} \le \frac{125\ \mu s}{2c}

For c=32c = 32: taccess≤1.95t_{access} \le 1.95 µs. Faster memory allows more channels per TSI.

Features

  • Strictly non-blocking: any input slot can go to any free output slot.
  • Gives a fixed delay of less than one frame per stage.
  • Size is limited by memory speed, so large exchanges combine T stages with S stages (TST).
  • Broadcasting is easy: several CM locations can hold the same SM address, so one input can feed several outputs.
  • 2081 Baisakh · 4 marks

Write a short note on ST switch.

Answer

An ST switch is a two-stage digital switch in which a space (S) stage is followed by a time (T) stage. The S stage is a time-multiplexed space switch that connects incoming TDM highways to outgoing highways; each outgoing highway has a time slot interchanger (TSI) that moves the sample to the required output slot.

 TDM in            S stage           T stage     TDM out
 I1 ------->+----------------+--->[TSI 1]---> O1
 I2 ------->|  time-shared   |--->[TSI 2]---> O2
 ..         |  crosspoint    |     ...         ..
 IN ------->|  matrix (N x N)|--->[TSI N]---> ON
            +----------------+
                  ^
            control memory (per output column)

Working: a sample arriving in slot ii on input highway IaI_a must go to slot jj on output highway ObO_b. In slot ii the S-stage control memory closes crosspoint (a,b)(a,b), so the sample enters TSI bb, which writes it and reads it out in slot jj.

Drawback (blocking): the space stage must move the sample in the same time slot ii in which it arrives. If another call from a different input already uses slot ii to reach output highway bb, the new call is blocked, even though slot jj on ObO_b is free. There is only one path for each connection, so blocking is high under heavy load. TS switches have the same problem the other way round.

Remedy: add a third stage to get several alternative paths, giving STS or TST switches. The TST form is preferred in practice (e.g. in most large digital exchanges) because time expansion is cheaper than space expansion.

  • 2080 Chaitra · 4 marks

Write a short note on Stored Program control (SPC) in digital exchange.

Answer

Stored Program Control (SPC) means that the exchange is controlled by a digital computer (processor) that runs a program stored in memory, instead of by hard-wired relay logic as in Strowger and crossbar exchanges. All digital exchanges use SPC.

What the program does

  • Scans subscriber lines and trunks to detect off-hook, on-hook and digits.
  • Analyses digits, finds a free path through the switching network and sets it up.
  • Sends ringing, tones and signaling messages, and supervises the call.
  • Records call data for charging (billing) and traffic statistics.
  • Runs maintenance, fault diagnosis and administration tasks.

Types

  • Centralised SPC: one main processor (usually duplicated for reliability) controls the whole exchange. Dual-processor modes are standby, synchronous duplex and load sharing.
  • Distributed SPC: control is shared among many smaller processors (line-group processors, call processors, etc.). It is more reliable and easier to expand; most modern exchanges use it.

Advantages

  • New services (call forwarding, call waiting, abbreviated dialling, conference calls) are added by changing software.
  • Easy change of subscriber data and routing through a terminal.
  • Automatic fault detection and remote maintenance.
  • Fast call set-up and support for common channel signaling (SS7).
  • Smaller size, lower power use and better reliability.

Disadvantages: high initial software cost, and a software fault can affect the whole exchange, so careful testing and processor redundancy are needed.

  • 2080 Bhadra · 4+4 marks

Explain the working principle of Digital Telephone Exchange. Calculate how many cross point used in three stage switching.

Answer

Working principle of a digital telephone exchange

A digital exchange switches speech as PCM samples (8-bit words, 8000 per second) using time-division techniques, under stored program control (SPC).

 Subscriber +-------+  +----------+  +-------+
 lines ---->| Line  |->| Digital  |->| Trunk |--> other
 (analog)   | units |<-| switching|<-| units |    exch.
            |BORSCHT|  | (TST)    |  |(E1)   |
            +-------+  +----------+  +-------+
                ^           ^            ^
            +-------------------------------+
            | Control processor (SPC)       |
            | + signaling                   |
            +-------------------------------+
  1. Line unit: each analogue line has a line card doing the BORSCHT functions: Battery feed, Over-voltage protection, Ringing, Supervision, Coding (codec: A/D and D/A), Hybrid (2-wire to 4-wire), Testing.
  2. Concentration and multiplexing: coded samples from many lines are multiplexed onto 32-channel (2.048 Mbps) PCM highways.
  3. Digital switching network: time switches (TSIs) and space switches (usually in TST form) move each sample from its input slot and highway to the required output slot and highway.
  4. Control: the processor detects off-hook, collects digits, finds a free path, writes the control memories of the T and S stages, sends ringing and tones, supervises and records charging data.
  5. Trunk and signaling units: connect to other exchanges over E1 links; signaling is by CAS or CCS (SS7).

Crosspoints in a three-stage switch

Take NN inputs and NN outputs, divided into groups of nn lines, with kk middle-stage arrays.

  • First stage: N/nN/n arrays, each n×kn \times k
  • Second stage: kk arrays, each (N/n)×(N/n)(N/n) \times (N/n)
  • Third stage: N/nN/n arrays, each k×nk \times n
Nx=Nn(nk)+k(Nn)2+Nn(kn)=2Nk+k(Nn)2\begin{aligned} N_x &= \frac{N}{n}(nk) + k\left(\frac{N}{n}\right)^2 + \frac{N}{n}(kn) \\ &= 2Nk + k\left(\frac{N}{n}\right)^2 \end{aligned}

For a strictly non-blocking (Clos) network, k=2n−1k = 2n - 1:

Nx=(2n−1)[2N+(Nn)2]N_x = (2n-1)\left[2N + \left(\frac{N}{n}\right)^2\right]

Example: N=100N = 100, n=10n = 10, k=2n−1=19k = 2n-1 = 19:

Nx=2(100)(19)+19(10010)2=3800+1900=5700\begin{aligned} N_x &= 2(100)(19) + 19\left(\frac{100}{10}\right)^2 \\ &= 3800 + 1900 = 5700 \end{aligned}

A single-stage 100×100100 \times 100 switch needs N2=10 000N^2 = 10\,000 crosspoints, so the three-stage design saves 43% and is still non-blocking.

Answer: Nx=2Nk+k(N/n)2N_x = 2Nk + k(N/n)^2; for N=100N = 100, n=10n = 10, k=19k = 19, Nx=5700N_x = 5700 crosspoints.

  • 2080 Bhadra · 3+5 marks

What is blocking and non-blocking switching system? At which condition digital switch work as no blocking? Explain with example.

Answer

Blocking and non-blocking switching

  • A blocking switch is one in which a call between a free input and a free output may still fail because no free internal path exists (internal links are busy with other calls). Multistage networks with too few middle-stage paths are blocking. Blocking probability is the chance that such a call is lost.
  • A non-blocking switch is one in which any free input can always be connected to any free output, whatever other calls are in progress. A single-stage N×NN \times N crossbar is non-blocking but needs N2N^2 crosspoints.
PointBlockingNon-blocking
Internal pathMay not be freeAlways free
Crosspoints / hardwareFewerMore
CostLowerHigher
Grade of serviceSmall loss allowedZero internal loss
Typical useLocal exchanges with light trafficTrunk/transit switches

Condition for a digital switch to be non-blocking

(a) Space-division three-stage (Clos) network. With nn inputs per first-stage array and kk middle arrays, the worst case is: the calling input's array already has n−1n-1 busy inputs using n−1n-1 different middle arrays, and the called output's array has n−1n-1 busy outputs using n−1n-1 other middle arrays. One more middle array is needed:

k≥2n−1k \ge 2n - 1

(b) TST digital switch. Each incoming link has cc time slots, and the time-multiplexed space stage has ll internal time slots. By the same reasoning (with time slots in place of middle arrays):

l≥2c−1l \ge 2c - 1

So a TST switch is non-blocking when the space stage runs at about twice the slot rate of the external links (time expansion).

(c) STS digital switch. With NN incoming TDM links and kk centre-stage TSIs: k≥2N−1k \ge 2N - 1.

(d) Single time switch (TSI). It is always non-blocking provided the memory is fast enough: taccess≤125 μs/2ct_{access} \le 125\ \mu s / 2c.

Example

A TST switch has E1 links with c=32c = 32 slots. For non-blocking operation, l=2(32)−1=63l = 2(32) - 1 = 63, so practical designs use 64 internal slots. With l=32l = 32 only (no expansion), and link occupancy 0.9 E, blocking probability is

PB=[1−(1−0.9)2]32=0.9932≈0.725P_B = \left[1 - (1 - 0.9)^2\right]^{32} = 0.99^{32} \approx 0.725

Hence the time expansion to l≥63l \ge 63 is what makes the switch non-blocking.

For a space switch example: N=200N = 200, n=10n = 10 needs k≥19k \ge 19 middle arrays for non-blocking operation.

  • 2080 Baisakh · 6 marks

Derive expressions for the blocking probability of both STS and TST switches. Show that the blocking probability of a TST switch is lower than that of a STS switch.

Answer

Blocking probability of multistage switches is found with Lee graphs: each path is drawn as links in series, parallel paths are independent, and each link is busy with probability p′p'.

STS switch

NN incoming TDM links enter an N×kN \times k space stage, then kk centre-stage TSIs, then a k×Nk \times N space stage.

 A o---(k parallel paths, each = 2 links)---o B
        input S -> TSI_i -> output S
  • Each path uses two internal links in series: input S → TSI and TSI → output S.
  • If pp is the occupancy of external links, the internal link occupancy is
p′=pβ,β=kN (space expansion)p' = \frac{p}{\beta}, \quad \beta = \frac{k}{N} \text{ (space expansion)}
  • A path is free only if both links are free: probability (1−p′)2(1-p')^2. Path busy: 1−(1−p′)21-(1-p')^2.
  • There are kk parallel paths (one per centre TSI); the call is blocked only if all are busy:
PB,STS=[1−(1−p′)2]kP_{B,STS} = \left[1 - (1 - p')^2\right]^{k}

TST switch

NN incoming links, each with cc slots, go to input TSIs; a time-multiplexed space stage with ll internal time slots; then output TSIs.

  • The alternative paths are the ll internal time slots of the space stage. Each path uses two links: input TSI → S (in slot xx) and S → output TSI (in slot xx).
  • Internal link occupancy:
p′=pα,α=lc (time expansion)p' = \frac{p}{\alpha}, \quad \alpha = \frac{l}{c} \text{ (time expansion)}
  • Same Lee graph with ll parallel paths:
PB,TST=[1−(1−p′)2]lP_{B,TST} = \left[1 - (1 - p')^2\right]^{l}

Why TST has lower blocking

The expressions have the same form; the blocking depends on the number of parallel paths (the exponent) and p′p'.

  1. In STS the exponent is kk, the number of centre TSIs. Each extra path needs a whole TSI and adds crosspoints to both space stages (2Nk2Nk), so kk stays small.
  2. In TST the exponent is ll, the number of internal time slots. It is equal to or greater than the frame size cc (32 to 1024 in practice) and is raised just by running the space stage and memories faster, which is cheap.
  3. So for a practical size, l≫kl \gg k and the TST result is far smaller.

Example: p=0.7p = 0.7 E, no expansion (p′=0.7p' = 0.7), 1−(1−0.7)2=0.911-(1-0.7)^2 = 0.91:

  • STS with N=k=16N = k = 16 links: PB=0.9116=0.221P_B = 0.91^{16} = 0.221
  • TST with c=l=32c = l = 32 slots: PB=0.9132=0.0489P_B = 0.91^{32} = 0.0489
  • TST with time expansion l=64l = 64 (p′=0.35p' = 0.35): PB=0.577564≈5.5×10−16P_B = 0.5775^{64} \approx 5.5 \times 10^{-16}

Hence TST gives lower blocking at lower cost, which is why large digital exchanges use TST (or TSST/TSSST) structures.

  • 2080 Baisakh · 6 marks

A TST network is used in a digital switch and the secondary multiplex contains 120 time slots. How many time slots would be included in the time multiplexed space stage for non-blocking operation? What would be the blocking probability if the time multiplexed space stage contained 120 time slots, 150 time slots, and 200 time slots? Assume channel occupancies of 0.6E and 0.9E.

Answer

Given: time slots per frame on the external (secondary multiplex) link, c=120c = 120; occupancy p=0.6p = 0.6 E and 0.90.9 E.

Non-blocking condition

A TST switch is strictly non-blocking when the number of space-stage time slots satisfies l≥2c−1l \ge 2c - 1:

l=2(120)−1=239 time slotsl = 2(120) - 1 = 239 \text{ time slots}

Blocking probability

Lee-graph formula for TST:

PB=[1−(1−p′)2]l,p′=p clP_B = \left[1 - (1 - p')^2\right]^{l}, \qquad p' = \frac{p\,c}{l}

Step 1: internal link occupancy p′p'

llp′p' (p = 0.6)p′p' (p = 0.9)
1200.6000.900
1500.4800.720
2000.3600.540

Step 2: path-busy probability q=1−(1−p′)2q = 1 - (1-p')^2

llqq (p = 0.6)qq (p = 0.9)
1200.84000.9900
1500.72960.9216
2000.59040.7884

Step 3: PB=q lP_B = q^{\,l}

For example, l=120l = 120, p=0.9p = 0.9:

PB=0.99120=0.299P_B = 0.99^{120} = 0.299

and l=150l = 150, p=0.6p = 0.6:

PB=0.7296150=2.90×10−21P_B = 0.7296^{150} = 2.90 \times 10^{-21}
llPBP_B at 0.6 EPBP_B at 0.9 E
1208.19×10−108.19 \times 10^{-10}0.2990.299
1502.90×10−212.90 \times 10^{-21}4.80×10−64.80 \times 10^{-6}
2001.70×10−461.70 \times 10^{-46}2.24×10−212.24 \times 10^{-21}

Comment: even without time expansion (l=120l = 120) the blocking is negligible at 0.6 E, but at 0.9 E it is about 30%. A modest time expansion (150 or 200 slots) brings it down to a negligible value, far fewer than the 239 slots needed for strictly non-blocking operation.

Answer: l=239l = 239 slots for non-blocking; PBP_B = 8.19×10−108.19 \times 10^{-10}, 2.90×10−212.90 \times 10^{-21}, 1.70×10−461.70 \times 10^{-46} (0.6 E) and 0.2990.299, 4.80×10−64.80 \times 10^{-6}, 2.24×10−212.24 \times 10^{-21} (0.9 E) for ll = 120, 150, 200.

  • 2078 Bhadra · 8 marks

Design a three-stage, 200×200 switch with division of input lines (N = 200) into groups with each group of n = 20 lines. Use k number of crossbars in the middle stage where k = 4. Also, redesign this three stage 200×200 switch using Clos criteria with a minimum number of cross points.

Answer

Three-stage network with NN lines, groups of nn, and kk middle arrays:

  • Stage 1: N/nN/n arrays of n×kn \times k
  • Stage 2: kk arrays of (N/n)×(N/n)(N/n) \times (N/n)
  • Stage 3: N/nN/n arrays of k×nk \times n
Nx=2Nk+k(Nn)2N_x = 2Nk + k\left(\frac{N}{n}\right)^2

Design 1: N=200N = 200, n=20n = 20, k=4k = 4

Number of groups N/n=200/20=10N/n = 200/20 = 10.

StageArraysSizeCrosspoints
11020×420 \times 410×80=80010 \times 80 = 800
2410×1010 \times 104×100=4004 \times 100 = 400
3104×204 \times 2010×80=80010 \times 80 = 800
Total2000
Nx=2(200)(4)+4(10)2=1600+400=2000N_x = 2(200)(4) + 4(10)^2 = 1600 + 400 = 2000
 20 in  [20x4] \                   / [4x20] 20 out
 20 in  [20x4] -->  4 x [10x10] -->  [4x20] 20 out
  ...     ...  /     middle      \    ...
 20 in  [20x4]  (10 arrays)          [4x20] 20 out

Each first-stage array has one link to each middle array; each middle array has one link to each third-stage array.

This needs only 5% of the 2002=40 000200^2 = 40\,000 crosspoints of a single-stage switch, but only 4 of the 20 inputs of a group can be connected at once, so it is a blocking (concentrating) design, since k=4<2n−1=39k = 4 < 2n - 1 = 39.

Design 2: Clos non-blocking with minimum crosspoints

For strict non-blocking, k=2n−1k = 2n - 1. Crosspoints are minimum when

n=N2=2002=10n = \sqrt{\frac{N}{2}} = \sqrt{\frac{200}{2}} = 10 k=2n−1=19,Nn=20010=20k = 2n - 1 = 19, \qquad \frac{N}{n} = \frac{200}{10} = 20
StageArraysSizeCrosspoints
12010×1910 \times 1920×190=380020 \times 190 = 3800
21920×2020 \times 2019×400=760019 \times 400 = 7600
32019×1019 \times 1020×190=380020 \times 190 = 3800
Total15 200

Check with the Clos minimum formula:

Nx,min=4N(2N−1)=4(200)(400−1)=800×19=15 200\begin{aligned} N_{x,min} &= 4N\left(\sqrt{2N} - 1\right) \\ &= 4(200)\left(\sqrt{400} - 1\right) = 800 \times 19 = 15\,200 \end{aligned}

Answer: Design 1 needs 2000 crosspoints (blocking). The Clos design (n=10n = 10, k=19k = 19, 20 outer arrays) needs 15 200 crosspoints and is strictly non-blocking, which is 38% of a 40 000-crosspoint single-stage switch.

  • 2078 Bhadra · 2+5 marks

What is multi stage switching? Explain TST switching with neat diagram and its blocking probability.

Answer

Multistage switching

Multistage switching builds a large switch from several stages of smaller switches connected by internal links, instead of one big N×NN \times N matrix. It cuts the number of crosspoints (or memory) needed, and gives several alternative paths for each call. The cost is a small chance of internal blocking unless enough paths are provided. Examples: three-stage space networks, STS, TST, TSST.

TST switching

A Time–Space–Time switch has an input time stage, a time-multiplexed space stage and an output time stage.

 Link 1 -->[TSI-A1]--+-----------+--[TSI-B1]--> Link 1
 Link 2 -->[TSI-A2]--| N x N     |--[TSI-B2]--> Link 2
   ...        ...    | space     |     ...
 Link N -->[TSI-AN]--| (TMS)     |--[TSI-BN]--> Link N
                     +-----------+
  c slots   in T      l internal    out T    c slots
                       slots

Working (example): connect slot 5 of link 1 to slot 20 of link N.

  1. The control processor searches for an internal slot xx that is free both on the output of TSI-A1 and on the input of TSI-BN (say x=12x = 12).
  2. Input T stage: TSI-A1 moves the sample from slot 5 to slot 12.
  3. Space stage: in slot 12 its control memory closes crosspoint (1, N), so the sample passes from highway 1 to highway N.
  4. Output T stage: TSI-BN moves the sample from slot 12 to slot 20.
  5. The return direction uses another slot, often x+l/2x + l/2 (mod ll), so one search serves both directions.

Since any internal slot xx can be used, there are ll alternative paths.

Blocking probability

Using a Lee graph: each path has two internal links in series (TSI-A → S and S → TSI-B), and there are ll paths in parallel. With external occupancy pp and time expansion α=l/c\alpha = l/c, the internal link occupancy is p′=p/αp' = p/\alpha:

PB=[1−(1−p′)2]lP_B = \left[1 - (1 - p')^2\right]^{l}
  • Non-blocking when l≥2c−1l \ge 2c - 1.
  • Example: c=l=32c = l = 32, p=0.6p = 0.6: PB=[1−0.42]32=0.8432≈3.8×10−3P_B = [1 - 0.4^2]^{32} = 0.84^{32} \approx 3.8 \times 10^{-3}.

Why it is preferred: time expansion is cheap (faster memory), the space stage is small, and the blocking is very low, so TST (and TSST) is used in most large digital exchanges.

  • 2076 Chaitra · 2+5 marks

What do you mean by stored program control (SPC)? Explain different modes of dual processor architecture used in an electronic switching system using centralized SPC.

Answer

Stored program control

Stored Program Control (SPC) is the control of an exchange by a processor running programs stored in memory. Call processing, routing, charging and maintenance are done in software, so new features can be added by changing the program.

Dual-processor architecture in centralised SPC

In centralised SPC one processor controls the whole exchange, so a failure would stop all calls. The processor is therefore duplicated. The two processors (P1 and P2) share or copy the memory and work in one of three modes.

  +------+   +------+
  |  P1  |   |  P2  |
  +--+---+   +---+--+
     |  +----+   |
     +--|Comp|---+     (comparator in
     |  +----+   |      sync duplex)
  +--+---+   +---+--+
  |  M1  |   |  M2  |   (memories)
  +------+   +------+
        \     /
     exchange hardware

1. Standby mode

  • One processor is active; the other is idle on standby.
  • The standby takes over when the active one fails.
  • The memory is shared or the active processor regularly copies its data to a secondary store, so the standby can reload the current call data.
  • Simple, but calls being set up at the time of failure may be lost.

2. Synchronous duplex mode

  • Both processors run the same program at the same time, each with its own memory, and get the same inputs.
  • A comparator checks their outputs after each step. If they agree, normal working continues.
  • On a mismatch, both run diagnostic programs; the faulty one is taken out of service and the healthy one continues alone.
  • Gives fast fault detection and no loss of calls, but transient faults may cause needless check routines.

3. Load-sharing mode

  • Each processor handles part of the traffic (e.g. calls are given to them alternately or at random).
  • They share a common memory and use an exclusion device (ED) so both do not seize the same resource at once.
  • If one fails, the other carries the full load (with possible reduced grade of service at the busy hour).
  • Gives higher call-handling capacity than standby or synchronous modes.

Availability

For one processor, A=MTBFMTBF+MTTRA = \dfrac{MTBF}{MTBF + MTTR}. For a dual system the mean time between system failures becomes

MTBFD=MTBF22 MTTRMTBF_D = \frac{MTBF^2}{2\,MTTR}

so unavailability falls from about MTTR/MTBFMTTR/MTBF to about 2 MTTR2/MTBF22\,MTTR^2/MTBF^2. This large gain is why all centralised SPC exchanges use duplicated processors.

  • 2076 Chaitra · 8 marks

Describe the working principle of TSI switch in sequential read, random write mode.

Answer

A time switch or TSI (Time Slot Interchanger) changes the time slot of a PCM sample within a TDM frame. In the sequential read, random write mode (also called input-associated control), the address at which each incoming sample is written is chosen by a control memory, and the speech memory is read out in order.

Main parts

  • Speech memory (SM): cc words of 8 bits, one per time slot.
  • Control memory (CM): cc words of log⁡2c\log_2 c bits. CM(ii) holds the SM address where the sample of input slot ii must be written.
  • Time-slot counter: a modulo-cc counter giving the current slot number.
  • Processor interface: loads CM entries when calls are set up or released.
 Input TDM                              Output TDM
 --------->+---------------------+---------->
           |   Speech memory     |
           |   (c x 8 bits)      |
           +---------------------+
  write addr ^             ^ read addr
   (random)  |             | (sequential)
      +------------+  +-------------+
      | Control    |<-| Time-slot   |
      | memory     |  | counter     |
      +------------+  +-------------+
           ^
   processor (call set-up)

Working principle

  1. Write phase (random): in input slot ii, the counter addresses CM(ii). The value stored there, say jj, is used as the write address, so the sample is stored in SM(jj).
  2. Read phase (sequential): the counter also gives the read address directly. In output slot jj, SM(jj) is read and sent out. So a sample written into SM(jj) leaves in output slot jj.
  3. To set up a call from input slot ii to output slot jj, the processor writes jj into CM(ii).
  4. One write and one read happen in every slot.

Example

Connect input TS2 to output TS9 in a 32-slot frame:

  • Processor sets CM(2) = 9.
  • In TS2 the sample is written into SM(9).
  • In TS9 the counter reads SM(9), so the sample appears in output slot 9.
  • Delay = 9−2=79 - 2 = 7 slots =7×3.9≈27.3= 7 \times 3.9 \approx 27.3 µs.

Timing

Slot time = 125/c125/c µs and two memory accesses per slot are needed:

taccess≤125 μs2ct_{access} \le \frac{125\ \mu s}{2c}

For 32 channels this is 1.95 µs. Maximum channels for a memory of access time tt: cmax=125 μs/2tc_{max} = 125\ \mu s / 2t.

Features

  • Strictly non-blocking for its own frame.
  • Delay is less than one frame (125 µs).
  • Broadcast is not possible in this mode, because one input sample is written into only one SM location. (In sequential-write, random-read mode, several outputs can read the same location.)
  • Useful at the input side of a TST switch, where each input slot has a fixed control entry.
  • 2076 Asoj · 7 marks

Derive the necessary conditions for a 3-stage network to be strictly non-blocking.

Answer

A three-stage network is strictly non-blocking if a free input can always be connected to a free output, however the existing calls are routed. The condition was derived by C. Clos (1953).

Network

  • NN inputs and NN outputs; groups of nn lines.
  • Stage 1: N/nN/n arrays of n×kn \times k; Stage 2: kk arrays of (N/n)×(N/n)(N/n) \times (N/n); Stage 3: N/nN/n arrays of k×nk \times n.
  • Each first-stage array has one link to every middle array; each middle array has one link to every third-stage array.
  input array A          output array B
  (n inputs)             (n outputs)
  n-1 busy -> use        n-1 busy -> use
  n-1 middle arrays      n-1 other middle
       \                      /
        \  need 1 more free  /
         +-- middle array --+
     total k >= (n-1)+(n-1)+1

Derivation of k≥2n−1k \ge 2n - 1

Consider a new call from a free input of first-stage array A to a free output of third-stage array B. Take the worst case:

  1. Array A has nn inputs; the other n−1n - 1 are busy. Each of those calls uses a different link from A, so they occupy n−1n - 1 middle arrays.
  2. Array B has nn outputs; the other n−1n - 1 are busy, using n−1n - 1 middle arrays.
  3. In the worst case, these two sets of middle arrays are all different, so 2(n−1)2(n - 1) middle arrays cannot be used.
  4. To connect A to B, at least one more middle array must be free:
k≥(n−1)+(n−1)+1=2n−1k \ge (n - 1) + (n - 1) + 1 = 2n - 1

This is the necessary and sufficient condition for strict non-blocking.

Crosspoints and optimum nn

Nx=2Nk+k(Nn)2=(2n−1)[2N+N2n2]N_x = 2Nk + k\left(\frac{N}{n}\right)^2 = (2n-1)\left[2N + \frac{N^2}{n^2}\right]

Differentiating and treating 2n−1≈2n2n - 1 \approx 2n for large nn:

dNxdn=4N+N22−2nn3≈4N−2N2n2=0nopt=N2\begin{aligned} \frac{dN_x}{dn} &= 4N + N^2\frac{2 - 2n}{n^3} \approx 4N - \frac{2N^2}{n^2} = 0 \\ n_{opt} &= \sqrt{\frac{N}{2}} \end{aligned}

Substituting:

Nx,min=4N(2N−1)N_{x,min} = 4N\left(\sqrt{2N} - 1\right)

Example

N=200N = 200: n=10n = 10, k=19k = 19, Nx,min=4(200)(20−1)=15 200N_{x,min} = 4(200)(20 - 1) = 15\,200, compared with N2=40 000N^2 = 40\,000 for a single-stage switch.

The same idea applies to digital switches: a TST switch is strictly non-blocking when the internal time slots satisfy l≥2c−1l \ge 2c - 1, and an STS switch when the centre TSIs satisfy k≥2N−1k \ge 2N - 1.

  • 2076 Asoj · 5 marks

In the case of a Time Slot Interchanger (TSI), compare the working mechanisms between Sequential Write and Random Read with Random Write and Sequential Read.

Answer

A TSI has a speech memory (SM) holding one PCM sample per time slot and a control memory (CM) set by the processor. The two modes differ in which access (write or read) the CM controls.

Sequential write, random read (output-associated control)

  • Input sample in slot ii is written into SM(ii), using the time-slot counter as address.
  • In output slot jj, CM(jj) gives the SM address to read; the sample of that input slot is sent out.
  • To connect input ii to output jj: set CM(jj) = ii.

Random write, sequential read (input-associated control)

  • In input slot ii, CM(ii) gives the SM address jj where the sample is written.
  • SM is read in order, so SM(jj) goes out in output slot jj.
  • To connect input ii to output jj: set CM(ii) = jj.
 SW-RR:  in --> SM[counter]      SM[CM[counter]] --> out
 RW-SR:  in --> SM[CM[counter]]  SM[counter]     --> out

Example: input TS4 to output TS10. SW-RR: sample stored in SM(4), CM(10) = 4. RW-SR: CM(4) = 10, sample stored in SM(10). Output is the same in both.

PointSequential write, random readRandom write, sequential read
Write addressFrom time-slot counterFrom control memory
Read addressFrom control memoryFrom time-slot counter
CM indexed byOutput slotInput slot
Control typeOutput-associatedInput-associated
Broadcast (one input to many outputs)PossibleNot possible
Typical useOutput stage of TSTInput stage of TST
DelayLess than one frameLess than one frame
Memory speed neededt≤125/2ct \le 125/2c µsSame

Both modes need the same memory size (c×8c \times 8 bits SM, c×log⁡2cc \times \log_2 c bits CM) and are non-blocking for one frame; they are often used together in a TST switch so that one control memory word can serve both directions.

  • 2075 Chaitra · 3+6 marks

Differentiate between TST and STS switch. Design a three stage switching system having 4 stage array of 5 input line and 6 second stage array. Also calculate the total number of cross points of the switching system.

Answer

Difference between TST and STS switches

PointTSTSTS
StructureTime – Space – TimeSpace – Time – Space
Outer stagesTSIs (memory)Time-multiplexed space switches
Centre stageTime-multiplexed space switchArray of TSIs
Number of alternative pathsll (internal time slots)kk (centre TSIs)
ExpansionTime expansion (cheap: faster memory)Space expansion (costly: more crosspoints and TSIs)
BlockingLower, [1−(1−p′)2]l[1-(1-p')^2]^lHigher, [1−(1−p′)2]k[1-(1-p')^2]^k
ControlPath search over time slotsPath search over centre TSIs
Best forLarge exchanges, heavy trafficSmall exchanges, light traffic

Three-stage design

Reading of the data: the first stage has 4 arrays with 5 input lines each, so N=4×5=20N = 4 \times 5 = 20 lines, n=5n = 5; the second stage has k=6k = 6 arrays.

  • Stage 1: 4 arrays of 5×65 \times 6
  • Stage 2: 6 arrays of 4×44 \times 4 (each middle array has one link from each of the 4 first-stage arrays)
  • Stage 3: 4 arrays of 6×56 \times 5
  5 in [5x6] \           / [6x5] 5 out
  5 in [5x6] --> 6 x [4x4] --> [6x5] 5 out
  5 in [5x6] -->  middle  --> [6x5] 5 out
  5 in [5x6] /           \ [6x5] 5 out
  (20 inputs)               (20 outputs)

Crosspoints

N1=4×(5×6)=120N2=6×(4×4)=96N3=4×(6×5)=120Nx=120+96+120=336\begin{aligned} N_1 &= 4 \times (5 \times 6) = 120 \\ N_2 &= 6 \times (4 \times 4) = 96 \\ N_3 &= 4 \times (6 \times 5) = 120 \\ N_x &= 120 + 96 + 120 = 336 \end{aligned}

Check: Nx=2Nk+k(N/n)2=2(20)(6)+6(4)2=240+96=336N_x = 2Nk + k(N/n)^2 = 2(20)(6) + 6(4)^2 = 240 + 96 = 336.

A single-stage 20×2020 \times 20 switch needs 400 crosspoints, so the saving is 64 crosspoints (16%). Since k=6<2n−1=9k = 6 < 2n - 1 = 9, this network is blocking; 9 middle arrays would be needed for strict non-blocking.

Answer: total crosspoints = 336.

  • 2075 Asoj · 4+4 marks

Describe technical structure of a telephone exchange. Compare TST and STS switch used in digital telephone exchange system.

Answer

Technical structure of a telephone exchange

A telephone exchange connects any subscriber line or trunk to any other on demand. It has the following main subsystems:

 Subscribers                         Other exchanges
   |                                       ^
 +------------+  +-------------+  +-------------+
 | Subscriber |->|  Switching  |->|  Trunk /    |
 | line units |<-|  network    |<-|  junction   |
 | (BORSCHT)  |  |  (T, S, TST)|  |  interface  |
 +------------+  +-------------+  +-------------+
      ^   ^             ^             ^
      |   +--------+    |    +--------+
      |   | Signal-|    |    |
      |   | ling   |<-->+<-->|  Control (SPC
      |   +--------+         |  processors)
      +--------------------> +---------------+
               O&M: charging, testing, admin
  1. Subscriber line interface: each line card does BORSCHT (battery feed, over-voltage protection, ringing, supervision, codec, hybrid, testing). Line concentrators combine many lightly used lines onto fewer switch ports.
  2. Switching network: makes the speech path. In digital exchanges it is a TST (or TSST) network of time switches and space switches.
  3. Trunk/junction interface: connects to other exchanges on E1 (2.048 Mbps) PCM links, with frame alignment and clock synchronisation.
  4. Signaling equipment: subscriber signaling (dial tone, DTMF receivers, ringing, busy tone) and inter-exchange signaling (CAS or CCS/SS7).
  5. Control unit: SPC processors that detect calls, analyse digits, find paths, set the switch, supervise calls and release them.
  6. Operation and maintenance: charging (billing) records, traffic measurement, alarms, testing and subscriber administration.
  7. Power and synchronisation: 48 V DC battery plant and a master clock.

Comparison of TST and STS switches

PointTST switchSTS switch
Stage orderTime – Space – TimeSpace – Time – Space
Centre stageSpace switch (TMS)TSIs
Alternative pathsll internal time slotskk centre TSIs
Blocking[1−(1−p/α)2]l[1-(1-p/\alpha)^2]^l, lower[1−(1−p/β)2]k[1-(1-p/\beta)^2]^k, higher
Expansion methodTime expansion, cheapSpace expansion, costly
Non-blocking conditionl≥2c−1l \ge 2c - 1k≥2N−1k \ge 2N - 1
Cost for large sizeLowerHigher
UseLarge digital exchanges (e.g. AXE, EWSD type)Small switches, light load

TST is preferred in practice because, for the same blocking, it needs less hardware and its paths are easy to find by searching time slots.

  • 2074 Asoj · 5+1+2 marks

Design space switch with following input trunks, output trunks and connection memory. How many virtual paths are required in this switch? Write down each output channels of output trunks in time t0 to t3, with respect to decode logic.
Input trunks and their channels:
Input trunkI1I2I3I4I5
ChannelsA4 A3 A2 A1B4 B3 B2 B1C4 C3 C2 C1D4 D3 D2 D1E4 E3 E2 E1
Connection memory (5 bits per output trunk, feeding the decode logic):
TimeO1O2O3O4
t00 0 1 0 00 1 0 0 01 0 0 0 00 0 0 1 0
t11 0 0 0 00 0 1 0 00 1 0 0 00 0 0 0 1
t20 0 0 0 10 0 0 1 00 0 1 0 01 0 0 0 0
t30 1 0 0 01 0 0 0 00 0 0 1 00 0 1 0 0

Answer

Assumptions: each 5-bit connection-memory word is a one-hot select for the decode logic, with the bits in the order I1 I2 I3 I4 I5 (left to right). Channel 1 of each trunk arrives first (in t0t_0), channel 2 in t1t_1, and so on, as the channels are written right-to-left ("A4 A3 A2 A1").

Design of the space switch

This is a time-multiplexed space switch with 5 input trunks (rows) and 4 output trunks (columns), with 4 time slots per frame.

          O1    O2    O3    O4
 I1 ----- x --- x --- x --- x
 I2 ----- x --- x --- x --- x
 I3 ----- x --- x --- x --- x
 I4 ----- x --- x --- x --- x
 I5 ----- x --- x --- x --- x
          |     |     |     |
       [CM1] [CM2] [CM3] [CM4]
       decode logic per column
       (4 words x 5 bits each)
  • Each output column has its own control memory with 4 words (one per time slot) of 5 bits.
  • In each time slot the decoder of a column closes the one crosspoint whose bit is 1, connecting that input trunk to the output trunk for that slot only.
  • The switch has 5×4=205 \times 4 = 20 crosspoints and 4×4=164 \times 4 = 16 control words.

Number of virtual paths

Each crosspoint can carry a different connection in each time slot:

virtual paths=5×4×4=80\text{virtual paths} = 5 \times 4 \times 4 = 80

In each time slot at most 4 connections exist (one per output), so in one frame at most 4×4=164 \times 4 = 16 of these paths are in use; the given memory uses all 16.

Decoding the connection memory

TimeO1O2O3O4
t0t_000100 → I301000 → I210000 → I100010 → I4
t1t_110000 → I100100 → I301000 → I200001 → I5
t2t_200001 → I500010 → I400100 → I310000 → I1
t3t_301000 → I210000 → I100010 → I400100 → I3

Output channels in t0t_0 to t3t_3

TimeO1O2O3O4
t0t_0C1B1A1D1
t1t_1A2C2B2E2
t2t_2E3D3C3A3
t3t_3B4A4D4C4

Written in the same style as the input trunks (latest on the left):

  • O1: B4 E3 A2 C1
  • O2: A4 D3 C2 B1
  • O3: D4 C3 B2 A1
  • O4: C4 A3 E2 D1

A space switch does not change the time slot of a sample: channel mm always leaves in time slot tm−1t_{m-1}. Changing slots needs a time switch, which is why practical exchanges use TST.

  • 2074 Asoj · 4 marks

Write a short note on TST and STS switches.

Answer

TST and STS are three-stage digital switches that combine time switches (TSIs) and time-multiplexed space switches (S) so that every call has many alternative paths. They solve the high blocking of two-stage ST and TS switches.

 TST:  link-->[T]-->[ S ]-->[T]-->link
 STS:  link-->[ S ]-->[T..T]-->[ S ]-->link

TST (Time–Space–Time)

  • Input TSI moves the sample from its external slot to a free internal slot xx; the space stage switches it in slot xx to the right output highway; the output TSI moves it to the required output slot.
  • Alternative paths = number of internal time slots ll.
  • Blocking: PB=[1−(1−p′)2]lP_B = [1-(1-p')^2]^l, p′=pc/lp' = pc/l; non-blocking if l≥2c−1l \ge 2c - 1.

STS (Space–Time–Space)

  • The input space stage sends the sample to one of kk centre TSIs; that TSI changes its slot; the output space stage sends it to the right outgoing link.
  • Alternative paths = number of centre TSIs kk.
  • Blocking: PB=[1−(1−p′)2]kP_B = [1-(1-p')^2]^k, p′=pN/kp' = pN/k; non-blocking if k≥2N−1k \ge 2N - 1.
PointTSTSTS
ExpansionTime (cheap)Space (costly)
BlockingLowerHigher
Best forLarge, busy exchangesSmall exchanges

TST is more widely used because time expansion only needs faster memory, giving lower blocking at lower cost.

  • 2074 Chaitra · 4+4 marks

What are the basic functions of a conventional exchange? Write economic and technical advantages of PCM switching compared to its analog switching.

Answer

Basic functions of a conventional exchange

  1. Attending (identification): continuously monitor all lines and detect a call request (off-hook) and the calling line.
  2. Information receiving: send dial tone and receive the dialled digits (pulse or DTMF).
  3. Information processing: analyse the digits to find the destination and route.
  4. Busy testing: check whether the called line or a trunk is free.
  5. Interconnection (switching): set up a speech path through the switching network between the calling and called line or trunk.
  6. Alerting: send ringing current to the called party and ring-back tone to the caller.
  7. Supervision: watch the call for answer and clear-down, and release the path when either party hangs up.
  8. Information sending: send address and line signals to other exchanges for outgoing calls.
  9. Charging and metering: record call duration and destination for billing.

Advantages of PCM (digital) switching over analog switching

Economic advantages

  • Shared crosspoints: one time-shared crosspoint or memory location serves many calls, so far fewer components are needed.
  • No codecs between exchanges in an integrated digital network: digital trunks are switched directly, removing costly A/D and D/A conversion at each exchange.
  • VLSI hardware: memories and logic chips are cheap, small and low power, so exchanges need less floor space, air conditioning and power.
  • Lower maintenance cost: no moving parts, self-diagnosis and remote maintenance by software.
  • Quick installation and easy expansion in modular units.

Technical advantages

  • No noise or loss accumulation: signals are regenerated, so quality does not depend on the number of switching stages or distance.
  • Constant transmission loss through the exchange, independent of path.
  • Less crosstalk and better speech quality.
  • Integration of transmission and switching on the same PCM format (2.048 Mbps E1).
  • Data and ISDN services carried directly, as everything is already digital.
  • Fast call set-up and easy use of common channel signaling (SS7).
  • SPC features: call waiting, forwarding, conference, itemised billing.
  • High reliability through duplicated processors and self-checking.

(Issues to note: need for network synchronisation, a codec and hybrid on every analogue line card, and echo control.)

  • 2073 Shrawan · 3+5 marks

What are advantages and issues of PCM switching when it compared with analog switching? Calculate and draw, how many cross points are found in three stages switching system, whereas 3 stages array of 4 input lines and 5 second stages array.

Answer

Advantages of PCM switching over analog switching

  • No accumulation of noise: digital signals are regenerated; quality does not depend on the number of switching stages.
  • Time sharing of hardware: one crosspoint or memory location serves many calls, so fewer components are needed.
  • Integration with PCM transmission: digital trunks are switched without A/D conversion at each exchange (integrated digital network).
  • Low cost, size and power using VLSI; easy maintenance and SPC features.
  • Data/ISDN services and fast common channel signaling.

Issues (problems) of PCM switching

  • Synchronisation: all exchanges must run on a common clock, otherwise frame slips occur.
  • Codec and hybrid per line: each analogue subscriber line needs BORSCHT functions, which costs more per line.
  • Delay and echo: buffering adds delay (up to a frame per T stage), and 2-wire/4-wire hybrids cause echo that may need cancellers.
  • Memory speed limits the size of a single time switch.
  • Software complexity and dependence on processors.

Crosspoints in the three-stage switch

Reading of the data: 3 first-stage arrays, each with 4 input lines, so N=3×4=12N = 3 \times 4 = 12, n=4n = 4; 5 second-stage arrays, k=5k = 5.

  • Stage 1: 3 arrays of 4×54 \times 5
  • Stage 2: 5 arrays of 3×33 \times 3
  • Stage 3: 3 arrays of 5×45 \times 4
 4 in [4x5] \             / [5x4] 4 out
 4 in [4x5] --> 5 x [3x3] --> [5x4] 4 out
 4 in [4x5] /    middle   \ [5x4] 4 out
 (12 inputs)                 (12 outputs)
 each outer array has 1 link to each
 of the 5 middle arrays
N1=3×(4×5)=60N2=5×(3×3)=45N3=3×(5×4)=60Nx=60+45+60=165\begin{aligned} N_1 &= 3 \times (4 \times 5) = 60 \\ N_2 &= 5 \times (3 \times 3) = 45 \\ N_3 &= 3 \times (5 \times 4) = 60 \\ N_x &= 60 + 45 + 60 = 165 \end{aligned}

Check: Nx=2Nk+k(N/n)2=2(12)(5)+5(3)2=120+45=165N_x = 2Nk + k(N/n)^2 = 2(12)(5) + 5(3)^2 = 120 + 45 = 165.

Comment: a single-stage 12×1212 \times 12 switch needs only 144 crosspoints, so for such a small NN the three-stage form gives no saving; it pays off only for large NN. Also k=5<2n−1=7k = 5 < 2n - 1 = 7, so the network is blocking.

Answer: total crosspoints = 165.

  • 2073 Chaitra · 8 marks

What is a space switch and time switch? How STS switch is differ than TST switch? Explain with telephone switching diagram.

Answer

Space switch

A space switch (S) connects physically separate input lines to output lines. In a digital exchange it is a time-multiplexed space switch: an N×MN \times M crosspoint (or multiplexer) matrix whose crosspoints are changed every time slot by a control memory, so one crosspoint carries different calls in different slots. It changes the highway of a sample but not its time slot.

Time switch

A time switch (T) or TSI changes the time slot of a sample on one TDM highway. It writes the incoming samples into a speech memory and reads them out in a different order under control of a control memory. It changes the slot but not the highway.

 Space switch:  I1 slot5 --> O3 slot5
 Time switch:   slot5 --> slot12 (same highway)

How STS differs from TST

 STS:
 links -->[ S ]-->[TSI 1]-->[ S ]--> links
                  [TSI 2]
                  [TSI k]
 TST:
 links -->[TSI]-->[  S  ]-->[TSI]--> links
          (N)   (l internal  (N)
                    slots)

STS working: in the external slot ii, the input S stage sends the sample to a free centre TSI mm. TSI mm moves it from slot ii to the required output slot jj. In slot jj the output S stage sends it to the correct outgoing link. Alternative paths = kk centre TSIs.

TST working: the input TSI moves the sample from slot ii to a free internal slot xx; in slot xx the space stage connects the input highway to the output highway; the output TSI moves it from slot xx to slot jj. Alternative paths = ll internal slots.

Telephone call example

Subscriber A is on slot 3 of highway 1; B is on slot 17 of highway 4.

  • TST: TSI-1 moves slot 3 → internal slot 9; S stage closes (1, 4) in slot 9; TSI-4 moves slot 9 → slot 17.
  • STS: in slot 3, S stage connects highway 1 to centre TSI 2; TSI 2 moves slot 3 → slot 17; in slot 17 the output S stage connects TSI 2 to highway 4.
PointSTSTST
Centre stageTSIsSpace switch
Pathskk TSIsll time slots
BlockingHigherLower
ExpansionSpace (costly)Time (cheap)
UseSmall switchesLarge exchanges
  • 2072 Chaitra · 8 marks

Explain digital switching system. Mention functions of switching system in telecommunication.

Answer

Digital switching system

A digital switching system is an exchange that switches voice and data in digital (PCM) form, using time-division switching under stored program control. Speech from analogue lines is sampled at 8 kHz, coded into 8-bit words and multiplexed onto 32-slot (2.048 Mbps) highways. The switching network then moves each word from its incoming slot and highway to the required outgoing slot and highway using time switches (T) and space switches (S), normally arranged as TST.

 Analog   +---------+  +------------+  +--------+  Digital
 lines -->| Line    |->| Switching  |->| Trunk  |--> trunks
          | card +  |  | network    |  | inter- |   (E1)
          | codec   |<-| (T-S-T)    |<-| face   |
          +---------+  +------------+  +--------+
               ^              ^             ^
               +------ Control processor ---+
                     (SPC) + signaling

Main parts: line interface (BORSCHT), concentrator/multiplexer, digital switching network, trunk interface, signaling units (DTMF receivers, SS7), control processor, and O&M system.

Features: no moving parts, time-shared hardware, regeneration of signals (no noise build-up), integration with digital transmission, and easy new services through software.

Functions of a switching system

  1. Attending: detect a call request (off-hook) on any line.
  2. Information receiving: give dial tone and receive dialled digits.
  3. Information processing: analyse digits to find the called party and route.
  4. Busy testing: check if the called line or outgoing trunk is free.
  5. Interconnection: set up a path through the switching network.
  6. Alerting: ring the called party; send ring-back tone to the caller.
  7. Supervision: watch for answer and clearing; release the path at the end.
  8. Information sending: send signaling to other exchanges for outgoing calls.
  9. Charging: record call details for billing.
  10. Operation and maintenance: fault detection, testing, traffic records, subscriber data changes.
  • 2072 Chaitra · 2+6 marks

What is multistage switching? Describe the STS switching with neat diagram and its blocking probabilities.

Answer

Multistage switching

Multistage switching builds a large switch from two or more stages of smaller switching arrays joined by internal links. It needs far fewer crosspoints than a single N×NN \times N matrix and gives alternative paths, but some blocking may occur unless the middle stage is large enough (e.g. k≥2n−1k \ge 2n - 1).

STS switching

A Space–Time–Space switch has an input time-multiplexed space stage, a centre stage of kk TSIs, and an output space stage.

 Link 1 -->+-------+  [TSI 1]  +-------+--> Link 1
 Link 2 -->| N x k |->[TSI 2]->| k x N |--> Link 2
   ...     | space |   ...     | space |      ...
 Link N -->+-------+  [TSI k]  +-------+--> Link N
   c slots  (in S)   centre T   (out S)

Working: to connect slot ii of link 1 to slot jj of link N:

  1. The controller looks for a centre TSI mm that is free in input slot ii (link from input S) and in output slot jj (link to output S).
  2. In slot ii, the input space stage closes crosspoint (1, mm), sending the sample to TSI mm.
  3. TSI mm stores the sample and reads it out in slot jj.
  4. In slot jj, the output space stage closes crosspoint (mm, N), sending it to outgoing link N.
  5. The reverse direction is set up the same way.

There are kk alternative paths, one through each centre TSI.

Blocking probability

Using a Lee graph: each path has two internal links in series (input S → TSI and TSI → output S), and kk paths are in parallel.

  • External link occupancy: pp. With space expansion β=k/N\beta = k/N, internal link occupancy is p′=p/βp' = p/\beta.
  • Probability one path is free: (1−p′)2(1-p')^2; busy: 1−(1−p′)21-(1-p')^2.
  • Blocking (all kk busy):
PB=[1−(1−p′)2]kP_B = \left[1 - (1 - p')^2\right]^{k}
  • Non-blocking when k≥2N−1k \ge 2N - 1.

Example: N=k=16N = k = 16, p=0.7p = 0.7: p′=0.7p' = 0.7, PB=(1−0.09)16=0.9116≈0.22P_B = (1 - 0.09)^{16} = 0.91^{16} \approx 0.22.

Remark: to reduce blocking, kk must be raised, which adds TSIs and crosspoints (space expansion). For large switches, TST is cheaper for the same blocking, so STS is used mainly in small switches.

  • 2072 Chaitra · 3+5 marks

What are the basic switching functions? Calculate and draw, how many cross points are found in three stage switching system, whereas 3 stage array of 4 input lines and 4 second stages array.

Answer

Basic switching functions

  1. Attending: detect a call request (off-hook) and identify the calling line.
  2. Information receiving: send dial tone and receive the dialled digits.
  3. Information processing: analyse the digits to find the called party and route.
  4. Busy testing: check that the called line or a trunk is free.
  5. Interconnection: set up a path through the switching network.
  6. Alerting: ring the called party and send ring-back tone to the caller.
  7. Supervision: detect answer and clear-down; release the path.
  8. Information sending: send signals to other exchanges for outgoing calls.
  9. Charging: record call details for billing.

Crosspoints in the three-stage switch

Reading of the data: 3 first-stage arrays with 4 input lines each, so N=3×4=12N = 3 \times 4 = 12, n=4n = 4; 4 second-stage arrays, k=4k = 4.

  • Stage 1: 3 arrays of 4×44 \times 4
  • Stage 2: 4 arrays of 3×33 \times 3
  • Stage 3: 3 arrays of 4×44 \times 4
 4 in [4x4] \             / [4x4] 4 out
 4 in [4x4] --> 4 x [3x3] --> [4x4] 4 out
 4 in [4x4] /    middle   \ [4x4] 4 out
 (12 inputs)                 (12 outputs)
 each outer array has one link to each
 of the 4 middle arrays
N1=3×(4×4)=48N2=4×(3×3)=36N3=3×(4×4)=48Nx=48+36+48=132\begin{aligned} N_1 &= 3 \times (4 \times 4) = 48 \\ N_2 &= 4 \times (3 \times 3) = 36 \\ N_3 &= 3 \times (4 \times 4) = 48 \\ N_x &= 48 + 36 + 48 = 132 \end{aligned}

Check with the formula: Nx=2Nk+k(N/n)2=2(12)(4)+4(3)2=96+36=132N_x = 2Nk + k(N/n)^2 = 2(12)(4) + 4(3)^2 = 96 + 36 = 132.

A single-stage 12×1212 \times 12 switch needs 122=14412^2 = 144 crosspoints, so the saving is 12 crosspoints. Since k=4<2n−1=7k = 4 < 2n - 1 = 7, the network is blocking; 7 middle arrays would make it strictly non-blocking.

Answer: total crosspoints = 132.

  • 2072 Kartik · 6+4 marks

Compare between TST and STS switch used in digital telephone exchange system. State the advantages and disadvantages of DTMF telephone set.

Answer

Comparison of TST and STS switches

Both are three-stage digital switches that give many alternative paths to reduce the blocking of two-stage ST/TS switches.

 TST: link-->[TSI]-->[ S ]-->[TSI]-->link
 STS: link-->[ S ]-->[TSI x k]-->[ S ]-->link
PointTSTSTS
Stage orderTime – Space – TimeSpace – Time – Space
Outer stagesTSIs (memories)Time-multiplexed space switches
Centre stageSpace switchkk TSIs
Alternative pathsll internal time slotskk centre TSIs
Blocking[1−(1−p′)2]l[1-(1-p')^2]^{l}, p′=pc/lp' = pc/l[1−(1−p′)2]k[1-(1-p')^2]^{k}, p′=pN/kp' = pN/k
Non-blocking ifl≥2c−1l \ge 2c - 1k≥2N−1k \ge 2N - 1
ExpansionTime expansion: faster memory, cheapSpace expansion: more TSIs and crosspoints, costly
Cost for large switchLowerHigher
ControlSearch for a free internal slotSearch for a free centre TSI
Typical useLarge exchanges, heavy trafficSmall switches, light traffic

TST gives lower blocking at lower cost, so most large digital exchanges use TST (or TSST).

DTMF telephone set

In Dual Tone Multi-Frequency dialling, each key sends two tones at once: one from the low group (697, 770, 852, 941 Hz) and one from the high group (1209, 1336, 1477, 1633 Hz). For example key "5" sends 770 Hz + 1336 Hz.

Advantages

  • Fast dialling: about 10 digits per second (≈ 50 ms tone + 50 ms gap) compared with about 1 s per digit for rotary pulse dialling.
  • Fewer errors: no pulse distortion; tones are easy to detect.
  • End-to-end signaling: tones pass through the speech path after the call is connected, so they can be used for IVR menus, phone banking, voice mail and remote control.
  • Extra keys (*, #, A–D) allow supplementary services.
  • Electronic, no moving parts, so the set is reliable and compact.
  • Works well with digital (SPC) exchanges.

Disadvantages

  • The exchange needs DTMF receivers (more costly than pulse counters), and enough of them for busy-hour traffic.
  • The set needs electronic tone generators and power from the line.
  • Talk-off: speech or music may imitate a valid tone pair and cause a false digit; receivers need guard times and twist limits.
  • Older electromechanical exchanges cannot use DTMF without converters.
  • Tones can be distorted by noise or line loss on poor lines.
  • 2071 Shrawan · 2+5 marks

What is the principle of time division switching? Describe the operation of time division space switch.

Answer

Principle of time division switching

In time division switching, the switching hardware is shared in time: each call is given a short time slot in a repeating frame and the hardware connects that call only during its slot. In PCM telephony the voice is sampled every 125 µs (8 kHz) and coded into 8 bits; 32 slots form one 125 µs frame. A crosspoint or memory location used by one call in slot 3 can be used by another call in slot 4. Two forms exist:

  • Time division space switching: a crosspoint matrix is shared in time (changes highway, keeps slot).
  • Time division time switching: samples are stored and read in a different order (changes slot, keeps highway).

Time division space switch (TMS)

A time-multiplexed space switch connects NN input TDM highways to MM output highways. It has an N×MN \times M crosspoint matrix (electronic gates or multiplexers) and a control memory for each output column (output-controlled) or each input row (input-controlled).

            O1      O2     ...   OM
 I1 ------- x ----- x ---------- x
 I2 ------- x ----- x ---------- x
 ...
 IN ------- x ----- x ---------- x
            |       |            |
         [CM 1]  [CM 2]   ...  [CM M]
     c words x log2(N) bits each
             ^
      time-slot counter (0..c-1)

Operation

  1. Each column control memory has cc words, one per time slot. Word tt holds the number of the input highway to connect to that output in slot tt.
  2. A time-slot counter steps through the slots. In slot tt it reads word tt of every control memory.
  3. A decoder for each column closes the one crosspoint named by the word, so the sample on that input highway passes to that output highway during slot tt.
  4. In the next slot new words are read and different crosspoints close. So each crosspoint is shared by up to cc calls per frame.
  5. The processor writes the control memories during call set-up and clears them at release.

Example: in slot 5, CM2 holds "3", so crosspoint (I3, O2) closes and the sample in slot 5 of I3 goes to slot 5 of O2. In slot 6, CM2 may hold "1", connecting I1 to O2.

Features

  • The sample keeps its time slot; only the highway changes. To change slots, a time stage is added (TS, ST, TST).
  • Control memory size: MM memories of c×⌈log⁡2N⌉c \times \lceil\log_2 N\rceil bits.
  • Crosspoint count is N×MN \times M, but each serves cc calls, so very large capacity.
  • Speed of the gates limits the number of slots per frame.
  • Alone it is blocking for slot changes; it is used as the middle stage of TST switches.
  • 2071 Shrawan · 2+6 marks

What do you mean by combination switch? Explain the working principle of 3-stage combination switch with its block diagram.

Answer

Combination switch

A combination switch is a digital switch built by combining time (T) stages and space (S) stages. A pure time switch is limited by memory speed and a pure space switch cannot change time slots, so the two are combined to get a large switch that can change both the highway and the time slot of a sample. Examples: TS, ST (two-stage), TST, STS (three-stage), TSST, TSSST.

Three-stage combination switch (TST)

The most common three-stage combination switch is Time–Space–Time.

         Input T       Space S        Output T
 HW1 --->[TSI A1]--+-------------+--[TSI B1]---> HW1
 HW2 --->[TSI A2]--|  N x N TMS  |--[TSI B2]---> HW2
  ...      ...     |  (control   |    ...
 HWN --->[TSI AN]--|  memory)    |--[TSI BN]---> HWN
                   +-------------+
  c slots each      l internal slots    c slots

Block functions

  • Input time stage: one TSI per incoming highway; moves a sample from its external slot to a chosen internal slot.
  • Space stage: a time-multiplexed N×NN \times N space switch; in each internal slot it connects any input highway to any output highway.
  • Output time stage: one TSI per outgoing highway; moves the sample from the internal slot to the required outgoing slot.
  • Control unit: finds a free internal slot and writes all three control memories.

Working principle

Connect slot 4 of highway 1 (caller A) to slot 22 of highway N (called B):

  1. The processor searches for an internal slot xx that is free at the output of TSI-A1 and at the input of TSI-BN, say x=10x = 10.
  2. Input T: TSI-A1 control memory set so the sample from slot 4 is read out in slot 10.
  3. Space: control memory of column N set to "1" for slot 10; the crosspoint (1, N) closes in slot 10.
  4. Output T: TSI-BN control memory set so the sample arriving in slot 10 is sent out in slot 22.
  5. The B→A direction uses another internal slot, often x+l/2x + l/2, so one search serves both directions.
  6. On release, the control memory entries are cleared.

Blocking

There are ll alternative paths (internal slots). With link occupancy pp and p′=pc/lp' = pc/l:

PB=[1−(1−p′)2]lP_B = \left[1 - (1 - p')^2\right]^{l}

Non-blocking when l≥2c−1l \ge 2c - 1.

Advantages: low blocking, cheap time expansion, small space stage, good for large exchanges. The other three-stage form, STS, uses space stages outside and TSIs in the middle, but needs more hardware for the same blocking.

  • 2071 Chaitra · 4+6 marks

What are principles of digital exchange? Describe non blocking switches with 3 stages switching matrix.

Answer

Principles of a digital exchange

  1. Digital (PCM) representation: speech is sampled at 8 kHz, quantised and coded into 8 bits (64 kbps per channel). 32 channels form a 2.048 Mbps E1 frame of 125 µs.
  2. Time division switching: switching hardware is shared in time. Time switches (TSIs) change the time slot of a sample; space switches change the highway. They are combined as TST, STS, TSST, etc.
  3. Stored program control: a processor runs software for call processing, routing, charging and maintenance.
  4. Line interface (BORSCHT): battery feed, over-voltage protection, ringing, supervision, coding, hybrid and testing for each analogue line.
  5. Digital trunks and CCS: inter-exchange trunks are PCM links, and signaling is usually SS7.
  6. Synchronisation: all exchanges work on a common clock to avoid slips.
 lines-->[Line units]-->[ T-S-T network ]-->[Trunks]
                ^              ^              ^
                +------[ SPC processor ]------+

Non-blocking switch with a three-stage matrix

A single N×NN \times N matrix is non-blocking but needs N2N^2 crosspoints. A three-stage network uses fewer.

Structure: NN inputs in groups of nn:

  • Stage 1: N/nN/n arrays of n×kn \times k
  • Stage 2: kk arrays of (N/n)×(N/n)(N/n) \times (N/n)
  • Stage 3: N/nN/n arrays of k×nk \times n
 n in [n x k] \               / [k x n] n out
 n in [n x k] --> k arrays  --> [k x n] n out
  ...         /  (N/n x N/n)  \    ...
Nx=2Nk+k(Nn)2N_x = 2Nk + k\left(\frac{N}{n}\right)^2

Clos condition for strict non-blocking: a new call from input array A to output array B is worst-hit when the other n−1n-1 inputs of A use n−1n-1 middle arrays and the other n−1n-1 outputs of B use n−1n-1 different middle arrays. One more is needed:

k≥2(n−1)+1=2n−1k \ge 2(n - 1) + 1 = 2n - 1

Minimum crosspoints: with k=2n−1k = 2n - 1, NxN_x is minimum at n=N/2n = \sqrt{N/2}, giving

Nx,min=4N(2N−1)N_{x,min} = 4N\left(\sqrt{2N} - 1\right)

Example: N=128N = 128: n=8n = 8, k=15k = 15, N/n=16N/n = 16:

Nx=2(128)(15)+15(16)2=3840+3840=7680N_x = 2(128)(15) + 15(16)^2 = 3840 + 3840 = 7680

A single-stage switch would need 1282=16 384128^2 = 16\,384 crosspoints.

Digital equivalent: in a TST switch the internal time slots act like middle arrays, so it is non-blocking when l≥2c−1l \ge 2c - 1 (e.g. c=32c = 32, l=63l = 63, so 64 internal slots are used).

  • 2070 Asar · 5 marks

What is Time (T) switch used in digital telephone exchange?

Answer

A time (T) switch, or Time Slot Interchanger (TSI), is the part of a digital exchange that moves a PCM sample from one time slot to another on a TDM highway. Since each subscriber occupies a fixed slot, swapping slots connects one subscriber to another.

 in (slots 0..c-1)                  out
 ------->[ Speech memory c x 8 ]------->
          ^ write       read ^
   [counter]          [Control memory]
                       (set by processor)

Parts

  • Speech memory (SM): cc locations of 8 bits, one per slot.
  • Control memory (CM): cc words of log⁡2c\log_2 c bits giving read or write addresses.
  • Time-slot counter and control logic.

Working (sequential write, random read)

  1. The sample in input slot ii is written into SM(ii).
  2. In output slot jj, CM(jj) holds ii, so SM(ii) is read and sent out in slot jj.
  3. Example: CM(7) = 3 connects input TS3 to output TS7, with a delay of 4 slots.

In the other mode (random write, sequential read), CM(ii) gives the write address and SM is read in order.

Speed limit: each slot needs a write and a read, so taccess≤125 μs/2ct_{access} \le 125\ \mu s / 2c. For c=32c = 32 this is 1.95 µs; a 50 ns memory can handle up to 125 μs/(2×50 ns)=1250125\ \mu s / (2 \times 50\ ns) = 1250 channels.

Features: non-blocking within its frame, delay under one frame, low cost (just memory). Since it serves only one highway, large exchanges combine T switches with space switches as TST.

  • 2070 Chaitra · 7 marks

Explain the role of Logic or digital electronics in upgrading the electromechanical switching system into digital switching system.

Answer

Electromechanical exchanges (Strowger step-by-step and crossbar) used relays, selectors and moving contacts both to make the speech path and to control the call. Logic and digital electronics replaced these parts step by step and turned them into digital SPC exchanges.

Role of logic / digital electronics

  1. Common control instead of direct control: in Strowger exchanges each selector was stepped directly by dial pulses. Electronic registers and markers made of logic circuits store the digits first and then decide the route, so any free path can be chosen (alternate routing, digit translation).
  2. Stored program control: hard-wired relay logic was replaced by a processor and memory. Call handling, routing tables and features became software, so new services (call waiting, forwarding, abbreviated dialling) need only a program change.
  3. Electronic scanning: line relays were replaced by scanners (multiplexers and gates) that test every line every few milliseconds for off-hook, digits and on-hook.
  4. Digit reception: pulse counters and DTMF receivers built with digital filters replaced rotary selectors.
  5. Electronic crosspoints: reed relays first, then semiconductor gates replaced mechanical crossbars, giving faster and wear-free switching.
  6. Time division switching: with PCM codecs, speech becomes 8-bit words. Memories (TSIs) and logic gate matrices (space switches) replace metallic crosspoints; one gate serves many calls in different slots.
  7. Digital transmission integration: exchanges connect directly to PCM E1 trunks, removing analogue/digital conversion between exchanges.
  8. Signaling: logic-based signaling units allowed common channel signaling (SS7) with fast message-based call set-up.
  9. Charging and administration: counters and memory replaced electromechanical meters; detailed call records are produced automatically.
  10. Maintenance: self-test logic, alarms and remote diagnostics reduce fault-finding time.

Results of the upgrade

PointElectromechanicalDigital (logic-based)
Switching elementRelays, selectorsGates, memories (TSI)
ControlHard-wiredStored program
Speed of set-upSecondsMilliseconds
Size and powerLarge, highSmall, low
MaintenanceFrequent, manualLow, automatic
New servicesDifficultSoftware update

Thus digital electronics is the basis of modern exchanges, giving lower cost, higher reliability and many more services.

  • 2070 Chaitra · 2+2+4 marks

What are the drawbacks of ST and TS switch and how are they solved by STS switch? Explain.

Answer

Drawbacks of ST and TS switches

A TS switch has a time stage followed by a space stage; an ST switch has a space stage followed by a time stage. Both are two-stage and give only one possible path for each connection, so they block heavily.

  • TS drawback: the input TSI moves the sample from slot ii to the wanted output slot jj on its own highway, and then the space stage must connect the highway to the output highway in slot jj. If another input highway already uses slot jj to reach the same output highway, the call is blocked, even though both A and B are free.
  • ST drawback: the space stage must connect input highway aa to output highway bb in the input slot ii. If slot ii is already used on output highway bb by a call from another input, the call is blocked. The output TSI could change the slot afterwards, but it never receives the sample.
 TS: [T] -> [S]   S must use slot j  (fixed)
 ST: [S] -> [T]   S must use slot i  (fixed)
 => one path only, blocking = P(that slot busy)

How STS solves this

An STS switch has an input space stage, a middle stage of kk TSIs and an output space stage.

 HW1 -->+------+  [TSI 1]  +------+--> HW1
 HW2 -->|N x k |->[TSI 2]->|k x N |--> HW2
 HWN -->+------+  [TSI k]  +------+--> HWN
  1. In input slot ii, the input S stage can send the sample to any free centre TSI mm (one with a free link in slot ii).
  2. TSI mm changes the slot from ii to jj.
  3. In slot jj, the output S stage connects TSI mm to output highway bb.
  4. The call is blocked only if all kk centre TSIs are busy in slot ii or slot jj.

So there are now kk alternative paths instead of one. Using a Lee graph with internal link occupancy p′=pN/kp' = pN/k:

PB=[1−(1−p′)2]kP_B = \left[1 - (1 - p')^2\right]^{k}

For p′=0.3p' = 0.3 and k=8k = 8: PB=0.518≈4.6×10−3P_B = 0.51^8 \approx 4.6 \times 10^{-3}, compared with about 0.30.3 for a single path. With k≥2N−1k \ge 2N - 1 the STS switch becomes strictly non-blocking.

(TST solves the same problem using ll internal time slots as alternative paths, and is usually cheaper.)

  • 2069 Chaitra · 2+6 marks

What do you mean by S (space) and T (time) switches? Show that 3-stage STS or TST network can minimize the switching problems associated with 2-stage ST or TS network with their working models.

Answer

S and T switches

  • Space (S) switch: a time-multiplexed crosspoint matrix that connects an input TDM highway to an output highway in a given time slot. It changes the highway but not the slot.
  • Time (T) switch / TSI: a speech memory with a control memory that writes samples and reads them in a different order. It changes the slot but not the highway.

Problem with two-stage ST and TS networks

To connect slot ii on highway aa to slot jj on highway bb, both the slot and the highway must change.

 TS: HWa --[T: i->j]--[S: a->b in slot j]-- HWb
 ST: HWa --[S: a->b in slot i]--[T: i->j]-- HWb
  • In TS, the space stage must use slot jj; if slot jj is already used by another input highway to reach bb, the call is blocked.
  • In ST, the space stage must use slot ii; if slot ii is busy into highway bb, the call is blocked.
  • There is only one path, so the blocking probability equals the chance that this particular link is busy (about pp). This is too high for an exchange.

Three-stage TST network

 HWa->[TSI-A: i->x]->[S: a->b in x]->[TSI-B: x->j]->HWb
  1. Input TSI moves the sample from slot ii to any free internal slot xx.
  2. Space stage connects aa to bb in slot xx.
  3. Output TSI moves it from xx to jj.

The controller can pick any of ll internal slots, so there are ll alternative paths.

PB,TST=[1−(1−p′)2]l,p′=pclP_{B,TST} = \left[1 - (1 - p')^2\right]^{l}, \quad p' = \frac{pc}{l}

Three-stage STS network

 HWa->[S: a->m in i]->[TSI m: i->j]->[S: m->b in j]->HWb
  1. Input space stage sends the sample in slot ii to any free centre TSI mm.
  2. TSI mm changes the slot from ii to jj.
  3. Output space stage connects TSI mm to highway bb in slot jj.

There are kk alternative paths (centre TSIs):

PB,STS=[1−(1−p′)2]k,p′=pNkP_{B,STS} = \left[1 - (1 - p')^2\right]^{k}, \quad p' = \frac{pN}{k}

Comparison (example)

For p′=0.3p' = 0.3: a two-stage network blocks with probability about 0.3 (one link busy). A three-stage network with 8 paths gives (1−0.72)8=0.518≈4.6×10−3(1 - 0.7^2)^8 = 0.51^8 \approx 4.6 \times 10^{-3}, and with 32 paths about 4.4×10−104.4 \times 10^{-10}.

NetworkPathsBlocking
TS / ST1High (≈ pp)
STSkkLow, non-blocking if k≥2N−1k \ge 2N-1
TSTllVery low, non-blocking if l≥2c−1l \ge 2c-1

So 3-stage STS or TST networks remove the single-path problem of 2-stage networks; TST is preferred because time expansion is cheaper than space expansion.

  • 2069 Bhadra (old course) · 6+6 marks

With a block and logical diagram, explain the working principle of space switch (S) used in digital switching system. Justify, why TST switch and why not TS or ST switch are used in digital switching system?

Answer

Space (S) switch in a digital switching system

A digital space switch is a time-multiplexed space switch (TMS). It connects NN incoming PCM highways to MM outgoing highways. Its crosspoints are electronic gates, and the connection pattern changes every time slot under the control of control memories. It changes the highway of a sample but keeps its time slot.

Block diagram

 Incoming          Crosspoint           Outgoing
 highways          matrix               highways
 HW1 ------->+--------------------+---> OHW1
 HW2 ------->|   N x M gates      |---> OHW2
  ...        |   (or M x N:1 MUX) |      ...
 HWN ------->+--------------------+---> OHWM
                ^    ^        ^
             [CM1] [CM2] ... [CMM]   c words each
                ^
         time-slot counter <--- clock (frame sync)
                ^
         call-control processor (writes CMs)

Logical diagram (one output column)

Each output is in effect an NN-to-1 multiplexer whose select lines come from its control memory.

  I1 --|AND|--+
  I2 --|AND|--+
   ...        +--[OR]--> output Oj
  IN --|AND|--+
        ^
   decoder (log2 N -> N lines)
        ^
   CMj word for current slot

Only the AND gate selected by the decoder is enabled in a slot, so only that input reaches the output.

Working principle

  1. Each control memory has cc words, one per slot; word tt of CMj_j holds the address of the input highway to connect to output jj in slot tt.
  2. The time-slot counter, synchronised to the frame, steps t=0…c−1t = 0 \dots c-1.
  3. In slot tt, every CM is read; each decoder enables one gate, so input IaI_a is connected to output OjO_j for the duration of slot tt (about 3.9 µs for a 32-slot frame).
  4. In the next slot different gates may be enabled, so each crosspoint is shared by up to cc calls.
  5. At call set-up the processor writes the input number into the right CM word; at release it clears it.

Example: CM2_2 word 5 = 3 → in slot 5, I3 is connected to O2. Word 6 = 1 → in slot 6, I1 is connected to O2.

Control memory size: M×c×⌈log⁡2N⌉M \times c \times \lceil\log_2 N\rceil bits (or with one-hot words, NN bits each).

Why TST and not TS or ST

Problem of two-stage TS and ST:

  • TS: the TSI moves the sample to the output slot jj, then the S stage must connect the highways in slot jj. If slot jj on the output highway is already taken by a call from another input highway, the call is blocked.
  • ST: the S stage must connect the highways in the input slot ii; if slot ii is already busy into that output highway, the call is blocked.
  • Only one path exists, so blocking is about equal to link occupancy pp (e.g. 0.3 to 0.7), far above an acceptable grade of service (e.g. 0.002).

TST gives many paths: the input TSI can move the sample to any free internal slot xx, the S stage switches in slot xx, and the output TSI moves it to slot jj. There are ll paths:

PB=[1−(1−p′)2]l,p′=pclP_B = \left[1 - (1 - p')^2\right]^{l}, \quad p' = \frac{pc}{l}

For example, c=l=32c = l = 32, p=0.5p = 0.5: PB=0.7532≈1.0×10−4P_B = 0.75^{32} \approx 1.0 \times 10^{-4}.

Why TST rather than STS:

  1. Paths in TST come from time slots; increasing ll (time expansion) needs only faster memory and logic, which is cheap.
  2. In STS, more paths need more centre TSIs and more crosspoints (space expansion), which is costly.
  3. Non-blocking needs only l≥2c−1l \ge 2c - 1, easily met (e.g. 64 internal slots for 32-slot links).
  4. The space stage is small (N×NN \times N highways), and the time stages are just memories, so TST is cheaper, more modular and suits large exchanges.
  5. Path search is simple: find one internal slot free on two links; the reverse path can use slot x+l/2x + l/2.

Hence practical digital exchanges use TST (or its extension TSST) instead of two-stage TS or ST switches.

  • 2068 Bhadra (old course) · 5+5+4 marks

Explain the working principle of time switch and space switch used in digital telephone exchange. What are their drawbacks and how are they solved? Explain.

Answer

Time switch (T)

A time switch or Time Slot Interchanger (TSI) changes the time slot of PCM samples on a TDM highway.

 in ---->[ Speech memory (c x 8 bits) ]----> out
           ^ write addr      read addr ^
     [slot counter]          [Control memory]
  • Speech memory (SM): cc words of 8 bits, one per slot. Control memory (CM): cc words of log⁡2c\log_2 c bits.
  • Sequential write, random read: sample in slot ii is written to SM(ii); in output slot jj, CM(jj) = ii gives the read address. So input slot ii appears in output slot jj.
  • Random write, sequential read: CM(ii) = jj gives the write address; SM is read in order.
  • Example: CM(7) = 3 connects TS3 to TS7 with a delay of 4 slots.

Space switch (S)

A time-multiplexed space switch connects NN input highways to MM output highways through a gate matrix. Each output column has a control memory with cc words; in slot tt the word selects which input is connected to that output. The slot stays the same; only the highway changes.

        O1  O2  ..  OM
  I1 ---x---x-------x
  I2 ---x---x-------x
  IN ---x---x-------x
        |   |       |
      [CM1][CM2]..[CMM]

Example: CM2_2(5) = 3 connects I3 to O2 during slot 5.

Drawbacks

Time switch

  1. Size limited by memory speed: two accesses per slot, so taccess≤125 μs/2ct_{access} \le 125\ \mu s/2c. A 50 ns memory allows at most 125 μs/(2×50 ns)=1250125\ \mu s/(2 \times 50\ ns) = 1250 channels, not enough for a large exchange.
  2. Works on one highway only; it cannot connect different highways.
  3. Adds a delay of up to one frame (125 µs).

Space switch

  1. Cannot change time slots: a caller in slot 3 cannot reach a called party in slot 17.
  2. Crosspoints grow as N×MN \times M; a big single matrix is costly.
  3. Gate speed limits the number of slots.

Two-stage TS or ST combination

  • Only one path per connection; the S stage must use a fixed slot (jj in TS, ii in ST). If that slot is already busy into the output highway, the call is blocked, so blocking is high (about pp).

How the drawbacks are solved

  1. Combine T and S: time stages change slots; space stages change highways. Together they connect any slot on any highway to any slot on any other highway.
  2. Use three stages (TST or STS) to give many alternative paths:
    • TST: ll internal slots as paths, PB=[1−(1−p′)2]lP_B = [1-(1-p')^2]^l, non-blocking for l≥2c−1l \ge 2c - 1.
    • STS: kk centre TSIs as paths, PB=[1−(1−p′)2]kP_B = [1-(1-p')^2]^k, non-blocking for k≥2N−1k \ge 2N - 1.
  3. Time expansion in TST (more internal slots than external) reduces blocking cheaply.
  4. Multistage space networks (Clos, k≥2n−1k \ge 2n - 1) reduce crosspoints while staying non-blocking.
  5. Larger structures such as TSST or TSSST for very large exchanges.
  6. Faster memory and parallel (wider) highways (serial-to-parallel conversion) allow more channels per TSI.

Example: with c=l=32c = l = 32 and p=0.5p = 0.5, a TST switch has PB=0.7532≈1.0×10−4P_B = 0.75^{32} \approx 1.0 \times 10^{-4}, compared with about 0.5 for a two-stage TS switch.

  • 2065 Magh (old course) · 2+6+6+6 marks

What is space switch? How does it work? Describe the working principle and drawbacks of two stage Space-Time switch. And also explain to solve the drawbacks problem.

Answer

What is a space switch?

A space switch connects physically separate input lines (or highways) to output lines. In analogue exchanges it is a metallic crossbar matrix with one crosspoint per call. In digital exchanges it is a time-multiplexed space switch (TMS): a matrix of electronic gates whose connections change every time slot, so each crosspoint is shared by many calls. It changes the highway of a PCM sample, not its time slot.

How does it work?

            O1     O2    ...   OM
 I1 ------- x ---- x --------- x
 I2 ------- x ---- x --------- x
 ...
 IN ------- x ---- x --------- x
            |      |           |
         [CM1]  [CM2]  ...  [CMM]
          c words x log2(N) bits
              ^
     slot counter (0..c-1), clock
              ^
       processor writes CMs
  1. There are NN input and MM output PCM highways, each with cc slots per 125 µs frame.
  2. Each output column has a control memory with cc words; word tt contains the number of the input to connect in slot tt.
  3. A time-slot counter steps through the slots. In each slot, every CM is read and a decoder enables one gate in its column (logically each column is an NN:1 multiplexer).
  4. The sample in slot tt of the chosen input passes to the same slot tt of the output.
  5. In the next slot other gates close, so one crosspoint can carry up to cc different calls per frame.
  6. The processor writes the CM word at call set-up and clears it at release.

Example: CM3_3(word 4) = 2 → in slot 4, I2 is connected to O3. CM3_3(word 5) = 6 → in slot 5, I6 is connected to O3.

Features: very fast, large capacity per crosspoint, but it cannot change the time slot, and the crosspoint count grows as N×MN \times M.

Two-stage Space–Time (ST) switch: working principle

In an ST switch, a time-multiplexed space stage is followed by a time stage (one TSI on each outgoing highway).

 HW1 -->+-------------+-->[TSI 1]--> OHW1
 HW2 -->|  N x N      |-->[TSI 2]--> OHW2
  ...   |  space      |     ...
 HWN -->|  stage (S)  |-->[TSI N]--> OHWN
        +-------------+
          ^ CM (S)        ^ CM (T)
          +---- processor -+

Connect caller A (slot ii on input highway aa) to called B (slot jj on output highway bb):

  1. Space stage: in slot ii, the S-stage control memory of column bb holds aa, so crosspoint (a,b)(a, b) closes and the sample of A goes onto the link to TSI bb, still in slot ii.
  2. Time stage: TSI bb writes the sample into its speech memory in slot ii and, under its control memory, reads it out in slot jj.
  3. The sample now appears in slot jj of output highway bb: A is connected to B.
  4. The reverse direction (B → A) is set up the same way through the other half of the switch.

Example: A in TS5 of HW1, B in TS20 of HW3. In TS5 crosspoint (1, 3) closes; TSI 3 moves the sample from TS5 to TS20.

Drawbacks of the two-stage ST switch

  1. Single path, high blocking: the space stage must use the caller's slot ii. If any other input highway already sends a call to highway bb in slot ii, the link from S to TSI bb is busy in slot ii, and the new call is blocked, even though B and slot jj are free. Blocking is about the occupancy of that link, PB≈pP_B \approx p (e.g. 0.5 at 0.5 E), far worse than the usual grade of service (0.002 to 0.01).
  2. No alternative routing: the controller has no choice of path, so a single busy link causes loss.
  3. Space stage cannot change slots; the slot change is only done at the end.
  4. Large exchanges need a big S matrix and fast TSIs.
  5. Same type of problem in TS switches, where the space stage must use the output slot jj.
 New call: HW2 slot i -> HW3 slot k
 Link S->TSI3 already used in slot i
 by HW1 -> BLOCKED (no other path)

Solving the drawbacks

The cure is to add a third stage so each call has many alternative paths.

1. STS switch (Space–Time–Space)

 HW -->[S: N x k]-->[TSI 1..k]-->[S: k x N]--> HW
  • In slot ii the input space stage can send the sample to any of kk centre TSIs that is free in slot ii; the TSI changes i→ji \to j; in slot jj the output space stage sends it to highway bb.
  • Paths = kk: PB=[1−(1−p′)2]kP_B = [1-(1-p')^2]^k, p′=pN/kp' = pN/k; non-blocking if k≥2N−1k \ge 2N - 1.

2. TST switch (Time–Space–Time)

 HW -->[TSI: i -> x]-->[S in slot x]-->[TSI: x -> j]--> HW
  • The input TSI moves the sample to any free internal slot xx; the space stage switches it in slot xx; the output TSI moves it to slot jj.
  • Paths = ll internal slots: PB=[1−(1−p′)2]lP_B = [1-(1-p')^2]^l, p′=pc/lp' = pc/l; non-blocking if l≥2c−1l \ge 2c - 1.

Numerical comparison (p=0.5p = 0.5, no expansion, c=l=32c = l = 32):

  • Two-stage ST: PB≈0.5P_B \approx 0.5
  • TST: PB=[1−0.52]32=0.7532≈1.0×10−4P_B = [1 - 0.5^2]^{32} = 0.75^{32} \approx 1.0 \times 10^{-4}

3. Why TST is chosen in practice

  • Time expansion (more internal slots) needs only faster memory; space expansion needs more TSIs and crosspoints.
  • For the same blocking, TST uses less hardware and is easier to control (search for one free internal slot; reverse path at x+l/2x + l/2).
  • Very large exchanges extend it to TSST or TSSST, using multistage space stages in the middle.

Thus the single-path blocking of ST (and TS) switches is removed by three-stage STS or, preferably, TST structures.

  • 2064 Poush (old course) · 1+4+1+10 marks

What is Time (T) Switch? How does it work? Is it possible to make a switching network in a digital exchange with the help of two stage Time-Space (TS) switch? If not, explain why and how this problem is solved.

Answer

What is a time (T) switch?

A time switch, or Time Slot Interchanger (TSI), is a digital switch that moves PCM samples from one time slot to another on a TDM highway. Since each subscriber has a fixed slot, interchanging slots connects two subscribers.

How does it work?

 TDM in                                TDM out
 ------->+------------------------+------->
         | Speech memory (SM)     |
         | c words x 8 bits       |
         +------------------------+
   write addr ^              ^ read addr
   +-----------+        +-----------+
   | slot      |------->| Control   |
   | counter   |        | memory CM |
   +-----------+        +-----------+
                              ^ processor
  • Sequential write, random read: in input slot ii the sample is written into SM(ii). In output slot jj, CM(jj) holds ii, so SM(ii) is read out in slot jj.
  • Random write, sequential read: CM(ii) = jj gives the write address; SM is read in order, so the sample leaves in slot jj.
  • Example: to connect TS3 to TS7 (sequential write), set CM(7) = 3. Delay = 4 slots ≈ 15.6 µs (slot = 3.9 µs).
  • Timing: one write and one read per slot, so taccess≤125 μs/2ct_{access} \le 125\ \mu s / 2c; for c=32c = 32, t≤1.95t \le 1.95 µs.

Is a two-stage TS switch possible?

Partly, but not as a practical exchange network. A TS switch can be built and it will work at very light load, but it suffers from heavy blocking, so it cannot give an acceptable grade of service.

 HW1 -->[TSI 1]--+-----------+--> OHW1
 HW2 -->[TSI 2]--|  N x N    |--> OHW2
  ...            |  space S  |     ...
 HWN -->[TSI N]--+-----------+--> OHWN

Working of TS: to connect A (slot ii, highway aa) to B (slot jj, highway bb):

  1. TSI aa moves A's sample from slot ii to slot jj (the slot B needs).
  2. In slot jj the space stage closes crosspoint (a,b)(a, b), sending the sample to highway bb.

Why the TS switch is not suitable

  1. Single fixed path: the space stage must use slot jj, the called party's slot. If another input highway is already sending a call to highway bb in slot jj, the crosspoint column for bb is busy in slot jj, and the new call is blocked, even though A and B are free.
  2. Also, if slot jj on the output of TSI aa is already used by another call from highway aa (to a different output highway), the call is blocked.
  3. The blocking probability is roughly the chance that one of these links is busy: with link occupancy pp,
PB=1−(1−p)2P_B = 1 - (1 - p)^2

For p=0.5p = 0.5: PB=0.75P_B = 0.75; for p=0.2p = 0.2: PB=0.36P_B = 0.36. Telephone networks need PBP_B of about 0.002 to 0.01. 4. There is no alternative path, so path search cannot help. 5. The same problem exists in the ST switch, where the space stage must use the caller's slot ii.

 Call 1: HW1 -> HW3 using slot 7   (S busy: col 3, slot 7)
 Call 2: HW2 -> HW3, B is in slot 7 -> BLOCKED

How the problem is solved

Add a third stage to create alternative paths: TST (or STS).

 HWa->[TSI-A: i->x]->[S: a->b in x]->[TSI-B: x->j]->HWb
  1. Input T stage: moves the sample from slot ii to any free internal slot xx, not just slot jj.
  2. Space stage: in slot xx connects highway aa to highway bb.
  3. Output T stage: moves the sample from slot xx to the wanted slot jj.
  4. The controller can choose among ll internal slots, so the call is blocked only if all ll are unusable.

Blocking with Lee graph (ll parallel paths, two links each, p′=pc/lp' = pc/l):

PB=[1−(1−p′)2]lP_B = \left[1 - (1 - p')^2\right]^{l}

Example: c=l=32c = l = 32, p=0.5p = 0.5:

PB=(1−0.25)32=0.7532≈1.0×10−4P_B = (1 - 0.25)^{32} = 0.75^{32} \approx 1.0 \times 10^{-4}

compared with 0.750.75 for the TS switch.

Non-blocking TST: with time expansion l≥2c−1l \ge 2c - 1 (e.g. 64 internal slots for 32-slot links), no internal blocking at all.

Other solutions

  • STS: space stages outside and kk TSIs in the middle; kk alternative paths; non-blocking if k≥2N−1k \ge 2N - 1. It needs more hardware than TST.
  • TSST / TSSST: for very large exchanges, the middle space stage is made multistage (Clos), keeping crosspoints low.
NetworkPaths per callBlocking at p=0.5p = 0.5, 32 slots
T onlyNot between highways—
TS / ST1≈ 0.75
TST32≈ 1.0×10−41.0 \times 10^{-4}
TST, l=63l = 63630 (non-blocking)

Hence digital exchanges use TST (time–space–time) networks rather than two-stage TS switches.

Questions from Old Question Collection (BEI EX 756) (IOE BEI IV/II Telecommunication (EX 756) papers, 2079 to 2081), Old Question Collection (EX 703) (IOE BEX IV/I Telecommunication (EX 703) papers, 2069 to 2081) and Old Questions (EX 703 and earlier) (IOE EX 703 papers 2069-2075 and older-course BEX IV/II papers 2064-2069). Answers are written for this site; check them against your class notes.

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