Crossbar switching

7/29/2019 Crossbar switching

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Switching

(definition) The establishment on demand, ofan individual connection from a desired inlet toa desired outlet within a set of inlets and outlets

for as long as is required for the transfer ofinformation.


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Intro to Switching

A switch routes a call based on a number system

e.g 1 302 369 6923 access code area code exchange code subscriber code


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Intro to Switching

LocalEnd Office

Tandem

Switch

TrunkGroup

Transit

End Office

Transit Concentrator

Transit

Local (line-to-line) switching

Transit (tandem)

Call distribution


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Switching System

Switching Matrix

Subscriber Lines

T

ru

n

k

s

Signaling

Control


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Essential Switch Functions

Interconnection Control

Alerting Attending

Information receiving Information sending

Busy Testing Supervising


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Functions of Switching Systems

Signalingmonitor line activitysend incoming information to control function

send control signals to outgoing lines

Controlprocess signaling and set-up/knock down connections

Switchingmake connections between input and output lines


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Basic Switch Requirements

A switch must be able to connect any incomingcall to one of a multitude of outgoing calls

A switch must have the ability to hold andterminate calls

A switch has to prevent new calls from intrudinginto circuits already in use


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General Switch Requirements

Speed of call setup should be kept short relativeto the call holding time

Grade of service should be high

.99 overall.95 busy-hour

HIGH availability!


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Switching Methods

Connectivity

Full: any input to any output

Blocking

Blocking: Possibility exists that call setup may faildue to insufficient switching resources

Non-blocking: If any input Ij and output Oj are free,they can be connected


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Time Division Switching

Time mapping of inputs and outputs

I1 I2 . . . . . . In

O1 O2 . . . . . . O36 . . . . . .


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Space Division Switching

Spatial mapping of inputs and outputs

Used primarily in analog switching systems

Space

I1

In

.

.

.

.

.

.

O1

Om


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Space-Time-Space Switching

I1 I2 . . . . . . In

O1 O2 . . . . . . O36 . . . . . .


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Single Stage Switches

Crosspoint switches

Complex - many crosspoints (ij)Poor utilization of crosspointsNot fault tolerant


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Multiple Stage Switches

Input connected to output via two or moresmaller switches

Crosspoints shared by several possibleconnections (potential for blocking)

Possible to provide multiple paths between any2 ports


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Three Stage Switch Matrix(Multiple Stage Switch example)


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Reducing Cross-point Complexity

Increase number of stages

Allow some blocking

Switch in more than one dimension


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Number of Cross-points

Non-blocking SwitchNbr of Lines 3-Stage 1-Stage

128 7,680 16,256

512 63,488 261,635

2K 516,096 4.2M

8K 4.2M 67M

32K 33M 1B128K 268M 17B

P C S


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The basic idea of

crossbar switching is toprovide a matrix of nm

sets of contacts with

only n+m activators or

less to select one of thenm sets of contacts.

This type of switching is

also known as

coordinate switching as

the switching contacts

are arranged in a xy-

plane.

Principal of Crossbar Switching

C f C b i h


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A set of horizontal and vertical wires (shown by solid

lines)

A set of horizontal and vertical contact points connected

to these wires. The contact point form pairs, each pair

consisting of a bank of three or four horizontal and a

corresponding bank of vertical contact points. A contactpoint pair acts as a crosspoint switch.

The contact points are mechanically mounted and

electrically insulated on a set of horizontal and verticalbars shown as dotted lines.

The bars, in turns are connected to a set of

electromagnets.

Components of Crossbar switch

W ki f C b S it h


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The crosspoint switches remains separated or open

when not in use. When an electromagnetic, say in thehorizontal direction, is energized, the bar attached to it

slightly rotates in such a way that the contact points

attached to the bar move closer to its facing contact

points make do not actually make any contact. Now if anelectromagnetic in the vertical direction is energized, the

corresponding bar rotates causing the contact points at

the intersection of the two bars to close. This happens

because the contact points move towards each other.

Working of Crossbar Switch

C t


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Cont.

As an example, if electromagnets M2 and M3/ are

energized, a contact is established at the crosspoint 6

such that the subscriber B is connected the subscriberC.

E i i f l t hi th i t


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Energizing sequence for latching the crosspoints

Let us consider a 66 crossbar schematic shown below.

C t


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Let us consider the establishment of the following

connections in sequence: A to C and B to E.

First the horizontal bar A is energized.

Then the vertical bar B is energized

The crosspoint AC is latched and the conversation

between A and C can now proceed.Suppose we now energize the horizontal bar of B to

establish the connection B-E, the crosspoint BC may

latch and B will be brought into the circuit of A-C. This is

prevented by an energizing sequence for latching thecrosspoints.

A crosspoint latches only if the horizontal bar isenergized first and then the vertical bar.

Cont.

C t


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Cont.

In order to establish the connection B-E, the vertical bar

E need to be energized after the horizontal bar is

energized.

In this case the crosspoint AE may latch as the

horizontal bar A has already been energized for

establishing the the connection A-C.

This should also be avoided and is done by de-

energizing the horizontal bar A after the crosspoint is

latched and making a suitable arrangement such thatthe latch is maintained even though the energisation in

the horizontal direction is withdrawn.

The crosspoint remains latched as long as the vertical

C t


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Cont.

The complete procedure for establishing a connection in

a crossbar switch:

1. energize horizontal bar

2. energize vertical bar

3. de-energize horizontal bar

Design parameters


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Design parameters

In a non-blocking crossbar configurations, there are N2

switching elements for N subscribers. When all thesubscribers are engaged, only N/2 switches are

actually used for connections.

Crossbar switch configurations


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Crossbar switch configurations

Different switch points are used to establish a

connection between two given subscribers dependingupon who initiate the call. For example when the

subscriber C wishes to call subscriber B, crosspoint

CB is energized. On the other hand when B initiates

the call to contact C, the switch BC is used. Bydesigning a suitable control mechanism, only one

switch may be used to establish a connection

between two subscribers, irrespective of which of

them initiates the call. The crosspoints in the diagonal

connect the inlets and the outlets of the same

subscriber. Hence they can also be eliminated.

Cont


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Cont.

Cont


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The crosspoints in the diagonal connect the inlets and

the outlets of the same subscriber. Hence they can

also be eliminated.

Cont.

Class work


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Class work

Calculate the number of switches in adiagonal crosspoint matrix if the

number of Subscribers is N.

Blocking Crossbar switch


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Blocking Crossbar switch

The diagonal crosspoint matrix is a nonblocking

configuration. The number of crosspoint switches can

be reduced significantly by designing blockingconfigurations.

The number of vertical

bars is less than the

number of subscribers.

The vertical bars

determines the number

of simultaneous callsthat can be out through

the switch.

Cont


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32

Cont.

Let a connection be required to be established between

the subscriber A and B. The sequence to be followed

in establishing the A-B circuit may be summarized as:

Energize horizontal bar A

Energize free vertical bar P

De-energize horizontal bar A

Energize horizontal bar B

Energize vertical bar P

De-energize horizontal bar B

Cont


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Cont.

Alternative Energizing sequence:

Energize horizontal A and B

Energize vertical P

De-energize horizontal A and B

The number of switches required is 2NK, where N is the

number of subscribers and K is the number of vertical

bars that are used to establish the connections.


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Definition:space-division switching

Space division switching was originally developed forthe analog environment and has been carried over into

the digital realm. A space division switching is one inwhich the signal paths are physically separate from oneanother (divided in space).


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Limitations of crossbar switch

The basic building block of the switch is crossbar switch in which ametallic cross-point or semiconductor gate is enabled or disabledby a control unit for the establishment of a physical path. But thecrossbar switch has a no. of limitations:

The no. of cross-points grows with the square of the no. ofattached stations. This is costly for a large switch.

The loss of a cross-point prevents connection between the twodevices whose lines intersect at that cross-point.

Cross-points are inefficiently utilized. Only a small fraction ofcross-points are engaged even when all devices are active.


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Contd.

To overcome these limitations of crossbarswitch, multiple stage switches are employed.Although a multistage network requires a more

complex control scheme, it has severaladvantages over a single stage switch.


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Single stage vs Multistage networks


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Single stage vs Multistage networks


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Two stage representation of N X N

network

Theorem:For any single stage network there exists an equivalent multistagenetwork.

So, N X N single stage network with capacity k can be realized by a twostage network of N X K and K X N stages.


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Cont.

Any of the N inlets can be connected to any ofthe K outputs of 1st stage.

Similarly, Any of the K inputs can be connectedto any of the N outputs of 2nd stage.

So, there are K alternative paths and 2NK

switching elements.

Any of the N inlets can be connected to any of the N outlets.


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Cont.

Each stage has NK switching elements

Assume only a fraction of the subscribers to be activeon an average

K can be equal to N/16 So, no. of switching elements,

S = 2NK = N2/8 --- (1)

Example:N = 1024, K = 64S = 131,027So, for large N, the switching matrix NxK may still be

difficult to realize practically.

T k i h l i l


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Two stage network with multiple

switching matrices

M inlets aredivided into rblocks of p

inlets. M = pr

N inlets aredivided into sblocks of qoutlets. N = qs


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Cont.

For full connectivity there must be at least one outletfrom each block in the 1st stage terminating as inlet onevery block of the 2nd stage.

So, block sizes are p x s and r x q respectively

So, S = psr + qrs --- (2) Putting values for M, N S = Ms + Nr --- (3) The number simultaneous calls in the network,

switching capacity, SC = rs --- (4)


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Cont. For rs connections to be simultaneously active, the s active

inputs in one block of the 1st stage must be uniformlydistributed across all the s blocks in the 2nd stage at the rateof one per block.

Blocking may occur in two conditions:

I. Calls are uniformly distributed (there are rs calls inprogress and (rs + 1)th calls arrives)

II. Calls are not uniformly distributed, there is a call inprogress from I-th block from the first stage to the J-

th block in the 2nd

stage and another call originates inthe I-th block destined to J-th block.


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The blocking probability

Let be the probability the a given inlet is active.Now, probability that an outlet at the I-th block is active is, =(p)/s

The probability that another inlet becomes active and seeks anoutlet other than the one which already active is given by(p - 1)/(s- 1)

Now, Probability that an ready active outlet is soughtPB = ((p)/s)[1(p-1)/(s-1)]

Substituting, p = M/r, we havePB = ((M)/rs)[1((M/r)-1)/(s-1)] --- (5)


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Discussion.

If s and r decrease then S can be minimized

But if we decrease s and r we are increasingblocking probability!

So, we have to choose values for s and r as smallas possible but giving sufficient links to providea reasonable grade of service.


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Cont.

If N > M, network is expanding traffic

If M > N, concentrating the traffic

If N = M, matrix size is uniform

i.e. r=s, p=q

So,

S = 2Nr --- (6)SC = r2 --- (7)


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Cont.

For square switching matrices as in standard ICs. p=r=s=q = N

Thus the network has N blocks each in the 1st

and 2nd

stages and each block is a square matrix of N X Ninlets and outlets.So,S= N N + N N = 2N N --- (8)SC = N X N = N --- (9)


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Cont.

In the two stage network discussed so far, thereis only one link between a block in the 1st stageand a block in the 2nd stage.

What will happen if this particular link failure?

Rise of severe blocking in the network!!

How can we improve this performance?


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Cont.

Increase number of links between the blocks of thestages.

Consider k links beings introduced between every 1st

and 2nd

stage pair. Design parameters for M = N are as follows:

p = q = N, s = r = kN

S = 2Nk N --- (10)SC = N --- (11)


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Cont.

In order to make the network non-blocking, must haveK = N

Now, S = 2N2 --- (12)

And, SC = N --- (13)

So, a two-stage non-blocking network requires twice

the number of switching elements as the single stagenon-blocking network.


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Three-stage

network

Switching matrices

Stage 1: p x s

Stage 2: r x r

Stage 3: s x p


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Cont.

What is the improvement here?

It has s alternatives from stage 1 to stage 3.

S= rps + sr2 + spr = 2Ns + sr2 =s(2N+r2) --- (14)

If we use square matrices in stage 1 and 3, thenp = s = (N/r)

S=2N2/r + Nr --- (15)


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Cont.

There is an optimum value for r which would minimizevalue of S. (equation 15)

To obtain this optimum value differentiate thisequation and set to zero.

dS/dr = -2N2/r2 + N = 0

=> r = (2N) Smin=2N (2N) and p = N/r = (N/2) --- (16)

Optimum ratio of the number of blocks to the numberof inputs per block is

r/p= (2N)/ (N/2) = 2 --- (17)


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Lees graph for 3 stage network

= probability of a link being busy = probability of a link is free

Now, = 1 -

If there are sparallel links, the blocking probability is theprobability that all the links are busy:PB =

s, QB = 1PB = 1 -s


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Cont.

When a series of s links are needed to complete aconnection, the blocking probability is easilydetermined as one minus the probability that they areavailable:

PB = 1()s=1(1)s

For a three stage network, there are two links in series

for every path and there are s parallel paths.Therefore,

PB = [1()2]

s= [1(1)2]s --- (18)


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Cont.

Ifis the probability that an inlet at first stage isbusy, then

=p/s= /k --- (19)

Now, substitute the value of in equation (18),we get the blocking probability for three-stageswitch as:

PB = [1(1 - /k)2]s --- (20)

k represents either space expansion orconcentration.


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Analysis

Look at equation (20), the term /k is the factor todecide the value for blocking probability.

If/k is small probability is low.

Ifis large then k must be large, i.e. if inlets are well loaded,we need an expanding first stage.

On the other hand, ifis small k may be small, i.e. if inputsare lightly loaded, the first stage may be a concentrating one.


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Three stage non-blocking


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Cont.

Multistage non-blocking and fullyavailable networks are known asClos networks.

3 stage switching network can bemade non-blocking.

How???

By providing adequate number of blocks

in 2nd stage, i.e. increase value ofs.


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Cont.

Worst situation can occur when the situations arise:1. (p-1) inlets in a blockI in 1st stage are busy

2. (p-1) inlets in a blockO in 3rd stage are busy

3. The (p-1) 2nd stage blocks, on which (p-1) outlets from

blockIare terminated, are different from the (p-1) 2nd stageblocks from which the links are established to the blockO.

4. The free inlet of blockIneeds to be terminated on the freeoutlet of blockO.


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Definition: time-division switching

Switching of time-division multiplexed (TDM) channels byshifting bits between time slots in a TDM frame.

Time-division multiplexing (TDM) is a type of digital or(rarely) analog multiplexing in which two or more signals or bitstreams are transferred apparently simultaneously as sub-channels in one communication channel, but physically are

taking turns on the channel. The time domain is divided intoseveral recurrent timeslots of fixed length, one for each sub-channel.

Statistical Time division


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Statistical Time-division

Multiplexing (STDM) STDM is an advanced version of TDM in which both the address of the

terminal and the data itself are transmitted together for better routing. UsingSTDM allows bandwidth to be split over 1 line.

If there is one 10MBit line coming into a building, STDM can be used toprovide 178 terminals with a dedicated 56k connection (178 * 56k = 9.96Mb).

A more common use however is to only grant the bandwidth when that muchis needed.

STDM does not reserve a time slot for each terminal, rather it assigns a slotwhen the terminal is requiring data to be sent or received.


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Preliminaries

8 kHz sampling rate -> a sample occurs in every125sec.

In this 125s sampling interval about 120sare unused !!!

How can we utilize this efficiently?

Establish a dynamic control mechanism, whereby aswitching element can be shared by a number ofsimultaneously active speech circuit.

Basic Time division space


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Basic Time division space

switching

N X N time division spaceswitch

Each inlet/outlet is a single

speech circuit correspondingto a subscriber line.

Speech is carried as eitherPAM or PCM samples.


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Two stage Equivalent

Matrices: (1st and 2nd stage)

N X 1 and 1 X N


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Cont.

If PAM samples are switched on then Analog timedivision switch

If PCM binary samples are switched on then Digitaltime division switch

Interconnection is through a bus

Number of simultaneous conversations

SC = 125/ts

ts is time insec to setup a connection and transfersample value.


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Cont.

Selection of inlet/outlet is controlleddynamically.

Simplest manner is to select in a cyclic manner

Cyclic control is organized using a modulo-Ncounter and a k-to-2kdecoder

Ceil(log2N) = k


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Cont.

All the inlets/outlets are scanned within 125sec, the switchingcapacity, SC, of the network is the same as number of inlets oroutlets in the system.

Switching is non-blocking

Lacks full availability -> Not possible to connect any inlet to anyoutlet.

How to obtain it full availability?

Make one of the controls (input or output), memory based

Figure at next slide

I d i


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Input driven

TDSS

A control memory on outputside.

modulo-N counter also acts as

memory address register(MAR)

Control memory has N wordscorresponding to N inlets and

has a width of ceil(log2N) bits.


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Cont.

For an active inlet i, the corresponding outlet address isis contained in the i-th location of the control memory.

Address is decoded by MDR (memory data register)

Then proper outlet is enabled. Then the sample value is transferred.

Thus any inlet i can be connected to any outlet kensuring full availability.


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Finally, Definition

Since a single switching element, the bus, isbeing time shared by N connections, all ofwhich can be active simultaneously, and aphysical connection is established between theinlet and outlet for the duration of the sampletransfer, the switching technique is known as

time division space switching


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Output controlled TDSS

Each location of

the control memoryis rigidly associatedwith a given outlet.


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Analysis

For both input and out controlled configurations, thenumber of inlets or outlets N is equal to the switchingcapacity SC.

N = SC = 125/(ti+ tm+ td+ tt)ti= time to increment modulo-N counter

tm= time to read control memory

td= time to decode address and select inlet or outlet

tt = time to transfer the sample value from inlet tooutlet


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Cont.

Number of switching elements:on the input side = N

on the output side = N

total = 2NSwitching capacity, SC = N

Traffic handling capacity,

TC = SC/(theoretically max load) = 1

Cost of the switching network = cost of the switching element + costof the control memory = 2N + N = 3N

Cost capacity index, CCI = SC/(cost per subscriber line)= N/ (3N/N) = N/3


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Discussion

Use of cyclic control in input or output-controlledswitches restricts the number of subscribers on thesystem rather than the switching capacity.

Thats why cyclic control demands all the lines to bescanned irrespective of whether they are active or not.

Practically, number of active subscriber is around 20%of the total.

G li d


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Generalized

TDSS

Control memory for controlling bothinlets and outlets.

Permits a larger number ofsubscribers than the switchingcapacity of the network.

Each word in control memory hastwo addresses: inlet & outlet

Control memory width is2*ceil(log

2N).


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Operation procedure of the switch

modulo-SC counter is updated at the clock rate

Control memory words are read one after another

Inlet address is used to connect the corresponding inlet

to the bus and so also the outlet address to connect theoutlet

The sample is then transferred from inlet to outlet

Next, clock updates the counter and the cycle of

operation is repeated.


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Cont.

Recall the switching capacity,

SC = 125/ts

ts=(ti+ tm+ td+ tt)

If time to read memory i.e. tmofts is dominatingfactor in the equation then it means that control

memory is busy through out the sampling intervalof 125s


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Cont.

Time slot is 125/Ms, duration of 1 sample. In 1 time slot N samples are switched.

The output is cyclically scanned.

1-to-M relationship between the outlets and control memory, i.e.

M location in control memory (CM) corresponding to eachoutlet.

CM has MN words.

Number of trunks can be supported, N = 125/(M*ts)

Where, ts is the switching time as earlier.

Cost of the switch, C = Number of switch + Number of memorywords = 2N + MN

E i


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Exercise

Calculate number of trunks that can be supported on a timemultiplexed space switch, given that

32 channels are multiplexed, control memory access time is 100

ns, bus switching and transfer time is 100 ns per transfer.

Solution:

Here, M = 32, ts =100 + 100 = 200ns

N = 125/(M*ts) = 125/(32 * 200 * 10-3) = 20 (approx.)


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Time Multiplexed Time Switch

Unlike time multiplexed space switches, itpermits time slot interchange (TSI) of samplevalues.

Such an operation necessarily implies a delaybetween the reception and the transmission of asample.

Illustration in next graph.


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M channels are multiplexed oneach trunk

The switch is organized insequential write/ random read

fashion Time slot duration, tTS= 125/M

Time slot clock runs at the timeslot rate

Time slot counter is incrementedby 1 at the end of each time slot.


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Cont.

Contents of counter provides location addressesfor the data memory and the control memory.

Data memory and control memory access take

place simultaneously at the starting of the timeslot.

Contents of the control memory are used as theaddress of the data memory and the data readout to the output trunk

Cont


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Cont.

Even if there is no time slot interchange, a sample isdelayed by a minimum of one time slot in action.

Depending on the output time slot, delay range is tTS to

MtTS microsec In the example given in figure 1st location in CM

contains value 1 which implies that the contents ofinput time slot 1 is switched to output time slot 1.

Delay for this sample is tTS microsec

i 2 i Th f i i l i


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Location 2 contains 7. Therefore, input time slot 7 isswitched to output time slot 2.

Delay for this sample: (M(7-2) + 1)tTSs=(M4)tTSs Location 3 contains 4. Therefore, input time slot 4 is

switched to output time slot 3.

Delay for this sample: (M(4-3) + 1)tTSs=MtTSs = 125s

There are two sequential memory access per time slot.So, time constraint may be stated as

tTS = 2tm, 125 = 2Mtm

C i i


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Cost estimation

No switching elements!

Cost is equal to the number of memorylocations.

There are M locations each in the control anddata memory

So, total cost is given by

C = 2M units

E i


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Exercise

Calculate maximum access time that can be permitted for thedata and control memories in TSI switch with a single input andsingle output trunk multiplexing 2500 channels. Estimate thecost of the switch and compare the result with a single stagespace division switch.

Solution:

tm =(125*103)/(2500*2) = 25 ns

C = 2 * 2500 = 5000 units

This switch is non-blocking and full available. An equivalentsingle stage space division switch uses a matrix of 2500 X 2500.So, cost is 6.25 million units

Cost advantage of time switch = (6.25*106)/5000 = 1250


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Combination Switches

Wh t t d f ?


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What we study so far?

Time division space switch do not provide fullavailability as they are not capable of time slotinterchange

Time slot interchange switch not capable ofswitching sample values across trunks without the helpof some space switching matrices.

So, here comes the idea of combining both techniques

and getting their advantages in a single platform. This iscombination switch.

C bi ti it h


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Combination switch

C td


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Contd.

Crossbar switching

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