7Sep/127

ISDN

ISDN (Integrated Services Digital Network) is a solution to bring digital telephony to the phone devices and then offer more flexibility and more throughput than analog telephony systems. We are going to study some protocols used in the ISDN architecture and we will understand the management and signaling exchanges made to establish a phone call. This exercise helps you to read ITU-T standards and understand the ISO Reference model. Even if ISDN is not so popular these days, it can be viewed as a first step to define GSM where some similar principles can be found, but in a more complex form due to the use of radio links.

The following scheme represents a simplified architecture of the ISDN network.

Terminal Equipment (TE) are connected through a bus to a Network Termination (NT) . The NT belongs and is managed by the telephony provider. NT manages the ISDN bus (addresses, right to talk,…) and adapt information sends the the TE to the provider network.

NT sends simultaneously two synchronous binary trains. The first train contains information sent by the NT to TE, and can be read only by TE. The other train allows communication from TE to NT. The NT send some synchronisation signals and TE can write bits at specific times.

Both trains include some channels:

– There is two B channels represented by B1 and B2, this channel may contains digitalized voice following a-law standard at 64 kbit/s. NT allows TE to use or not these B channels.

– A D channel, which will be used either to carry management frames to allocate addresses, signalisation frames to establish or receive phone call.

– E bits are also present on the binary train from NT to TE. It contains a copy of bits received in the D by the NT

– other bits (s or x) are for synchronisation or for other purposes like terminal management and are out of the scope of this exercise.

Question 1: What is the throughput of each binary train ?

Question 2: What is each channel B throughput ?

Question 3: What is channel D throughput ?

The access to B channels is managed by the NT. One TE wanting to access channel B to establish a phone call must ask the NT the right to talk. For that it sends a request on the D channel which allows a random access.

Question 4: Can we have collisions (two TE sending at the same time) in this architecture? Is it possible on channel D ? Is it possible on channel B ?

Question 5: how E bits can be used to detect collisions?

Lap-D protocol

The binary information sent on the D channel can be viewed as a frame, close to a subset of the HDLC standard. The next figure shows this frame.

Address field is divided into two parts. SAPI (Service Access Point Identifier) field contains the L3 protocol number. For instance:

  • 0 for signalisation protocol (i.e. messages between NT and TE to establish and close telephone calls). The very long Q.931 recommendation describes this mechanisms. We will just study a short part of this document.
  • 16 for data (for instance to send X.25 packets)
  • 63 for network management; for instance to allocate layer 2 addresses (TEI)  to TE. Figure 13 and Table 8 (page 37 of the english version) of the Q.921 standard give a overview of this protocol.

TEI (Terminal End point Identifier} field contains either the TE address. Value 127 means a broadcast frame when send by the NT and an unknown address when sent by the TE.

Details of the control field are available on Table 5 (page 15 of the english version) the Q.921 recommandation.

Question 6: Draw the protocol stack and indicate SAP numbers

Question 7: On how many bytes are coded U, I and S type frames ?

The following exchange has been captured by a network analyser. The analyser do not shows flags or bits added to avoid flag into the PDU. CRC field is shown.

Canal D : TE -> NT
fc ff 03 0f a8 45 01 ff 20 7d
Canal D : NT -> TE
fe ff 03 0f a8 45 02 87 e8 a3

Question 8: What are the values for the field SAPI, TEI in the first frame. To which upper layer entity the SDU must be delivered ? What is the frame type ?

Question 9: Looking at the L3 protocol PDU, discover the goal of this exchange, what type of packets are exchanged ?

When a user composes a phone number on a TE, we have the following exchange.

Canal D : TE -> NT
00 87 7f 78 8c
Canal D : NT -> TE
00 87 73 14 46

Question 10: what is the SAPI value ?

Question 12: What is the goal of this exchange ?

The rest of the exchange is the following:

TE -> NT
00 87 00 00 08 01 01 05 04 03 80 90 a3 18 01 83
9e 24 01 80 6d 06 80 50 39 39 39 39 70 05 80 33
36 39 39 7d 02 91 81 7e 01 04 c6 3a
NT -> TE
00 87 01 02 fd 46
NT -> TE
02 87 00 02 08 01 81 0d 18 01 89 85 5a
TE -> NT
02 87 01 02 8b 7f
TE -> NT
00 87 02 02 08 01 01 7d 08 03 87 e2 0d 14 01 01
36 c2 6Â
NT -> TE
00 87 01 04 cb 23
NT -> TE
02 87 02 04 08 01 81 02 18 01 89 37 6e
TE -> NT
02 87 01 04 bd 1a ‡ó
NT -> TE
02 87 04 04 08 01 81 03 08 02 87 fe 1e 02 82 81
40 99
TE -> NT
02 87 01 06 af 39
NT -> TE
02 87 06 04 08 01 81 07 5c a2
TE -> NT
02 87 01 08 d1 d0
NT -> TE
02 87 01 05 34 0b
TE -> NT
02 87 01 09 58 c1
NT -> TE
02 87 01 05 34 0b
TE -> NT
02 87 01 09 58 c1
NT -> TE
02 87 01 05 34 0b
TE -> NT
02 87 01 09 58 c1
NT -> TE
02 87 01 05 34 0b
TE -> NT
02 87 01 09 58 c1
NT -> TE
02 87 01 05 34 0b ‡4
TE -> TNR
02 87 01 09 58 c1 ‡XÁ
TNR -> Equipement
02 87 01 05 34 0b ‡4
Equipement -> TNR
02 87 01 09 58 c1 ‡XÁ
NT -> TE
02 87 01 05 34 0b ‡4
TE -> TNR
02 87 01 09 58 c1 ‡XÁ
Equipement -> TNR
00 87 04 08 08 01 01 45 08 02 87 90 6c c9 ‡E ‡lÉ
TNR -> Equipement
02 87 01 05 34 0b ‡4
TNR -> Equipement
00 87 01 06 d9 00 ‡Ù
Equipement -> TNR
02 87 01 09 58 c1 ‡XÁ
TNR -> Equipement
02 87 08 06 08 01 81 4d 28 60 ‡M (`
Equipement -> TNR
02 87 01 0a c3 f3
TE -> NT
00 87 06 0a 08 01 01 5a 27 cd
02 87 06 04 08 01 81 07 5c a2
TE -> NT
02 87 01 08 d1 d0
NT -> TE
02 87 01 05 34 0b
TE -> NT
02 87 01 09 58 c1
NT -> TE
02 87 01 05 34 0b
TE -> NT
02 87 01 09 58 c1
NT -> TE
02 87 01 05 34 0b
TE -> NT
02 87 01 09 58 c1
NT -> TE
02 87 01 05 34 0b
TE -> NT
02 87 01 09 58 c1
NT -> TE
02 87 01 05 34 0b
TE -> NT
02 87 01 09 58 c1
NT -> TE
02 87 01 05 34 0b
TE -> NT
02 87 01 09 58 c1
NT -> TE
02 87 01 05 34 0b
TE -> NT
02 87 01 09 58 c1
TE -> NT
00 87 04 08 08 01 01 45 08 02 87 90 6c c9
NT -> TE
02 87 01 05 34 0b
NT -> TE
00 87 01 06 d9 00
TE -> NT
02 87 01 09 58 c1
NT -> TE
02 87 08 06 08 01 81 4d 28 60
TE -> NT
02 87 01 0a c3 f3
TE -> NT
00 87 06 0a 08 01 01 5a 27 cd

Question 12: Draw the time diagram for this exchange, without representing RR frames, but when it is possible display information regarding the signalisation protocol

Question 13: Why do we have a very long exchange of RR frames ?

This content is published under the Attribution-Noncommercial-No Derivative Works 3.0 Unported license.

7 Responses to “ISDN”

  1. shahid says:

    Question 1: What is the throughput of each binary train ?In 250 microsecond there are 48 bits.So, in (250*4=)1ms there are (48*4=) 192 bitsSo, in (1000ms=) 1 second there are 192 Kbit/secondAnswer: 192Kbit/secondQuestion 2: What is each channel B throughput ?—-Let’s consider B1.In 250 microsecond there are 16 bits of B1 So, in (250*4=)1ms there are (16*4=) 64 bitsSo, in (1000ms=) 1 second there are 64 Kbit/secondSame result for B2. So, each B channel has 
    64 Kbit/second .Answer: 64 Kbit/second Question 3: What is channel D throughput ?In above way, 16Kbit/secondAnswer: 16 Kbit/second 

  2. admin says:

    Question 4: Can we have collisions (two TE sending at the same time) in this architecture? Is it possible on channel D ? Is it possible on channel B ?


    We can have collisions on channel D when a node talks to the NT. On the other way this is not possible since only then NT can talk.

    On channel B there is no risk of collision, since the right to talk is given by the NT. So a TE will use a Channel B only if it is authorized by the NT.
  3. admin says:

    Question 5: how E bits can be used to detect collisions?

    Since the TE has an echo of the bit it sends, if the bit E does not reflect the bit it has sent, it means, either a transmission error or another TE has sent another bit at the same time. 
  4. admin says:

    Question 6: Draw the protocol stack and indicate SAP numbers


                                 +—+ +—–+ +———+
                                 |SIG| |DATA | | MNGT. |
                                 | 0 | | 16  | | 63      | 
                                 +||-+-+-||–+-+-||——+
                                 |  LAP-D                |
    +———–+ +————+ +———————–+
    | Channel B1| | Channel B2 | |   Channel D           |
    +———–+-+————+-+———————–+
    |        Binary Train                                |
    +—————————————————-+
  5. admin says:

    Question 7: On how many bytes are coded U, I and S type frames ?

    I and S frame headers are coded on two bytes (since counters are on 7 bits). U frames are coded on 1 byte.
  6. admin says:

    Canal D : TE -> NT
    fc ff 03 0f a8 45 01 ff 20 7d
    Canal D : NT -> TE
    fe ff 03 0f a8 45 02 87 e8 a3

    Question 8: What are the values for the field SAPI, TEI in the first frame. To which upper layer entity the SDU must be delivered ? What is the frame type ? Question 9: Looking at the L3 protocol PDU, discover the goal of this exchange, what type of packets are exchanged ?

    SAPI=63

    TEI=127

    This is Management frame. They are used to assign an address to the requesting TE. The 127 TEI is used as an unknown address when the TEI send the request and as a broadcast address when the NT replies. A random number inside the PDU allows the requesting node to match the answer with the query.

  7. admin says:

    Question 10: what is the SAPI value ?

    Signalization

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