Telecom Professional Qualification study materials
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Telecom Professional Qualification: Transmission, Switching, and Transport Technologies

Telecom Professional Qualification · Work through your exam, one topic at a time.

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Transmission, Switching, and Transport Technologies

This domain covers how bits are carried and cross-connected: multiplexing hierarchies, the PDH/SDH/OTN rate ladders, protection schemes, and the switching modes that decide how a connection is routed. Numeric recall matters here - wrong rate constants lose marks quickly.

Multiplexing and the rate ladders

  • TDM: an E1 is 2.048 Mbit/s, built from 32 timeslots of 64 kbit/s. TS0 carries framing and CRC, TS16 carries signalling, leaving 30 voice channels (30B+D). A T1 is 1.544 Mbit/s from 24 × 64 kbit/s plus framing - do not mix the two.
  • PDH is plesiochronous, so higher-order multiplexing needs bit stuffing and demultiplexing is cascaded. SDH is synchronous: STM-1 = 155.52 Mbit/s, STM-4 = 622.08, STM-16 = 2,488.32, STM-64 = 9,953.28, STM-256 ≈ 39.8 Gbit/s. Pointers absorb timing wander.
  • A VC-4 carries 63 × 2.048 Mbit/s = 129.024 Mbit/s of payload inside the STM-1 frame.
  • WDM: DWDM channels sit on the ITU-T G.694.1 grid anchored at 193.1 THz, spaced 100 or 50 GHz; CWDM uses 20 nm spacing.
  • OTN wraps clients into ODUk/OTUk with strong FEC: OTU1 = 2.666, OTU2 = 10.709, OTU4 = 111.81 Gbit/s.

Switching and protection

  • Circuit, packet and cell (ATM) switching; MPLS forwards on labels rather than longest-prefix match.
  • 1+1 protection bridges permanently to both paths and selects at the receiver - hitless and simple, but capacity is duplicated. 1:1 keeps a standby path and allows extra traffic; 1:N shares one standby across N working paths.
  • MSP ring protection (SDH) versus SNCP path protection; control-plane restoration is slower but more capacity-efficient.
  • Erlang B models blocking for loss systems (circuit groups); Erlang C models queueing delay.

Worked example

An STM-16 link: 16 × 155.52 = 2,488.32 Mbit/s. Dividing naively by 2.048 gives about 1,214 E1s, but the real answer is 16 VC-4 × 63 E1 = 1,008 E1s (2,064 Mbit/s of payload), because section/line and path overhead consume the rest. Always check whether the question asks for line rate capacity or mapped payload.

Traps that cost marks

  • Confusing E1 (2.048) with T1 (1.544), or STM-1 with a rounded 155 Mbit/s when the question needs two decimals.
  • Confusing 1+1 (permanent bridge) with 1:1 (standby, extra traffic allowed).
  • Forgetting that OTU rates exceed the client rate because of FEC and overhead, so a 10G client maps into OTU2 at 10.709.