Telecom Basics
This chapter is the foundation of the whole course. The concepts here (dB, Erlang, switching, multiplexing) are used in almost every later chapter.
1.1Basic Structure of a Telecommunication System
Every communication system has five parts: Source → Transmitter → Channel (medium) → Receiver → Destination. Noise is added in the channel and damages the signal. The medium can be a wire (copper), optical fibre or air (radio).
Figure 1.1: Block diagram of a communication system
1.2Analog vs Digital and PCM
Voice is basically analog. To send it over a digital network there are three steps: Sampling → Quantization → Encoding (this is called PCM).
Nyquist Sampling Theorem
If the maximum frequency of a signal is fmax, the sampling rate must be fs ≥ 2 fmax, otherwise aliasing occurs.
Example: Bit rate of a voice channel
Voice band ≈ 0.3–3.4 kHz; assume 4 kHz. Sampling = 2 × 4 = 8 kHz (8000 samples/second). If each sample is 8 bits:
Bit rate = 8000 × 8 = 64 kbps (this is DS0)
One E1 line has 32 timeslots: 32 × 64 kbps = 2.048 Mbps (30 voice + 1 frame sync + 1 signalling). T1 = 24 × 64 + 8 kbps = 1.544 Mbps.1.3Decibel (dB) — the Most Needed Maths in Telecom
Signal levels vary over a very wide range (milliwatts to kilowatts), so a logarithmic scale is used.
dB = 10 log10(P2 / P1) | dBm = 10 log10(P / 1 mW) | dBW = 10 log10(P / 1 W)
| Power ratio | dB | Memory trick |
|---|---|---|
| 2 times | +3 dB | ×2 = +3 dB, ÷2 = −3 dB |
| 10 times | +10 dB | ×10 = +10 dB |
| 100 times | +20 dB | ×100 = +20 dB |
| 1000 times | +30 dB | 1 W = 30 dBm (because 1 W = 1000 mW) |
| 4 times | +6 dB | 3+3 |
| 5 times | ≈ +7 dB | 10 − 3 |
Rules: dBm + dB = dBm (gain/loss can be added and subtracted); dBm − dBm = dB; dBm + dBm cannot be added directly (first convert to mW, then add). For voltage/current use 20 log.
Example: How many dBm is 40 W?
40 W = 40,000 mW → 10 log(40000) = 10 × 4.602 = 46.02 dBm. Shortcut: 40 = 4×10 → 6 + 10 = 16 dBW → 16 + 30 = 46 dBm ✔
Example: Adding powers
Adding −30 dBm and −30 dBm? Each is 1 µW; total 2 µW → −30 + 3 = −27 dBm (not −60).
1.4Noise and Shannon Capacity
Thermal noise floor: N = kTB. At room temperature (290 K):
N (dBm) = −174 + 10 log10(BHz) + NF
SNR (dB) = Prx (dBm) − N (dBm)
SNR (dB) = Prx (dBm) − N (dBm)
Here −174 dBm/Hz is the noise density and NF is the Noise Figure of the receiver.
Shannon-Hartley Theorem
The maximum error-free data rate of a channel:
C = B · log2(1 + SNR) [bps]
Here SNR is the linear ratio, not dB. C increases if B or SNR is increased — that is why 5G needs wider bandwidth and MIMO.Example: Shannon capacity of a 20 MHz channel
SNR = 15 dB → linear = 101.5 = 31.62.
C = 20×106 × log2(1 + 31.62) = 20×106 × log2(32.62) = 20×106 × 5.03 ≈ 100.6 Mbps.
Noise floor: −174 + 10 log(20×106) = −174 + 73.01 = −100.99 dBm; with NF = 7 dB it is −93.99 ≈ −94 dBm.
C = 20×106 × log2(1 + 31.62) = 20×106 × log2(32.62) = 20×106 × 5.03 ≈ 100.6 Mbps.
Noise floor: −174 + 10 log(20×106) = −174 + 73.01 = −100.99 dBm; with NF = 7 dB it is −93.99 ≈ −94 dBm.
1.5Switching: Circuit vs Packet
| Topic | Circuit switching (PSTN, 2G CS) | Packet switching (IP, 4G/5G) |
|---|---|---|
| Path | A dedicated path/resources are allocated at the start of the call | Each packet is routed independently |
| Resource use | Resources stay held even when silent (wastage) | Statistical multiplexing — efficient |
| Delay | Fixed and low, no jitter | Variable delay and jitter (QoS needed) |
| Failure | Call drops if the path breaks | Finds an alternative path |
| Suitable for | Traditional voice | Data, VoIP/VoLTE, video |
| Example protocols | SS7/ISUP | IP, SIP, GTP |
1.6Multiplexing and Transport
- FDM — divides by frequency (radio/TV channels).
- TDM — divides by time slots (E1, SDH, GSM).
- WDM — lights of different wavelengths on the same fibre (CWDM/DWDM).
| SDH level | Bit rate | Remember |
|---|---|---|
| STM-1 | 155.52 Mbps | = 63 × E1 (2.048) |
| STM-4 | 622.08 Mbps | 4 × STM-1 |
| STM-16 | 2.488 Gbps | 16 × STM-1 |
| STM-64 | 9.953 Gbps | 64 × STM-1 |
Optical fibre: Single-mode (SMF, core ≈ 9 µm, long distance; 1310/1550 nm) and Multi-mode (MMF, core 50/62.5 µm, short distance). Attenuation is lowest at 1550 nm (~0.2 dB/km). Optical link budget: Prx = Ptx − (α·L + connector loss + splice loss).
1.7OSI and TCP/IP Models
Figure 1.2: OSI vs TCP/IP layers
- TCP — reliable, connection-oriented (data/HTTP); UDP — fast, no reliability (voice RTP, GTP-U); SCTP — for signalling (S1AP, NGAP, Diameter).
Example: IPv4 subnetting
192.168.10.0/26 → subnet mask 255.255.255.192; host bits = 32 − 26 = 6; total addresses = 26 = 64; usable hosts = 64 − 2 = 62. Network = .0, broadcast = .63, range .1–.62. IPv6 = 128 bits; IPv6 is important in the mobile core because IPv4 is running out.
1.8Signalling Protocols — Who Is Where
| Protocol | Function | Where used |
|---|---|---|
| SS7 (MTP, SCCP, ISUP, TCAP/MAP) | Traditional signalling | 2G/3G CS core, with the HLR (MAP) |
| Diameter | AAA, subscriber data, charging | 4G: S6a, Gx, Rx; IMS: Cx |
| SIP | Session control | IMS/VoLTE |
| GTP (GTP-C / GTP-U) | Tunnelling (control / user plane) | S11, S5/S8, S1-U, N3 |
| S1AP / NGAP | RAN–core signalling | S1-MME / N2 |
| NAS (EMM/ESM, 5GMM/5GSM) | UE–core signalling | UE ↔ MME/AMF |
| PFCP | Control-user plane separation | N4 (SMF–UPF), Sxa/Sxb |
| HTTP/2 + JSON | Service-based interface | 5GC SBA |
1.9Network Performance (QoS and KPI)
- Throughput, Latency (delay), Jitter (variation in delay), Packet loss.
- CSSR (Call Setup Success Rate), CDR/DCR (Call/Drop Rate), HOSR (Handover Success Rate), RRC/E-RAB Setup Success Rate, PRB utilization, site Availability.
- For voice: delay < 150 ms (one way), loss < 1%; MOS scale 1–5 (4+ is good).
Example: Erlang — the unit of traffic
1 Erlang = one channel busy 100% of the time.
A = λ × h
λ = number of calls per hour, h = average call duration (in hours). 1200 calls per hour, average 3 minutes → A = 1200 × (3/60) = 60 Erlang. (Detailed in Chapter 3.)Practice (MCQ)
1. What is the bit rate if a 4 kHz voice signal is PCM coded (8 bits/sample)?
Answer: (b) — fs = 8 kHz, 8 bits → 64 kbps (DS0).
2. How many dBm is 1 W of power?
Answer: (c) — 1 W = 1000 mW → 10 log 1000 = 30 dBm.
3. −30 dBm + −30 dBm = ?
Answer: (b) — Double power = +3 dB.
4. In the Shannon capacity formula, in what form is SNR used?
Answer: (b) — C = B log₂(1+SNR), SNR is linear.
5. What is the bit rate of an E1 line?
Answer: (b) — 32 × 64 kbps = 2.048 Mbps. 1.544 is T1.
6. What is the bit rate of STM-1?
Answer: (b) — STM-1 = 155.52 Mbps ≈ 63 E1.
7. Which is a feature of circuit switching?
Answer: (a) — In circuit switching, resources are allocated for the call in advance.
8. What is the approximate thermal noise floor at 20 MHz bandwidth (NF = 0)?
Answer: (b) — −174 + 73 = −101 dBm.
9. Which protocol is commonly used on the S1-MME interface?
Answer: (b) — S1-MME = S1AP/SCTP.
10. How many usable hosts are in a /26 subnet?
Answer: (b) — 2⁶ − 2 = 62.
11. In which multiplexing are lights of different colours sent on the same fibre?
Answer: (c) — WDM = Wavelength Division Multiplexing.
12. According to Nyquist, what is the minimum rate to sample 20 kHz audio?
Answer: (b) — fs ≥ 2 fmax = 40 kHz.