Communication Systems

Every communication system moves information from a source to a destination through a noisy channel. Learn the block diagram (source, transmitter, channel, receiver), analog vs digital communication, frequency bands and wavelength, bandwidth, decibels and dBm, noise (thermal kTB, noise figure), SNR and receiver sensitivity.

Radio, TV, mobile phones, Wi-Fi, satellites and fibre all follow the same model defined by Claude Shannon in 1948. Understanding signal power, noise and bandwidth is the foundation for every modulation and coding technique that follows.

An analogy

Communication is like talking across a noisy room: you can speak louder (power), slower (bandwidth/rate), or more clearly (modulation and coding) to be understood.

The block diagram

Source (voice, data) → transducer → transmitter (modulation, amplification) → channel (wire, fibre, air) with noise and distortion → receiver (filtering, amplification, demodulation) → destination.

Spectrum

LF/MF/HF (long-distance radio), VHF/UHF (FM, TV, mobile), microwave (Wi-Fi 2.4/5/6 GHz, 5G mid-band, satellites), mmWave (5G FR2, radar), optical (fibre). Higher frequency → more bandwidth but shorter range and line-of-sight propagation.

Noise and SNR

Thermal noise sets a floor proportional to bandwidth. Every receiver stage adds noise (noise figure). SNR at the detector determines analog quality or digital bit-error rate.

Analog vs digital

Analog systems degrade gradually with noise; digital systems regenerate signals, use error correction and compression, and share channels efficiently — which is why almost all modern systems are digital, with analog RF front-ends.

Formulas

• Wavelength: λ = c / f
• dBm: P_dBm = 10 log₁₀(P / 1 mW)
• Thermal noise: N = kTB → −174 dBm/Hz + 10 log₁₀ B (290 K)
• Receiver sensitivity: S_min = −174 + 10 log B + NF + SNR_req (dBm)

Worked example: Receiver sensitivity

Given: B = 1 MHz, NF = 5 dB, required SNR = 10 dB

Find: Sensitivity.

• Noise floor = −174 + 60 = −114 dBm
• + NF 5 + SNR 10 = −99 dBm

Answer: −99 dBm

Worked example: Wavelength

Given: 2.4 GHz Wi-Fi

Find: λ.

• λ = 3 × 10⁸/2.4 × 10⁹ = 0.125 m

Answer: 12.5 cm (quarter-wave antenna ≈ 3.1 cm)

Common mistakes

• Wrong: Adding dB values of powers directly. — Right: Convert to linear, add, convert back.
• Wrong: Ignoring bandwidth in noise. — Right: Noise power grows with 10 log B.

Applications

• Mobile networks: 4G/5G link design.
• Broadcasting: AM/FM/DAB radio and TV.
• IoT: LoRa, BLE, NB-IoT sensitivity trade-offs.

Key takeaways

• Source → Tx → channel → Rx → destination.
• λ = c/f; dBm = 10 log(P/1 mW).
• Noise floor −174 + 10 log B + NF.
• Digital systems dominate modern comms.

Going further

• Friis noise formula: Cascaded noise figure.
• Fading channels: Multipath, Rayleigh and Rician fading.
• Spectrum regulation: ITU and national allocations.