Telemetry and wireless sensor networks

Telemetry systems (4–20 mA, FM/FM, PCM frames) and wireless sensor networks: topologies, standards, duty-cycled energy budgets and radio link budgets.

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Why it matters

Measurements are useless if they cannot reach the people and controllers who act on them. Telemetry carries readings from pipelines, substations, wells, satellites, rockets and patients to a remote station; wireless sensor networks (WSNs) let hundreds of battery-powered nodes monitor a plant, a farm or a structure without running cables. Designing them means combining everything in this subject — modulation, multiplexing, noise, link budgets — with the hard constraint of battery energy.

Key ideas

Telemetry system. Transducer → signal conditioning → multiplexer (TDM or FDM) → modulator/transmitter → channel (wire, power line, radio, satellite, optical fibre) → receiver/demodulator → demultiplexer → display, recorder or control system. Supervisory control and data acquisition (SCADA) adds commands in the reverse direction.

Classification by the signal used.

  • Voltage and current telemetry over wires: the 4–20 mA current loop is the industrial standard because loop current is unaffected by wire resistance, a live zero (4 mA) distinguishes a broken wire (0 mA) from a zero reading, and the loop can power the transmitter. HART superimposes FSK (1200 Hz for 1, 2200 Hz for 0) on the 4–20 mA loop for digital data.
  • Frequency telemetry: the measurand sets a frequency, which is immune to amplitude errors over long lines.
  • Radio telemetry: FM/FM (sensor signals frequency-modulate subcarriers, which are FDM-combined and then frequency-modulate the RF carrier — IRIG standard subcarrier channels) for analog data, and PCM/FM or PCM/PSK for digital data in TDM frames with a synchronisation word.
  • Optical-fibre telemetry for EMI-heavy or hazardous areas.

PCM telemetry frames. Each frame contains one word per channel (or several for fast channels — super-commutation) plus a frame-sync pattern. Bit rate = (bits per frame) × (frames per second). Sampling at about 5 times the signal bandwidth is common in practice to ease reconstruction filtering.

Wireless sensor networks. A node has a sensor, a microcontroller with ADC, a low-power radio, memory and a power source (battery, often with energy harvesting). Nodes report to a sink or gateway, which connects to the plant network or cloud.

  • Topologies: star (simple, low latency, limited range), tree/cluster, and mesh (multi-hop routing around failed nodes and obstacles, self-healing, at the cost of relay energy and latency).
  • Standards: IEEE 802.15.4 physical/MAC layer (2.4 GHz, 250 kbit/s) underlies Zigbee, WirelessHART and ISA100.11a (industrial process, with time-synchronised channel hopping for reliability); Bluetooth Low Energy for short range; LoRaWAN and NB-IoT for kilometre-range, low-rate links.
  • Energy: the radio dominates consumption, so nodes sleep most of the time and wake briefly (duty cycling). Average current I_avg = D·I_active + (1 − D)·I_sleep, where D is the duty cycle, and battery life ≈ capacity/I_avg. Data aggregation in the network and event-driven reporting save further energy.
  • Other design issues: time synchronisation, localisation, security (encryption, authentication), interference in crowded 2.4 GHz bands, and latency for control loops.

Radio link budget. Received power P_r(dBm) = P_t + G_t + G_r − L_path − L_other. In free space, path loss FSPL = 20·log₁₀(4πdf/c) dB, rising 6 dB for each doubling of distance or frequency; indoors and in plants, obstacles make the loss grow faster (path-loss exponent 3–4). The link works if P_r exceeds receiver sensitivity by a fade margin.

Formulas

R_b = (N·n + b_sync)·f_frame R_b bit rate (bit/s), N channels, n bits per word, b_sync sync bits per frame, f_frame frames per second.

I_avg = D·I_active + (1 − D)·I_sleep, D = t_on/T D duty cycle, t_on active time per period T (s), currents (A).

t_life ≈ C/I_avg C battery capacity (A·h), t_life (h); ignores self-discharge and temperature effects.

FSPL(dB) = 20·log₁₀(4πdf/c) d distance (m), f frequency (Hz), c = 3 × 10⁸ m/s.

P_r = P_t + G_t + G_r − L_path (all in dB/dBm) P_t transmit power (dBm), G_t, G_r antenna gains (dBi).

E = P·t Energy (J) from power (W) and time (s).

Worked examples

Example 1 (standard). A PCM telemetry system carries 32 channels, each sampled at 500 samples/s and encoded with 12 bits. Each frame adds a 16-bit sync word. Find the bits per frame and the bit rate.

  1. Bits per frame = 32 × 12 + 16 = 400 bits.
  2. One sample per channel per frame, so f_frame = 500 frames/s.
  3. R_b = 400 × 500 = 200 000 bit/s. Answer: 400 bits per frame, 200 kbit/s.

Example 2 (GATE level). A 2.4 GHz WSN node transmits 0 dBm through a 2 dBi antenna to a gateway 100 m away (2 dBi antenna, free space). Receiver sensitivity is −95 dBm. The node wakes every 10 s for 10 ms drawing 20 mA, and sleeps at 5 µA, from a 2000 mA·h battery. Find the link margin and the battery life.

  1. FSPL = 20·log₁₀(4π × 100 × 2.4 × 10⁹/3 × 10⁸) = 20·log₁₀(10 053) = 80.0 dB.
  2. P_r = 0 + 2 + 2 − 80.0 = −76.0 dBm; margin = −76.0 − (−95) = 19.0 dB.
  3. D = 0.01/10 = 0.001; I_avg = 0.001 × 20 mA + 0.999 × 5 µA = 20.0 µA + 5.0 µA = 25.0 µA.
  4. t_life = 2000 mA·h/0.025 mA = 80 000 h ≈ 9.1 years. Answer: margin ≈ 19 dB; life ≈ 80 000 h (≈ 9 years) before self-discharge — note that the sleep current is a fifth of the total.

Example 3 (energy). Fifty nodes each draw 0.5 mW continuously for 24 h. Find the total energy.

  1. Total power = 50 × 0.5 mW = 25 mW.
  2. E = 0.025 W × 86 400 s = 2160 J. Answer: 2160 J.

Common mistakes

  • Ignoring sleep current; at low duty cycles it can dominate battery life.
  • Adding gains and losses in linear units and dB together. Work entirely in dB/dBm.
  • Forgetting the sync word or overhead bits when computing PCM telemetry bit rate.
  • Assuming free-space loss indoors; real plant links need larger fade margins.
  • Thinking a 4–20 mA loop's zero is 0 mA; 4 mA is the live zero, and 0 mA indicates a fault.

For GATE IN

  • 4–20 mA loop calculations (current for a given measurand, live zero, loop resistance).
  • PCM/TDM telemetry frame and bit-rate numericals; FM/FM subcarrier bandwidths.
  • Duty-cycle and battery-life estimates; energy in joules.
  • Free-space path loss and simple link budgets.
  • Conceptual questions on WSN topologies, protocols and energy saving.

Quick check

  1. A 4–20 mA transmitter spans 0–200 °C. What current corresponds to 50 °C?
  2. Doubling the distance in free space changes path loss by how much?
  3. Why does HART not disturb the 4–20 mA reading?
  4. Which topology lets data route around a failed node? Answers: 1. 8 mA. 2. +6 dB. 3. Its FSK signal is AC with zero average, so the DC loop current is unchanged. 4. Mesh.

Interview questions

All Communication and Optical Instrumentation interview questions

Try answering each one aloud before you open it.

  1. 1.What is telemetry in the context of communication systems?Concept

    Telemetry is the process of recording and transmitting the readings of an instrument. In communication systems, it involves the collection of data from remote or inaccessible points and transmitting it to a receiving station for monitoring, analysis, and control.

  2. 2.Explain the basic components of a wireless sensor network (WSN).Concept

    A wireless sensor network (WSN) typically consists of sensor nodes, a communication network, and a base station. Sensor nodes are equipped with sensors to monitor physical or environmental conditions, such as temperature or pressure. These nodes communicate wirelessly with each other and with the base station, which collects and processes the data.

  3. 3.How does a telemetry system differ from a traditional data communication system?Concept

    A telemetry system is specifically designed for remote monitoring and control, often involving sensors and data acquisition from inaccessible locations. Traditional data communication systems focus on the transmission of data between two or more points, without necessarily involving remote sensing or control. Telemetry systems often include additional components like sensors and data loggers.

  4. 4.Why are wireless sensor networks used in environmental monitoring?Application

    Wireless sensor networks are used in environmental monitoring because they can be deployed in remote and harsh environments without the need for physical infrastructure. They provide real-time data collection and can cover large areas, making them ideal for monitoring environmental parameters like temperature, humidity, and pollution levels.

  5. 5.What happens if a sensor node in a WSN fails?Application

    If a sensor node in a WSN fails, the network's ability to collect data from that specific location is compromised. However, many WSNs are designed with redundancy and self-healing capabilities, allowing other nodes to compensate for the failed node by rerouting data through alternative paths.

  6. 6.Explain how data is transmitted in a telemetry system.Concept

    In a telemetry system, data is collected from sensors and transmitted to a remote receiver. This transmission can occur via various communication channels, such as radio waves, satellite links, or the internet. The data is often encoded and modulated to ensure it can be accurately received and decoded at the destination.

  7. 7.What are the advantages of using optical communication in telemetry systems?Application

    Optical communication in telemetry systems offers high bandwidth, low signal attenuation, and immunity to electromagnetic interference. These advantages make it suitable for transmitting large volumes of data over long distances with high reliability and speed.

  8. 8.Calculate the data rate required for a telemetry system that transmits 1000 sensor readings per second, with each reading being 16 bits.Numerical

    Data rate = Number of readings per second × Bits per reading = 1000 readings/second × 16 bits/reading = 16000 bits/second or 16 kbps.

  9. 9.What is the role of a base station in a wireless sensor network?Concept

    The base station in a wireless sensor network acts as a central point for data collection and processing. It receives data from sensor nodes, processes it, and may transmit it to other networks or systems for further analysis. The base station often has more processing power and storage capacity than individual sensor nodes.

  10. 10.If a telemetry system uses a satellite link with a latency of 500 ms, how does this affect real-time data monitoring?Application

    A latency of 500 ms in a satellite link can introduce a noticeable delay in real-time data monitoring. This delay may affect time-sensitive applications, where immediate data processing and response are critical. However, for many telemetry applications, such a delay is manageable and does not significantly impact overall system performance.

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