14.3 Fundamentals of Electronic Communications

14.3 Fundamentals of Electronic Communications

Fundamentals of Electronic Communications shares topic D of the Electrical Engineering subject with Electronics 1 and 2 in the PRC Enhanced Table of Specifications (PRBEE Resolution No. 40, s. 2024), the pair carrying 2.25% of the whole examination and 5 of the 100 Electrical Engineering items. The 2011 syllabus listed the same content as Principles of Communication. For a practising REE the material is not academic: SCADA telemetry, teleprotection signalling, distribution automation and smart-meter backhaul are all communication systems.

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1. Elements of a Communication System

$\text{Information source} \rightarrow \text{Transmitter} \rightarrow \text{Channel} \rightarrow \text{Receiver} \rightarrow \text{Destination}$

with noise entering primarily in the channel. Modulation — impressing a low-frequency message onto a high-frequency carrier — exists for four reasons: it allows practical antenna sizes, it permits frequency-division multiplexing of many users, it improves noise immunity, and it matches the signal to the channel.

Antenna length is tied to wavelength, and this is why baseband transmission is impractical:

$\lambda = \frac{c}{f}, \qquad c = 3 \times 10^{8} \text{ m/s}$

A half-wave antenna for a 3 kHz voice signal would be 50 km long; at 100 MHz it is 1.5 m.

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2. Amplitude Modulation

The modulation index for a carrier of amplitude $V_c$ and message amplitude $V_m$:

$m = \frac{V_m}{V_c} = \frac{V_{\max} - V_{\min}}{V_{\max} + V_{\min}}$

Overmodulation ($m > 1$) clips the envelope and generates severe distortion and splatter into adjacent channels.

$\text{Bandwidth} = 2f_{m(\max)}$

$P_t = P_c\left(1 + \frac{m^{2}}{2}\right), \qquad P_{SB} = P_c\frac{m^{2}}{2}$

At $m = 1$, $P_t = 1.5P_c$: two thirds of the transmitted power sits in the carrier, which conveys no information at all. That inefficiency motivates single sideband (SSB), which suppresses the carrier and one sideband, halving bandwidth and concentrating all power in the intelligence.

$I_t = I_c\sqrt{1 + \frac{m^{2}}{2}}$

For several simultaneous modulating signals, the effective index is $m_t = \sqrt{m_1^{2} + m_2^{2} + \cdots}$.

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3. Angle Modulation

Carson's rule gives the practical bandwidth:

$BW = 2\left(\Delta f + f_m\right) = 2f_m\left(m_f + 1\right)$

FM trades bandwidth for noise immunity: a limiter strips amplitude noise before demodulation, which AM cannot do. Narrowband FM ($m_f < 1$) occupies about the same bandwidth as AM; wideband FM broadcasting uses $\Delta f = 75$ kHz with $f_m = 15$ kHz, giving $BW = 2(75 + 15) = 180$ kHz within the 200 kHz channel allocation.

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4. Decibels and Noise

$dB = 10\log_{10}\frac{P_2}{P_1} = 20\log_{10}\frac{V_2}{V_1} \quad \text{(equal impedances)}$

Absolute references: dBm is referred to 1 mW, dBW to 1 W, so 0 dBm = −30 dBW. Cascaded stage gains and losses in decibels simply add.

Thermal noise power in a bandwidth $B$:

$P_n = kTB, \qquad k = 1.38 \times 10^{-23} \text{ J/K}$

$NF_{dB} = 10\log_{10}\frac{(S/N)_{in}}{(S/N)_{out}}$

Friis' formula shows why the first amplifier dominates the system noise figure:

$F_{total} = F_1 + \frac{F_2 - 1}{G_1} + \frac{F_3 - 1}{G_1G_2} + \cdots$

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5. Sampling, PCM and Digital Modulation

Nyquist sampling theorem:

$f_s \ge 2f_{\max}$

Sampling below this rate causes aliasing, in which high-frequency content masquerades irreversibly as low-frequency content. Telephone-quality speech band-limited to 3.4 kHz is sampled at 8 kHz and quantised to 8 bits, giving the familiar 64 kbps PCM channel.

$\text{Quantisation levels} = 2^{n}, \qquad \text{Bit rate} = f_s \times n$

$SQNR_{dB} \approx 6.02n + 1.76$

Each additional bit buys about 6 dB of signal-to-quantisation-noise ratio.

Shannon-Hartley capacity:

$C = B\log_{2}\left(1 + \frac{S}{N}\right) \quad \text{bits/s}$

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6. Transmission Media and Utility Communications

Fibre's immunity to electromagnetic interference is decisive in substations, where ground-potential rise during a fault can destroy metallic communication circuits and endanger personnel. OPGW — optical ground wire — combines the shield-wire function with an embedded fibre bundle, which is why it has become standard on new transmission lines.

Modern substation communications run IEC 61850 over Ethernet, with GOOSE messaging replacing hard-wired interlocking, and DNP3 or IEC 60870-5-101/104 carrying SCADA telemetry to the control centre.

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Solved Board Exam Examples

Example 1: AM Power Distribution

An AM transmitter has an unmodulated carrier power of 5 kW and is modulated to 80%. Find the total transmitted power and the power in each sideband.

Solution.

$P_t = P_c\left(1 + \frac{m^{2}}{2}\right) = 5\left(1 + \frac{0.64}{2}\right) = 5(1.32) = \boxed{6.6 \text{ kW}}$

$P_{SB(\text{total})} = 6.6 - 5.0 = 1.6 \text{ kW} \implies P_{\text{each}} = \boxed{0.8 \text{ kW}}$

Even at 80% modulation, 76% of the transmitted power is spent on a carrier that conveys nothing.

Example 2: FM Bandwidth by Carson's Rule

An FM signal has a maximum deviation of 25 kHz and a highest modulating frequency of 10 kHz. Find the modulation index and the bandwidth.

Solution.

$m_f = \frac{\Delta f}{f_m} = \frac{25}{10} = 2.5$

$BW = 2(\Delta f + f_m) = 2(25 + 10) = \boxed{70 \text{ kHz}}$

An AM signal carrying the same 10 kHz message would need only 20 kHz — the bandwidth price FM pays for its noise immunity.

Example 3: PCM Bit Rate for a SCADA Voice Channel

An analog channel band-limited to 3.4 kHz is sampled at the standard telephony rate and quantised to 256 levels. Determine the sampling rate, bits per sample and resulting bit rate.

Solution. The Nyquist minimum is $2(3.4) = 6.8$ kHz; the standard practical rate is $f_s = 8$ kHz, which leaves guard band for the anti-aliasing filter.

$n = \log_{2}(256) = 8 \text{ bits per sample}$

$\text{Bit rate} = 8000 \times 8 = \boxed{64\,000 \text{ bps} = 64 \text{ kbps}}$

The signal-to-quantisation-noise ratio is approximately $6.02(8) + 1.76 = 49.9$ dB.