Engineer Radio Electronic Communication common mistakes
Engineer Radio Electronic Communication · Understand the traps and how to avoid them.
Treating free-space path loss as a loss that depends on frequency in the same way as a physical dissipative loss, and forgetting the 20·log(4π/c) constant when computing it by hand.
Candidates memorize 'FSPL increases with frequency' but never derive it, so they mix up the wavelength form (20·log(4πd/λ)) with the frequency form (32.44 + 20·log d_km + 20·log f_MHz) and lose the unit conversions. The 32.44 constant is often misremembered as 32, 20·log(4π/c) evaluated incorrectly, or applied with d in metres.
What to do instead
Derive once from first principles: FSPL = 20·log(4πd/λ) = 20·log(4πdf/c). With d in km and f in MHz this becomes 32.44 + 20·log(d_km) + 20·log(f_MHz). Practise five numeric problems changing units each time, and always sanity-check: doubling distance adds 6 dB, doubling frequency adds 6 dB.
Confusing antenna gain, directivity, and effective aperture, and applying the Friis equation without converting EIRP correctly or accounting for impedance mismatch and cable loss.
Candidates treat gain and directivity as synonyms and forget that gain includes efficiency. They also plug transmit power in dBW against receive power in dBm, or forget that EIRP = P_t + G_t − L_cable, producing answers off by 30 dB or by the cable loss.
What to do instead
Write the link budget line by line with units attached: P_t (dBm) − L_tx + G_tx = EIRP; then − FSPL + G_rx − L_rx = P_rx. Memorize G = η·D and A_eff = G·λ²/(4π). Do three full link budgets end-to-end, converting every term to dBm/dB before summing.
Mixing up dBm, dBW, dBμV, and dBμV/m when converting field strength to received voltage, and forgetting the 107 dB constant in the E(dBμV/m) to V(dBμV) relation.
The field-to-voltage conversion E(dBμV/m) = V(dBμV) + 20·log f(MHz) + 77.2 (for a 50 Ω system, or 107 for a half-wave dipole with 0 dBi reference) is a formula candidates half-remember. They also apply the 50 Ω impedance factor to a 75 Ω system or forget the antenna factor entirely.
What to do instead
Build a one-page conversion sheet: 0 dBm = −30 dBW = 107 dBμV (50 Ω). Derive the field-strength constant from P = E²/(120π)·A_eff once, then use it. In every problem state the reference impedance and antenna factor before substituting numbers.
Miscalculating the noise floor by forgetting to add 10·log(B) in the right units or by omitting the receiver noise figure, then reporting sensitivity without the required SNR.
Candidates use kTB = −174 dBm/Hz but then multiply by bandwidth in Hz incorrectly (using kHz without the 30 dB correction) or drop the noise figure term. They also confuse sensitivity (minimum detectable signal at a given SNR) with the noise floor itself.
What to do instead
Use the template: N = −174 + 10·log(B_Hz) + NF (dBm); sensitivity = N + required SNR. Check with a worked example: 200 kHz bandwidth gives −174 + 53 = −121 dBm noise floor, +8 dB NF = −113 dBm, +10 dB SNR = −103 dBm. Always state bandwidth in Hz before taking the log.
Applying Carson's rule and Bessel-function sideband analysis as if they were interchangeable, or computing FM deviation without distinguishing peak deviation from RMS deviation.
Candidates memorize B ≈ 2(Δf + f_m) but then use it for narrowband cases where B ≈ 2·f_m applies, or they confuse the modulation index β = Δf/f_m with the deviation ratio. In Bessel analysis they forget that the carrier can vanish at certain β values (J0 = 0).
What to do instead
State the regime first: if β < 0.3 use narrowband B ≈ 2·f_m; otherwise Carson B ≈ 2(Δf + f_m). Memorize that J0(β) = 0 at β ≈ 2.405, 5.52, 8.65. Practise one problem each way and check the answer against the 98% power bandwidth definition.
Confusing the roles of the mixer, local oscillator, and IF filter in a superheterodyne receiver, and mispredicting image frequency and reciprocal mixing effects.
Candidates can state the block diagram but cannot compute f_image = f_LO + f_IF (high-side injection) or f_LO − f_IF (low-side), and they forget that image rejection depends on the RF filter before the mixer, not the IF filter after it.
What to do instead
Draw the spectrum with f_RF, f_LO, f_IF, and f_image marked. Use the rule: for high-side injection f_image = f_RF + 2·f_IF. Work three problems with different IFs and note how a higher IF improves image rejection but worsens adjacent-channel selectivity, motivating double conversion.
Treating QAM and PSK constellations as having the same Eb/N0 versus BER curve, or forgetting the 3 dB penalty of coherent versus non-coherent detection and the coding gain from FEC.
Candidates memorize '16-QAM needs about 4 dB more than QPSK for the same BER' as a number without understanding that it comes from the distance between constellation points at equal average energy. They then apply it to the wrong comparison (e.g., 16-QAM versus 16-PSK) or forget to subtract coding gain when FEC is used.
What to do instead
Anchor on the Shannon limit and the waterfall curves: QPSK and 16-QAM differ by roughly 3–4 dB at BER 10⁻⁵ for equal average energy. Derive the minimum distance for square QAM (d = √(6E_s/(M−1))). For FEC, subtract the coding gain (e.g., 5–7 dB for rate-1/2 convolutional with Viterbi) explicitly and state the coding rate.
Assuming a spurious emission or harmonic is acceptable because it is below the fundamental, without checking it against the absolute limit in dBc or the regulatory mask (e.g., −36 dBm below 1 GHz, −30 dBm above).
Candidates apply a relative dBc rule of thumb (e.g., 'harmonics at −40 dBc are fine') when the standard specifies an absolute power limit in dBm at the antenna port, often independent of the carrier power for low-power devices. They also forget that the limit varies with frequency band and service.
What to do instead
Always identify the applicable standard and its absolute limit before judging compliance. For EN 300 220 / 300 328 style limits, memorize the tiered absolute values (−36 dBm below 1 GHz, −30 dBm above 1 GHz for general spurious, with tighter limits in restricted bands). Convert dBc to dBm using the actual carrier power, not the rated maximum.
Reading a spectrum analyzer trace without correcting for resolution bandwidth, detector mode, and reference level, and treating the displayed noise floor as the true noise floor.
The displayed average noise level (DANL) changes with RBW: halving RBW lowers the displayed noise by 3 dB. Candidates compare traces taken at different RBW settings, or use a peak detector and mistake the peak of the noise for the average, or set the reference level too high and compress the front end.
What to do instead
Normalize every measurement: noise floor in dBm/Hz = displayed level − 10·log(RBW_Hz). Always record RBW, VBW, detector (peak/average/RMS), and reference level. Verify linearity by checking the attenuator setting; a 10 dB attenuator increase should drop the displayed level by 10 dB if the analyzer is not compressing.
Confusing the ITU Radio Regulations allocation status (primary, secondary, permitted) with the national assignment, and misapplying the 'secondary service must not cause harmful interference' rule.
Candidates learn the ITU Table of Frequency Allocations as if it were directly binding on operators, and they treat secondary allocations as unusable or as having equal rights. They also forget that national administrations may impose additional constraints and that the RR footnotes often change the status for specific countries.
What to do instead
Separate the layers: ITU RR Article 5 allocation (global/regional) → national table of allocations → licence conditions. Memorize the Article 1 definitions of primary and secondary service and the Article 15 interference principle. For any band question, state the allocation status, the relevant footnote, and the national implementation before answering.
Treating a cellular or trunked radio network as a set of independent links, ignoring frequency reuse distance, co-channel interference, and the Erlang-based capacity model.
Candidates can compute a single link budget but not the cluster size N = i² + ij + j², the reuse distance D = R·√(3N), or the co-channel carrier-to-interference ratio C/I = (1/6)·(D/R)^γ for a hexagonal layout. They also confuse offered traffic with carried traffic and forget the Erlang B blocking formula assumptions.
What to do instead
Memorize the reuse geometry: N = i² + ij + j² (N = 3, 4, 7, 9, 12...), D/R = √(3N), and C/I ≈ (1/6)(3N)^(γ/2). For capacity, use Erlang B for blocked-calls-cleared systems and Erlang C for queued systems, and always state the blocking target (e.g., 2%). Work one example with N = 7 and γ = 4 to get C/I ≈ 18 dB.