4.4 Communications
Key Takeaways
- Analog modulation varies a carrier: AM changes amplitude, FM changes frequency, and PM changes phase; FM and PM are angle-modulation methods that trade bandwidth for noise immunity.
- Standard double-sideband AM bandwidth equals twice the highest message frequency, BW = 2·fm, while FM bandwidth follows Carson's rule, BW ≈ 2(Δf + fm).
- The Shannon–Hartley theorem gives the maximum error-free channel capacity C = B·log2(1 + S/N) in bits per second, where B is bandwidth in hertz and S/N is the linear signal-to-noise ratio.
- Signal-to-noise ratio in decibels is SNR(dB) = 10·log10(S/N); to use Shannon's formula you must first convert dB back to the linear power ratio S/N = 10^(SNR_dB/10).
- Digital modulation (ASK, FSK, PSK, QAM) encodes bits onto a carrier with bits/symbol = log2 M; higher-order schemes pack more bits per symbol but need higher SNR, and multiplexing (FDM, TDM, CDM) shares one channel among many signals.
Modulating a carrier
The Communications area is 5–8 questions and is highly formula-driven, which makes it a reliable point source if you know where the equations live in the NCEES FE Reference Handbook. Modulation impresses a low-frequency message signal m(t) onto a high-frequency carrier c(t) = Ac·cos(2πfc·t) so it can be transmitted efficiently and share spectrum with other signals. The three classic analog methods each change a different carrier property:
- Amplitude modulation (AM) varies the carrier amplitude in proportion to the message.
- Frequency modulation (FM) varies the carrier instantaneous frequency.
- Phase modulation (PM) varies the carrier phase.
FM and PM are together called angle modulation: they hold amplitude constant and encode information in the angle, which makes them far more resistant to additive amplitude noise than AM — the reason FM radio sounds cleaner than AM. The cost is bandwidth. For AM, the modulation index μ = Am/Ac must satisfy μ ≤ 1 to avoid overmodulation and distortion, and the transmitted power splits between the carrier (which carries no information) and the two sidebands (which do), explaining why DSB and especially SSB schemes that suppress the carrier are more power-efficient than standard AM.
Bandwidth
Bandwidth questions reduce to a few rules. For standard double-sideband AM, with a message band-limited to fm, the transmission bandwidth is twice the highest message frequency:
BW(AM) = 2 · fm
For FM, the bandwidth is estimated with Carson's rule, where Δf is the peak frequency deviation:
BW(FM) ≈ 2(Δf + fm) = 2·fm·(β + 1)
where β = Δf/fm is the FM modulation index. Single-sideband (SSB) transmits only one sideband, halving AM bandwidth to fm.
| Scheme | Carrier property varied | Bandwidth |
|---|---|---|
| AM (DSB) | Amplitude | 2·fm |
| SSB | Amplitude (one sideband) | fm |
| FM | Frequency | ≈ 2(Δf + fm) |
| PM | Phase | Depends on deviation and fm |
Worked bandwidth: An FM signal has peak deviation Δf = 75 kHz (broadcast FM) and message fm = 15 kHz. Carson's rule gives BW ≈ 2(75 + 15) = 180 kHz — far wider than the 2·15 = 30 kHz an AM signal of the same message would use, illustrating FM's bandwidth-for-noise trade.
Channel capacity and SNR
The headline result of information theory is the Shannon–Hartley theorem, giving the maximum error-free data rate (channel capacity) over a band-limited noisy channel:
C = B · log2(1 + S/N)
Here C is capacity in bits per second, B is the channel bandwidth in hertz, and S/N is the linear signal-to-noise power ratio. Capacity grows linearly with bandwidth but only logarithmically with SNR — doubling bandwidth doubles capacity, while doubling SNR barely changes it.
The most common trap is the signal-to-noise ratio (SNR) unit. SNR is usually quoted in decibels (a power ratio, so 10·log10):
SNR(dB) = 10 · log10(S/N) → S/N = 10^(SNR_dB/10)
Worked capacity: A channel has B = 4 kHz and SNR = 30 dB. Convert first: S/N = 10^(30/10) = 10³ = 1000. Then C = 4000·log2(1 + 1000) = 4000·log2(1001). Since log2(1001) ≈ 9.97, C ≈ 39,900 bps ≈ 40 kbps. Leaving SNR in dB is the classic error and gives a nonsensical answer.
Digital modulation and multiplexing
Digital systems map bits onto a carrier. The basic schemes mirror the analog ones:
- ASK (amplitude-shift keying) — bits set amplitude levels.
- FSK (frequency-shift keying) — bits select among frequencies.
- PSK (phase-shift keying) — bits set carrier phase (BPSK = 1 bit/symbol, QPSK = 2 bits/symbol).
- QAM (quadrature amplitude modulation) — combines amplitude and phase; 16-QAM carries 4 bits/symbol, 64-QAM carries 6.
The bits per symbol for an M-ary scheme is bits/symbol = log2 M, so 16-QAM (M = 16) → log2 16 = 4 bits/symbol. The symbol rate (baud) and bit rate relate by bit rate = baud × log2 M. Higher-order schemes pack more bits per symbol but demand higher SNR to keep the constellation points distinguishable, so designers trade spectral efficiency against power/SNR. Multiplexing shares one physical channel among many signals: FDM (frequency-division) assigns separate frequency bands, TDM (time-division) assigns time slots, and CDM/CDMA (code-division) assigns orthogonal codes so users overlap in both time and frequency.
Noise, link basics, and data-rate trade
The enemy of any link is noise, dominated by thermal noise with power N = kTB, where k is Boltzmann's constant, T the system noise temperature, and B the bandwidth — note that widening B lets in more noise, partly offsetting the capacity gain. The bit error rate (BER) falls as the energy-per-bit to noise-density ratio Eb/N0 rises, which is the digital analog of SNR.
A useful framing: Shannon sets the ceiling C = B·log2(1 + S/N) that no scheme can beat error-free, and practical modulation/coding choices decide how close to that ceiling a link operates. For the FE, the takeaways are the relationships — capacity grows with bandwidth and (logarithmically) with SNR, higher-order modulation needs more SNR, and angle modulation buys noise immunity at the price of bandwidth — applied to the handbook formulas rather than memorized derivations.
A communication channel has a bandwidth of 4 kHz and a signal-to-noise ratio of 30 dB. Using the Shannon–Hartley theorem, the maximum channel capacity is approximately:
An FM signal has a peak frequency deviation of 5 kHz and a maximum message frequency of 3 kHz. Using Carson's rule, the approximate transmission bandwidth is:
Which statement about FM (frequency modulation) compared with AM (amplitude modulation) is correct for the FE exam?