Communications

A. Basic modulation/demodulation concepts

Analog Modulation

AM and FM/PM, modulation index, Carson-rule bandwidth, and power distribution — the formula family behind nearly every analog-comms question on the FE.

Modulation moves a low-frequency message

m(t)

up to a carrier frequency

f_c

so it can radiate from an antenna of practical size, share a channel with other signals, and trade bandwidth for noise immunity. The FE Communications topic leans hard on a small set of relationships: how a message rides on a carrier (amplitude vs. angle), how wide the resulting spectrum is, and how transmitted power splits between carrier and sidebands. The FE Reference Handbook lays these out compactly under “Amplitude Modulation,” “Angle Modulation,” and “Frequency Modulation (FM)” — know each symbol, because the exam supplies the formula but not the judgment to use it.

Conventional AM and the modulation index

Standard (double-sideband, large-carrier) AM transmits a carrier plus a scaled copy of the message:

x_{AM}(t)=A_c\bigl[1+a\,m_n(t)\bigr]\cos(2\pi f_c t)

, where

m_n(t)=m(t)/\max|m(t)|

is the normalized message (so

|m_n|\le 1

) and

a

is the modulation index. For a single-tone message,

a

is the ratio of message-induced amplitude swing to carrier amplitude, often read straight off an envelope as

a=(E_{max}-E_{min})/(E_{max}+E_{min})

. Keep

a\le 1

: at

a>1

the envelope crosses zero, overmodulation distorts the signal, and a simple envelope detector fails.

x_{AM}(t)=A_c\bigl[1+a\,m_n(t)\bigr]\cos(2\pi f_c t),\qquad a=\frac{E_{max}-E_{min}}{E_{max}+E_{min}}

AM power and efficiency

The carrier in conventional AM carries no information — it is overhead that an envelope detector needs. For single-tone modulation at index

a

, total power is

P_t=P_c\bigl(1+a^2/2\bigr)

, the two sidebands together carry

P_c\,a^2/2

, and the efficiency (fraction of power in the message) is

\eta=a^2/(2+a^2)

. Even at full

a=1

the best you can do is

\eta=33.3\%

; at

a=0.7

only about

19.7\%

of transmitted power is information. That waste is why SSB exists.

P_t=P_c\!\left(1+\frac{a^2}{2}\right),\qquad \eta=\frac{a^2}{2+a^2}

DSB-SC and single sideband (SSB)

Double-sideband suppressed-carrier,

x_{DSB}(t)=A_c\,m(t)\cos(2\pi f_c t)

, deletes the carrier and puts all power in the sidebands — but it must be demodulated synchronously (a coherent local oscillator, often via a Costas loop) because there is no envelope to follow. SSB goes further and transmits only one sideband, halving the occupied bandwidth from

2W

to

W

and cutting power further. The handbook gives the upper/lower-sideband spectra; the practical takeaways are the bandwidths: AM and DSB occupy

2W

, SSB occupies

W

, where

W

is the message bandwidth.

x_{DSB}(t)=A_c\,m(t)\cos(2\pi f_c t),\qquad B_{DSB}=2W,\quad B_{SSB}=W

Angle modulation: FM and PM

Angle modulation writes the carrier as

x_{Ang}(t)=A_c\cos\bigl(2\pi f_c t+\phi(t)\bigr)

and varies the angle rather than the amplitude. In phase modulation (PM) the phase deviation tracks the message directly,

\phi(t)=k_P\,m(t)

; in frequency modulation (FM) the instantaneous-frequency deviation tracks the message, so the phase is the integral,

\phi(t)=k_F\!\int m(\lambda)\,d\lambda

. The instantaneous frequency is

f_i(t)=f_c+\tfrac{1}{2\pi}\,d\phi/dt

, and the peak frequency deviation

\Delta f

is the most-deviated excursion from

f_c

. Because the amplitude is constant, FM/PM resist amplitude noise — the basis of FM’s wideband SNR advantage.

\text{PM: } \phi(t)=k_P\,m(t),\qquad \text{FM: } \phi(t)=k_F\!\int_{-\infty}^{t}\! m(\lambda)\,d\lambda

FM deviation ratio and Carson's rule

The frequency-deviation ratio is

D=\Delta f/W=k_F\,\max|m(t)|/(2\pi W)

, where

W

is the message bandwidth. (For single-tone FM this

D

is the modulation index

\beta

.) An angle-modulated signal has theoretically infinite bandwidth, so we use the 98%-power bandwidth: for narrowband FM (

D\ll 1

),

B\approx 2W

; for wideband FM (

D>1

), Carson’s rule gives

B\approx 2(D+1)W=2(\Delta f+W)

. Carson’s rule is the single most-tested FM result — commit it to memory in both the

D

form and the

(\Delta f+W)

form.

D=\frac{\Delta f}{W},\qquad B\approx 2(D+1)\,W=2(\Delta f+W)\quad(\text{Carson, }D>1)

Demodulation and noise intuition

Each scheme has a matched detector. Conventional AM (carrier present,

a\le 1

) recovers the message with a cheap envelope detector; DSB-SC and SSB need a synchronous (coherent) demodulator that multiplies by a phase-locked carrier replica and low-pass filters. FM uses a discriminator (a frequency-to-voltage device) or a phase-locked loop. The payoff of spending bandwidth: wideband FM’s output SNR improves with

D^2

, so trading channel width for noise immunity is governed by the deviation ratio — a qualitative result the FE expects you to recognize even though it rarely asks the full SNR formula.

Exam strategy

Sort the problem first: amplitude or angle. For AM, decide whether the carrier is present (conventional,

2W

, envelope-detectable, efficiency

\le 33\%

) or suppressed (DSB

2W

/ SSB

W

, synchronous detection). For FM, compute

D=\Delta f/W

, then choose

B\approx 2W

if

D\ll 1

or Carson

B\approx 2(D+1)W

otherwise — and remember

2(\Delta f+W)

is the same thing. Read carefully whether a quoted frequency is the peak deviation

\Delta f

, the message bandwidth

W

, or the modulating frequency

f_m

; mixing them is the classic Carson trap.

Key equations

Conventional AM signal

x_{AM}(t)=A_c\bigl[1+a\,m_n(t)\bigr]\cos(2\pi f_c t)

$a$ = modulation index ( $\le 1$ to avoid overmodulation); $m_n(t)=m(t)/\max|m(t)|$ is the normalized message; $A_c$ , $f_c$ = carrier amplitude and frequency.

Modulation index from envelope

a=\frac{E_{max}-E_{min}}{E_{max}+E_{min}}

Reads $a$ directly off the maximum and minimum envelope amplitudes of a single-tone AM waveform (dimensionless).

AM total power

P_t=P_c\!\left(1+\frac{a^2}{2}\right)

Single-tone AM. $P_c$ = unmodulated carrier power (W); the two sidebands together carry $P_c a^2/2$ .

AM efficiency

\eta=\frac{a^2}{2+a^2}

Fraction of transmitted power in the sidebands (information). Maximum $33.3\%$ at $a=1$ .

DSB-SC signal

x_{DSB}(t)=A_c\,m(t)\cos(2\pi f_c t)

Suppressed carrier; all power in sidebands; needs synchronous (coherent) demodulation. Bandwidth $2W$ .

Bandwidths of AM variants

B_{AM}=B_{DSB}=2W,\qquad B_{SSB}=W

$W$ = message bandwidth (Hz). SSB occupies half the spectrum of DSB/AM.

Angle-modulated carrier

x_{Ang}(t)=A_c\cos\!\bigl(2\pi f_c t+\phi(t)\bigr)

$\phi(t)$ = phase deviation set by the message. PM: $\phi=k_P m(t)$ ; FM: $\phi=k_F\!\int m\,d\lambda$ .

Instantaneous frequency (FM)

f_i(t)=f_c+\frac{1}{2\pi}\frac{d\phi}{dt}

Peak excursion of $f_i$ from $f_c$ is the peak frequency deviation $\Delta f$ (Hz).

FM deviation ratio

D=\frac{\Delta f}{W}

$\Delta f$ = peak frequency deviation (Hz); $W$ = message bandwidth (Hz). For single tone, $D$ equals the modulation index $\beta$ .

Carson's rule (wideband FM)

B\approx 2(D+1)\,W=2(\Delta f+W)

98%-power bandwidth for $D>1$ . Reduces to $B\approx 2W$ for narrowband FM ( $D\ll 1$ ).

Worked examples

AM transmitted power and efficiency

Problem. A conventional AM transmitter has an unmodulated carrier power of $50\,\text{W}$ and is single-tone modulated to $a=0.70$ . Find the total transmitted power, the total sideband power, and the modulation efficiency.

Solution. Total power:

P_t=P_c(1+a^2/2)=50\,(1+0.49/2)=50\,(1.245)=62.3\,\text{W}

. Sideband power:

P_{sb}=P_c\,a^2/2=50\,(0.245)=12.3\,\text{W}

(split equally,

6.13\,\text{W}

each). Efficiency:

\eta=a^2/(2+a^2)=0.49/2.49=0.197=19.7\%

. Sanity check:

P_{sb}/P_t=12.25/62.25=0.197

matches

\eta

, and the carrier still holds the remaining

50\,\text{W}

— consistent.

P_t=50\!\left(1+\tfrac{0.7^2}{2}\right)=62.3\ \text{W},\qquad \eta=\frac{0.49}{2.49}=19.7\%

Carson-rule bandwidth of an FM broadcast signal

Problem. A commercial FM station uses a peak frequency deviation of $\Delta f=75\,\text{kHz}$ for an audio message bandwidth $W=15\,\text{kHz}$ . Find the deviation ratio and the transmission bandwidth by Carson’s rule.

Solution. Deviation ratio:

D=\Delta f/W=75/15=5

(wideband,

D>1

). Carson’s rule:

B\approx 2(D+1)W=2(6)(15)=180\,\text{kHz}

. Cross-check via the

(\Delta f+W)

form:

B\approx 2(75+15)=2(90)=180\,\text{kHz}

— agrees, confirming the

200\,\text{kHz}

FM channel spacing easily contains it.

D=\frac{75}{15}=5,\qquad B\approx 2(5+1)(15)=180\ \text{kHz}

Power saved by switching AM to SSB

Problem. A message of bandwidth $W=4\,\text{kHz}$ is sent first as conventional AM at $a=1$ and then as SSB. Compare occupied bandwidths and the fraction of useful (message) power in each.

Solution. Conventional AM: bandwidth

B=2W=8\,\text{kHz}

; efficiency

\eta=1/(2+1)=33.3\%

, so only one-third of transmitted power carries the message. SSB: bandwidth

B=W=4\,\text{kHz}

— half the spectrum — and with no carrier and one sideband removed, essentially

100\%

of transmitted power is information. Sanity check: SSB halves bandwidth and roughly triples power efficiency versus full AM — the standard motivation for SSB in spectrum-limited links.

B_{AM}=2W=8\ \text{kHz},\ \eta=33.3\%;\qquad B_{SSB}=W=4\ \text{kHz}

Common pitfalls

•Confusing the modulating frequency $f_m$ with the message bandwidth $W$ in Carson’s rule. For multitone/baseband messages, $W$ is the highest message frequency — use it, not a single tone.
•Allowing $a>1$ . Overmodulation makes the envelope cross zero, distorts the recovered audio, and breaks envelope detection; keep $a\le 1$ for conventional AM.
•Forgetting the carrier power when computing AM efficiency. The carrier consumes the majority of transmitted power; $\eta=a^2/(2+a^2)$ already accounts for that — don’t divide sideband power by sideband power.
•Using $2W$ for wideband FM. Narrowband ( $D\ll 1$ ) gives $B\approx 2W$ ; wideband ( $D>1$ ) requires Carson $2(D+1)W$ — check $D$ before choosing.
•Treating $D$ (deviation ratio) and the single-tone modulation index $\beta$ as always identical. They coincide for a single tone but $D$ uses the message bandwidth $W$ in general.
•Trying to envelope-detect DSB-SC or SSB. With no transmitted carrier the envelope is meaningless; these require a synchronous (coherent) demodulator.

References

NCEES FE Reference Handbook — Electrical and Computer Engineering: Amplitude Modulation, Angle Modulation, Frequency Modulation (FM)
Haykin, Communication Systems — AM/FM derivations, Carson's rule, SSB
Lathi & Ding, Modern Digital and Analog Communication Systems — modulation index, efficiency, FM bandwidth

B. Fourier transforms/Fourier series

Fourier Series and Transforms

Decompose periodic and aperiodic signals into spectra, apply transform pairs and theorems, and use Parseval to track power — the analysis backbone of communications problems.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

C. Multiplexing

Multiplexing

Share one channel among many signals by frequency (FDM/WDM), time (TDM), or code (CDMA) — with the guard bands, framing overhead, and capacity bookkeeping the FE tests.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

D. Digital communications

Digital Communications

From sampling and PCM through ASK/FSK/PSK/QAM to the Nyquist and Shannon limits — the bit-rate, bandwidth, and error-rate relationships the FE tests most.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders