FE and PE Exam

FE Electrical and Computer · areas 13 and 11

Communications and electromagnetics

Communications is 5–8 items on the E&C specification and Electromagnetics adds 4–6 items. One page holds both: modulation and channels on one side, the fields and transmission lines that carry the signal on the other.

Updated Oct 29, 2026 · Not affiliated with NCEES.

What Communications covers

Four subtopics in the July 2020 specification:

  • Modulation and demodulation: AM, FM and PCM.
  • Fourier transforms and series: spectra of periodic and pulsed signals.
  • Multiplexing: time division, frequency division, code division.
  • Digital communications: bit rate, symbol rate, noise and capacity.

16 communications and electromagnetics problems

0 / 16 answered · 0 correct
Question 1 of 16 · Electromagnetics · 4–6 items

A lossless transmission line with characteristic impedance 50 Ω is terminated in a 100 Ω load. What is the voltage reflection coefficient at the load?

Given
  • Characteristic impedance Z0 = 50 Ω
  • Load impedance ZL = 100 Ω
FindReflection coefficient Γ at the load

Solution

Γ = (ZL − Z0)/(ZL + Z0)

Γ = (100 − 50)/(100 + 50) = 0.333

Γ = (ZL − Z0)/(ZL + Z0) = (100 − 50)/(100 + 50) = 0.333.
This divides by ZL alone, (ZL − Z0)/ZL; the denominator is the sum of the impedances.
This is ZL/(ZL + Z0), half the transmission coefficient, not Γ.
This is the transmission coefficient τ = 2ZL/(ZL + Z0) = 1 + Γ.
Question 2 of 16 · Electromagnetics · 4–6 items

The electric field intensity at a distance of 2 m from a point charge in free space is 900 V/m. What is the magnitude of the charge? Use ε0 = 8.854 × 10−12 F/m.

Given
  • Distance r = 2 m
  • Electric field E = 900 V/m
  • Permittivity ε0 = 8.854 × 10−12 F/m
FindCharge Q (µC)

Solution

E = Q/(4πε0r2) → Q = 4πε0r2E

Q = 4π × 8.854 × 10−12 × (2 m)2 × 900 V/m = 0.40 µC

This drops the 4π: Q = Eε0r2 instead of E·4πε0r2.
This uses r instead of r2; the field of a point charge falls as 1/r2.
Q = E·4πε0r2 = 900 × 4π × 8.854 × 10−12 × 22 = 4.0 × 10−7 C = 0.40 µC.
This uses r3, the falloff of a dipole field, not of a single point charge.
Question 3 of 16 · Electromagnetics · 4–6 items

An infinite straight wire carries a current of 10 A. What is the magnitude of the magnetic field intensity H at a perpendicular distance of 0.5 m from the wire?

Given
  • Current I = 10 A
  • Distance r = 0.5 m
FindMagnetic field intensity H (A/m)

Solution

∮H·dl = I → H = I/(2πr)

H = 10 A/(2π × 0.5 m) = 3.18 A/m

This is I/r = 10/0.5 = 20.00 A/m, dropping the 2π circumference of the Amperian loop.
This is I/(2r), the field at the center of a circular loop, not near a straight wire.
This is I/(πr), as if 0.5 m were the diameter of the Amperian loop.
Ampère's law around a circle of radius r: H = I/(2πr) = 10/(2π × 0.5) = 3.18 A/m.
Question 4 of 16 · Electromagnetics · 4–6 items

A uniform plane wave in free space has an electric field amplitude of 30 V/m. What is the magnetic field amplitude H? The intrinsic impedance of free space is 377 Ω.

Given
  • Electric field amplitude E = 30 V/m
  • Intrinsic impedance η0 = 377 Ω
FindMagnetic field amplitude H (A/m)

Solution

η0 = E/H → H = E/η0

H = 30 V/m/377 Ω = 0.0796 A/m

H = E/η0 = 30/377 = 0.0796 A/m.
This uses η0 = 120 Ω, dropping the π from 120π ≈ 377 Ω: 30/120 = 0.250 A/m.
This is E2/η0, which is a power density in W/m2, not a field.
This multiplies E by η0 instead of dividing.
Question 5 of 16 · Electromagnetics · 4–6 items

An electromagnetic wave propagates in a conductor with conductivity σ = 5.8 × 107 S/m at 1 GHz. Using δ = √(2/(ωμσ)) with μ = 4π × 10−7 H/m, what is the skin depth?

Given
  • Conductivity σ = 5.8 × 107 S/m
  • Frequency f = 1 GHz
  • Permeability μ = 4π × 10−7 H/m
FindSkin depth δ (µm)

Solution

δ = √(2/(ωμσ)), ω = 2πf

δ = √(2/(2π × 109 × 4π × 10−7 × 5.8 × 107)) = 2.09 × 10−6 m

This drops the 2 under the square root: √(1/(ωμσ)).
δ = √(2/(ωμσ)) = √(2/(2π × 109 × 4π × 10−7 × 5.8 × 107)) = 2.09 µm.
This uses f instead of ω = 2πf, which makes δ larger by √(2π).
This is the skin depth at 1 MHz; 1 GHz is 1000 times higher, so δ is √1000 times smaller.
Question 6 of 16 · Communications · 5–8 items

A rectangular pulse of width 2 ms is applied to a system. The first null in its frequency spectrum occurs at approximately what frequency?

Given
  • Pulse width T = 2 ms
FindFirst spectral null (Hz)

Solution

X(f) = AT sinc(fT), nulls at f = n/T

f1 = 1/T = 1/(2 × 10−3 s) = 500 Hz

This is 1/(2T), as if the pulse width were 4 ms; the first null of a pulse of width T sits at 1/T.
A rectangular pulse of width T has a sinc spectrum with nulls at multiples of 1/T: 1/0.002 s = 500 Hz.
This is 2/T, the second null, or the first null of a 1 ms pulse.
This is 2π/T = 3142 rad/s, an angular frequency read as hertz.
Question 7 of 16 · Communications · 5–8 items

A pulse code modulation (PCM) system samples an analog signal at 8 kHz with 8 bits per sample. What is the bit rate of the PCM output?

Given
  • Sampling rate fs = 8 kHz
  • Bits per sample n = 8
FindPCM bit rate (kbps)

Solution

Rb = fs · n

Rb = 8000 samples/s × 8 bits/sample = 64 kbps

This divides the sampling rate by the bits per sample; each sample adds 8 bits, so the rates multiply.
This counts one bit per sample, the sampling rate alone.
This multiplies by a 4 kHz signal bandwidth; the 8 kHz given is already the sampling rate.
Bit rate = fs × n = 8000 samples/s × 8 bits = 64,000 bit/s = 64 kbps.
Question 8 of 16 · Communications · 5–8 items

Four channels, each requiring 4 kHz bandwidth, are multiplexed using frequency division multiplexing (FDM) with 1 kHz guard bands between channels. What is the total bandwidth required?

Given
  • Number of channels N = 4
  • Channel bandwidth B = 4 kHz
  • Guard band = 1 kHz
FindTotal bandwidth (kHz)

Solution

Btotal = N·B + (N − 1)·Bg

Btotal = 4 × 4 kHz + 3 × 1 kHz = 19 kHz

This counts only the four 4 kHz channels and leaves out the guard bands.
This adds only one guard band; four channels in a row need three gaps between them.
Four channels need three guard bands between neighbors: 4 × 4 + 3 × 1 = 19 kHz.
This adds one guard band per channel (four); only the gaps between adjacent channels need a guard band.
Question 9 of 16 · Communications · 5–8 items

An amplitude modulated (AM) signal with 100% modulation carries 1000 W in the carrier. What is the total power in the modulated signal?

Given
  • Modulation index m = 1 (100 %)
  • Carrier power Pc = 1000 W
FindTotal transmitted power Pt (W)

Solution

Pt = Pc(1 + m2/2)

Pt = 1000 W × (1 + 12/2) = 1500 W

This is the power in the two sidebands alone, Pcm2/2; the carrier is still transmitted.
This is the carrier alone; the two sidebands add power on top of it.
This adds only one sideband, Pcm2/4; both sidebands together carry Pcm2/2.
Pt = Pc(1 + m2/2) = 1000 × (1 + 0.5) = 1500 W.
Question 10 of 16 · Communications · 5–8 items

A radio link transmits 1.0 W. The transmit antenna gain is 8 dBi, the receive antenna gain is 10 dBi, the free-space path loss is 100 dB, and the feeder and connector loss at the transmitter is 2 dB. The receiver input noise power is −100 dBm. What are the received signal power and the input signal-to-noise ratio?

Given
  • Transmit power Pt = 1.0 W
  • Transmit antenna gain Gt = 8 dBi
  • Receive antenna gain Gr = 10 dBi
  • Free-space path loss Lfs = 100 dB
  • Feeder loss Lf = 2 dB
  • Receiver noise power N = −100 dBm
FindReceived power (dBm) and SNR (dB)

Solution

Pt = 10 log10(1000 mW/1 mW) = +30 dBm

Pr = Pt + Gt + Gr − Lfs − Lf = 30 + 8 + 10 − 100 − 2 = −54 dBm

SNR = Pr − N = −54 dBm − (−100 dBm) = 46 dB

Pr = 30 + 8 + 10 − 100 − 2 = −54 dBm (1 W = +30 dBm), and SNR = −54 − (−100) = 46 dB.
This takes 1 W as 0 dBm; 0 dBW = 1 W = +30 dBm.
This leaves out the 2 dB feeder loss at the transmitter.
This subtracts the two antenna gains as if they were losses.
Question 11 of 16 · Communications · 5–8 items

A frequency modulation (FM) transmitter has a frequency deviation of 75 kHz and modulates a 15 kHz audio signal. What is the modulation index?

Given
  • Frequency deviation Δf = 75 kHz
  • Modulating frequency fm = 15 kHz
FindModulation index β

Solution

β = Δf/fm = 75 kHz/15 kHz = 5

β = Δf/fm = 75 kHz/15 kHz = 5.
This is Δf − fm in kHz; the index is a ratio, not a difference.
This is Δf + fm in kHz, half of Carson's bandwidth.
This is Carson's bandwidth 2(Δf + fm) in kHz, not the modulation index.
Question 12 of 16 · Communications · 5–8 items

A periodic signal has a fundamental frequency of 100 Hz. In its Fourier series representation, at what frequencies will the harmonics appear?

Solution

A signal with period T0 = 1/f0 contains only frequencies nf0:

fn = n × 100 Hz, n = 1, 2, 3, …

A 50 Hz component would make the period 20 ms, contradicting the 100 Hz fundamental.
Harmonics sit at integer multiples of the fundamental, nf0 with n = 1, 2, 3, …
The 0 Hz term is the DC (average) component, not a harmonic; harmonics start at the fundamental.
Odd harmonics only describe signals with half-wave symmetry, such as a square wave; a general periodic signal has all of them.
Question 13 of 16 · Communications · 5–8 items

A communication system has a signal power of 100 mW and a noise power of 0.1 mW. What is the signal-to-noise ratio (SNR) in decibels?

Given
  • Signal power Ps = 100 mW
  • Noise power Pn = 0.1 mW
FindSNR (dB)

Solution

SNR = 10 log10(Ps/Pn)

SNR = 10 log10(100 mW/0.1 mW) = 10 log10(1000) = 30 dB

This is log10(1000) without the factor 10 in the decibel definition.
This is 10 log10(100), the signal power in milliwatts read as if it were the ratio; the 0.1 mW noise power was never divided in.
SNR = 10 log10(100 mW/0.1 mW) = 10 log10(1000) = 30 dB.
This is 20 log10(1000), the voltage-ratio formula applied to a power ratio.
Question 14 of 16 · Communications · 5–8 items

An 8-bit PCM system has a quantization step size of 0.05 V. What is the full-scale voltage range of the system?

Given
  • Number of bits n = 8
  • Quantization step q = 0.05 V
FindFull-scale voltage range (V)

Solution

q = VFS/2n → VFS = 2n · q

VFS = 28 × 0.05 V = 12.80 V

This is 29 × 0.05, one bit too many.
An n-bit quantizer has 2n levels: 28 × 0.05 V = 256 × 0.05 = 12.80 V.
This is 27 × 0.05, one bit short.
This is n × q = 8 × 0.05: the bit count was used where the number of levels, 2n, belongs.
Question 15 of 16 · Communications · 5–8 items

A wireless provider has channels 5.0 MHz wide and measures an SNR of 20.0 dB on each channel at the receiver. Using the Shannon capacity C = B log2(1 + SNR), what is the theoretical maximum throughput per channel, and what is the minimum number of non-overlapping channels needed for an aggregate throughput of 150 Mbps?

Given
  • Channel bandwidth B = 5.0 MHz
  • SNR = 20.0 dB
  • Required aggregate throughput = 150 Mbps
FindCapacity per channel (Mbps) and minimum number of channels

Solution

SNR = 1020/10 = 100

C = B log2(1 + SNR) = 5.0 MHz × log2(101) = 33.3 Mbps

N = ⌈150/33.3⌉ = ⌈4.5⌉ = 5

The capacity is right, but 150/33.3 = 4.5 was rounded down; four channels carry only 133 Mbps.
This doubles the capacity, mixing the Nyquist 2B signaling rate into the Shannon formula.
SNR = 102 = 100; C = 5.0 MHz × log2(101) = 33.3 Mbps; 150/33.3 = 4.5, rounded up to 5 channels.
This puts the 20 dB value straight into the formula (SNR = 20); it must first be converted to the ratio 100.
Question 16 of 16 · Communications · 5–8 items

A QPSK link carries 1.00 Mbps. The received signal power is −80.0 dBm, and the noise power measured in the symbol bandwidth (0.5 MHz) is −100.0 dBm. What are Eb/N0 and the approximate bit-error rate from BER ≈ Q(√(2Eb/N0))? Use Q(10) ≈ 7.6 × 10−24.

Given
  • Bit rate Rb = 1.00 Mbps (QPSK, symbol rate 0.5 Msymbol/s)
  • Received signal power Pr = −80.0 dBm
  • Noise power N = −100.0 dBm in the symbol bandwidth B = 0.5 MHz
  • Q(10) ≈ 7.6 × 10−24
FindEb/N0 (dB) and bit-error rate

Solution

SNR = −80 dBm − (−100 dBm) = 20 dB = 100

Eb/N0 = SNR · B/Rb = 100 × 0.5 MHz/1.0 Mbps = 50 = 16.99 dB

BER ≈ Q(√(2 × 50)) = Q(10) ≈ 7.6 × 10−24

Eb/N0 is right, but the dB number 16.99 went into Q(√(2Eb/N0)) unconverted; the formula needs the ratio 50.
This sets Eb/N0 equal to the SNR, ignoring the B/Rb = 0.5 factor of QPSK's 2 bits per symbol.
This multiplies the SNR by Rb/B = 2 instead of B/Rb = 0.5, inverting the bandwidth correction.
SNR = −80 − (−100) = 20 dB = 100. With B = Rs = 0.5 MHz, Eb/N0 = 100 × 0.5/1.0 = 50 = 16.99 dB, and BER ≈ Q(√100) = Q(10) ≈ 7.6 × 10−24.

Answer every question to see your score

Modulation and channels

Much of the area is bandwidth and power bookkeeping. FM bandwidth is estimated by Carson’s rule:

BFM ≈ 2(Δf + fm)

Shannon’s limit sets channel capacity: it grows in proportion to bandwidth but only with the logarithm of S/N, so each extra bit per second per hertz costs about twice the power ratio.

In AM the carrier carries no information, so the sidebands are what the modulation index buys; overmodulation (index above 1) distorts the envelope. PCM is sampling plus quantization: bits per sample set both the resolution and the bit rate.

Electromagnetics

A separate area of the specification, 4–6 items, with three subtopics:

  • Electrostatics and magnetostatics: fields of charges and currents, with vector analysis.
  • Electrodynamics: Maxwell’s equations and wave propagation.
  • Transmission lines: high-frequency lines, reflection and matching.

For lines, three relations do most of the work:

Γ = (ZL − Z0)/(ZL + Z0) VSWR = (1 + |Γ|)/(1 − |Γ|) Z0 = √(Zin·ZL) (quarter-wave)

Radiation protection, the safety side of fields, is a subtopic of Ethics and Professional Practice on the FE practice test.

Common traps in communications and fields

  • Bits or bauds. A symbol can carry several bits; bit rate = symbol rate × bits per symbol.
  • SNR in Shannon. The formula takes S/N as a power ratio. Convert from dB first.
  • Carson’s rule inputs. It takes the peak deviation and the highest modulating frequency, not the carrier.
  • Wavelength in a medium. λ = c/f holds in free space; in a dielectric the speed drops by √εr, and so does the wavelength.
  • |Γ| or VSWR. A matched line has Γ = 0 and VSWR = 1; VSWR never falls below 1.

Sources: Haykin, Communication Systems; Hayt and Buck, Engineering Electromagnetics.

Other E&C areas

Communications, quick answers

Is electromagnetics on the FE electrical exam?

Yes, as its own area: 4–6 questions on static fields, Maxwell’s equations, wave propagation and high-frequency transmission lines.

How many communications questions are on the FE electrical exam?

5–8, by the July 2020 specification.

More modulation problems, on your phone

FE Electrical and Computer questions, answers and worked solutions in a study app.

Download on theApp StoreAndroid: practice in the browser Get the FE Electrical and Computer app

Not affiliated with NCEES.