FE and PE Exam

FE Electrical and Computer · areas 7 and 8

Linear systems and signal processing

Two areas of the specification share this page because they share the mathematics: Linear Systems (5–8 items) and Signal Processing (5–8 items). Together that is 10 to 16 questions, more than Power Systems alone.

Updated Oct 29, 2026 · Not affiliated with NCEES.

What Linear Systems covers

Four subtopics in the July 2020 specification:

  • Frequency and transient response: how a system reacts to a step and to a sinusoid.
  • Resonance: RLC circuits, quality factor and bandwidth.
  • Laplace transforms: forward and inverse, including partial fractions.
  • Transfer functions: poles, zeros, gain and stability.

18 linear systems and signal processing problems

0 / 18 answered · 0 correct
Question 1 of 18 · Signal Processing · 5–8 items

A digital filter is described by y[n] = 0.5x[n] + 0.5x[n − 1]. What is its transfer function H(z)?

Solution

Y(z) = 0.5X(z) + 0.5z−1X(z) → H(z) = Y/X = 0.5(1 + z−1)

This puts the delay in the denominator, which describes a recursive filter with y[n − 1] feedback.
This drops the 0.5 coefficient on x[n].
This flips the sign of the delayed term; the difference equation adds x[n − 1].
Each one-sample delay becomes z−1: Y = 0.5X + 0.5z−1X, so H(z) = 0.5(1 + z−1).
Question 2 of 18 · Signal Processing · 5–8 items

A digital filter has the transfer function H(z) = 1/(1 − 0.8·z−1). Where is the pole located in the z-plane?

Given
  • H(z) = 1/(1 − 0.8z−1)
FindPole location in the z-plane

Solution

H(z) = 1/(1 − 0.8z−1) = z/(z − 0.8) → pole at z = 0.8

This solves 1 − 0.8z = 0, treating z−1 as z.
The sign is flipped: 1 − 0.8z−1 = 0 gives z = +0.8.
H(z) = z/(z − 0.8), so the pole is at z = 0.8, inside the unit circle.
H(z) = z/(z − 0.8) has a zero at z = 0; the pole comes from the denominator.
Question 3 of 18 · Signal Processing · 5–8 items

A signal is sampled at 8 kHz. What is the highest frequency that can be represented without aliasing?

Given
  • Sampling rate fs = 8 kHz
FindHighest unaliased frequency (kHz)

Solution

fmax = fs/2 = 8 kHz/2 = 4 kHz

This is fs/4, half the actual limit.
The Nyquist frequency is fs/2 = 8/2 = 4 kHz.
This is the sampling rate itself; a component at fs aliases to DC.
This is twice the sampling rate, far above the Nyquist limit.
Question 4 of 18 · Signal Processing · 5–8 items

A low-pass Butterworth filter has a cutoff frequency of 1 kHz and rolls off at 40 dB/decade. What is its order?

Given
  • Cutoff frequency fc = 1 kHz
  • Roll-off = 40 dB/decade
FindFilter order

Solution

roll-off = 20n dB/decade → n = 40/20 = 2

This divides 40 by 6, the per-pole slope in dB per octave, while the slope given is per decade.
This divides by 10 dB/decade; a single pole gives 20 dB/decade.
Each order adds 20 dB/decade of roll-off: 40/20 = 2.
A first-order filter rolls off at 20 dB/decade.
Question 5 of 18 · Signal Processing · 5–8 items

Which classic analog filter family is defined by a maximally flat passband magnitude response?

Solution

Butterworth: |H(jω)|2 = 1/(1 + (ω/ωc)2n), maximally flat passband.

Type I trades passband ripple for a steeper roll-off.
Elliptic filters have ripple in both the passband and the stopband in exchange for the sharpest transition.
Bessel filters are designed for maximally flat group delay (linear phase); their magnitude droops earlier.
The Butterworth response is maximally flat: no ripple, and its magnitude derivatives at ω = 0 vanish to the highest order.
Question 6 of 18 · Signal Processing · 5–8 items

A first-order low-pass filter has a cutoff frequency of 2 kHz. What is the magnitude response at 2 kHz?

Solution

|H(jf)| = 1/√(1 + (f/fc)2)

|H(jfc)| = 1/√2 = 0.707

This is 1 − 0.707, the drop from the passband gain, not the magnitude itself.
This is 1/√(1 + 22), the magnitude at twice the cutoff, as if f/fc were 2.
This is the −6 dB point; the cutoff is defined at −3 dB.
At f = fc, |H| = 1/√(1 + 1) = 0.707 (−3 dB).
Question 7 of 18 · Signal Processing · 5–8 items

A signal has a bandwidth of 5 kHz. If it is sampled at exactly the Nyquist rate, what is the spacing between spectral replicas in the frequency domain?

Given
  • Signal bandwidth B = 5 kHz
  • Sampling at the Nyquist rate
FindSpacing of spectral replicas (kHz)

Solution

fs = 2B = 10 kHz; replicas at nfs → spacing 10 kHz

This is the bandwidth, which is also fs/2, the folding frequency; replicas repeat every fs.
At the Nyquist rate fs = 2B = 10 kHz, and the spectrum repeats every fs.
This is fs + B, the upper edge of the first replica, not the spacing between replicas.
This is 2fs, the spacing of every second replica.
Question 8 of 18 · Signal Processing · 5–8 items

A signal band-limited to 6 kHz is sampled at 32 kHz and then decimated by 4 (one of every four samples kept). What is the maximum cutoff of the anti-aliasing low-pass filter applied before decimation?

Given
  • Signal bandwidth = 6 kHz
  • Sampling rate fs = 32 kHz
  • Decimation factor M = 4
FindMaximum anti-aliasing filter cutoff (kHz)

Solution

fs,new = fs/M = 32 kHz/4 = 8 kHz

fc,max = fs,new/2 = 4 kHz

After decimation by 4 the rate is 32/4 = 8 kHz, so components must stay below 8/2 = 4 kHz.
This keeps the full 6 kHz signal band, which would alias at the 8 kHz output rate.
This is the new sampling rate itself; the cutoff must be half of it.
This is the Nyquist frequency of the original 32 kHz rate; after decimation the rate is only 8 kHz.
Question 9 of 18 · Signal Processing · 5–8 items

A signal sampled at 8 kHz is analyzed with a 64-point DFT after windowing. What are the approximate null-to-null main-lobe widths for a rectangular window and for a Hamming window?

Given
  • Sampling rate fs = 8 kHz
  • DFT length N = 64
FindMain-lobe width, rectangular and Hamming (Hz)

Solution

Δf = fs/N = 8000/64 = 125 Hz

rectangular: 2Δf = 250 Hz; Hamming: 4Δf = 500 Hz

Bin spacing fs/N = 8000/64 = 125 Hz; the main lobe spans 2 bins for the rectangular window (250 Hz) and 4 bins for Hamming (500 Hz).
These are half-widths (fs/N and 2fs/N); the null-to-null width is twice that.
These are twice the null-to-null widths, as if counting 4 and 8 bins.
The Hamming window lowers sidelobes at the cost of a main lobe twice as wide as the rectangular one.
Question 10 of 18 · Linear Systems · 5–8 items

A system has the transfer function H(s) = 1/(s2 + 2s + 5). What is the damping ratio of this second-order system?

Given
  • Transfer function H(s) = 1/(s2 + 2s + 5)
FindDamping ratio ζ

Solution

s2 + 2ζωns + ωn2 = s2 + 2s + 5

ωn = √5 = 2.236 rad/s, ζ = 2/(2ωn) = 0.447

This divides 2 by 2ωn2 = 10; the coefficient equals 2ζωn with ωn = √5.
This divides 2 by ωn2 = 5, dropping both the 2 and the square root.
ωn2 = 5 → ωn = 2.236 rad/s; 2ζωn = 2 → ζ = 2/(2 × 2.236) = 0.447.
This is 2/√5: the factor 2 in 2ζωn was left out.
Question 11 of 18 · Linear Systems · 5–8 items

A series RLC circuit has R = 60 Ω, L = 0.2 H, and C = 50 µF. What is the damped natural frequency ωd of its underdamped step response?

Given
  • Resistance R = 60 Ω
  • Inductance L = 0.2 H
  • Capacitance C = 50 µF
FindDamped natural frequency ωd (rad/s)

Solution

ω0 = 1/√(LC) = 1/√(0.2 H × 50 × 10−6 F) = 316.2 rad/s, α = R/(2L) = 60/(2 × 0.2) = 150 s−1

ωd = √(ω02 − α2) = ω0√(1 − ζ2) = √(316.22 − 1502) = 278.4 rad/s

This is the decay rate α = R/(2L), the real part of the roots, not the oscillation frequency.
ω0 = 1/√(LC) = 316.2 rad/s and α = R/(2L) = 150 s−1; ωd = √(ω02 − α2) = √(100,000 − 22,500) = 278.4 rad/s.
This is the undamped frequency 1/√(LC); damping lowers the ringing frequency below it.
This adds α2 under the root instead of subtracting it.
Question 12 of 18 · Linear Systems · 5–8 items

A series RLC circuit has R = 10 Ω, L = 50 mH, and C = 100 µF. What is the resonant frequency?

Given
  • Resistance R = 10 Ω
  • Inductance L = 50 mH
  • Capacitance C = 100 µF
FindResonant frequency f0 (Hz)

Solution

f0 = 1/(2π√(LC))

f0 = 1/(2π√(0.050 H × 100 × 10−6 F)) = 71.2 Hz

f0 = 1/(2π√(0.050 × 100 × 10−6)) = 1/(2π × 2.236 × 10−3) = 71.2 Hz.
This is 1/(π√(LC)), dropping the 2 in 2π.
This is ω0 = 1/√(LC) in rad/s, not hertz.
This is 1/(2πLC), with the square root left off.
Question 13 of 18 · Linear Systems · 5–8 items

Given Y(s) = 5/[s(s + 2)2], what is the time-domain function y(t) for t ≥ 0?

Given
  • Y(s) = 5/(s(s + 2)2)
FindTime response y(t)

Solution

Y(s) = A/s + B/(s + 2) + C/(s + 2)2

A = 5/(s + 2)2|s=0 = 5/4, C = 5/s|s=−2 = −5/2, B = −A = −5/4

y(t) = 5/4 − (5/4)e−2t − (5/2)te−2t, t ≥ 0

The e−2t coefficient must be −5/4 so that y(0) = 0; −5/2 gives y(0) = −5/4.
Partial fractions: 5/4·(1/s) − 5/4·1/(s + 2) − 5/2·1/(s + 2)2; the last term inverts to te−2t.
This drops the step term 5/4 from the pole at s = 0, so y(t) would decay to zero instead of settling at 5/4.
The repeated-pole coefficient is 5/(s)|s=−2 = −5/2, not −5/4.
Question 14 of 18 · Linear Systems · 5–8 items

A linear system has the transfer function H(s) = 5/(s + 10). What is the DC gain of this system?

Given
  • Transfer function H(s) = 5/(s + 10)
FindDC gain

Solution

H(0) = 5/(0 + 10) = 0.5

This is the pole magnitude, which sets the bandwidth, not the gain.
This is the numerator alone; at s = 0 the denominator equals 10.
This is 10/5, the ratio inverted; H(0) puts the 5 over the 10.
DC gain = H(0) = 5/(0 + 10) = 0.5.
Question 15 of 18 · Linear Systems · 5–8 items

A first-order system has a transfer function H(s) = 20/(s + 4). What is the time constant of this system?

Given
  • Transfer function H(s) = 20/(s + 4)
FindTime constant τ (s)

Solution

H(s) = K/(s + a) → τ = 1/a

τ = 1/4 = 0.25 s

This is the DC gain 20/4 = 5, which is dimensionless, not a time.
This is the pole magnitude a = 4 s−1; the time constant is its reciprocal.
Write H(s) = 5/(s/4 + 1): the time constant is 1/4 = 0.25 s.
This is 1/20, using the gain constant in place of the pole.
Question 16 of 18 · Linear Systems · 5–8 items

A system has a transfer function with poles at s = −2 + j3 and s = −2 − j3. Is this system stable?

Solution

Re(s) = −2 < 0 for both poles → asymptotically stable.

Both poles have real part −2 < 0, so the response decays: a damped oscillation at 3 rad/s.
Instability needs a pole with a positive real part; both poles here have real part −2.
For an LTI system the pole locations are enough to decide stability.
Marginal stability means simple poles on the jω axis; these poles are to the left of it.
Question 17 of 18 · Linear Systems · 5–8 items

The transfer function of a system is H(s) = (s + 3)/(s2 + 3s + 2). What are the poles of this system?

Solution

s2 + 3s + 2 = (s + 1)(s + 2) = 0 → s = −1, −2

The zero is at s = −3; there is no pole-zero cancellation.

This counts the zero at −3 as a pole; poles come from the denominator only.
s2 + 3s + 2 = (s + 1)(s + 2), so the poles are −1 and −2; the numerator root −3 is a zero.
These are the roots of s2 + 3s with the constant term dropped.
The signs are flipped: s + 1 = 0 gives s = −1.
Question 18 of 18 · Linear Systems · 5–8 items

A series RLC circuit at resonance has R = 50 Ω, L = 10 mH, and C = 10 µF. What is the quality factor Q?

Given
  • Resistance R = 50 Ω
  • Inductance L = 10 mH
  • Capacitance C = 10 µF
FindQuality factor Q

Solution

Q = ω0L/R = (1/R)√(L/C)

Q = (1/50 Ω)√(0.010 H/10 × 10−6 F) = 31.6/50 = 0.632

Q = (1/R)√(L/C) = (1/50)√(0.010/10 × 10−6) = 31.6/50 = 0.632.
This is R√(C/L) = 1/Q, the reciprocal of the quality factor.
This is L/(RC) = 0.010/(50 × 10 × 10−6), a mix of the series and parallel forms that is not even dimensionless.
This is √(L/C), the characteristic impedance in ohms, before dividing by R.

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Reading a transfer function

Start from H(s) and factor the denominator: the poles say whether the system is stable (all in the left half plane), how fast it settles (the real part) and whether it rings (an imaginary part). For a second-order system, compare with the standard form:

H(s) = K·ωn2/(s2 + 2ζωns + ωn2)

and read ζ and ωn straight off the coefficients. Inverse Laplace problems are partial fractions plus a transform table; repeated poles produce t·e−at terms that are easy to drop.

Signal processing

Signal Processing is its own area in the specification, 5–8 items, with three subtopics:

  • Sampling: the Nyquist theorem and aliasing.
  • Analog filters: low-pass, high-pass, band-pass; Butterworth, Chebyshev and the like by passband shape and roll-off.
  • Digital filters: difference equations and Z-transforms; FIR filters have no feedback terms, IIR filters do.

The bridge between the two halves of this page is z = esT: the left half of the s-plane maps inside the unit circle, so a digital filter is stable when all its poles lie inside |z| = 1.

Common traps in linear systems and signals

  • Natural or damped frequency. A ringing step response oscillates at ωd = ωn√(1 − ζ2), not at ωn.
  • Final-value theorem. It gives the steady state only if the system is stable; on an unstable system it returns a number that means nothing.
  • Nyquist rate or Nyquist frequency. The rate is the minimum sampling rate, twice the highest signal frequency; the frequency is half the sampling rate, the highest one you can represent.
  • Aliasing. A tone above half the sampling rate does not vanish; it folds back and shows up at |f − k·fs|.
  • Decibels. A voltage ratio takes 20·log; a power ratio takes 10·log.
  • Hz or rad/s. Cutoff and resonant frequencies come out of formulas in rad/s; divide by 2π before comparing with a frequency in hertz.

Sources: Oppenheim and Willsky, Signals and Systems; the transform tables in any linear systems text.

Other E&C areas

Linear systems and signals, quick answers

Is signal processing on the FE electrical exam?

Yes, as its own area: 5–8 questions on sampling, analog filters and digital filters.

Do I need Z-transforms for the FE?

Yes. The specification lists them under digital filters, together with difference equations.

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