Control Systems is 6–9 items on the E&C specification. It is the feedback half of the transfer-function story: what closing a loop does to stability, accuracy and speed, read from block diagrams, Bode plots and pole locations.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Electrical and Computer
items · Control Systems6–9
problems on this page15
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What the area covers
Four subtopics in the July 2020 specification; the methods after each colon are our reading:
Block diagrams: feedforward and feedback, reduced to one transfer function.
Bode plots: magnitude and phase, gain and phase margins.
Response and stability: open-loop and closed-loop response; Routh-Hurwitz and pole location.
Controller performance: steady-state error, settling time, overshoot.
15 control systems problems
0 / 15 answered · 0 correct
Question 1 of 15 · Control Systems · 6–9 items
A second-order system has a damping ratio of 0.5 and natural frequency of 4 rad/s. What is the percent overshoot for a step input?
This swaps the roles of ζ and √(1 − ζ2) in the exponent: e−π√(1 − ζ2)/ζ.
This drops the square root in the denominator: e−πζ/(1 − ζ2).
PO = 100 e−πζ/√(1 − ζ2) = 100 e−π(0.5)/0.866 = 16.3 %; ωn does not affect overshoot.
This keeps only e−πζ and drops the √(1 − ζ2) term.
Question 2 of 15 · Control Systems · 6–9 items
The open-loop transfer function of a unity-feedback system is G(s) = 20/[s(s + 5)]. What is the velocity error constant Kv?
Given
Open-loop transfer function G(s) = 20/(s(s + 5))
Unity feedback
Find
Velocity error constant Kv (s−1)
Solution
Kv = lims→0 s·G(s)
Kv = lims→0 20/(s + 5) = 20/5 = 4 s−1
This divides the other way, 5/20 = 0.25; Kv is the gain divided by the pole value, 20/5.
Kv = lims→0 sG(s) = 20/(0 + 5) = 4 s−1.
This takes the numerator gain and forgets the (s + 5) factor, which equals 5 at s = 0.
This multiplies 20 by 5 instead of dividing.
Question 3 of 15 · Control Systems · 6–9 items
A Bode plot shows that the gain crossover frequency is 10 rad/s and the phase there is −150°. What is the phase margin?
Given
Gain crossover frequency = 10 rad/s
Phase at gain crossover = −150°
Find
Phase margin (°)
Solution
PM = 180° + ∠GH(jωgc)
PM = 180° + (−150°) = 30°
This is −180° − (−150°), the subtraction done in reverse; the margin is how far the phase sits above −180°, which is +30°.
PM = 180° + ∠G(jωgc) = 180° − 150° = 30°.
This is the magnitude of the phase itself, not its distance from −180°.
This is 180° − (−150°): the phase was subtracted instead of added.
Question 4 of 15 · Control Systems · 6–9 items
For the characteristic polynomial s4 + 2s3 + (3 + K)s2 + 4s + 5 = 0, what range of the real parameter K gives all roots negative real parts? Use the Routh-Hurwitz criterion.
Unity DC gain needs a pole at the origin (Type 1); this loop is Type 0 and keeps a steady-state error.
This is the open-loop DC gain G(0); closing the loop divides it by 1 + G(0).
This is the numerator constant alone, ignoring the denominator value of 25 at s = 0.
Question 6 of 15 · Control Systems · 6–9 items
A unity-feedback loop has the open-loop transfer function G(s) = 10/[s(s − 2)]. What is the steady-state error for a unit ramp input?
Solution
1 + G(s) = 0 → s2 − 2s + 10 = 0 → s = 1 ± j3
Right-half-plane roots: the closed loop is unstable and ess does not exist; check stability before using Kv.
The error constant formula assumes a stable closed loop; here it gives a meaningless number.
A pole at the origin is what gives Type 1 its finite ramp error; it does not prevent a steady state.
Dropping the sign does not rescue the formula: the final value theorem needs a stable closed loop first.
A stable Type 1 loop follows a ramp with a finite error 1/Kv; zero ramp error needs Type 2.
The characteristic equation s2 − 2s + 10 = 0 has roots 1 ± j3 in the right half-plane, so the error grows and the final value theorem does not apply.
Question 7 of 15 · Control Systems · 6–9 items
A measured disturbance D reaches the output through Gd(s) = 2/(5s + 1). A feedforward controller Gff sends D to the plant input, and the plant is Gp(s) = 4/(10s + 1). What Gff cancels the disturbance exactly?
Consider the state-space system ẋ = Ax + Bu with A = [[0, 1], [−2, −3]] and B = [0; 1]. (a) Is the system controllable? (b) If so, find the state feedback u = −Kx, K = [k1 k2], that places the closed-loop poles at s = −2 ± j1 (characteristic polynomial s2 + 4s + 5).
The controllability matrix [B AB] = [[0, 1], [1, −3]] has rank 2; A − BK gives s2 + (3 + k2)s + (2 + k1), so k2 = 1 and k1 = 3.
[B AB] has determinant −1 ≠ 0, so the pair is controllable and a gain exists.
The two gains are swapped: k1 sets the constant term and k2 the s term.
The sign of k1 is flipped; with u = −Kx the constant term becomes 2 + k1 = 5, so k1 = +3.
Question 9 of 15 · Control Systems · 6–9 items
For the open-loop transfer function G(s) = K/[s(s + 1)(s + 3)(s + 6)] with unity negative feedback, what are the number of root-locus asymptotes, their angles, and their centroid?
Given
Open-loop G(s) = K/[s(s + 1)(s + 3)(s + 6)]
Feedback = unity negative
Find
Number, angles and centroid of asymptotes
Solution
n − m = 4 − 0 = 4 asymptotes
θk = (2k + 1)·180°/4 = 45°, 135°, 225°, 315°
σa = (∑p − ∑z)/(n − m) = (0 − 1 − 3 − 6)/4 = −2.5
This forgets the pole at s = 0, leaving three poles; G(s) has four poles and no zeros.
The centroid here is the sum of the poles, not divided by n − m = 4; the angles are also off.
This uses k·360°/4; asymptote angles are (2k + 1)·180°/(n − m), which never include 0° for negative feedback.
Question 10 of 15 · Control Systems · 6–9 items
Design a full-order state observer for ẋ = Ax + Bu, y = Cx with A = [[0, 1], [−2, −3]], B = [0; 1], and C = [1, 0]. What observer gain L (a column vector) places the observer error eigenvalues at s = −5 and s = −6?
These are the desired eigenvalue magnitudes, not gains; L comes from matching the polynomial coefficients.
Question 11 of 15 · Control Systems · 6–9 items
A control system has a settling time (2% criterion) of 2 seconds. If the system is approximated as first-order, what is the time constant?
Given
Settling time Ts = 2 s (2% criterion)
Find
Time constant τ (s)
Solution
Ts (2 %) ≈ 4τ
τ = 2 s/4 = 0.50 s
This multiplies by 4 instead of dividing.
This is the settling time itself; the time constant is a quarter of it.
This divides by 3, the factor for the 5 % settling criterion; the stem uses 2 %.
For a first-order system, 2 % settling takes about 4τ: τ = Ts/4 = 2/4 = 0.50 s.
Question 12 of 15 · Control Systems · 6–9 items
For the open-loop transfer function G(s) = K/[s(s + 2)(s + 6)] with unity negative feedback, where is the real-axis breakaway point between the poles at 0 and −2, and what is K at that point?
Given
Open-loop G(s) = K/[s(s + 2)(s + 6)]
Feedback = unity negative
Find
Breakaway point s and gain K there
Solution
K = −s(s + 2)(s + 6) = −(s3 + 8s2 + 12s)
dK/ds = 0 → 3s2 + 16s + 12 = 0 → s = −0.903 (on the locus) or −4.431
K = 0.903 × 1.097 × 5.097 = 5.05
This is the other root of dK/ds = 0, but the real axis between −2 and −6 is not on the locus.
This takes the midpoint of the poles at 0 and −2; with a third pole at −6 pulling on it, the breakaway point shifts toward the origin.
The point is right, but K was taken as s(s + 2)(s + 6) without the minus sign from 1 + KG = 0.
dK/ds = 0 gives 3s2 + 16s + 12 = 0, s = −0.903 or −4.431; only −0.903 lies on the locus segment between 0 and −2, and K = 0.903 × 1.097 × 5.097 = 5.05.
Question 13 of 15 · Control Systems · 6–9 items
A second-order underdamped system has poles at s = −2 ± j3. What is the natural frequency ωn?
Given
Poles s = −2 ± j3
Find
Natural frequency ωn (rad/s)
Solution
ωn = √(σ2 + ωd2)
ωn = √(22 + 32) = 3.61 rad/s
This is σ = ζωn, the real part of the poles.
This is the damped frequency ωd, the imaginary part, which is smaller than ωn.
ωn is the distance of the poles from the origin: √(22 + 32) = √13 = 3.61 rad/s.
Question 14 of 15 · Control Systems · 6–9 items
For the unity-feedback system with open-loop G(s) = K/[s(s + 4)(s + 6)], is there a real positive gain K that places a pair of closed-loop poles at s = −2.5 ± j4.330 (ζ = 0.5, ωn = 5 rad/s)?
Given
Open-loop transfer function G(s) = K/(s(s + 4)(s + 6))
K = −s(s + 4)(s + 6) ≈ 60.0 + j112.6 (complex) → no real positive K
Discarding the imaginary part of a complex K is not allowed; a valid gain must be real and positive.
None of the open-loop poles (0, −4, −6) is at −2.5 ± j4.33, so the denominator is not zero there.
The magnitude condition gives |s||s + 4||s + 6| ≈ 128, but it applies only after the angle condition is met.
The angles from s, s + 4 and s + 6 add to about 242°, not an odd multiple of 180°, so K = −s(s + 4)(s + 6) ≈ 60 + j113 is complex.
Question 15 of 15 · Control Systems · 6–9 items
In a PID controller, which component provides a response proportional to the rate of change of the error signal?
Solution
u(t) = Kpe + Ki∫e dt + Kd de/dt
The I term responds to the accumulated (integrated) error over time.
Feed-forward acts on the reference or a measured disturbance, not on the error; it is not one of the three PID terms.
The D term outputs Kd·de/dt, proportional to the error's rate of change.
The P term responds to the present size of the error, not its rate of change.
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The relations that do most of the work
A negative-feedback loop with forward path G and feedback H closes to
T(s) = G(s)/(1 + G(s)H(s))
and its stability is decided by the roots of 1 + GH = 0. For a second-order closed loop, the time specs follow from ζ and ωn:
%OS = 100·e−ζπ/√(1 − ζ2) ts ≈ 4/(ζωn) (2% band)
Steady-state error depends on the system type, the number of free integrators (poles at s = 0) in the open loop, and on the error constants Kp, Kv and Ka. For Routh-Hurwitz, count the sign changes in the first column: each one is a right-half-plane pole.
Common traps in control systems
Open loop or closed loop. Margins and error constants come from the open-loop G·H; step response and overshoot belong to the closed loop.
System type. A type 0 system has a finite error to a step and infinite error to a ramp; each added integrator moves that one input up.
Gain margin units. A margin read as a ratio and one read in dB are different numbers: GMdB = 20·log(GM).
Where each margin is read. Phase margin at the gain crossover (|GH| = 1), gain margin at the phase crossover (−180°).
Overshoot depends on ζ alone. Changing ωn speeds the response up but leaves percent overshoot where it was.
Sources: Ogata, Modern Control Engineering; Nise, Control Systems Engineering.
How it connects to other areas
Transfer functions, poles and Laplace transforms are set up in Linear Systems. In this hub, control theory as such (block-diagram algebra, Routh, margins, error by system type) is owned here. Process loops, such as tuning a PID on a chemical process, cascade control and valve fail position, are the process control part of FE Chemical; sensors and measurement uncertainty are FE Mechanical measurements.
How many control systems questions are on the FE electrical exam?
6–9 questions in the current outline, a little more than Linear Systems alone.
What is the difference between gain margin and phase margin?
Gain margin is how much the loop gain can rise before instability, read at the frequency where the phase is −180°; phase margin is the extra phase lag the loop can take, read where the gain is 1 (0 dB).
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