Control Engineering 1: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

1Formulas and methods

Stability: Routh–Hurwitz & root locus

Stability
all roots of the characteristic polynomial in the left half-plane
Routh array, s¹ entry of a cubic
b1 = (a2a1 − a3a0)/a2
Right half-plane roots
number of sign changes in the first column
Cubic s³ + a₂s² + a₁s + a₀
stable when all a > 0 and a₂·a₁ > a₀
K/(s(s + a)(s + b))
Kmax = ab(a + b), ω = √(ab)
Characteristic equation
1 + K·G(s) = 0
Asymptotes
σa = (Σp − Σz)/(n − m), θ = (2q + 1)·180°/(n − m)
Break-away and break-in points
dK/ds = 0
Magnitude condition
K = 1/|G(s)|
Angle of departure
θd = 180° − Σ∠(from other poles) + Σ∠(from zeros)

Method

  1. Write the closed-loop characteristic polynomial: the denominator of 1 + K·G(s) = 0, multiplied out.
  2. Put the coefficients alternately into the first two rows of the Routh array and build the remaining rows.
  3. Count the sign changes in the first column: each one is a pole in the right half-plane. With K in the array, demand that every first-column entry stays positive.
  4. For a root locus, mark the poles and zeros, the real-axis segments, the asymptotes from the centroid and the break points where dK/ds = 0. Use K = 1/|G(s)| for the gain at a point.

Steady-state error & PID control

System type
number of poles of G(s) at s = 0
Error constants
Kp = lim G(s), Kv = lim s·G(s), Ka = lim s²·G(s) (s → 0)
Step A
ess = A/(1 + Kp)
Ramp A·t
ess = A/Kv
Parabola A·t²/2
ess = A/Ka
PID controller
C(s) = Kp + Ki/s + Kd·s
Second-order target
s² + 2ζωns + ωn²
Overshoot and timing
%OS = 100·e−ζπ/√(1−ζ²), tp = π/ωd, ts ≈ 4/(ζωn)

Method

  1. Count the poles of the open loop C(s)·G(s) at s = 0: that is the system type.
  2. Work out the error constant that matches the input: Kp for a step, Kv for a ramp, Ka for a parabola.
  3. Apply ess = A/(1 + Kp), A/Kv or A/Ka, or solve it for the gain you need.
  4. For a damping or speed specification, write the closed-loop characteristic polynomial and match it with s² + 2ζω_n·s + ω_n².

2Practice problems

1.Laplace transformsMedium
t (s)f(t)t = 0.5f(0.5) = ?

Find f(t) for F(s) = 60/(s(s + 2)(s + 5)) by partial fractions, and evaluate f(0.5).

F(s) = 60/(s(s + 2)(s + 5))t = 0.5 s
f(0.5) =
2.Laplace transformsHard
ReIm5

A system is described by the differential equation y″ + 6y′ + 34y = u′ + 3u, with input u(t) and output y(t). Find its transfer function G(s) = Y(s)/U(s) (zero initial conditions) and evaluate the impulse response g(t) at t = 0.25 s. Find g(0.25).

y″ + 6y′ + 34y = u′ + 3ut = 0.25 s
g(0.25) =
3.Block diagramsMedium
R(s)+18K+10s + 2G(s)C(s)3H₁−−

The plant G(s) = 10/(s + 2) has an inner (minor) negative feedback loop with gain H₁ = 3. A controller gain K = 18 and an outer unity negative feedback loop close the system. Find the closed-loop pole p.

G(s) = 10/(s + 2)H₁ = 3K = 18
p =
4.Block diagramsHard
R(s)+Kcontroller+D(s)+5s + 10plantC(s)−

A proportional controller K drives the plant G(s) = 5/(s + 10) in a unity negative feedback loop. A disturbance step of size 0.5 enters at the plant input while R = 0. Find the smallest gain K that keeps the steady-state output at or below 0.0125.

G(s) = 5/(s + 10)D = 0.5css ≤ 0.0125
K =
5.Step responseMedium
t (s)y(t)2?

A system has the transfer function G(s) = 168/(s² + 15s + 84) and receives a unit step input. Find the percent overshoot %OS of the step response.

G(s) = 168/(s² + 15s + 84)
%OS = %
6.Step responseHard
R(s)+300s(s + 8)G(s)C(s)−

A unity negative feedback loop has G(s) = 300/(s(s + 8)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 300/(s(s + 8))
ts = s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1(s + 3)(s + 4)(s + 5)C(s)−

A unity-feedback loop has L(s) = K/((s + 3)(s + 4)(s + 5)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/((s + 3)(s + 4)(s + 5))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−71j1j2−2

For the loop L(s) = K/(s(s + 5)(s + 6)), find the gain K that puts a closed-loop pole at s = −2.

L(s) = K/(s(s + 5)(s + 6))s = −2
K =
9.Steady-state error & PID controlMedium
R(s)+K(s + 4)(s + 10)C(s)−

A unity-feedback loop has G(s) = K/((s + 4)(s + 10)). Find the gain K that gives a steady-state error of 25 % for a step input.

G(s) = K/((s + 4)(s + 10))ess = 25 %
K =
10.Steady-state error & PID controlHard
r(t) = 1t+1 + 2/s4(s + 5)(s + 10)C(s)−

A PI controller C(s) = 1 + 2/s drives the plant G(s) = 4/((s + 5)(s + 10)) in a unity-feedback loop. Find the steady-state error ess for the ramp input r(t) = 1t.

C(s) = 1 + 2/sG(s) = 4/((s + 5)(s + 10))r(t) = 1t
ess =
11.Bode plots & marginsMedium
vin​L = 5 mHC = 10 µFR = 2 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 2 Ω, with L = 5 mH and C = 10 µF. Find the −3 dB bandwidth BW in Hz.

R = 2 ΩL = 5 mHC = 10 µF
BW = Hz
12.Bode plots & marginsHard
|H| (dB)0.010.010.10.1111010100100−80−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 1.2/(s(s + 2)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 1.2/(s(s + 2))
ωc = rad/s

3Solutions

1.Laplace transforms

Answer: f(0.5) = 2.65

Laplace transform: F(s) = ∫₀∞ f(t)·e−st dt turns derivatives into multiplications by s: ℒ{f′} = sF(s) − f(0), ℒ{f″} = s²F(s) − s·f(0) − f′(0). Standard pairs: 1 ↔ 1/s, e−at ↔ 1/(s + a), t·e−at ↔ 1/(s + a)², sin ωt ↔ ω/(s² + ω²), cos ωt ↔ s/(s² + ω²), and a shift s → s + σ multiplies the time function by e−σt. To invert a rational F(s), split it into partial fractions: the residue of a simple pole p is A = [(s − p)·F(s)] at s = p. Final value theorem: f(∞) = lims→0 s·F(s), valid only when all poles of s·F(s) lie in the left half-plane. Initial value theorem: f(0⁺) = lims→∞ s·F(s).

Partial fractions

F(s) = A/s + B/(s + 2) + C/(s + 5)
A = 60/(2·5) = 6
B = 60/((−2)·3) = −10
C = 60/((−5)·(−3)) = 4

Inverse transform

f(t) = 6 − 10·e−2t + 4·e−5t

Evaluate

f(0.5) = 6 − 10·e−2·0.5 + 4·e−5·0.5 = 2.65

Answer

f(0.5) = 2.65
2.Laplace transforms

Answer: g(0.25) = 0.1489

Laplace transform: F(s) = ∫₀∞ f(t)·e−st dt turns derivatives into multiplications by s: ℒ{f′} = sF(s) − f(0), ℒ{f″} = s²F(s) − s·f(0) − f′(0). Standard pairs: 1 ↔ 1/s, e−at ↔ 1/(s + a), t·e−at ↔ 1/(s + a)², sin ωt ↔ ω/(s² + ω²), cos ωt ↔ s/(s² + ω²), and a shift s → s + σ multiplies the time function by e−σt. To invert a rational F(s), split it into partial fractions: the residue of a simple pole p is A = [(s − p)·F(s)] at s = p. Final value theorem: f(∞) = lims→0 s·F(s), valid only when all poles of s·F(s) lie in the left half-plane. Initial value theorem: f(0⁺) = lims→∞ s·F(s).

Transfer function

G(s) = Y(s)/U(s) = (s + 3)/(s² + 6s + 34)

Partial fractions

s² + 6s + 34 = (s + 3)² + 5²
G(s) = 1·(s + 3)/((s + 3)² + 5²) + (0/5)·5/((s + 3)² + 5²)

Impulse response

g(t) = e−3t·(cos 5t)
g(0.25) = 0.1489

Answer

g(0.25) = 0.1489
3.Block diagrams

Answer: p = −212

Block-diagram algebra: blocks in series multiply (G₁·G₂), blocks in parallel add (G₁ + G₂), and a loop with forward path G and feedback path H reduces to T = G/(1 + G·H) for negative feedback and G/(1 − G·H) for positive feedback. Reduce inner loops first, then work outwards. The closed-loop poles are the roots of the characteristic equation 1 + G(s)H(s) = 0. For a stable loop, a step input of size A gives the steady-state output A·T(0); with unity feedback the steady-state error to a step is A/(1 + Kp) with Kp = G(0).

Inner loop

Gᵢ = G/(1 + G·H₁) = 10/(s + 2 + 10·3) = 10/(s + 32)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 180/(s + 32 + 180) = 180/(s + 212)

Result

p = −212

Answer

p = −212
4.Block diagrams

Answer: K = 38

Block-diagram algebra: blocks in series multiply (G₁·G₂), blocks in parallel add (G₁ + G₂), and a loop with forward path G and feedback path H reduces to T = G/(1 + G·H) for negative feedback and G/(1 − G·H) for positive feedback. Reduce inner loops first, then work outwards. The closed-loop poles are the roots of the characteristic equation 1 + G(s)H(s) = 0. For a stable loop, a step input of size A gives the steady-state output A·T(0); with unity feedback the steady-state error to a step is A/(1 + Kp) with Kp = G(0).

Disturbance response

css = D·G(0)/(1 + K·G(0)) = 0.5·5/(10 + 5K)

Solve for K

10 + 5K = 0.5·5/0.0125 = 200
K = (200 − 10)/5 = 38

Answer

K = 38
5.Step response

Answer: %OS = 1.142 %

Step response: a first-order system G(s) = K/(τs + 1) answers a step of size U with y(t) = K·U·(1 − e−t/τ): it reaches 63.2 % of the final value K·U after one time constant τ and 98 % after 4τ. A second-order system G(s) = K·ωₙ²/(s² + 2ζωₙs + ωₙ²) has the poles s = −ζωₙ ± jωd with ωd = ωₙ√(1 − ζ²). For 0 < ζ < 1 the step response overshoots: percent overshoot %OS = 100·e−ζπ/√(1−ζ²), peak time tp = π/ωd, and the 2 % settling time ts ≈ 4/(ζωₙ). Compare the denominator with s² + 2ζωₙs + ωₙ² to read off ωₙ and ζ. For a mass–spring–damper ωₙ = √(k/m) and ζ = c/(2√(km)); for a series RLC circuit ωₙ = 1/√(LC) and ζ = (R/2)·√(C/L).

Compare with the standard form

The denominator must match s² + 2ζωₙs + ωₙ².

ωₙ² = 84 → ωₙ = 9.165 rad/s
2ζωₙ = 15 → ζ = 15/(2·9.165) = 0.8183

Result

%OS = 100·e−ζπ/√(1−ζ²) = 100·e−4.473 = 1.142 %

Answer

%OS = 1.142 %
6.Step response

Answer: ts = 1 s

Step response: a first-order system G(s) = K/(τs + 1) answers a step of size U with y(t) = K·U·(1 − e−t/τ): it reaches 63.2 % of the final value K·U after one time constant τ and 98 % after 4τ. A second-order system G(s) = K·ωₙ²/(s² + 2ζωₙs + ωₙ²) has the poles s = −ζωₙ ± jωd with ωd = ωₙ√(1 − ζ²). For 0 < ζ < 1 the step response overshoots: percent overshoot %OS = 100·e−ζπ/√(1−ζ²), peak time tp = π/ωd, and the 2 % settling time ts ≈ 4/(ζωₙ). Compare the denominator with s² + 2ζωₙs + ωₙ² to read off ωₙ and ζ. For a mass–spring–damper ωₙ = √(k/m) and ζ = c/(2√(km)); for a series RLC circuit ωₙ = 1/√(LC) and ζ = (R/2)·√(C/L).

Closed-loop transfer function

T(s) = 300/(s² + 8s + 300): ωₙ = 17.32 rad/s, ζ = 0.2309

Settling time

ζωₙ = 8/2 = 4
ts ≈ 4/(ζωₙ) = 1 s

Answer

ts = 1 s

For this loop ζωₙ = a/2 whatever K is: raising K adds overshoot but does not shorten the settling time.

7.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 504

Routh–Hurwitz criterion: a linear system is stable when every root of its characteristic polynomial lies in the left half of the s-plane. Write the coefficients alternately into the first two rows of the Routh array. Each entry of a new row comes from the two rows above it; for a cubic a₃s³ + a₂s² + a₁s + a₀ the s¹ entry is b₁ = (a₂·a₁ − a₃·a₀)/a₂. The number of sign changes in the first column equals the number of roots in the right half-plane. With a gain K in the coefficients, every first-column entry must stay positive: the gain that makes the s¹ entry zero is the stability limit, and the s² row then gives the frequency of the sustained oscillation.

Characteristic polynomial

s³ + 12s² + 47s + 60 + K = 0

Routh array

s³: 1, 47
s²: 12, 60 + K
s¹: (12·47 − 60 − K)/12
s⁰: 60 + K

Condition

The s¹ entry must stay positive.

K < 12·47 − 60 = 504

Answer

Kmax = 504

At the limit the loop oscillates at ω = √47 = 6.856 rad/s.

8.Stability: Routh–Hurwitz & root locus

Answer: K = 24

Root locus: as the gain K goes from 0 to ∞, the closed-loop poles, the roots of 1 + K·G(s) = 0, move from the open-loop poles to the open-loop zeros or to infinity. A point on the real axis belongs to the locus when the number of real poles and zeros to its right is odd. The n − m branches that go to infinity follow asymptotes at the angles (2q + 1)·180°/(n − m), which start from the centroid σₐ = (Σ poles − Σ zeros)/(n − m). Branches leave or join the real axis where dK/ds = 0. The angle condition ∠G(s) = 180° fixes the shape, and the magnitude condition K = 1/|G(s)| gives the gain at any point of the locus.

Magnitude condition

K = |s|·|s + 5|·|s + 6|

Evaluate at the point

K = 2·3·4 = 24

Answer

K = 24

Check: s(s + 5)(s + 6) + K is zero at s = −2.

9.Steady-state error & PID control

Answer: K = 120

Steady-state error: in a unity-feedback loop with open-loop transfer function G(s) the error is E(s) = R(s)/(1 + G(s)), and by the final value theorem ess = lims→0 s·E(s). The system type is the number of poles of G(s) at s = 0. The error constants are Kp = lim G(s), Kv = lim s·G(s) and Ka = lim s²·G(s), all for s → 0. A step of size A gives ess = A/(1 + Kp), a ramp A·t gives A/Kv and a parabola A·t²/2 gives A/Ka. A type-0 loop follows a step with a finite error, a type-1 loop follows it exactly and a ramp with a finite error.

Required error constant

1/(1 + Kp) = 0.25 → Kp = 3

Gain

K/40 = 3 → K = 120

Answer

K = 120

Check the closed-loop stability as well: for this second-order loop any K > 0 is stable.

10.Steady-state error & PID control

Answer: ess = 6.25

P, PI and PID control: the controller C(s) = Kp + Ki/s + Kd·s acts on the error. A proportional gain makes the loop faster and reduces the error of a type-0 plant, but never removes it. The integral term adds a pole at s = 0: the system type rises by one, and a step is followed without error. The derivative term adds damping. To choose gains, write the closed-loop characteristic polynomial and compare it with s² + 2ζωns + ωn². Then %OS = 100·e−ζπ/√(1−ζ²), tp = π/ωd with ωd = ωn√(1 − ζ²), and ts ≈ 4/(ζωn).

Velocity constant

s·C(s) = 1s + 2 → 2 as s → 0.

Kv = 2·4/50 = 0.16

Ramp error

ess = 1/0.16 = 6.25

Answer

ess = 6.25

The PI controller removes the step error completely; a ramp is followed with a constant lag.

11.Bode plots & margins

Answer: BW = 63.66 Hz

Frequency response: substitute s = jω in the transfer function. The magnitude in decibels is 20·log₁₀|H(jω)| and the phase is the sum of the angles of the numerator factors minus those of the denominator factors. A first-order RC or RL filter has its corner (−3 dB) frequency at ωc = 1/(RC) or R/L, and fc = ωc/(2π); a low-pass filter gives |H| = 1/√(1 + (ω/ωc)²) and φ = −atan(ω/ωc). In the asymptotic (straight-line) Bode plot each pole bends the magnitude down by 20 dB/decade at its corner frequency, each zero bends it up by 20 dB/decade, and an integrator 1/s gives −20 dB/decade at all frequencies. Gain crossover ωc: |L(jω)| = 1 (0 dB); phase margin PM = 180° + ∠L(jωc). Phase crossover ω180: ∠L(jω) = −180°; gain margin GM = −20·log₁₀|L(jω180)| dB. The closed loop is stable when both margins are positive.

Resonant frequency

ω₀ = 1/√(LC) = 1/√(0.005·1 × 10⁻⁵) = 4472 rad/s
f₀ = ω₀/(2π) = 711.8 Hz

Bandwidth

BW = R/L = 2/0.005 = 400 rad/s = 63.66 Hz

Answer

BW = 63.66 Hz
12.Bode plots & margins

Answer: ωc = 0.5765 rad/s

Frequency response: substitute s = jω in the transfer function. The magnitude in decibels is 20·log₁₀|H(jω)| and the phase is the sum of the angles of the numerator factors minus those of the denominator factors. A first-order RC or RL filter has its corner (−3 dB) frequency at ωc = 1/(RC) or R/L, and fc = ωc/(2π); a low-pass filter gives |H| = 1/√(1 + (ω/ωc)²) and φ = −atan(ω/ωc). In the asymptotic (straight-line) Bode plot each pole bends the magnitude down by 20 dB/decade at its corner frequency, each zero bends it up by 20 dB/decade, and an integrator 1/s gives −20 dB/decade at all frequencies. Gain crossover ωc: |L(jω)| = 1 (0 dB); phase margin PM = 180° + ∠L(jωc). Phase crossover ω180: ∠L(jω) = −180°; gain margin GM = −20·log₁₀|L(jω180)| dB. The closed loop is stable when both margins are positive.

Gain crossover

ω⁴ + 4ω² − 1.44 = 0
ω² = (−4 + √(16 + 4·1.44))/2 = 0.3324
ωc = 0.5765 rad/s

Answer

ωc = 0.5765 rad/s

ExamProblems · examproblems.com/courses/control-engineering-1