Control Engineering 1: cram pack

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

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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
ReIm−224

Find f(t) for F(s) = 4/(s² + 6s + 25) by completing the square, and evaluate f(0.1).

F(s) = 4/(s² + 6s + 25)t = 0.1 s
f(0.1) =
2.Laplace transformsHard
t (s)y(t)t = 0.5y(0.5) = ?

A system obeys y″ + 5y′ + 6y = 18·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 5 for t ≥ 0. Use the Laplace transform to find the output y(0.5).

y″ + 5y′ + 6y = 18·u(t)u(t) = 5y(0) = 0, y′(0) = 0t = 0.5 s
y(0.5) =
3.Block diagramsMedium
R(s)+11K+5s + 1G(s)C(s)0.5H₁−−

The plant G(s) = 5/(s + 1) has an inner (minor) negative feedback loop with gain H₁ = 0.5. A controller gain K = 11 and an outer unity negative feedback loop close the system. Find the closed-loop DC gain T(0).

G(s) = 5/(s + 1)H₁ = 0.5K = 11
T(0) =
4.Block diagramsHard
R(s)+14s + 3G₁+10s + 1G₂C(s)4H₁−0.2H₂−

Reduce the block diagram with G₁ = 14/(s + 3), G₂ = 10/(s + 1), an inner negative feedback H₁ = 4 around G₂ and an outer negative feedback H₂ = 0.2. Find the DC gain T(0) of C(s)/R(s).

G₁ = 14/(s + 3)G₂ = 10/(s + 1)H₁ = 4H₂ = 0.2
T(0) =
5.Step responseMedium
t (s)y(t)1

A system has the transfer function G(s) = 90/(s² + 4s + 90) and receives a unit step input. Find the damping ratio ζ.

G(s) = 90/(s² + 4s + 90)
ζ =
6.Step responseHard
k = 800 N/mc = 16 N·s/mm = 8 kgF = 100 Nx(t)

A mass m = 8 kg hangs on a spring k = 800 N/m with a viscous damper c = 16 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 100 N is suddenly applied. Find the time tp at which the displacement is largest.

m = 8 kgk = 800 N/mc = 16 N·s/mF = 100 N
tp = s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+21601s(s + 8)(s + 10)C(s)−

The loop in the diagram has negative feedback and L(s) = 2160/(s(s + 8)(s + 10)). How many closed-loop poles N lie in the right half-plane?

L(s) = 2160/(s(s + 8)(s + 10))
N =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8j2j4

For L(s) = K/((s + 7)(s² + 2s + 17)), find the angle of departure θd of the root locus from the complex pole in the upper half-plane, measured from the positive real axis.

L(s) = K/((s + 7)(s² + 2s + 17))
θd = °
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 10 % for a step input.

G(s) = K/((s + 4)(s + 10))ess = 10 %
K =
10.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s5s + 2C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 5/(s + 2) in a unity-feedback loop. The closed loop must have ωₙ = 6 rad/s and ζ = 0.7. Find the integral gain Ki.

G(s) = 5/(s + 2)ωₙ = 6 rad/sζ = 0.7
Ki =
11.Bode plots & marginsMedium
|H| (dB)0.010.010.10.1111010100100−20020406080ω (rad/s)φ (°)ω (rad/s)−90−450ω = 0.5

Find the phase ∠G(jω) of G(s) = 42(s + 2)/(s(s + 3)) at ω = 0.5 rad/s, in degrees.

G(s) = 42(s + 2)/(s(s + 3))ω = 0.5 rad/s
∠G(jω) = °
12.Bode plots & marginsHard
|H| (dB)0.10.111101010010010001000−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​

A unity-feedback loop has the open-loop transfer function L(s) = 116/(s(s + 4)). Find the phase margin PM in degrees.

L(s) = 116/(s(s + 4))
PM = °

3Solutions

1.Laplace transforms

Answer: f(0.1) = 0.2885

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).

Complete the square

s² + 6s + 25 = (s + 3)² + 16 = (s + 3)² + 4²

Match the numerator

F(s) = (4/4)·4/((s + 3)² + 4²)

Inverse transform

f(t) = e−3t·(sin 4t)
f(0.1) = e−3·0.1·(sin(4·0.1)) = 0.2885

Answer

f(0.1) = 0.2885
2.Laplace transforms

Answer: y(0.5) = 5.139

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).

Transform the equation

Y(s) = 90/(s(s + 2)(s + 3))

Partial fractions

A = 90/(2·3) = 15
B = 90/((−2)·1) = −45
C = 90/((−3)·(−1)) = 30

Output

y(t) = 15 − 45·e−2t + 30·e−3t
y(0.5) = 5.139

Answer

y(0.5) = 5.139
3.Block diagrams

Answer: T(0) = 0.9402

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) = 5/(s + 1 + 5·0.5) = 5/(s + 3.5)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 55/(s + 3.5 + 55) = 55/(s + 58.5)

Result

T(0) = 55/58.5 = 0.9402

Answer

T(0) = 0.9402
4.Block diagrams

Answer: T(0) = 0.9272

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 = G₂/(1 + G₂H₁) = 10/(s + 41)

Outer loop

T(s) = G₁Gᵢ/(1 + G₁GᵢH₂) = 140/((s + 3)(s + 41) + 28)

DC gain

T(0) = 140/(3·41 + 28) = 140/151 = 0.9272

Answer

T(0) = 0.9272
5.Step response

Answer: ζ = 0.2108

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 = 90 → ωₙ = 9.487 rad/s
2ζωₙ = 4 → ζ = 4/(2·9.487) = 0.2108

Result

ζ = 0.2108

Answer

ζ = 0.2108
6.Step response

Answer: tp = 0.3157 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).

Natural frequency and damping ratio

ωₙ = √(k/m) = √(800/8) = 10 rad/s
ζ = c/(2√(km)) = 16/(2·√(800·8)) = 0.1

Result

ωd = 9.95 rad/s
tp = π/ωd = 0.3157 s

Answer

tp = 0.3157 s
7.Stability: Routh–Hurwitz & root locus

Answer: N = 2

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 + 18s² + 80s + 2160 = 0

Routh array

s³: 1, 80
s²: 18, 2160
s¹: −40
s⁰: 2160

Sign changes in the first column

1 → 18 → −40 → 2160
N = 2

Answer

N = 2

The gain is above the limit K = 1440, so a complex pair has crossed into the right half-plane.

8.Stability: Routh–Hurwitz & root locus

Answer: θd = 56.31 °

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.

Pole locations

s² + 2s + 17 = 0 → s = −1 ± j4
s = −7

Angles from the other poles

∠(−1 + j4 + 7) = atan2(4, 6) = 33.69°
∠(−1 + j4 − (−1 − j4)) = ∠(j8) = 90°

Angle of departure

θd = 180° − 33.69° − 90° = 56.31°

Answer

θd = 56.31°
9.Steady-state error & PID control

Answer: K = 360

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) = 0.1 → Kp = 9

Gain

K/40 = 9 → K = 360

Answer

K = 360

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

10.Steady-state error & PID control

Answer: Ki = 7.2

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).

Characteristic polynomial

1 + C(s)G(s) = 0, multiplied by s(s + 2).

s² + (2 + 5Kp)s + 5Ki

Target polynomial

s² + 2·0.7·6s + 6² = s² + 8.4s + 36

Match the coefficients

5Ki = 36 → Ki = 7.2

Answer

Ki = 7.2

The integral term makes the loop type 1, so a step is followed without steady-state error.

11.Bode plots & margins

Answer: ∠G(jω) = −85.43 °

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.

Add the angles

∠G = atan(0.5/2) − 90° − atan(0.5/3) = 14.04° − 90° − 9.462°

Result

∠G(j0.5) = −85.43°

Answer

∠G(jω) = −85.43°
12.Bode plots & margins

Answer: PM = 21.03 °

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 + 16ω² − 13456 = 0
ω² = (−16 + √(256 + 4·13456))/2 = 108.3
ωc = 10.41 rad/s

Phase margin

∠L(jωc) = −90° − atan(10.41/4) = −159°
PM = 180° + ∠L = 21.03°

Answer

PM = 21.03°

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