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−5−10−15−20−255

Find f(t) for F(s) = (s + 26)/(s² + 8s + 52) by completing the square, and evaluate f(0.2).

F(s) = (s + 26)/(s² + 8s + 52)t = 0.2 s
f(0.2) =
2.Laplace transformsHard
t (s)y(t)t = 0.1y(0.1) = ?

A system obeys y″ + 7y′ + 12y = 60·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 10 for t ≥ 0. Use the Laplace transform to find the output y(0.1).

y″ + 7y′ + 12y = 60·u(t)u(t) = 10y(0) = 0, y′(0) = 0t = 0.1 s
y(0.1) =
3.Block diagramsMedium
R(s)+39s + 2G(s)C(s)0.1H−

A plant G(s) = 39/(s + 2) is controlled in a negative feedback loop whose sensor has the gain H = 0.1. A step of size 10 is applied at R. Find the steady-state output css.

G(s) = 39/(s + 2)H = 0.1R = 10
css =
4.Block diagramsHard
R(s)+Ks(s + 9)(s + 10)G(s)C(s)−

The plant G(s) = K/(s(s + 9)(s + 10)) is in a unity negative feedback loop. At the largest stable gain the loop oscillates. Find the frequency of oscillation ω in rad/s.

G(s) = K/(s(s + 9)(s + 10))unity negative feedback
ω = rad/s
5.Step responseMedium
t (s)y(t)2

A system has the transfer function G(s) = 76/(s² + 5s + 38) and receives a unit step input. Find the natural frequency ωn in rad/s.

G(s) = 76/(s² + 5s + 38)
ωn = rad/s
6.Step responseHard
R(s)+180s(s + 16)G(s)C(s)−

A unity negative feedback loop has G(s) = 180/(s(s + 16)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 180/(s(s + 16))
ts = s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 6)(s + 8)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 6)(s + 8)). The gain is raised until the closed loop is on the edge of stability. Find the frequency ω of the sustained oscillation.

L(s) = K/(s(s + 6)(s + 8))
ω = rad/s
8.Stability: Routh–Hurwitz & root locusHard
ReIm−0.5−1−1.5−2−2.5−3−3.5−4−4.5−50.5j0.5j1−1.5

For the loop L(s) = K/(s(s + 2)(s + 4)), find the gain K that puts a closed-loop pole at s = −1.5.

L(s) = K/(s(s + 2)(s + 4))s = −1.5
K =
9.Steady-state error & PID controlMedium
R(s)+K(s + 1)(s + 10)C(s)−

A unity-feedback loop has G(s) = K/((s + 1)(s + 10)). Find the gain K that gives a steady-state error of 4 % for a step input.

G(s) = K/((s + 1)(s + 10))ess = 4 %
K =
10.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s5s + 4C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 5/(s + 4) in a unity-feedback loop. The closed loop must have ωₙ = 5 rad/s and ζ = 0.7. Find the proportional gain Kp.

G(s) = 5/(s + 4)ωₙ = 5 rad/sζ = 0.7
Kp =
11.Bode plots & marginsMedium
vin​L = 20 mHC = 10 µFR = 20 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 20 Ω, with L = 20 mH and C = 10 µF. Find the centre (resonant) frequency f0 in Hz.

R = 20 ΩL = 20 mHC = 10 µF
f0 = Hz
12.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−80−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−270−180−900ω180​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 210/(s(s + 7)(s + 8)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 210/(s(s + 7)(s + 8))
ω180 = rad/s

3Solutions

1.Laplace transforms

Answer: f(0.2) = 1.698

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 + 8s + 52 = (s + 4)² + 36 = (s + 4)² + 6²

Match the numerator

F(s) = 1·(s + 4)/((s + 4)² + 6²) + (22/6)·6/((s + 4)² + 6²)

Inverse transform

f(t) = e−4t·(cos 6t + 3.667·sin 6t)
f(0.2) = e−4·0.2·(cos(6·0.2) + 3.667·sin(6·0.2)) = 1.698

Answer

f(0.2) = 1.698
2.Laplace transforms

Answer: y(0.1) = 2.384

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) = 600/(s(s + 3)(s + 4))

Partial fractions

A = 600/(3·4) = 50
B = 600/((−3)·1) = −200
C = 600/((−4)·(−1)) = 150

Output

y(t) = 50 − 200·e−3t + 150·e−4t
y(0.1) = 2.384

Answer

y(0.1) = 2.384
3.Block diagrams

Answer: css = 66.1

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).

Closed-loop transfer function

T(s) = G/(1 + G·H) = 39/(s + 2 + 3.9) = 39/(s + 5.9)

Result

css = R·T(0) = 10·39/5.9 = 66.1

Answer

css = 66.1
4.Block diagrams

Answer: ω = 9.487 rad/s

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).

Characteristic equation

s³ + 19s² + 90s + K = 0

Routh array

s³ | 1 90
s² | 19 K
s¹ | (19·90 − K)/19
s⁰ | K

Stability limit

All entries of the first column must be positive.

Kmax = 19·90 = 1710

Frequency of oscillation

19·(jω)² + 1710 = 0 → ω² = 1710/19 = 90
ω = 9.487 rad/s

Answer

ω = 9.487 rad/s
5.Step response

Answer: ωn = 6.164 rad/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).

Compare with the standard form

The denominator must match s² + 2ζωₙs + ωₙ².

ωₙ² = 38 → ωₙ = 6.164 rad/s
2ζωₙ = 5 → ζ = 5/(2·6.164) = 0.4056

Result

ωₙ = 6.164 rad/s

Answer

ωn = 6.164 rad/s
6.Step response

Answer: ts = 0.5 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) = 180/(s² + 16s + 180): ωₙ = 13.42 rad/s, ζ = 0.5963

Settling time

ζωₙ = 16/2 = 8
ts ≈ 4/(ζωₙ) = 0.5 s

Answer

ts = 0.5 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: ω = 6.928 rad/s

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.

Routh array

s³ + 14s² + 48s + K = 0
s³: 1, 48
s²: 14, K
s¹: (14·48 − K)/14
s⁰: K

Critical gain

K = 14·48 = 672

Auxiliary equation

At the critical gain the s¹ row is zero; the s² row gives the poles on the imaginary axis.

14s² + 672 = 0 → s = ±j√48
ω = 6.928 rad/s

Answer

ω = 6.928 rad/s
8.Stability: Routh–Hurwitz & root locus

Answer: K = 1.875

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 + 2|·|s + 4|

Evaluate at the point

K = 1.5·0.5·2.5 = 1.875

Answer

K = 1.875

Check: s(s + 2)(s + 4) + K is zero at s = −1.5.

9.Steady-state error & PID control

Answer: K = 240

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) = 0.04 → Kp = 24

Gain

K/10 = 24 → K = 240

Answer

K = 240

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

10.Steady-state error & PID control

Answer: Kp = 0.6

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 + 4).

s² + (4 + 5Kp)s + 5Ki

Target polynomial

s² + 2·0.7·5s + 5² = s² + 7s + 25

Match the coefficients

4 + 5Kp = 7 → Kp = 0.6

Answer

Kp = 0.6

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

11.Bode plots & margins

Answer: f0 = 355.9 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/√(0.02·1 × 10⁻⁵) = 2236 rad/s
f₀ = ω₀/(2π) = 355.9 Hz

Answer

f0 = 355.9 Hz
12.Bode plots & margins

Answer: ω180 = 7.483 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.

Phase crossover

(ω/7)·(ω/8) = 1 → ω180 = √(7·8) = 7.483 rad/s

Answer

ω180 = 7.483 rad/s

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