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
t (s)f(t)t = 0.8f(0.8) = ?

Find f(t) for F(s) = (2s + 23)/(s² + 4s + 4) and evaluate f(0.8).

F(s) = (2s + 23)/(s² + 4s + 4)t = 0.8 s
f(0.8) =
2.Laplace transformsHard
t (s)y(t)t = 0.3y(0.3) = ?

A system obeys y″ + 5y′ + 4y = 16·u(t), with y(0) = 3 and y′(0) = 0. At t = 0 a step input u(t) = 2 is applied. Use the Laplace transform to find y(0.3).

y″ + 5y′ + 4y = 16·u(t)y(0) = 3y′(0) = 0u(t) = 2t = 0.3 s
y(0.3) =
3.Block diagramsMedium
R(s)+12K+6s + 4G(s)C(s)3H₁−−

The plant G(s) = 6/(s + 4) has an inner (minor) negative feedback loop with gain H₁ = 3. A controller gain K = 12 and an outer unity negative feedback loop close the system. Find the closed-loop pole p.

G(s) = 6/(s + 4)H₁ = 3K = 12
p =
4.Block diagramsHard
R(s)+Kcontroller+D(s)+3s + 8plantC(s)−

A proportional controller K drives the plant G(s) = 3/(s + 8) in a unity negative feedback loop. A disturbance step of size 2 enters at the plant input while R = 0. Find the smallest gain K that keeps the steady-state output at or below 0.015.

G(s) = 3/(s + 8)D = 2css ≤ 0.015
K =
5.Step responseMedium
t (s)y(t)2?

A system has the transfer function G(s) = 42/(s² + 5s + 21) and receives a unit step input. Find the percent overshoot %OS of the step response.

G(s) = 42/(s² + 5s + 21)
%OS = %
6.Step responseHard
R(s)+165s(s + 6)G(s)C(s)−

A unity negative feedback loop has G(s) = 165/(s(s + 6)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 165/(s(s + 6))
ts = s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 1)(s + 10)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 1)(s + 10)). 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 + 1)(s + 10))
ω = rad/s
8.Stability: Routh–Hurwitz & root locusHard
ReIm−5j5

The root locus of L(s) = K/((s + 3)(s + 5)(s + 6)) is sketched. Find the frequency ω at which the locus crosses the imaginary axis.

L(s) = K/((s + 3)(s + 5)(s + 6))
ω = rad/s
9.Steady-state error & PID controlMedium
R(s)+K(s + 2)(s + 5)C(s)−

A unity-feedback loop has G(s) = K/((s + 2)(s + 5)). Find the gain K that gives a steady-state error of 25 % for a step input.

G(s) = K/((s + 2)(s + 5))ess = 25 %
K =
10.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s10s + 1C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 10/(s + 1) in a unity-feedback loop. The closed loop must have ωₙ = 5 rad/s and ζ = 0.8. Find the proportional gain Kp.

G(s) = 10/(s + 1)ωₙ = 5 rad/sζ = 0.8
Kp =
11.Bode plots & marginsMedium
vin​L = 1 mHC = 4.7 µFR = 5 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 5 Ω, with L = 1 mH and C = 4.7 µF. Find the quality factor Q.

R = 5 ΩL = 1 mHC = 4.7 µF
Q =
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) = 10/(s(s + 2)(s + 3)). Find the gain margin GM in dB.

L(s) = 10/(s(s + 2)(s + 3))
GM = dB

3Solutions

1.Laplace transforms

Answer: f(0.8) = 3.473

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).

Factor the denominator

s² + 4s + 4 = (s + 2)²

Partial fractions

2s + 23 = 2(s + 2) + 19
F(s) = 2/(s + 2) + 19/(s + 2)²

Inverse transform

f(t) = 2·e−2t + 19·t·e−2t
f(0.8) = (2 + 19·0.8)·e−2·0.8 = 3.473

Answer

f(0.8) = 3.473
2.Laplace transforms

Answer: y(0.3) = 3.563

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

(s² + 5s + 4)·Y(s) = 3s + 15 + 32/s
Y(s) = (3s² + 15s + 32)/(s(s + 1)(s + 4))

Partial fractions

A = 32/(1·4) = 8
B = 20/((−1)·3) = −6.667
C = 20/((−4)·(−3)) = 1.667

Result

y(t) = 8 − 6.667·e−t + 1.667·e−4t
y(0.3) = 3.563

Answer

y(0.3) = 3.563

Check: y(0) = A + B + C = 3.

3.Block diagrams

Answer: p = −94

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) = 6/(s + 4 + 6·3) = 6/(s + 22)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 72/(s + 22 + 72) = 72/(s + 94)

Result

p = −94

Answer

p = −94
4.Block diagrams

Answer: K = 130.7

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)) = 2·3/(8 + 3K)

Solve for K

8 + 3K = 2·3/0.015 = 400
K = (400 − 8)/3 = 130.7

Answer

K = 130.7
5.Step response

Answer: %OS = 12.94 %

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 = 21 → ωₙ = 4.583 rad/s
2ζωₙ = 5 → ζ = 5/(2·4.583) = 0.5455

Result

%OS = 100·e−ζπ/√(1−ζ²) = 100·e−2.045 = 12.94 %

Answer

%OS = 12.94 %
6.Step response

Answer: ts = 1.333 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) = 165/(s² + 6s + 165): ωₙ = 12.85 rad/s, ζ = 0.2335

Settling time

ζωₙ = 6/2 = 3
ts ≈ 4/(ζωₙ) = 1.333 s

Answer

ts = 1.333 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 = 3.162 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 + 11s² + 10s + K = 0
s³: 1, 10
s²: 11, K
s¹: (11·10 − K)/11
s⁰: K

Critical gain

K = 11·10 = 110

Auxiliary equation

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

11s² + 110 = 0 → s = ±j√10
ω = 3.162 rad/s

Answer

ω = 3.162 rad/s
8.Stability: Routh–Hurwitz & root locus

Answer: ω = 7.937 rad/s

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.

Critical gain

s³ + 14s² + 63s + 90 + K = 0
s¹: (14·63 − 90 − K)/14 = 0 → K = 792

Crossing frequency

14s² + 90 + 792 = 14s² + 882 = 0 → s² = −63
ω = √63 = 7.937 rad/s

Answer

ω = 7.937 rad/s
9.Steady-state error & PID control

Answer: K = 30

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/10 = 3 → K = 30

Answer

K = 30

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 = 0.7

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 + 1).

s² + (1 + 10Kp)s + 10Ki

Target polynomial

s² + 2·0.8·5s + 5² = s² + 8s + 25

Match the coefficients

1 + 10Kp = 8 → Kp = 0.7

Answer

Kp = 0.7

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

11.Bode plots & margins

Answer: Q = 2.917

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/√(0.001·4.7 × 10⁻⁶) = 14586 rad/s
f₀ = ω₀/(2π) = 2322 Hz

Quality factor

BW = R/L = 5/0.001 = 5000 rad/s = 795.8 Hz
Q = ω₀/BW = 14586/5000 = 2.917

Answer

Q = 2.917
12.Bode plots & margins

Answer: GM = 9.542 dB

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/2)·(ω/3) = 1 → ω180 = √(2·3) = 2.449 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 4)·√(ω² + 9)) = K/(2·3·(2 + 3)) = K/30

Gain margin

GM = 20·log₁₀(30/10) = 9.542 dB

Answer

GM = 9.542 dB

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