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−1−2−3−4−5−6−7−8−91

F(s) = (4s + 15)/(s² + 11s + 24). Its inverse transform has the form f(t) = A·e−3t + B·e−8t. Find the coefficient A by partial fractions.

F(s) = (4s + 15)/(s² + 11s + 24)
A =
2.Laplace transformsHard
t (s)y(t)t = 0.6y(0.6) = ?

Solve y″ + 7y′ + 10y = 0 with y(0) = −8 and y′(0) = 4 using the Laplace transform, and find y(0.6).

y″ + 7y′ + 10y = 0y(0) = −8y′(0) = 4t = 0.6 s
y(0.6) =
3.Block diagramsMedium
R(s)+5(s + 7)(s + 1)(s + 6)G(s)C(s)−

A unity negative feedback loop has G(s) = 5(s + 7)/((s + 1)(s + 6)). A step of size 5 is applied at the reference. Find the steady-state error ess.

G(s) = 5(s + 7)/((s + 1)(s + 6))R = 5
ess =
4.Block diagramsHard
R(s)+Kcontroller+D(s)+1s + 1plantC(s)−

A proportional controller K drives the plant G(s) = 1/(s + 1) in a unity negative feedback loop. A disturbance step of size 10 enters at the plant input while R = 0. Find the smallest gain K that keeps the steady-state output at or below 0.5.

G(s) = 1/(s + 1)D = 10css ≤ 0.5
K =
5.Step responseMedium
t (s)y(t)1

A system has the transfer function G(s) = 23/(s² + 2s + 23) and receives a unit step input. Find the natural frequency ωn in rad/s.

G(s) = 23/(s² + 2s + 23)
ωn = rad/s
6.Step responseHard
R(s)+Ks(s + 10)G(s)C(s)−

A unity negative feedback loop has G(s) = K/(s(s + 10)). Find the gain K that gives a step response with 5 % overshoot.

G(s) = K/(s(s + 10))%OS = 5 %
K =
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K = 601s(s + 4)(s + 5)C(s)−

The loop in the diagram runs with K = 60 and L(s) = K/(s(s + 4)(s + 5)). By what factor m can the gain be raised before the closed loop becomes unstable?

L(s) = K/(s(s + 4)(s + 5))K = 60
m =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10−12−14−16j2j4

The root locus of L(s) = K(s + 13)/(s(s + 7)) leaves the real axis between s = 0 and s = −7, circles round and joins the real axis again to the left of the zero. Find the break-away point sb.

L(s) = K(s + 13)/(s(s + 7))
sb =
9.Steady-state error & PID controlMedium
r(t) = 1t+Ks(s + 5)C(s)−

A unity-feedback loop with G(s) = K/(s(s + 5)) must follow the ramp r(t) = 1t with a steady-state error of 0.1. Find the gain K.

G(s) = K/(s(s + 5))r(t) = 1tess = 0.1
K =
10.Steady-state error & PID controlHard
R(s)+K = 810s(s + 5)C(s)−

A proportional controller K = 8 drives the plant G(s) = 10/(s(s + 5)) in a unity-feedback loop. Find the percent overshoot %OS of the step response.

G(s) = 10/(s(s + 5))K = 8
%OS = %
11.Bode plots & marginsMedium
vin​L = 5 mHC = 0.1 µFR = 50 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 50 Ω, with L = 5 mH and C = 0.1 µF. Find the quality factor Q.

R = 50 ΩL = 5 mHC = 0.1 µF
Q =
12.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−80−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

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

L(s) = 2/(s(s + 1))
ωc = rad/s

3Solutions

1.Laplace transforms

Answer: A = 0.6

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 + 11s + 24 = (s + 3)(s + 8)
F(s) = A/(s + 3) + B/(s + 8)

Cover-up rule

Multiply by (s + 3) and set s = −3: the B-term vanishes.

A = (4s + 15)/(s + 8) |s = −3 = 3/5 = 0.6

Answer

A = 0.6
(B = 3.4, so f(t) = 0.6·e−3t + 3.4·e−8t)
2.Laplace transforms

Answer: y(0.6) = −3.415

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

s²Y − (−8)s − 4 + 7(sY − (−8)) + 10Y = 0
Y(s) = (−8s − 52)/((s + 2)(s + 5))

Partial fractions

A = −36/3 = −12
B = −12/(−3) = 4

Result

y(t) = −12·e−2t + 4·e−5t
y(0.6) = −3.415

Answer

y(0.6) = −3.415

Check: y(0) = A + B = −8.

3.Block diagrams

Answer: ess = 0.7317

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).

Position error constant

Kₚ = G(0) = 5·7/(1·6) = 5.833

Steady-state error

ess = A/(1 + Kₚ) = 5/(1 + 5.833) = 0.7317

Answer

ess = 0.7317
4.Block diagrams

Answer: K = 19

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)) = 10·1/(1 + 1K)

Solve for K

1 + 1K = 10·1/0.5 = 20
K = (20 − 1)/1 = 19

Answer

K = 19
5.Step response

Answer: ωn = 4.796 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 = 23 → ωₙ = 4.796 rad/s
2ζωₙ = 2 → ζ = 2/(2·4.796) = 0.2085

Result

ωₙ = 4.796 rad/s

Answer

ωn = 4.796 rad/s
6.Step response

Answer: K = 52.49

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) = K/(s² + 10s + K)

Damping ratio from the overshoot

ζ = −ln(0.05)/√(π² + ln²(0.05)) = 0.6901

Gain

ωₙ = 10/(2ζ) = 7.245 rad/s
K = ωₙ² = 52.49

Answer

K = 52.49
7.Stability: Routh–Hurwitz & root locus

Answer: m = 3

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 + 9s² + 20s + K = 0
s³: 1, 20
s²: 9, K
s¹: (9·20 − K)/9
s⁰: K

Largest stable gain

Kmax = 9·20 = 180

Factor

m = 180/60 = 3

Answer

m = 3

This factor is the gain margin as a ratio: 9.542 dB.

8.Stability: Routh–Hurwitz & root locus

Answer: sb = −4.168

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.

Gain along the real axis

K(s) = −(s² + 7s)/(s + 13)

Stationary points

Differentiate with the quotient rule and set the numerator to zero.

(2s + 7)(s + 13) − (s² + 7s) = s² + 26s + 91 = 0
s = −13 ± √78 = −4.168, −21.83

Break-away point

sb = −4.168

Answer

sb = −4.168

Between the two break points the locus is a circle centred on the zero, with radius √78 = 8.832.

9.Steady-state error & PID control

Answer: K = 50

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 velocity constant

Kv = A/ess = 1/0.1 = 10

Gain

K = 5·Kv = 50

Answer

K = 50

A large gain makes this loop very lightly damped: ζ = 0.3536. That conflict is why compensators are used.

10.Steady-state error & PID control

Answer: %OS = 40.07 %

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

s(s + 5) + 8·10 = s² + 5s + 80

ωₙ and ζ

ωₙ = √80 = 8.944 rad/s
ζ = 5/(2·8.944) = 0.2795

Overshoot

%OS = 100·e−0.2795π/√(1 − 0.2795²) = 40.07 %

Answer

%OS = 40.07 %
11.Bode plots & margins

Answer: Q = 4.472

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 × 10⁻⁷) = 44721 rad/s
f₀ = ω₀/(2π) = 7118 Hz

Quality factor

BW = R/L = 50/0.005 = 10000 rad/s = 1592 Hz
Q = ω₀/BW = 44721/10000 = 4.472

Answer

Q = 4.472
12.Bode plots & margins

Answer: ωc = 1.25 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 + 1ω² − 4 = 0
ω² = (−1 + √(1 + 4·4))/2 = 1.562
ωc = 1.25 rad/s

Answer

ωc = 1.25 rad/s

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