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 = 1f(1) = ?

Find f(t) for F(s) = (2s + 39)/(s² + 4s + 4) and evaluate f(1).

F(s) = (2s + 39)/(s² + 4s + 4)t = 1 s
f(1) =
2.Laplace transformsHard
t (s)y(t)t = 0.1y(0.1) = ?

Solve y″ + 7y′ + 10y = 0 with y(0) = 9 and y′(0) = −9 using the Laplace transform, and find y(0.1).

y″ + 7y′ + 10y = 0y(0) = 9y′(0) = −9t = 0.1 s
y(0.1) =
3.Block diagramsMedium
R(s)+K(s + 2)(s + 3)G(s)C(s)−

The plant G(s) = K/((s + 2)(s + 3)) is in a unity negative feedback loop. Find the gain K for which the closed loop is critically damped (two equal real poles).

G(s) = K/((s + 2)(s + 3))unity negative feedback
K =
4.Block diagramsHard
R(s)+15s + 4G₁+4s + 8G₂C(s)0.5H₁−0.5H₂−

Reduce the block diagram with G₁ = 15/(s + 4), G₂ = 4/(s + 8), an inner negative feedback H₁ = 0.5 around G₂ and an outer negative feedback H₂ = 0.5. Find the DC gain T(0) of C(s)/R(s).

G₁ = 15/(s + 4)G₂ = 4/(s + 8)H₁ = 0.5H₂ = 0.5
T(0) =
5.Step responseMedium
t (s)y(t)1ts​ = ?

A system has the transfer function G(s) = 91/(s² + 4s + 91) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 91/(s² + 4s + 91)
ts = s
6.Step responseHard
k = 500 N/mc = 28 N·s/mm = 2.5 kgF = 50 Nx(t)

A mass m = 2.5 kg hangs on a spring k = 500 N/m with a viscous damper c = 28 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 50 N is suddenly applied. Find the time tp at which the displacement is largest.

m = 2.5 kgk = 500 N/mc = 28 N·s/mF = 50 N
tp = s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 1)(s + 2)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 1)(s + 2)). Find the largest gain Kmax for which the closed loop is stable.

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

The root locus of L(s) = K/(s(s + 4)(s + 6)) leaves the real axis between s = 0 and s = −4. Find the break-away point sb.

L(s) = K/(s(s + 4)(s + 6))
sb =
9.Steady-state error & PID controlMedium
r(t) = 10 + 4t+2(s + 1)s(s + 4)(s + 10)C(s)−

The input r(t) = 10 + 4t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 2(s + 1)/(s(s + 4)(s + 10)). Find the steady-state error ess.

G(s) = 2(s + 1)/(s(s + 4)(s + 10))r(t) = 10 + 4t
ess =
10.Steady-state error & PID controlHard
R(s)+K = 81s(s + 4)C(s)−

A proportional controller K = 8 drives the plant G(s) = 1/(s(s + 4)) in a unity-feedback loop. Find the peak time tp of the step response.

G(s) = 1/(s(s + 4))K = 8
tp = s
11.Bode plots & marginsMedium
vin​L = 50 mHC = 0.1 µFR = 20 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 20 Ω, with L = 50 mH and C = 0.1 µF. Find the −3 dB bandwidth BW in Hz.

R = 20 ΩL = 50 mHC = 0.1 µF
BW = 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) = 106/(s(s + 3)(s + 7)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 106/(s(s + 3)(s + 7))
ω180 = rad/s

3Solutions

1.Laplace transforms

Answer: f(1) = 5.007

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 + 39 = 2(s + 2) + 35
F(s) = 2/(s + 2) + 35/(s + 2)²

Inverse transform

f(t) = 2·e−2t + 35·t·e−2t
f(1) = (2 + 35·1)·e−2·1 = 5.007

Answer

f(1) = 5.007
2.Laplace transforms

Answer: y(0.1) = 8.005

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 − 9s − (−9) + 7(sY − 9) + 10Y = 0
Y(s) = (9s + 54)/((s + 2)(s + 5))

Partial fractions

A = 36/3 = 12
B = 9/(−3) = −3

Result

y(t) = 12·e−2t − 3·e−5t
y(0.1) = 8.005

Answer

y(0.1) = 8.005

Check: y(0) = A + B = 9.

3.Block diagrams

Answer: K = 0.25

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

1 + K/((s + 2)(s + 3)) = 0 → s² + 5s + 6 + K = 0

Repeated root

A quadratic s² + βs + γ has a double root when β² = 4γ.

5² = 4(6 + K) → K = 25/4 − 6 = 0.25

Answer

K = 0.25
Both closed-loop poles are then at s = −2.5.
4.Block diagrams

Answer: T(0) = 0.8571

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) = 4/(s + 10)

Outer loop

T(s) = G₁Gᵢ/(1 + G₁GᵢH₂) = 60/((s + 4)(s + 10) + 30)

DC gain

T(0) = 60/(4·10 + 30) = 60/70 = 0.8571

Answer

T(0) = 0.8571
5.Step response

Answer: ts = 2 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 = 91 → ωₙ = 9.539 rad/s
2ζωₙ = 4 → ζ = 4/(2·9.539) = 0.2097

Result

ζωₙ = 2 → ts ≈ 4/2 = 2 s

Answer

ts = 2 s
6.Step response

Answer: tp = 0.2419 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) = √(500/2.5) = 14.14 rad/s
ζ = c/(2√(km)) = 28/(2·√(500·2.5)) = 0.396

Result

ωd = 12.99 rad/s
tp = π/ωd = 0.2419 s

Answer

tp = 0.2419 s
7.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 6

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 equation

s³ + 3s² + 2s + K = 0

Routh array

s³: 1, 2
s²: 3, K
s¹: (3·2 − K)/3
s⁰: K

Conditions

Every entry of the first column must be positive.

K > 0
3·2 − K > 0 → K < 6

Answer

Kmax = 6

At K = 6 the closed loop oscillates at ω = √2 = 1.414 rad/s.

8.Stability: Routh–Hurwitz & root locus

Answer: sb = −1.569

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) = −s(s + 4)(s + 6) = −(s³ + 10s² + 24s)

Stationary points

dK/ds = −(3s² + 20s + 24) = 0
s = (−20 ± √(400 − 288))/6 = −1.569, −5.097

Pick the point on the locus

The locus lies on the real axis between 0 and −4; the other root is not on the locus.

sb = −1.569

Answer

sb = −1.569
9.Steady-state error & PID control

Answer: ess = 80

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.

Velocity constant

Kv = lim s·G(s) = 0.05

Step part

Kp = ∞ → estep = 0

Ramp error

ess = 4/0.05 = 80

Answer

ess = 80
10.Steady-state error & PID control

Answer: tp = 1.571 s

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 + 4) + 8·1 = s² + 4s + 8

ωₙ and ζ

ωₙ = √8 = 2.828 rad/s
ζ = 4/(2·2.828) = 0.7071

Peak time

ωd = 2.828·√(1 − 0.7071²) = 2 rad/s
tp = π/2 = 1.571 s

Answer

tp = 1.571 s
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/√(0.05·1 × 10⁻⁷) = 14142 rad/s
f₀ = ω₀/(2π) = 2251 Hz

Bandwidth

BW = R/L = 20/0.05 = 400 rad/s = 63.66 Hz

Answer

BW = 63.66 Hz
12.Bode plots & margins

Answer: ω180 = 4.583 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/3)·(ω/7) = 1 → ω180 = √(3·7) = 4.583 rad/s

Answer

ω180 = 4.583 rad/s

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