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

F(s) = (s + 17)/(s² + 11s + 30). Its inverse transform has the form f(t) = A·e−5t + B·e−6t. Find the coefficient A by partial fractions.

F(s) = (s + 17)/(s² + 11s + 30)
A =
2.Laplace transformsHard
t (s)y(t)t = 0.6y(0.6) = ?

A system obeys y″ + 3y′ + 2y = 8·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 1 for t ≥ 0. Use the Laplace transform to find the output y(0.6).

y″ + 3y′ + 2y = 8·u(t)u(t) = 1y(0) = 0, y′(0) = 0t = 0.6 s
y(0.6) =
3.Block diagramsMedium
R(s)+17K+8s + 7G(s)C(s)3H₁−−

The plant G(s) = 8/(s + 7) has an inner (minor) negative feedback loop with gain H₁ = 3. A controller gain K = 17 and an outer unity negative feedback loop close the system. Find the closed-loop DC gain T(0).

G(s) = 8/(s + 7)H₁ = 3K = 17
T(0) =
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)3ts​ = ?

A system has the transfer function G(s) = 15/(s² + s + 5) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 15/(s² + s + 5)
ts = s
6.Step responseHard
k = 400 N/mc = 4 N·s/mm = 1 kgF = 20 Nx(t)

A mass m = 1 kg hangs on a spring k = 400 N/m with a viscous damper c = 4 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 20 N is suddenly applied. Find the damping ratio ζ.

m = 1 kgk = 400 N/mc = 4 N·s/mF = 20 N
ζ =
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 10s(s + 1)(s + 4)C(s)−

A unity-feedback loop has L(s) = K(s + 10)/(s(s + 1)(s + 4)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K(s + 10)/(s(s + 1)(s + 4))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
ReImζ = 0.5−2−4−6j2j4j6

The root locus of L(s) = K/(s(s + 6)) is shown. Find the gain K for which the closed-loop poles have the damping ratio ζ = 0.5.

L(s) = K/(s(s + 6))ζ = 0.5
K =
9.Steady-state error & PID controlMedium
R(s)+K(s + 5)(s + 10)C(s)−

A unity-feedback loop has G(s) = K/((s + 5)(s + 10)). Find the gain K that gives a steady-state error of 25 % for a step input.

G(s) = K/((s + 5)(s + 10))ess = 25 %
K =
10.Steady-state error & PID controlHard
R(s)+K = 402s(s + 4)C(s)−

A proportional controller K = 40 drives the plant G(s) = 2/(s(s + 4)) in a unity-feedback loop. Find the peak time tp of the step response.

G(s) = 2/(s(s + 4))K = 40
tp = s
11.Bode plots & marginsMedium
vin​C = 22 nFR = 47 kΩvout​

The RC high-pass filter shown has R = 47 kΩ, C = 22 nF. A sinusoidal input of 460 Hz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 47 kΩC = 22 nFf = 460 Hz
φ = °
12.Bode plots & marginsHard
|H| (dB)11101010010010001000−80−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​

A unity-feedback loop has the open-loop transfer function L(s) = 880/(s(s + 20)). Find the phase margin PM in degrees.

L(s) = 880/(s(s + 20))
PM = °

3Solutions

1.Laplace transforms

Answer: A = 12

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 + 30 = (s + 5)(s + 6)
F(s) = A/(s + 5) + B/(s + 6)

Cover-up rule

Multiply by (s + 5) and set s = −5: the B-term vanishes.

A = (s + 17)/(s + 6) |s = −5 = 12/1 = 12

Answer

A = 12
(B = −11, so f(t) = 12·e−5t − 11·e−6t)
2.Laplace transforms

Answer: y(0.6) = 0.8143

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) = 8/(s(s + 1)(s + 2))

Partial fractions

A = 8/(1·2) = 4
B = 8/((−1)·1) = −8
C = 8/((−2)·(−1)) = 4

Output

y(t) = 4 − 8·e−t + 4·e−2t
y(0.6) = 0.8143

Answer

y(0.6) = 0.8143
3.Block diagrams

Answer: T(0) = 0.8144

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) = 8/(s + 7 + 8·3) = 8/(s + 31)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 136/(s + 31 + 136) = 136/(s + 167)

Result

T(0) = 136/167 = 0.8144

Answer

T(0) = 0.8144
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: ts = 8 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 = 5 → ωₙ = 2.236 rad/s
2ζωₙ = 1 → ζ = 1/(2·2.236) = 0.2236

Result

ζωₙ = 0.5 → ts ≈ 4/0.5 = 8 s

Answer

ts = 8 s
6.Step response

Answer: ζ = 0.1

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) = √(400/1) = 20 rad/s
ζ = c/(2√(km)) = 4/(2·√(400·1)) = 0.1

Result

ζ = 0.1

Answer

ζ = 0.1
7.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 4

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

s(s + 1)(s + 4) + K(s + 10) = s³ + 5s² + (4 + K)s + 10K

Routh array

s³: 1, 4 + K
s²: 5, 10K
s¹: (5·(4 + K) − 10K)/5
s⁰: 10K

Condition

20 − 5K > 0
K < 20/5 = 4

Answer

Kmax = 4

With the zero to the right of s = −5 the loop would stay stable for every K > 0.

8.Stability: Routh–Hurwitz & root locus

Answer: K = 36

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.

Compare with the standard form

s² + 6s + K = s² + 2ζωₙs + ωₙ²
2·0.5·ωₙ = 6 → ωₙ = 6 rad/s

Gain

K = ωₙ² = 6² = 36

Answer

K = 36

The closed-loop poles sit where the ζ line crosses the vertical branch at s = −3.

9.Steady-state error & PID control

Answer: K = 150

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/50 = 3 → K = 150

Answer

K = 150

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

10.Steady-state error & PID control

Answer: tp = 0.3604 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) + 40·2 = s² + 4s + 80

ωₙ and ζ

ωₙ = √80 = 8.944 rad/s
ζ = 4/(2·8.944) = 0.2236

Peak time

ωd = 8.944·√(1 − 0.2236²) = 8.718 rad/s
tp = π/8.718 = 0.3604 s

Answer

tp = 0.3604 s
11.Bode plots & margins

Answer: φ = 18.5 °

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.

Corner frequency

ωc = 1/(RC) = 1/(47000·2.2 × 10⁻⁸) = 967.1 rad/s
fc = 153.9 Hz → f/fc = 460/153.9 = 2.989

Phase

φ = 90° − atan(2.989) = 18.5°

Answer

φ = 18.5°
12.Bode plots & margins

Answer: PM = 37.04 °

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 + 400ω² − 774400 = 0
ω² = (−400 + √(160000 + 4·774400))/2 = 702.4
ωc = 26.5 rad/s

Phase margin

∠L(jωc) = −90° − atan(26.5/20) = −143°
PM = 180° + ∠L = 37.04°

Answer

PM = 37.04°

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