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−2−4−62

Find f(t) for F(s) = (5s + 33)/(s² + 8s + 25) by completing the square, and evaluate f(0.5).

F(s) = (5s + 33)/(s² + 8s + 25)t = 0.5 s
f(0.5) =
2.Laplace transformsHard
t (s)y(t)t = 0.25y(0.25) = ?

Solve y″ + 6y′ + 5y = 0 with y(0) = −5 and y′(0) = −14 using the Laplace transform, and find y(0.25).

y″ + 6y′ + 5y = 0y(0) = −5y′(0) = −14t = 0.25 s
y(0.25) =
3.Block diagramsMedium
R(s)+16K+7s + 7G(s)C(s)0.5H₁−−

The plant G(s) = 7/(s + 7) has an inner (minor) negative feedback loop with gain H₁ = 0.5. A controller gain K = 16 and an outer unity negative feedback loop close the system. Find the closed-loop DC gain T(0).

G(s) = 7/(s + 7)H₁ = 0.5K = 16
T(0) =
4.Block diagramsHard
R(s)+K(s + 1)(s + 4)(s + 6)G(s)C(s)−

The plant G(s) = K/((s + 1)(s + 4)(s + 6)) 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 + 1)(s + 4)(s + 6))unity negative feedback
ω = rad/s
5.Step responseMedium
t (s)y(t)1?

A system has the transfer function G(s) = 39/(s² + 7s + 39) and receives a unit step input. Find the percent overshoot %OS of the step response.

G(s) = 39/(s² + 7s + 39)
%OS = %
6.Step responseHard
R(s)+115s(s + 12)G(s)C(s)−

A unity negative feedback loop has G(s) = 115/(s(s + 12)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 115/(s(s + 12))
ts = s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1(s + 4)(s + 5)(s + 6)C(s)−

A unity-feedback loop has L(s) = K/((s + 4)(s + 5)(s + 6)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/((s + 4)(s + 5)(s + 6))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−71j1j2sb​

The root locus of L(s) = K/((s + 4)(s + 6)) is shown. Find the gain K at the break-away point sb, where the two closed-loop poles meet on the real axis.

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

A unity-feedback loop with G(s) = K/(s(s + 1)) must follow the ramp r(t) = 1t with a steady-state error of 0.5. Find the gain K.

G(s) = K/(s(s + 1))r(t) = 1tess = 0.5
K =
10.Steady-state error & PID controlHard
R(s)+K = 410s(s + 10)C(s)−

A proportional controller K = 4 drives the plant G(s) = 10/(s(s + 10)) in a unity-feedback loop. Find the peak time tp of the step response.

G(s) = 10/(s(s + 10))K = 4
tp = s
11.Bode plots & marginsMedium
|H| (dB)0.10.111101010010010001000−40−200204060ω (rad/s)φ (°)ω (rad/s)−135−90−450ω = 30

Find the phase ∠G(jω) of G(s) = 27(s + 9)/(s(s + 4)) at ω = 30 rad/s, in degrees.

G(s) = 27(s + 9)/(s(s + 4))ω = 30 rad/s
∠G(jω) = °
12.Bode plots & marginsHard
|H| (dB)0.10.111101010010010001000−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​

A unity-feedback loop has the open-loop transfer function L(s) = 416/(s(s + 8)). Find the phase margin PM in degrees.

L(s) = 416/(s(s + 8))
PM = °

3Solutions

1.Laplace transforms

Answer: f(0.5) = 0.6329

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 + 25 = (s + 4)² + 9 = (s + 4)² + 3²

Match the numerator

F(s) = 5·(s + 4)/((s + 4)² + 3²) + (13/3)·3/((s + 4)² + 3²)

Inverse transform

f(t) = e−4t·(5·cos 3t + 4.333·sin 3t)
f(0.5) = e−4·0.5·(5·cos(3·0.5) + 4.333·sin(3·0.5)) = 0.6329

Answer

f(0.5) = 0.6329
2.Laplace transforms

Answer: y(0.25) = −6.232

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 − (−5)s − (−14) + 6(sY − (−5)) + 5Y = 0
Y(s) = (−5s − 44)/((s + 1)(s + 5))

Partial fractions

A = −39/4 = −9.75
B = −19/(−4) = 4.75

Result

y(t) = −9.75·e−t + 4.75·e−5t
y(0.25) = −6.232

Answer

y(0.25) = −6.232

Check: y(0) = A + B = −5.

3.Block diagrams

Answer: T(0) = 0.9143

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) = 7/(s + 7 + 7·0.5) = 7/(s + 10.5)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 112/(s + 10.5 + 112) = 112/(s + 122.5)

Result

T(0) = 112/122.5 = 0.9143

Answer

T(0) = 0.9143
4.Block diagrams

Answer: ω = 5.831 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 + 11s² + 34s + 24 + K = 0

Routh array

s³ | 1 34
s² | 11 24 + K
s¹ | (11·34 − 24 − K)/11
s⁰ | 24 + K

Stability limit

All entries of the first column must be positive.

Kmax = 11·34 − 24 = 350

Frequency of oscillation

11·(jω)² + 24 + 350 = 0 → ω² = 374/11 = 34
ω = 5.831 rad/s

Answer

ω = 5.831 rad/s
5.Step response

Answer: %OS = 11.93 %

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 = 39 → ωₙ = 6.245 rad/s
2ζωₙ = 7 → ζ = 7/(2·6.245) = 0.5604

Result

%OS = 100·e−ζπ/√(1−ζ²) = 100·e−2.126 = 11.93 %

Answer

%OS = 11.93 %
6.Step response

Answer: ts = 0.6667 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) = 115/(s² + 12s + 115): ωₙ = 10.72 rad/s, ζ = 0.5595

Settling time

ζωₙ = 12/2 = 6
ts ≈ 4/(ζωₙ) = 0.6667 s

Answer

ts = 0.6667 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: Kmax = 990

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

s³ + 15s² + 74s + 120 + K = 0

Routh array

s³: 1, 74
s²: 15, 120 + K
s¹: (15·74 − 120 − K)/15
s⁰: 120 + K

Condition

The s¹ entry must stay positive.

K < 15·74 − 120 = 990

Answer

Kmax = 990

At the limit the loop oscillates at ω = √74 = 8.602 rad/s.

8.Stability: Routh–Hurwitz & root locus

Answer: K = 1

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.

Break-away point

dK/ds = 0 for K = −(s + 4)(s + 6) gives 2s + 10 = 0.

sb = −5

Magnitude condition

K = |−5 + 4|·|−5 + 6| = 1·1 = 1

Answer

K = 1

For this gain the closed loop is critically damped (ζ = 1).

9.Steady-state error & PID control

Answer: K = 2

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 = 1/0.5 = 2

Gain

K = 1·Kv = 2

Answer

K = 2

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: tp = 0.8112 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 + 10) + 4·10 = s² + 10s + 40

ωₙ and ζ

ωₙ = √40 = 6.325 rad/s
ζ = 10/(2·6.325) = 0.7906

Peak time

ωd = 6.325·√(1 − 0.7906²) = 3.873 rad/s
tp = π/3.873 = 0.8112 s

Answer

tp = 0.8112 s
11.Bode plots & margins

Answer: ∠G(jω) = −99.1 °

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.

Add the angles

∠G = atan(30/9) − 90° − atan(30/4) = 73.3° − 90° − 82.41°

Result

∠G(j30) = −99.1°

Answer

∠G(jω) = −99.1°
12.Bode plots & margins

Answer: PM = 22.18 °

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 + 64ω² − 173056 = 0
ω² = (−64 + √(4096 + 4·173056))/2 = 385.2
ωc = 19.63 rad/s

Phase margin

∠L(jωc) = −90° − atan(19.63/8) = −157.8°
PM = 180° + ∠L = 22.18°

Answer

PM = 22.18°

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