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 = 0.6f(0.6) = ?

Find f(t) for F(s) = (2s + 7)/(s² + 4s + 4) and evaluate f(0.6).

F(s) = (2s + 7)/(s² + 4s + 4)t = 0.6 s
f(0.6) =
2.Laplace transformsHard
t (s)y(t)t = 0.1y(0.1) = ?

A system obeys y″ + 5y′ + 6y = 24·u(t), with y(0) = 3 and y′(0) = 0. At t = 0 a step input u(t) = 5 is applied. Use the Laplace transform to find y(0.1).

y″ + 5y′ + 6y = 24·u(t)y(0) = 3y′(0) = 0u(t) = 5t = 0.1 s
y(0.1) =
3.Block diagramsMedium
R(s)+15s + 6G(s)C(s)0.25H−

A plant G(s) = 15/(s + 6) is controlled in a negative feedback loop whose sensor has the gain H = 0.25. Find the time constant τ of the closed loop.

G(s) = 15/(s + 6)H = 0.25
τ = s
4.Block diagramsHard
R(s)+38s(s + 3)G(s)C(s)2H−

The plant G(s) = 38/(s(s + 3)) is in a negative feedback loop with sensor gain H = 2. The closed-loop poles are complex. Find their imaginary part ωd (the positive value), in rad/s.

G(s) = 38/(s(s + 3))H = 2
ωd = rad/s
5.Step responseMedium
t (s)y(t)1ts​ = ?

A system has the transfer function G(s) = 60/(s² + 9s + 60) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 60/(s² + 9s + 60)
ts = s
6.Step responseHard
R(s)+235s(s + 8)G(s)C(s)−

A unity negative feedback loop has G(s) = 235/(s(s + 8)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 235/(s(s + 8))
ts = s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 9s(s + 3)(s + 4)C(s)−

A unity-feedback loop has L(s) = K(s + 9)/(s(s + 3)(s + 4)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K(s + 9)/(s(s + 3)(s + 4))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10j2

The root locus of L(s) = K(s + 9)/(s(s + 6)) leaves the real axis between s = 0 and s = −6, circles round and joins the real axis again to the left of the zero. Find the break-away point sb.

L(s) = K(s + 9)/(s(s + 6))
sb =
9.Steady-state error & PID controlMedium
R(s)+K = 42.52s + 1C(s)−

A process G(s) = 2.5/(2s + 1) is controlled by a proportional gain K = 4 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.

G(s) = 2.5/(2s + 1)K = 4
ess = %
10.Steady-state error & PID controlHard
R(s)+K = 105s(s + 10)C(s)−

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

G(s) = 5/(s(s + 10))K = 10
%OS = %
11.Bode plots & marginsMedium
|H| (dB)0.10.111101010010010001000−40−200204060ω (rad/s)φ (°)ω (rad/s)−135−90−450ω = 1

Find the phase ∠G(jω) of G(s) = 46(s + 15)/(s(s + 14)) at ω = 1 rad/s, in degrees.

G(s) = 46(s + 15)/(s(s + 14))ω = 1 rad/s
∠G(jω) = °
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) = K/(s(s + 4)(s + 9)). Find the gain K that gives a gain margin of 6 dB.

L(s) = K/(s(s + 4)(s + 9))GM = 6 dB
K =

3Solutions

1.Laplace transforms

Answer: f(0.6) = 1.145

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 + 7 = 2(s + 2) + 3
F(s) = 2/(s + 2) + 3/(s + 2)²

Inverse transform

f(t) = 2·e−2t + 3·t·e−2t
f(0.6) = (2 + 3·0.6)·e−2·0.6 = 1.145

Answer

f(0.6) = 1.145
2.Laplace transforms

Answer: y(0.1) = 3.433

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 + 6)·Y(s) = 3s + 15 + 120/s
Y(s) = (3s² + 15s + 120)/(s(s + 2)(s + 3))

Partial fractions

A = 120/(2·3) = 20
B = 102/((−2)·1) = −51
C = 102/((−3)·(−1)) = 34

Result

y(t) = 20 − 51·e−2t + 34·e−3t
y(0.1) = 3.433

Answer

y(0.1) = 3.433

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

3.Block diagrams

Answer: τ = 0.1026 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).

Closed-loop transfer function

T(s) = G/(1 + G·H) = 15/(s + 6 + 3.75) = 15/(s + 9.75)

Result

τ = 1/9.75 = 0.1026 s

Answer

τ = 0.1026 s
4.Block diagrams

Answer: ωd = 8.588 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

s(s + 3) + 38·2 = 0 → s² + 3s + 76 = 0

Roots

s = −1.5 ± j·√(76 − 2.25) = −1.5 ± j8.588
ωd = 8.588 rad/s

Answer

ωd = 8.588 rad/s
5.Step response

Answer: ts = 0.8889 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 = 60 → ωₙ = 7.746 rad/s
2ζωₙ = 9 → ζ = 9/(2·7.746) = 0.5809

Result

ζωₙ = 4.5 → ts ≈ 4/4.5 = 0.8889 s

Answer

ts = 0.8889 s
6.Step response

Answer: ts = 1 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) = 235/(s² + 8s + 235): ωₙ = 15.33 rad/s, ζ = 0.2609

Settling time

ζωₙ = 8/2 = 4
ts ≈ 4/(ζωₙ) = 1 s

Answer

ts = 1 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 = 42

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 + 3)(s + 4) + K(s + 9) = s³ + 7s² + (12 + K)s + 9K

Routh array

s³: 1, 12 + K
s²: 7, 9K
s¹: (7·(12 + K) − 9K)/7
s⁰: 9K

Condition

84 − 2K > 0
K < 84/2 = 42

Answer

Kmax = 42

With the zero to the right of s = −7 the loop would stay stable for every K > 0.

8.Stability: Routh–Hurwitz & root locus

Answer: sb = −3.804

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 + 6s)/(s + 9)

Stationary points

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

(2s + 6)(s + 9) − (s² + 6s) = s² + 18s + 54 = 0
s = −9 ± √27 = −3.804, −14.2

Break-away point

sb = −3.804

Answer

sb = −3.804

Between the two break points the locus is a circle centred on the zero, with radius √27 = 5.196.

9.Steady-state error & PID control

Answer: ess = 9.091 %

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.

Position constant

Kp = 4·2.5 = 10

Step error

ess = 100/(1 + 10) = 9.091 %

Answer

ess = 9.091 %
10.Steady-state error & PID control

Answer: %OS = 4.321 %

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) + 10·5 = s² + 10s + 50

ωₙ and ζ

ωₙ = √50 = 7.071 rad/s
ζ = 10/(2·7.071) = 0.7071

Overshoot

%OS = 100·e−0.7071π/√(1 − 0.7071²) = 4.321 %

Answer

%OS = 4.321 %
11.Bode plots & margins

Answer: ∠G(jω) = −90.27 °

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(1/15) − 90° − atan(1/14) = 3.814° − 90° − 4.086°

Result

∠G(j1) = −90.27°

Answer

∠G(jω) = −90.27°
12.Bode plots & margins

Answer: K = 234.6

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/4)·(ω/9) = 1 → ω180 = √(4·9) = 6 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 16)·√(ω² + 81)) = K/(4·9·(4 + 9)) = K/468

Required gain

468/K = 106/20 = 1.995
K = 468/1.995 = 234.6

Answer

K = 234.6

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