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 = 0.3f(0.3) = ?

Find f(t) for F(s) = 50/(s(s + 1)(s + 5)) by partial fractions, and evaluate f(0.3).

F(s) = 50/(s(s + 1)(s + 5))t = 0.3 s
f(0.3) =
2.Laplace transformsHard
ReIm−22

A system is described by the differential equation y″ + 4y′ + 13y = 16u, with input u(t) and output y(t). Find its transfer function G(s) = Y(s)/U(s) (zero initial conditions) and evaluate the impulse response g(t) at t = 0.25 s. Find g(0.25).

y″ + 4y′ + 13y = 16ut = 0.25 s
g(0.25) =
3.Block diagramsMedium
R(s)+6(s + 6)(s + 7)(s + 8)G(s)C(s)−

A unity negative feedback loop has G(s) = 6(s + 6)/((s + 7)(s + 8)). A step of size 1 is applied at the reference. Find the steady-state error ess.

G(s) = 6(s + 6)/((s + 7)(s + 8))R = 1
ess =
4.Block diagramsHard
R(s)+16s + 1G₁+2s + 6G₂C(s)0.5H₁−0.1H₂−

Reduce the block diagram with G₁ = 16/(s + 1), G₂ = 2/(s + 6), an inner negative feedback H₁ = 0.5 around G₂ and an outer negative feedback H₂ = 0.1. Find the DC gain T(0) of C(s)/R(s).

G₁ = 16/(s + 1)G₂ = 2/(s + 6)H₁ = 0.5H₂ = 0.1
T(0) =
5.Step responseMedium
t (s)y(t)0.5?

A system has the transfer function G(s) = 45.5/(s² + 8s + 91) and receives a unit step input. Find the percent overshoot %OS of the step response.

G(s) = 45.5/(s² + 8s + 91)
%OS = %
6.Step responseHard
+−24 VR = 42 ΩL = 50 mHC = 10 µFvC​

A series RLC circuit with R = 42 Ω, L = 50 mH and C = 10 µF is connected to a 24 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the time tp of the first voltage peak, in ms.

R = 42 ΩL = 50 mHC = 10 µFV = 24 V
tp = ms
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 1(s − 2)(s + 5)C(s)−

The plant G(s) = (s + 1)/((s − 2)(s + 5)) has an unstable pole at s = 2. A proportional gain K closes a unity-feedback loop around it. Find the smallest gain Kmin that makes the closed loop stable.

G(s) = (s + 1)/((s − 2)(s + 5))
Kmin =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10−12j2

A unity-feedback loop has L(s) = K/(s(s + 4)(s + 10)). Find the point σa where the asymptotes of the root locus cross the real axis.

L(s) = K/(s(s + 4)(s + 10))
σa =
9.Steady-state error & PID controlMedium
R(s)+K(s + 1)(s + 5)C(s)−

A unity-feedback loop has G(s) = K/((s + 1)(s + 5)). Find the gain K that gives a steady-state error of 4 % for a step input.

G(s) = K/((s + 1)(s + 5))ess = 4 %
K =
10.Steady-state error & PID controlHard
R(s)+K = 85s(s + 5)C(s)−

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

G(s) = 5/(s(s + 5))K = 8
%OS = %
11.Bode plots & marginsMedium
vin​R = 2.2 kΩC = 2.2 nFvout​

The RC low-pass filter shown has R = 2.2 kΩ, C = 2.2 nF. A sinusoidal input of 6.6 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 2.2 kΩC = 2.2 nFf = 6.6 kHz
φ = °
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) = 40/(s(s + 2)(s + 5)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 40/(s(s + 2)(s + 5))
ω180 = rad/s

3Solutions

1.Laplace transforms

Answer: f(0.3) = 1.298

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).

Partial fractions

F(s) = A/s + B/(s + 1) + C/(s + 5)
A = 50/(1·5) = 10
B = 50/((−1)·4) = −12.5
C = 50/((−5)·(−4)) = 2.5

Inverse transform

f(t) = 10 − 12.5·e−t + 2.5·e−5t

Evaluate

f(0.3) = 10 − 12.5·e−0.3 + 2.5·e−5·0.3 = 1.298

Answer

f(0.3) = 1.298
2.Laplace transforms

Answer: g(0.25) = 2.205

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).

Transfer function

G(s) = Y(s)/U(s) = 16/(s² + 4s + 13)

Partial fractions

s² + 4s + 13 = (s + 2)² + 3²
G(s) = (16/3)·3/((s + 2)² + 3²)

Impulse response

g(t) = e−2t·(5.333·sin 3t)
g(0.25) = 2.205

Answer

g(0.25) = 2.205
3.Block diagrams

Answer: ess = 0.6087

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) = 6·6/(7·8) = 0.6429

Steady-state error

ess = A/(1 + Kₚ) = 1/(1 + 0.6429) = 0.6087

Answer

ess = 0.6087
4.Block diagrams

Answer: T(0) = 3.137

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) = 2/(s + 7)

Outer loop

T(s) = G₁Gᵢ/(1 + G₁GᵢH₂) = 32/((s + 1)(s + 7) + 3.2)

DC gain

T(0) = 32/(1·7 + 3.2) = 32/10.2 = 3.137

Answer

T(0) = 3.137
5.Step response

Answer: %OS = 23.43 %

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 = 8 → ζ = 8/(2·9.539) = 0.4193

Result

%OS = 100·e−ζπ/√(1−ζ²) = 100·e−1.451 = 23.43 %

Answer

%OS = 23.43 %
6.Step response

Answer: tp = 2.326 ms

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 = 1/√(LC) = 1/√(0.05·1 × 10⁻⁵) = 1414 rad/s
ζ = (R/2)·√(C/L) = (42/2)·√(1 × 10⁻⁵/0.05) = 0.297

Result

ωd = 1350 rad/s
tp = π/ωd = 2.326 ms

Answer

tp = 2.326 ms
7.Stability: Routh–Hurwitz & root locus

Answer: Kmin = 10

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 − 2)(s + 5) + K(s + 1) = s² + (3 + K)s + (1K − 10)

Conditions

For s² + c₁s + c₀ all roots lie in the left half-plane exactly when c₁ > 0 and c₀ > 0.

3 + K > 0 → K > −3
1K − 10 > 0 → K > 10
Kmin = 10

Answer

Kmin = 10
8.Stability: Routh–Hurwitz & root locus

Answer: σa = −4.667

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.

Poles and zeros

poles: 0, −4, −10
n − m = 3 − 0 = 3

Centroid

σa = (−14)/3 = −4.667

Answer

σa = −4.667

The 3 asymptotes leave the centroid at angles of 60°, 180°, 300°.

9.Steady-state error & PID control

Answer: K = 120

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) = 0.04 → Kp = 24

Gain

K/5 = 24 → K = 120

Answer

K = 120

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

10.Steady-state error & PID control

Answer: %OS = 25.87 %

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) + 8·5 = s² + 5s + 40

ωₙ and ζ

ωₙ = √40 = 6.325 rad/s
ζ = 5/(2·6.325) = 0.3953

Overshoot

%OS = 100·e−0.3953π/√(1 − 0.3953²) = 25.87 %

Answer

%OS = 25.87 %
11.Bode plots & margins

Answer: φ = −11.35 °

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/(2200·2.2 × 10⁻⁹) = 206612 rad/s
fc = 32883 Hz → f/fc = 6600/32883 = 0.2007

Phase

φ = − atan(0.2007) = −11.35°

Answer

φ = −11.35°
12.Bode plots & margins

Answer: ω180 = 3.162 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/2)·(ω/5) = 1 → ω180 = √(2·5) = 3.162 rad/s

Answer

ω180 = 3.162 rad/s

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