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 = 0.2f(0.2) = ?

Find f(t) for F(s) = 30/(s(s + 4)(s + 6)) by partial fractions, and evaluate f(0.2).

F(s) = 30/(s(s + 4)(s + 6))t = 0.2 s
f(0.2) =
2.Laplace transformsHard
t (s)y(t)t = 0.2y(0.2) = ?

A system obeys y″ + 11y′ + 30y = 30·u(t), with y(0) = 6 and y′(0) = 0. At t = 0 a step input u(t) = 5 is applied. Use the Laplace transform to find y(0.2).

y″ + 11y′ + 30y = 30·u(t)y(0) = 6y′(0) = 0u(t) = 5t = 0.2 s
y(0.2) =
3.Block diagramsMedium
R(s)+17s + 2G(s)C(s)2H−

A plant G(s) = 17/(s + 2) is controlled in a negative feedback loop whose sensor has the gain H = 2. Find the time constant τ of the closed loop.

G(s) = 17/(s + 2)H = 2
τ = s
4.Block diagramsHard
R(s)+K(s + 2)(s + 3)(s + 5)G(s)C(s)−

The plant G(s) = K/((s + 2)(s + 3)(s + 5)) 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 + 2)(s + 3)(s + 5))unity negative feedback
ω = rad/s
5.Step responseMedium
t (s)y(t)3

A system has the transfer function G(s) = 201/(s² + 10s + 67) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.

G(s) = 201/(s² + 10s + 67)
σ =
6.Step responseHard
+−5 VR = 181 ΩL = 500 mHC = 22 µFvC​

A series RLC circuit with R = 181 Ω, L = 500 mH and C = 22 µF is connected to a 5 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the undamped natural frequency ωn in rad/s.

R = 181 ΩL = 500 mHC = 22 µFV = 5 V
ωn = rad/s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 11s(s + 3)(s + 4)C(s)−

A unity-feedback loop has L(s) = K(s + 11)/(s(s + 3)(s + 4)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K(s + 11)/(s(s + 3)(s + 4))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−0.5−1−1.5−2−2.5−3−3.5−4−4.5−50.5j0.5j1−1

For the loop L(s) = K/(s(s + 3)(s + 4)), find the gain K that puts a closed-loop pole at s = −1.

L(s) = K/(s(s + 3)(s + 4))s = −1
K =
9.Steady-state error & PID controlMedium
R(s)+K = 200.510s + 1C(s)−

A process G(s) = 0.5/(10s + 1) is controlled by a proportional gain K = 20 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.

G(s) = 0.5/(10s + 1)K = 20
ess = %
10.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s10s + 1C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 10/(s + 1) in a unity-feedback loop. The closed loop must have ωₙ = 8 rad/s and ζ = 0.6. Find the proportional gain Kp.

G(s) = 10/(s + 1)ωₙ = 8 rad/sζ = 0.6
Kp =
11.Bode plots & marginsMedium
vin​L = 20 mHC = 0.47 µFR = 10 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 10 Ω, with L = 20 mH and C = 0.47 µF. Find the −3 dB bandwidth BW in Hz.

R = 10 ΩL = 20 mHC = 0.47 µF
BW = Hz
12.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−270−180−900ω180​

A unity-feedback loop has the open-loop transfer function L(s) = 636/(s(s + 7)(s + 9)). Find the gain margin GM in dB.

L(s) = 636/(s(s + 7)(s + 9))
GM = dB

3Solutions

1.Laplace transforms

Answer: f(0.2) = 0.318

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 + 4) + C/(s + 6)
A = 30/(4·6) = 1.25
B = 30/((−4)·2) = −3.75
C = 30/((−6)·(−2)) = 2.5

Inverse transform

f(t) = 1.25 − 3.75·e−4t + 2.5·e−6t

Evaluate

f(0.2) = 1.25 − 3.75·e−4·0.2 + 2.5·e−6·0.2 = 0.318

Answer

f(0.2) = 0.318
2.Laplace transforms

Answer: y(0.2) = 5.701

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 + 11s + 30)·Y(s) = 6s + 66 + 150/s
Y(s) = (6s² + 66s + 150)/(s(s + 5)(s + 6))

Partial fractions

A = 150/(5·6) = 5
B = −30/((−5)·1) = 6
C = −30/((−6)·(−1)) = −5

Result

y(t) = 5 + 6·e−5t − 5·e−6t
y(0.2) = 5.701

Answer

y(0.2) = 5.701

Check: y(0) = A + B + C = 6.

3.Block diagrams

Answer: τ = 0.02778 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) = 17/(s + 2 + 34) = 17/(s + 36)

Result

τ = 1/36 = 0.02778 s

Answer

τ = 0.02778 s
4.Block diagrams

Answer: ω = 5.568 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 + 10s² + 31s + 30 + K = 0

Routh array

s³ | 1 31
s² | 10 30 + K
s¹ | (10·31 − 30 − K)/10
s⁰ | 30 + K

Stability limit

All entries of the first column must be positive.

Kmax = 10·31 − 30 = 280

Frequency of oscillation

10·(jω)² + 30 + 280 = 0 → ω² = 310/10 = 31
ω = 5.568 rad/s

Answer

ω = 5.568 rad/s
5.Step response

Answer: σ = −5

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 = 67 → ωₙ = 8.185 rad/s
2ζωₙ = 10 → ζ = 10/(2·8.185) = 0.6108

Result

σ = −ζωₙ = −10/2 = −5

Answer

σ = −5
6.Step response

Answer: ωn = 301.5 rad/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 = 1/√(LC) = 1/√(0.5·2.2 × 10⁻⁵) = 301.5 rad/s
ζ = (R/2)·√(C/L) = (181/2)·√(2.2 × 10⁻⁵/0.5) = 0.6003

Result

ωₙ = 301.5 rad/s

Answer

ωn = 301.5 rad/s
7.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 21

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 + 11) = s³ + 7s² + (12 + K)s + 11K

Routh array

s³: 1, 12 + K
s²: 7, 11K
s¹: (7·(12 + K) − 11K)/7
s⁰: 11K

Condition

84 − 4K > 0
K < 84/4 = 21

Answer

Kmax = 21

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: K = 6

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.

Magnitude condition

K = |s|·|s + 3|·|s + 4|

Evaluate at the point

K = 1·2·3 = 6

Answer

K = 6

Check: s(s + 3)(s + 4) + K is zero at s = −1.

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 = 20·0.5 = 10

Step error

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

Answer

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

Answer: Kp = 0.86

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

1 + C(s)G(s) = 0, multiplied by s(s + 1).

s² + (1 + 10Kp)s + 10Ki

Target polynomial

s² + 2·0.6·8s + 8² = s² + 9.6s + 64

Match the coefficients

1 + 10Kp = 9.6 → Kp = 0.86

Answer

Kp = 0.86

The integral term makes the loop type 1, so a step is followed without steady-state error.

11.Bode plots & margins

Answer: BW = 79.58 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/√(0.02·4.7 × 10⁻⁷) = 10314 rad/s
f₀ = ω₀/(2π) = 1642 Hz

Bandwidth

BW = R/L = 10/0.02 = 500 rad/s = 79.58 Hz

Answer

BW = 79.58 Hz
12.Bode plots & margins

Answer: GM = 4 dB

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/7)·(ω/9) = 1 → ω180 = √(7·9) = 7.937 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 49)·√(ω² + 81)) = K/(7·9·(7 + 9)) = K/1008

Gain margin

GM = 20·log₁₀(1008/636) = 4 dB

Answer

GM = 4 dB

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