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−1−2−3−4−5−61

F(s) = (6s + 36)/(s² + 5s + 6). Its inverse transform has the form f(t) = A·e−2t + B·e−3t. Find the coefficient A by partial fractions.

F(s) = (6s + 36)/(s² + 5s + 6)
A =
2.Laplace transformsHard
t (s)y(t)t = 0.2y(0.2) = ?

A system obeys y″ + 10y′ + 24y = 120·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(0.2).

y″ + 10y′ + 24y = 120·u(t)u(t) = 1y(0) = 0, y′(0) = 0t = 0.2 s
y(0.2) =
3.Block diagramsMedium
R(s)+K(s + 4)(s + 5)G(s)C(s)−

The plant G(s) = K/((s + 4)(s + 5)) is in a unity negative feedback loop. Find the gain K for which the closed loop is critically damped (two equal real poles).

G(s) = K/((s + 4)(s + 5))unity negative feedback
K =
4.Block diagramsHard
R(s)+9controller+D(s)+5s + 6plantC(s)−

A proportional controller K = 9 drives the plant G(s) = 5/(s + 6) in a unity negative feedback loop. A disturbance D enters at the plant input (it is added to the controller output). A reference step of size 2 and a disturbance step of size 0.5 are applied together. Find the steady-state output css.

K = 9G(s) = 5/(s + 6)D = 0.5R = 2
css =
5.Step responseMedium
t (s)y(t)1?

A system has the transfer function G(s) = 93/(s² + 14s + 93) and receives a unit step input. Find the percent overshoot %OS of the step response.

G(s) = 93/(s² + 14s + 93)
%OS = %
6.Step responseHard
+−24 VR = 126 ΩL = 100 mHC = 1 µFvC​

A series RLC circuit with R = 126 Ω, L = 100 mH and C = 1 µF is connected to a 24 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the peak capacitor voltage vC,max.

R = 126 ΩL = 100 mHC = 1 µFV = 24 V
vC,max = V
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 1)(s + 5)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 1)(s + 5)). The gain is raised until the closed loop is on the edge of stability. Find the frequency ω of the sustained oscillation.

L(s) = K/(s(s + 1)(s + 5))
ω = rad/s
8.Stability: Routh–Hurwitz & root locusHard
ReIm−0.5−1−1.5−2−2.5−3−3.5−40.5j0.5j1

The root locus of L(s) = K/(s(s + 2)(s + 3)) leaves the real axis between s = 0 and s = −2. Find the break-away point sb.

L(s) = K/(s(s + 2)(s + 3))
sb =
9.Steady-state error & PID controlMedium
r(t) = 1t+Ks(s + 5)C(s)−

A unity-feedback loop with G(s) = K/(s(s + 5)) must follow the ramp r(t) = 1t with a steady-state error of 0.2. Find the gain K.

G(s) = K/(s(s + 5))r(t) = 1tess = 0.2
K =
10.Steady-state error & PID controlHard
R(s)+K = 164s(s + 10)C(s)−

A proportional controller K = 16 drives the plant G(s) = 4/(s(s + 10)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 4/(s(s + 10))K = 16
ts = s
11.Bode plots & marginsMedium
vin​L = 5 mHC = 10 µFR = 10 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 10 Ω, with L = 5 mH and C = 10 µF. Find the centre (resonant) frequency f0 in Hz.

R = 10 ΩL = 5 mHC = 10 µF
f0 = Hz
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) = 144/(s(s + 8)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 144/(s(s + 8))
ωc = rad/s

3Solutions

1.Laplace transforms

Answer: A = 24

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 + 5s + 6 = (s + 2)(s + 3)
F(s) = A/(s + 2) + B/(s + 3)

Cover-up rule

Multiply by (s + 2) and set s = −2: the B-term vanishes.

A = (6s + 36)/(s + 3) |s = −2 = 24/1 = 24

Answer

A = 24
(B = −18, so f(t) = 24·e−2t − 18·e−3t)
2.Laplace transforms

Answer: y(0.2) = 1.272

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) = 120/(s(s + 4)(s + 6))

Partial fractions

A = 120/(4·6) = 5
B = 120/((−4)·2) = −15
C = 120/((−6)·(−2)) = 10

Output

y(t) = 5 − 15·e−4t + 10·e−6t
y(0.2) = 1.272

Answer

y(0.2) = 1.272
3.Block diagrams

Answer: K = 0.25

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

1 + K/((s + 4)(s + 5)) = 0 → s² + 9s + 20 + K = 0

Repeated root

A quadratic s² + βs + γ has a double root when β² = 4γ.

9² = 4(20 + K) → K = 81/4 − 20 = 0.25

Answer

K = 0.25
Both closed-loop poles are then at s = −4.5.
4.Block diagrams

Answer: css = 1.814

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).

Transfer functions

C/D = G/(1 + K·G) = 5/(s + 6 + 45)
C/R = K·G/(1 + K·G) = 45/(s + 6 + 45)

Steady state

css = 2·45/51 + 0.5·5/51 = 1.814

Answer

css = 1.814

A larger K reduces the effect of the disturbance, but never to zero with a proportional controller.

5.Step response

Answer: %OS = 3.632 %

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 = 93 → ωₙ = 9.644 rad/s
2ζωₙ = 14 → ζ = 14/(2·9.644) = 0.7259

Result

%OS = 100·e−ζπ/√(1−ζ²) = 100·e−3.315 = 3.632 %

Answer

%OS = 3.632 %
6.Step response

Answer: vC,max = 36.67 V

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/√(0.1·1 × 10⁻⁶) = 3162 rad/s
ζ = (R/2)·√(C/L) = (126/2)·√(1 × 10⁻⁶/0.1) = 0.1992

Result

%OS = 52.8 %
vC,max = 24·(1 + 0.528) = 36.67 V

Answer

vC,max = 36.67 V
7.Stability: Routh–Hurwitz & root locus

Answer: ω = 2.236 rad/s

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.

Routh array

s³ + 6s² + 5s + K = 0
s³: 1, 5
s²: 6, K
s¹: (6·5 − K)/6
s⁰: K

Critical gain

K = 6·5 = 30

Auxiliary equation

At the critical gain the s¹ row is zero; the s² row gives the poles on the imaginary axis.

6s² + 30 = 0 → s = ±j√5
ω = 2.236 rad/s

Answer

ω = 2.236 rad/s
8.Stability: Routh–Hurwitz & root locus

Answer: sb = −0.7847

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) = −s(s + 2)(s + 3) = −(s³ + 5s² + 6s)

Stationary points

dK/ds = −(3s² + 10s + 6) = 0
s = (−10 ± √(100 − 72))/6 = −0.7847, −2.549

Pick the point on the locus

The locus lies on the real axis between 0 and −2; the other root is not on the locus.

sb = −0.7847

Answer

sb = −0.7847
9.Steady-state error & PID control

Answer: K = 25

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 = 1/0.2 = 5

Gain

K = 5·Kv = 25

Answer

K = 25

A large gain makes this loop very lightly damped: ζ = 0.5. That conflict is why compensators are used.

10.Steady-state error & PID control

Answer: ts = 0.8 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) + 16·4 = s² + 10s + 64

ωₙ and ζ

ωₙ = √64 = 8 rad/s
ζ = 10/(2·8) = 0.625

Settling time

ζωₙ = 10/2 = 5
ts ≈ 4/5 = 0.8 s

Answer

ts = 0.8 s

The settling time does not depend on K here: raising K only makes the response oscillate faster.

11.Bode plots & margins

Answer: f0 = 711.8 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/√(0.005·1 × 10⁻⁵) = 4472 rad/s
f₀ = ω₀/(2π) = 711.8 Hz

Answer

f0 = 711.8 Hz
12.Bode plots & margins

Answer: ωc = 10.75 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.

Gain crossover

ω⁴ + 64ω² − 20736 = 0
ω² = (−64 + √(4096 + 4·20736))/2 = 115.5
ωc = 10.75 rad/s

Answer

ωc = 10.75 rad/s

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