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 = 0.5f(0.5) = ?

Find f(t) for F(s) = 30/(s(s + 2)(s + 5)) by partial fractions, and evaluate f(0.5).

F(s) = 30/(s(s + 2)(s + 5))t = 0.5 s
f(0.5) =
2.Laplace transformsHard
t (s)y(t)t = 1.2y(1.2) = ?

A system obeys y″ + 7y′ + 6y = 9·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 5 for t ≥ 0. Use the Laplace transform to find the output y(1.2).

y″ + 7y′ + 6y = 9·u(t)u(t) = 5y(0) = 0, y′(0) = 0t = 1.2 s
y(1.2) =
3.Block diagramsMedium
R(s)+K(s + 2)(s + 8)G(s)C(s)−

The plant G(s) = K/((s + 2)(s + 8)) 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 + 2)(s + 8))unity negative feedback
K =
4.Block diagramsHard
R(s)+64s(s + 5)G(s)C(s)1H−

The plant G(s) = 64/(s(s + 5)) is in a negative feedback loop with sensor gain H = 1. The closed-loop poles are complex. Find their imaginary part ωd (the positive value), in rad/s.

G(s) = 64/(s(s + 5))H = 1
ωd = rad/s
5.Step responseMedium
t (s)y(t)1

A system has the transfer function G(s) = 53/(s² + 3s + 53) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.

G(s) = 53/(s² + 3s + 53)
σ =
6.Step responseHard
R(s)+10s(s + 4)G(s)C(s)−

A unity negative feedback loop has G(s) = 10/(s(s + 4)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 10/(s(s + 4))
ts = s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 2)(s + 3)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 2)(s + 3)). 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 + 2)(s + 3))
ω = rad/s
8.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10−12j2

The root locus of L(s) = K(s + 10)/(s(s + 2)) leaves the real axis between s = 0 and s = −2, circles round and joins the real axis again to the left of the zero. Find the break-away point sb.

L(s) = K(s + 10)/(s(s + 2))
sb =
9.Steady-state error & PID controlMedium
r(t) = 5t+10(s + 1)s(s + 4)(s + 5)C(s)−

A ramp r(t) = 5t is applied to a unity-feedback loop with G(s) = 10(s + 1)/(s(s + 4)(s + 5)). Find the steady-state error ess.

G(s) = 10(s + 1)/(s(s + 4)(s + 5))r(t) = 5t
ess =
10.Steady-state error & PID controlHard
R(s)+K = 501s(s + 8)C(s)−

A proportional controller K = 50 drives the plant G(s) = 1/(s(s + 8)) in a unity-feedback loop. Find the percent overshoot %OS of the step response.

G(s) = 1/(s(s + 8))K = 50
%OS = %
11.Bode plots & marginsMedium
vin​L = 100 mHC = 2.2 µFR = 20 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 20 Ω, with L = 100 mH and C = 2.2 µF. Find the centre (resonant) frequency f0 in Hz.

R = 20 ΩL = 100 mHC = 2.2 µF
f0 = Hz
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 + 3)(s + 8)). Find the gain K that gives a gain margin of 3 dB.

L(s) = K/(s(s + 3)(s + 8))GM = 3 dB
K =

3Solutions

1.Laplace transforms

Answer: f(0.5) = 1.325

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 + 2) + C/(s + 5)
A = 30/(2·5) = 3
B = 30/((−2)·3) = −5
C = 30/((−5)·(−3)) = 2

Inverse transform

f(t) = 3 − 5·e−2t + 2·e−5t

Evaluate

f(0.5) = 3 − 5·e−2·0.5 + 2·e−5·0.5 = 1.325

Answer

f(0.5) = 1.325
2.Laplace transforms

Answer: y(1.2) = 4.79

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) = 45/(s(s + 1)(s + 6))

Partial fractions

A = 45/(1·6) = 7.5
B = 45/((−1)·5) = −9
C = 45/((−6)·(−5)) = 1.5

Output

y(t) = 7.5 − 9·e−t + 1.5·e−6t
y(1.2) = 4.79

Answer

y(1.2) = 4.79
3.Block diagrams

Answer: K = 9

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 + 2)(s + 8)) = 0 → s² + 10s + 16 + K = 0

Repeated root

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

10² = 4(16 + K) → K = 100/4 − 16 = 9

Answer

K = 9
Both closed-loop poles are then at s = −5.
4.Block diagrams

Answer: ωd = 7.599 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 + 5) + 64·1 = 0 → s² + 5s + 64 = 0

Roots

s = −2.5 ± j·√(64 − 6.25) = −2.5 ± j7.599
ωd = 7.599 rad/s

Answer

ωd = 7.599 rad/s
5.Step response

Answer: σ = −1.5

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 = 53 → ωₙ = 7.28 rad/s
2ζωₙ = 3 → ζ = 3/(2·7.28) = 0.206

Result

σ = −ζωₙ = −3/2 = −1.5

Answer

σ = −1.5
6.Step response

Answer: ts = 2 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) = 10/(s² + 4s + 10): ωₙ = 3.162 rad/s, ζ = 0.6325

Settling time

ζωₙ = 4/2 = 2
ts ≈ 4/(ζωₙ) = 2 s

Answer

ts = 2 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: ω = 2.449 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 + 5s² + 6s + K = 0
s³: 1, 6
s²: 5, K
s¹: (5·6 − K)/5
s⁰: K

Critical gain

K = 5·6 = 30

Auxiliary equation

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

5s² + 30 = 0 → s = ±j√6
ω = 2.449 rad/s

Answer

ω = 2.449 rad/s
8.Stability: Routh–Hurwitz & root locus

Answer: sb = −1.056

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 + 2s)/(s + 10)

Stationary points

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

(2s + 2)(s + 10) − (s² + 2s) = s² + 20s + 20 = 0
s = −10 ± √80 = −1.056, −18.94

Break-away point

sb = −1.056

Answer

sb = −1.056

Between the two break points the locus is a circle centred on the zero, with radius √80 = 8.944.

9.Steady-state error & PID control

Answer: ess = 10

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.

Velocity constant

Kv = lim s·G(s) = 0.5

Ramp error

ess = 5/0.5 = 10

Answer

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

Answer: %OS = 11.59 %

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 + 8) + 50·1 = s² + 8s + 50

ωₙ and ζ

ωₙ = √50 = 7.071 rad/s
ζ = 8/(2·7.071) = 0.5657

Overshoot

%OS = 100·e−0.5657π/√(1 − 0.5657²) = 11.59 %

Answer

%OS = 11.59 %
11.Bode plots & margins

Answer: f0 = 339.3 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/√(0.1·2.2 × 10⁻⁶) = 2132 rad/s
f₀ = ω₀/(2π) = 339.3 Hz

Answer

f0 = 339.3 Hz
12.Bode plots & margins

Answer: K = 186.9

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/3)·(ω/8) = 1 → ω180 = √(3·8) = 4.899 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 9)·√(ω² + 64)) = K/(3·8·(3 + 8)) = K/264

Required gain

264/K = 103/20 = 1.413
K = 264/1.413 = 186.9

Answer

K = 186.9

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