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) = (6s + 33)/(s² + 4s + 4) and evaluate f(0.6).

F(s) = (6s + 33)/(s² + 4s + 4)t = 0.6 s
f(0.6) =
2.Laplace transformsHard
t (s)y(t)t = 0.6y(0.6) = ?

A system obeys y″ + 7y′ + 6y = 36·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(0.6).

y″ + 7y′ + 6y = 36·u(t)u(t) = 5y(0) = 0, y′(0) = 0t = 0.6 s
y(0.6) =
3.Block diagramsMedium
R(s)+11s + 8G(s)C(s)0.25H−

A plant G(s) = 11/(s + 8) 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) = 11/(s + 8)H = 0.25
τ = s
4.Block diagramsHard
R(s)+10controller+D(s)+3s + 8plantC(s)−

A proportional controller K = 10 drives the plant G(s) = 3/(s + 8) in a unity negative feedback loop. A disturbance D enters at the plant input (it is added to the controller output). With R = 0, a disturbance step of size 0.5 is applied. Find the steady-state output css.

K = 10G(s) = 3/(s + 8)D = 0.5R = 0
css =
5.Step responseMedium
t (s)y(t)1

A system has the transfer function G(s) = 25/(s² + 7s + 25) and receives a unit step input. Find the natural frequency ωn in rad/s.

G(s) = 25/(s² + 7s + 25)
ωn = rad/s
6.Step responseHard
k = 100 N/mc = 6 N·s/mm = 2 kgF = 20 Nx(t)

A mass m = 2 kg hangs on a spring k = 100 N/m with a viscous damper c = 6 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 20 N is suddenly applied. Find the damping ratio ζ.

m = 2 kgk = 100 N/mc = 6 N·s/mF = 20 N
ζ =
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 7s(s + 1)(s + 2)C(s)−

A unity-feedback loop has L(s) = K(s + 7)/(s(s + 1)(s + 2)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K(s + 7)/(s(s + 1)(s + 2))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4j2

The root locus of L(s) = K/((s + 1)(s + 2)(s + 3)) is sketched. Find the frequency ω at which the locus crosses the imaginary axis.

L(s) = K/((s + 1)(s + 2)(s + 3))
ω = rad/s
9.Steady-state error & PID controlMedium
r(t) = 10t+Ks(s + 2)C(s)−

A unity-feedback loop with G(s) = K/(s(s + 2)) must follow the ramp r(t) = 10t with a steady-state error of 0.2. Find the gain K.

G(s) = K/(s(s + 2))r(t) = 10tess = 0.2
K =
10.Steady-state error & PID controlHard
r(t) = 5t+2 + 1/s4(s + 1)(s + 10)C(s)−

A PI controller C(s) = 2 + 1/s drives the plant G(s) = 4/((s + 1)(s + 10)) in a unity-feedback loop. Find the steady-state error ess for the ramp input r(t) = 5t.

C(s) = 2 + 1/sG(s) = 4/((s + 1)(s + 10))r(t) = 5t
ess =
11.Bode plots & marginsMedium
vin​L = 10 mHC = 2.2 µFR = 5 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 5 Ω, with L = 10 mH and C = 2.2 µF. Find the centre (resonant) frequency f0 in Hz.

R = 5 ΩL = 10 mHC = 2.2 µF
f0 = Hz
12.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−40−200204060ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​

A unity-feedback loop has the open-loop transfer function L(s) = 120/(s(s + 10)). Find the phase margin PM in degrees.

L(s) = 120/(s(s + 10))
PM = °

3Solutions

1.Laplace transforms

Answer: f(0.6) = 5.602

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

6s + 33 = 6(s + 2) + 21
F(s) = 6/(s + 2) + 21/(s + 2)²

Inverse transform

f(t) = 6·e−2t + 21·t·e−2t
f(0.6) = (6 + 21·0.6)·e−2·0.6 = 5.602

Answer

f(0.6) = 5.602
2.Laplace transforms

Answer: y(0.6) = 10.41

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) = 180/(s(s + 1)(s + 6))

Partial fractions

A = 180/(1·6) = 30
B = 180/((−1)·5) = −36
C = 180/((−6)·(−5)) = 6

Output

y(t) = 30 − 36·e−t + 6·e−6t
y(0.6) = 10.41

Answer

y(0.6) = 10.41
3.Block diagrams

Answer: τ = 0.09302 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) = 11/(s + 8 + 2.75) = 11/(s + 10.75)

Result

τ = 1/10.75 = 0.09302 s

Answer

τ = 0.09302 s
4.Block diagrams

Answer: css = 0.03947

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) = 3/(s + 8 + 30)

Steady state

css = 0.5·3/38 = 0.03947

Answer

css = 0.03947

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

5.Step response

Answer: ωn = 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).

Compare with the standard form

The denominator must match s² + 2ζωₙs + ωₙ².

ωₙ² = 25 → ωₙ = 5 rad/s
2ζωₙ = 7 → ζ = 7/(2·5) = 0.7

Result

ωₙ = 5 rad/s

Answer

ωn = 5 rad/s
6.Step response

Answer: ζ = 0.2121

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 = √(k/m) = √(100/2) = 7.071 rad/s
ζ = c/(2√(km)) = 6/(2·√(100·2)) = 0.2121

Result

ζ = 0.2121

Answer

ζ = 0.2121
7.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 1.5

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 + 1)(s + 2) + K(s + 7) = s³ + 3s² + (2 + K)s + 7K

Routh array

s³: 1, 2 + K
s²: 3, 7K
s¹: (3·(2 + K) − 7K)/3
s⁰: 7K

Condition

6 − 4K > 0
K < 6/4 = 1.5

Answer

Kmax = 1.5

With the zero to the right of s = −3 the loop would stay stable for every K > 0.

8.Stability: Routh–Hurwitz & root locus

Answer: ω = 3.317 rad/s

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.

Critical gain

s³ + 6s² + 11s + 6 + K = 0
s¹: (6·11 − 6 − K)/6 = 0 → K = 60

Crossing frequency

6s² + 6 + 60 = 6s² + 66 = 0 → s² = −11
ω = √11 = 3.317 rad/s

Answer

ω = 3.317 rad/s
9.Steady-state error & PID control

Answer: K = 100

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 = 10/0.2 = 50

Gain

K = 2·Kv = 100

Answer

K = 100

A large gain makes this loop very lightly damped: ζ = 0.1. That conflict is why compensators are used.

10.Steady-state error & PID control

Answer: ess = 12.5

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).

Velocity constant

s·C(s) = 2s + 1 → 1 as s → 0.

Kv = 1·4/10 = 0.4

Ramp error

ess = 5/0.4 = 12.5

Answer

ess = 12.5

The PI controller removes the step error completely; a ramp is followed with a constant lag.

11.Bode plots & margins

Answer: f0 = 1073 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/√(0.01·2.2 × 10⁻⁶) = 6742 rad/s
f₀ = ω₀/(2π) = 1073 Hz

Answer

f0 = 1073 Hz
12.Bode plots & margins

Answer: PM = 48.19 °

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 + 100ω² − 14400 = 0
ω² = (−100 + √(10000 + 4·14400))/2 = 80
ωc = 8.944 rad/s

Phase margin

∠L(jωc) = −90° − atan(8.944/10) = −131.8°
PM = 180° + ∠L = 48.19°

Answer

PM = 48.19°

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