Control Engineering 2: Feedback Control Design: cram pack

Formula sheets, the key method steps for every topic and 10 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.Step responseMedium
t (s)y(t)1ts​ = ?

A system has the transfer function G(s) = 45/(s² + 9s + 45) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 45/(s² + 9s + 45)
ts = s
2.Step responseHard
k = 500 N/mc = 9 N·s/mm = 2 kgF = 20 Nx(t)

A mass m = 2 kg hangs on a spring k = 500 N/m with a viscous damper c = 9 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 largest displacement xmax in mm.

m = 2 kgk = 500 N/mc = 9 N·s/mF = 20 N
xmax = mm
3.Bode plots & marginsMedium
vin​L = 20 mHC = 4.7 µFR = 20 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 20 Ω, with L = 20 mH and C = 4.7 µF. Find the −3 dB bandwidth BW in Hz.

R = 20 ΩL = 20 mHC = 4.7 µF
BW = Hz
4.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) = 762/(s(s + 6)(s + 10)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 762/(s(s + 6)(s + 10))
ω180 = rad/s
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 1(s − 4)(s + 5)C(s)−

The plant G(s) = (s + 1)/((s − 4)(s + 5)) has an unstable pole at s = 4. 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 − 4)(s + 5))
Kmin =
6.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6j2j4j6

The root locus of L(s) = K/((s + 2)(s + 3)(s + 5)) is sketched. Find the frequency ω at which the locus crosses the imaginary axis.

L(s) = K/((s + 2)(s + 3)(s + 5))
ω = rad/s
7.Steady-state error & PID controlMedium
R(s)+K = 10.55s + 1C(s)−

A process G(s) = 0.5/(5s + 1) is controlled by a proportional gain K = 1 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/(5s + 1)K = 1
ess = %
8.Steady-state error & PID controlHard
r(t) = 2t+1 + 10/s20(s + 5)(s + 10)C(s)−

A PI controller C(s) = 1 + 10/s drives the plant G(s) = 20/((s + 5)(s + 10)) in a unity-feedback loop. Find the steady-state error ess for the ramp input r(t) = 2t.

C(s) = 1 + 10/sG(s) = 20/((s + 5)(s + 10))r(t) = 2t
ess =
9.Block diagramsMedium
R(s)+13K+1s + 4G(s)C(s)2H₁−−

The plant G(s) = 1/(s + 4) has an inner (minor) negative feedback loop with gain H₁ = 2. A controller gain K = 13 and an outer unity negative feedback loop close the system. Find the closed-loop DC gain T(0).

G(s) = 1/(s + 4)H₁ = 2K = 13
T(0) =
10.Block diagramsHard
R(s)+16s(s + 5)G(s)C(s)1H−

The plant G(s) = 16/(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) = 16/(s(s + 5))H = 1
ωd = rad/s

3Solutions

1.Step response

Answer: ts = 0.8889 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 = 45 → ωₙ = 6.708 rad/s
2ζωₙ = 9 → ζ = 9/(2·6.708) = 0.6708

Result

ζωₙ = 4.5 → ts ≈ 4/4.5 = 0.8889 s

Answer

ts = 0.8889 s
2.Step response

Answer: xmax = 65.46 mm

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) = √(500/2) = 15.81 rad/s
ζ = c/(2√(km)) = 9/(2·√(500·2)) = 0.1423

Result

xst = F/k = 20/500 m = 40 mm
%OS = 63.66 %
xmax = 40·(1 + 0.6366) = 65.46 mm

Answer

xmax = 65.46 mm
3.Bode plots & margins

Answer: BW = 159.2 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 × 10⁻⁶) = 3262 rad/s
f₀ = ω₀/(2π) = 519.1 Hz

Bandwidth

BW = R/L = 20/0.02 = 1000 rad/s = 159.2 Hz

Answer

BW = 159.2 Hz
4.Bode plots & margins

Answer: ω180 = 7.746 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/6)·(ω/10) = 1 → ω180 = √(6·10) = 7.746 rad/s

Answer

ω180 = 7.746 rad/s
5.Stability: Routh–Hurwitz & root locus

Answer: Kmin = 20

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 − 4)(s + 5) + K(s + 1) = s² + (1 + K)s + (1K − 20)

Conditions

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

1 + K > 0 → K > −1
1K − 20 > 0 → K > 20
Kmin = 20

Answer

Kmin = 20
6.Stability: Routh–Hurwitz & root locus

Answer: ω = 5.568 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 + 10s² + 31s + 30 + K = 0
s¹: (10·31 − 30 − K)/10 = 0 → K = 280

Crossing frequency

10s² + 30 + 280 = 10s² + 310 = 0 → s² = −31
ω = √31 = 5.568 rad/s

Answer

ω = 5.568 rad/s
7.Steady-state error & PID control

Answer: ess = 66.67 %

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 = 1·0.5 = 0.5

Step error

ess = 100/(1 + 0.5) = 66.67 %

Answer

ess = 66.67 %
8.Steady-state error & PID control

Answer: ess = 0.5

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) = 1s + 10 → 10 as s → 0.

Kv = 10·20/50 = 4

Ramp error

ess = 2/4 = 0.5

Answer

ess = 0.5

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

9.Block diagrams

Answer: T(0) = 0.6842

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 = G/(1 + G·H₁) = 1/(s + 4 + 1·2) = 1/(s + 6)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 13/(s + 6 + 13) = 13/(s + 19)

Result

T(0) = 13/19 = 0.6842

Answer

T(0) = 0.6842
10.Block diagrams

Answer: ωd = 3.122 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) + 16·1 = 0 → s² + 5s + 16 = 0

Roots

s = −2.5 ± j·√(16 − 6.25) = −2.5 ± j3.122
ωd = 3.122 rad/s

Answer

ωd = 3.122 rad/s

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