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)2

A system has the transfer function G(s) = 54/(s² + 6s + 27) and receives a unit step input. Find the natural frequency ωn in rad/s.

G(s) = 54/(s² + 6s + 27)
ωn = rad/s
2.Step responseHard
k = 100 N/mc = 18 N·s/mm = 20 kgF = 10 Nx(t)

A mass m = 20 kg hangs on a spring k = 100 N/m with a viscous damper c = 18 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 10 N is suddenly applied. Find the damping ratio ζ.

m = 20 kgk = 100 N/mc = 18 N·s/mF = 10 N
ζ =
3.Bode plots & marginsMedium
vin​R = 1 kΩC = 2.2 nFvout​

The RC low-pass filter shown has R = 1 kΩ, C = 2.2 nF. A sinusoidal input of 18 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 1 kΩC = 2.2 nFf = 18 kHz
φ = °
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) = 434/(s(s + 6)(s + 7)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 434/(s(s + 6)(s + 7))
ω180 = rad/s
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 4)(s + 5)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 4)(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 + 4)(s + 5))
ω = rad/s
6.Stability: Routh–Hurwitz & root locusHard
ReIm−0.5−1−1.5−2−2.5−3−3.5−40.5j0.5j1sb​

The root locus of L(s) = K/((s + 1)(s + 3)) is shown. Find the gain K at the break-away point sb, where the two closed-loop poles meet on the real axis.

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

A unity-feedback loop has G(s) = K/((s + 1)(s + 2)). Find the gain K that gives a steady-state error of 2 % for a step input.

G(s) = K/((s + 1)(s + 2))ess = 2 %
K =
8.Steady-state error & PID controlHard
R(s)+25 + Kd​s10s(s + 2)C(s)−

A PD controller C(s) = 25 + Kd·s drives the plant G(s) = 10/(s(s + 2)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.9.

G(s) = 10/(s(s + 2))Kp = 25ζ = 0.9
Kd = s
9.Block diagramsMedium
R(s)+3s + 4G(s)C(s)0.1H−

A plant G(s) = 3/(s + 4) is controlled in a negative feedback loop whose sensor has the gain H = 0.1. A step of size 2 is applied at R. Find the steady-state output css.

G(s) = 3/(s + 4)H = 0.1R = 2
css =
10.Block diagramsHard
R(s)+9s + 2G₁+10s + 6G₂C(s)4H₁−0.1H₂−

Reduce the block diagram with G₁ = 9/(s + 2), G₂ = 10/(s + 6), an inner negative feedback H₁ = 4 around G₂ and an outer negative feedback H₂ = 0.1. Find the DC gain T(0) of C(s)/R(s).

G₁ = 9/(s + 2)G₂ = 10/(s + 6)H₁ = 4H₂ = 0.1
T(0) =

3Solutions

1.Step response

Answer: ωn = 5.196 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 = 27 → ωₙ = 5.196 rad/s
2ζωₙ = 6 → ζ = 6/(2·5.196) = 0.5774

Result

ωₙ = 5.196 rad/s

Answer

ωn = 5.196 rad/s
2.Step response

Answer: ζ = 0.2012

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) = √(100/20) = 2.236 rad/s
ζ = c/(2√(km)) = 18/(2·√(100·20)) = 0.2012

Result

ζ = 0.2012

Answer

ζ = 0.2012
3.Bode plots & margins

Answer: φ = −13.97 °

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.

Corner frequency

ωc = 1/(RC) = 1/(1000·2.2 × 10⁻⁹) = 454545 rad/s
fc = 72343 Hz → f/fc = 18000/72343 = 0.2488

Phase

φ = − atan(0.2488) = −13.97°

Answer

φ = −13.97°
4.Bode plots & margins

Answer: ω180 = 6.481 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/7) = 1 → ω180 = √(6·7) = 6.481 rad/s

Answer

ω180 = 6.481 rad/s
5.Stability: Routh–Hurwitz & root locus

Answer: ω = 4.472 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 + 9s² + 20s + K = 0
s³: 1, 20
s²: 9, K
s¹: (9·20 − K)/9
s⁰: K

Critical gain

K = 9·20 = 180

Auxiliary equation

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

9s² + 180 = 0 → s = ±j√20
ω = 4.472 rad/s

Answer

ω = 4.472 rad/s
6.Stability: Routh–Hurwitz & root locus

Answer: K = 1

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.

Break-away point

dK/ds = 0 for K = −(s + 1)(s + 3) gives 2s + 4 = 0.

sb = −2

Magnitude condition

K = |−2 + 1|·|−2 + 3| = 1·1 = 1

Answer

K = 1

For this gain the closed loop is critically damped (ζ = 1).

7.Steady-state error & PID control

Answer: K = 98

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 error constant

1/(1 + Kp) = 0.02 → Kp = 49

Gain

K/2 = 49 → K = 98

Answer

K = 98

Check the closed-loop stability as well: for this second-order loop any K > 0 is stable.

8.Steady-state error & PID control

Answer: Kd = 2.646 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 + 2) + 10(25 + Kds) = s² + (2 + 10Kd)s + 250

Natural frequency

ωₙ = √250 = 15.81 rad/s

Derivative gain

2 + 10Kd = 2·0.9·15.81 = 28.46
Kd = 2.646 s

Answer

Kd = 2.646 s

The derivative term adds damping without changing ωₙ.

9.Block diagrams

Answer: css = 1.395

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) = 3/(s + 4 + 0.3) = 3/(s + 4.3)

Result

css = R·T(0) = 2·3/4.3 = 1.395

Answer

css = 1.395
10.Block diagrams

Answer: T(0) = 0.8911

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 = G₂/(1 + G₂H₁) = 10/(s + 46)

Outer loop

T(s) = G₁Gᵢ/(1 + G₁GᵢH₂) = 90/((s + 2)(s + 46) + 9)

DC gain

T(0) = 90/(2·46 + 9) = 90/101 = 0.8911

Answer

T(0) = 0.8911

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