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)2ts​ = ?

A system has the transfer function G(s) = 198/(s² + 15s + 99) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 198/(s² + 15s + 99)
ts = s
2.Step responseHard
t (s)y(t)1tp​ = 1 s2 %

The step response of a second-order system without zeros shows 2 % overshoot, and the first peak occurs at t = 1 s. Find the natural frequency ωn.

%OS = 2 %tp = 1 s
ωn = rad/s
3.Bode plots & marginsMedium
vin​L = 20 mHC = 10 µFR = 5 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 5 Ω, with L = 20 mH and C = 10 µF. Find the quality factor Q.

R = 5 ΩL = 20 mHC = 10 µF
Q =
4.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) = 192/(s(s + 4)(s + 8)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 192/(s(s + 4)(s + 8))
ω180 = rad/s
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 3)(s + 5)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 3)(s + 5)). Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/(s(s + 3)(s + 5))
Kmax =
6.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8j2j4

For L(s) = K/((s + 7)(s² + 2s + 17)), find the angle of departure θd of the root locus from the complex pole in the upper half-plane, measured from the positive real axis.

L(s) = K/((s + 7)(s² + 2s + 17))
θd = °
7.Steady-state error & PID controlMedium
R(s)+K(s + 2)(s + 4)C(s)−

A unity-feedback loop has G(s) = K/((s + 2)(s + 4)). Find the gain K that gives a steady-state error of 5 % for a step input.

G(s) = K/((s + 2)(s + 4))ess = 5 %
K =
8.Steady-state error & PID controlHard
R(s)+K2s(s + 10)C(s)−

A proportional controller K drives the plant G(s) = 2/(s(s + 10)) in a unity-feedback loop. Find the gain K that gives the closed loop a damping ratio ζ = 0.5.

G(s) = 2/(s(s + 10))ζ = 0.5
K =
9.Block diagramsMedium
R(s)+9K+4s + 1G(s)C(s)3H₁−−

The plant G(s) = 4/(s + 1) has an inner (minor) negative feedback loop with gain H₁ = 3. A controller gain K = 9 and an outer unity negative feedback loop close the system. Find the closed-loop pole p.

G(s) = 4/(s + 1)H₁ = 3K = 9
p =
10.Block diagramsHard
R(s)+K(s + 5)(s + 6)(s + 8)G(s)C(s)−

The plant G(s) = K/((s + 5)(s + 6)(s + 8)) is in a unity negative feedback loop. At the largest stable gain the loop oscillates. Find the frequency of oscillation ω in rad/s.

G(s) = K/((s + 5)(s + 6)(s + 8))unity negative feedback
ω = rad/s

3Solutions

1.Step response

Answer: ts = 0.5333 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 = 99 → ωₙ = 9.95 rad/s
2ζωₙ = 15 → ζ = 15/(2·9.95) = 0.7538

Result

ζωₙ = 7.5 → ts ≈ 4/7.5 = 0.5333 s

Answer

ts = 0.5333 s
2.Step response

Answer: ωn = 5.017 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).

Damping ratio

ζ = −ln(0.02)/√(π² + ln²(0.02)) = 3.912/√(9.87 + 15.3) = 0.7797

Natural frequency

ωd = π/tp = π/1 = 3.142 rad/s
ωₙ = ωd/√(1 − ζ²) = 3.142/0.6261 = 5.017 rad/s

Answer

ωn = 5.017 rad/s
3.Bode plots & margins

Answer: Q = 8.944

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/√(0.02·1 × 10⁻⁵) = 2236 rad/s
f₀ = ω₀/(2π) = 355.9 Hz

Quality factor

BW = R/L = 5/0.02 = 250 rad/s = 39.79 Hz
Q = ω₀/BW = 2236/250 = 8.944

Answer

Q = 8.944
4.Bode plots & margins

Answer: ω180 = 5.657 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/4)·(ω/8) = 1 → ω180 = √(4·8) = 5.657 rad/s

Answer

ω180 = 5.657 rad/s
5.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 120

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 equation

s³ + 8s² + 15s + K = 0

Routh array

s³: 1, 15
s²: 8, K
s¹: (8·15 − K)/8
s⁰: K

Conditions

Every entry of the first column must be positive.

K > 0
8·15 − K > 0 → K < 120

Answer

Kmax = 120

At K = 120 the closed loop oscillates at ω = √15 = 3.873 rad/s.

6.Stability: Routh–Hurwitz & root locus

Answer: θd = 56.31 °

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.

Pole locations

s² + 2s + 17 = 0 → s = −1 ± j4
s = −7

Angles from the other poles

∠(−1 + j4 + 7) = atan2(4, 6) = 33.69°
∠(−1 + j4 − (−1 − j4)) = ∠(j8) = 90°

Angle of departure

θd = 180° − 33.69° − 90° = 56.31°

Answer

θd = 56.31°
7.Steady-state error & PID control

Answer: K = 152

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) = 0.05 → Kp = 19

Gain

K/8 = 19 → K = 152

Answer

K = 152

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

8.Steady-state error & PID control

Answer: K = 50

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) + 2K = s² + 10s + 2K

Natural frequency

2·0.5·ωₙ = 10 → ωₙ = 10 rad/s

Gain

2K = ωₙ² = 100 → K = 50

Answer

K = 50
9.Block diagrams

Answer: p = −49

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) = 4/(s + 1 + 4·3) = 4/(s + 13)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 36/(s + 13 + 36) = 36/(s + 49)

Result

p = −49

Answer

p = −49
10.Block diagrams

Answer: ω = 10.86 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

s³ + 19s² + 118s + 240 + K = 0

Routh array

s³ | 1 118
s² | 19 240 + K
s¹ | (19·118 − 240 − K)/19
s⁰ | 240 + K

Stability limit

All entries of the first column must be positive.

Kmax = 19·118 − 240 = 2002

Frequency of oscillation

19·(jω)² + 240 + 2002 = 0 → ω² = 2242/19 = 118
ω = 10.86 rad/s

Answer

ω = 10.86 rad/s

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