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)0.5

A system has the transfer function G(s) = 20.5/(s² + 9s + 41) and receives a unit step input. Find the damping ratio ζ.

G(s) = 20.5/(s² + 9s + 41)
ζ =
2.Step responseHard
k = 100 N/mc = 9 N·s/mm = 2.5 kgF = 20 Nx(t)

A mass m = 2.5 kg hangs on a spring k = 100 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 damping ratio ζ.

m = 2.5 kgk = 100 N/mc = 9 N·s/mF = 20 N
ζ =
3.Bode plots & marginsMedium
vin​C = 100 nFR = 47 kΩvout​

The RC high-pass filter shown has R = 47 kΩ, C = 100 nF. A sinusoidal input of 140 Hz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 47 kΩC = 100 nFf = 140 Hz
φ = °
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) = 506/(s(s + 7)(s + 9)). Find the gain margin GM in dB.

L(s) = 506/(s(s + 7)(s + 9))
GM = dB
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+K = 8.41s(s + 3)(s + 4)C(s)−

The loop in the diagram runs with K = 8.4 and L(s) = K/(s(s + 3)(s + 4)). By what factor m can the gain be raised before the closed loop becomes unstable?

L(s) = K/(s(s + 3)(s + 4))K = 8.4
m =
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 + 1)(s + 4)C(s)−

A unity-feedback loop has G(s) = K/((s + 1)(s + 4)). Find the gain K that gives a steady-state error of 10 % for a step input.

G(s) = K/((s + 1)(s + 4))ess = 10 %
K =
8.Steady-state error & PID controlHard
R(s)+2 + Kd​s4s(s + 3)C(s)−

A PD controller C(s) = 2 + Kd·s drives the plant G(s) = 4/(s(s + 3)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.707.

G(s) = 4/(s(s + 3))Kp = 2ζ = 0.707
Kd = s
9.Block diagramsMedium
R(s)+13K+4s + 1G(s)C(s)5H₁−−

The plant G(s) = 4/(s + 1) has an inner (minor) negative feedback loop with gain H₁ = 5. 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) = 4/(s + 1)H₁ = 5K = 13
T(0) =
10.Block diagramsHard
R(s)+18controller+D(s)+1s + 10plantC(s)−

A proportional controller K = 18 drives the plant G(s) = 1/(s + 10) in a unity negative feedback loop. A disturbance D enters at the plant input (it is added to the controller output). A reference step of size 2 and a disturbance step of size 10 are applied together. Find the steady-state output css.

K = 18G(s) = 1/(s + 10)D = 10R = 2
css =

3Solutions

1.Step response

Answer: ζ = 0.7028

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 = 41 → ωₙ = 6.403 rad/s
2ζωₙ = 9 → ζ = 9/(2·6.403) = 0.7028

Result

ζ = 0.7028

Answer

ζ = 0.7028
2.Step response

Answer: ζ = 0.2846

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) = √(100/2.5) = 6.325 rad/s
ζ = c/(2√(km)) = 9/(2·√(100·2.5)) = 0.2846

Result

ζ = 0.2846

Answer

ζ = 0.2846
3.Bode plots & margins

Answer: φ = 13.6 °

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/(47000·1 × 10⁻⁷) = 212.8 rad/s
fc = 33.86 Hz → f/fc = 140/33.86 = 4.134

Phase

φ = 90° − atan(4.134) = 13.6°

Answer

φ = 13.6°
4.Bode plots & margins

Answer: GM = 5.986 dB

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/7)·(ω/9) = 1 → ω180 = √(7·9) = 7.937 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 49)·√(ω² + 81)) = K/(7·9·(7 + 9)) = K/1008

Gain margin

GM = 20·log₁₀(1008/506) = 5.986 dB

Answer

GM = 5.986 dB
5.Stability: Routh–Hurwitz & root locus

Answer: m = 10

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 + 7s² + 12s + K = 0
s³: 1, 12
s²: 7, K
s¹: (7·12 − K)/7
s⁰: K

Largest stable gain

Kmax = 7·12 = 84

Factor

m = 84/8.4 = 10

Answer

m = 10

This factor is the gain margin as a ratio: 20 dB.

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 = 36

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) = 0.1 → Kp = 9

Gain

K/4 = 9 → K = 36

Answer

K = 36

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 = 0.2498 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 + 3) + 4(2 + Kds) = s² + (3 + 4Kd)s + 8

Natural frequency

ωₙ = √8 = 2.828 rad/s

Derivative gain

3 + 4Kd = 2·0.707·2.828 = 3.999
Kd = 0.2498 s

Answer

Kd = 0.2498 s

The derivative term adds damping without changing ωₙ.

9.Block diagrams

Answer: T(0) = 0.7123

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 + 4·5) = 4/(s + 21)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 52/(s + 21 + 52) = 52/(s + 73)

Result

T(0) = 52/73 = 0.7123

Answer

T(0) = 0.7123
10.Block diagrams

Answer: css = 1.643

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) = 1/(s + 10 + 18)
C/R = K·G/(1 + K·G) = 18/(s + 10 + 18)

Steady state

css = 2·18/28 + 10·1/28 = 1.643

Answer

css = 1.643

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

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