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)1

A system has the transfer function G(s) = 44/(s² + 2s + 44) and receives a unit step input. Find the damping ratio ζ.

G(s) = 44/(s² + 2s + 44)
ζ =
2.Step responseHard
+−24 VR = 103 ΩL = 50 mHC = 4.7 µFvC​

A series RLC circuit with R = 103 Ω, L = 50 mH and C = 4.7 µF is connected to a 24 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the time tp of the first voltage peak, in ms.

R = 103 ΩL = 50 mHC = 4.7 µFV = 24 V
tp = ms
3.Bode plots & marginsMedium
vin​L = 50 mHC = 4.7 µFR = 50 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 50 Ω, with L = 50 mH and C = 4.7 µF. Find the quality factor Q.

R = 50 ΩL = 50 mHC = 4.7 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) = 54/(s(s + 6)(s + 7)). Find the gain margin GM in dB.

L(s) = 54/(s(s + 6)(s + 7))
GM = dB
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+K = 881s(s + 3)(s + 8)C(s)−

The loop in the diagram runs with K = 88 and L(s) = K/(s(s + 3)(s + 8)). By what factor m can the gain be raised before the closed loop becomes unstable?

L(s) = K/(s(s + 3)(s + 8))K = 88
m =
6.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10−12j2

The root locus of L(s) = K(s + 11)/(s(s + 8)) leaves the real axis between s = 0 and s = −8, circles round and joins the real axis again to the left of the zero. Find the break-away point sb.

L(s) = K(s + 11)/(s(s + 8))
sb =
7.Steady-state error & PID controlMedium
R(s)+K(s + 4)(s + 5)C(s)−

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

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

A proportional controller K drives the plant G(s) = 2/(s(s + 5)) in a unity-feedback loop. Find the gain K that gives the closed loop a damping ratio ζ = 0.8.

G(s) = 2/(s(s + 5))ζ = 0.8
K =
9.Block diagramsMedium
R(s)+36s(s + 3)G(s)C(s)−

A unity negative feedback loop has G(s) = 36/(s(s + 3)). The reference is a ramp r(t) = 1·t. Find the steady-state error ess.

G(s) = 36/(s(s + 3))r(t) = 1·t
ess =
10.Block diagramsHard
R(s)+9controller+D(s)+6s + 1plantC(s)−

A proportional controller K = 9 drives the plant G(s) = 6/(s + 1) 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 5 is applied. Find the steady-state output css.

K = 9G(s) = 6/(s + 1)D = 5R = 0
css =

3Solutions

1.Step response

Answer: ζ = 0.1508

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 = 44 → ωₙ = 6.633 rad/s
2ζωₙ = 2 → ζ = 2/(2·6.633) = 0.1508

Result

ζ = 0.1508

Answer

ζ = 0.1508
2.Step response

Answer: tp = 1.758 ms

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 = 1/√(LC) = 1/√(0.05·4.7 × 10⁻⁶) = 2063 rad/s
ζ = (R/2)·√(C/L) = (103/2)·√(4.7 × 10⁻⁶/0.05) = 0.4993

Result

ωd = 1787 rad/s
tp = π/ωd = 1.758 ms

Answer

tp = 1.758 ms
3.Bode plots & margins

Answer: Q = 2.063

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/√(0.05·4.7 × 10⁻⁶) = 2063 rad/s
f₀ = ω₀/(2π) = 328.3 Hz

Quality factor

BW = R/L = 50/0.05 = 1000 rad/s = 159.2 Hz
Q = ω₀/BW = 2063/1000 = 2.063

Answer

Q = 2.063
4.Bode plots & margins

Answer: GM = 20.1 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/6)·(ω/7) = 1 → ω180 = √(6·7) = 6.481 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 36)·√(ω² + 49)) = K/(6·7·(6 + 7)) = K/546

Gain margin

GM = 20·log₁₀(546/54) = 20.1 dB

Answer

GM = 20.1 dB
5.Stability: Routh–Hurwitz & root locus

Answer: m = 3

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 + 11s² + 24s + K = 0
s³: 1, 24
s²: 11, K
s¹: (11·24 − K)/11
s⁰: K

Largest stable gain

Kmax = 11·24 = 264

Factor

m = 264/88 = 3

Answer

m = 3

This factor is the gain margin as a ratio: 9.542 dB.

6.Stability: Routh–Hurwitz & root locus

Answer: sb = −5.255

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.

Gain along the real axis

K(s) = −(s² + 8s)/(s + 11)

Stationary points

Differentiate with the quotient rule and set the numerator to zero.

(2s + 8)(s + 11) − (s² + 8s) = s² + 22s + 88 = 0
s = −11 ± √33 = −5.255, −16.74

Break-away point

sb = −5.255

Answer

sb = −5.255

Between the two break points the locus is a circle centred on the zero, with radius √33 = 5.745.

7.Steady-state error & PID control

Answer: K = 480

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) = 0.04 → Kp = 24

Gain

K/20 = 24 → K = 480

Answer

K = 480

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 = 4.883

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 + 5) + 2K = s² + 5s + 2K

Natural frequency

2·0.8·ωₙ = 5 → ωₙ = 3.125 rad/s

Gain

2K = ωₙ² = 9.766 → K = 4.883

Answer

K = 4.883
9.Block diagrams

Answer: ess = 0.08333

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).

Velocity error constant

Kᵥ = lims→0 s·G(s) = 36/3 = 12

Steady-state error

ess = A/Kᵥ = 1/12 = 0.08333

Answer

ess = 0.08333
10.Block diagrams

Answer: css = 0.5455

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) = 6/(s + 1 + 54)

Steady state

css = 5·6/55 = 0.5455

Answer

css = 0.5455

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

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