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) = 83/(s² + 10s + 83) and receives a unit step input. Find the natural frequency ωn in rad/s.

G(s) = 83/(s² + 10s + 83)
ωn = rad/s
2.Step responseHard
k = 200 N/mc = 34 N·s/mm = 4 kgF = 20 Nx(t)

A mass m = 4 kg hangs on a spring k = 200 N/m with a viscous damper c = 34 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 time tp at which the displacement is largest.

m = 4 kgk = 200 N/mc = 34 N·s/mF = 20 N
tp = s
3.Bode plots & marginsMedium
vin​R = 220 ΩL = 10 mHvout​

The RL high-pass filter shown has R = 220 Ω, L = 10 mH. A sinusoidal input of 7 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 220 ΩL = 10 mHf = 7 kHz
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) = 24/(s(s + 2)(s + 3)). Find the gain margin GM in dB.

L(s) = 24/(s(s + 2)(s + 3))
GM = dB
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 3)(s + 10)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 3)(s + 10)). Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/(s(s + 3)(s + 10))
Kmax =
6.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−71j1j2

A unity-feedback loop has L(s) = K/((s + 6)(s + 3)(s + 4)). Find the point σa where the asymptotes of the root locus cross the real axis.

L(s) = K/((s + 6)(s + 3)(s + 4))
σa =
7.Steady-state error & PID controlMedium
r(t) = 5 + 5t+4s(s + 1)C(s)−

The input r(t) = 5 + 5t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 4/(s(s + 1)). Find the steady-state error ess.

G(s) = 4/(s(s + 1))r(t) = 5 + 5t
ess =
8.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s10s + 2C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 10/(s + 2) in a unity-feedback loop. The closed loop must have ωₙ = 5 rad/s and ζ = 0.8. Find the proportional gain Kp.

G(s) = 10/(s + 2)ωₙ = 5 rad/sζ = 0.8
Kp =
9.Block diagramsMedium
R(s)+17K+3s + 1G(s)C(s)1H₁−−

The plant G(s) = 3/(s + 1) has an inner (minor) negative feedback loop with gain H₁ = 1. A controller gain K = 17 and an outer unity negative feedback loop close the system. Find the closed-loop DC gain T(0).

G(s) = 3/(s + 1)H₁ = 1K = 17
T(0) =
10.Block diagramsHard
R(s)+18controller+D(s)+8s + 9plantC(s)−

A proportional controller K = 18 drives the plant G(s) = 8/(s + 9) 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 0.5 are applied together. Find the steady-state output css.

K = 18G(s) = 8/(s + 9)D = 0.5R = 2
css =

3Solutions

1.Step response

Answer: ωn = 9.11 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 = 83 → ωₙ = 9.11 rad/s
2ζωₙ = 10 → ζ = 10/(2·9.11) = 0.5488

Result

ωₙ = 9.11 rad/s

Answer

ωn = 9.11 rad/s
2.Step response

Answer: tp = 0.5559 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).

Natural frequency and damping ratio

ωₙ = √(k/m) = √(200/4) = 7.071 rad/s
ζ = c/(2√(km)) = 34/(2·√(200·4)) = 0.601

Result

ωd = 5.651 rad/s
tp = π/ωd = 0.5559 s

Answer

tp = 0.5559 s
3.Bode plots & margins

Answer: φ = 26.57 °

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 = R/L = 220/0.01 = 22000 rad/s
fc = 3501 Hz → f/fc = 7000/3501 = 1.999

Phase

φ = 90° − atan(1.999) = 26.57°

Answer

φ = 26.57°
4.Bode plots & margins

Answer: GM = 1.938 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/2)·(ω/3) = 1 → ω180 = √(2·3) = 2.449 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 4)·√(ω² + 9)) = K/(2·3·(2 + 3)) = K/30

Gain margin

GM = 20·log₁₀(30/24) = 1.938 dB

Answer

GM = 1.938 dB
5.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 390

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 + 13s² + 30s + K = 0

Routh array

s³: 1, 30
s²: 13, K
s¹: (13·30 − K)/13
s⁰: K

Conditions

Every entry of the first column must be positive.

K > 0
13·30 − K > 0 → K < 390

Answer

Kmax = 390

At K = 390 the closed loop oscillates at ω = √30 = 5.477 rad/s.

6.Stability: Routh–Hurwitz & root locus

Answer: σa = −4.333

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.

Poles and zeros

poles: −6, −3, −4
n − m = 3 − 0 = 3

Centroid

σa = (−13)/3 = −4.333

Answer

σa = −4.333

The 3 asymptotes leave the centroid at angles of 60°, 180°, 300°.

7.Steady-state error & PID control

Answer: ess = 1.25

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.

Velocity constant

Kv = lim s·G(s) = 4

Step part

Kp = ∞ → estep = 0

Ramp error

ess = 5/4 = 1.25

Answer

ess = 1.25
8.Steady-state error & PID control

Answer: Kp = 0.6

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

1 + C(s)G(s) = 0, multiplied by s(s + 2).

s² + (2 + 10Kp)s + 10Ki

Target polynomial

s² + 2·0.8·5s + 5² = s² + 8s + 25

Match the coefficients

2 + 10Kp = 8 → Kp = 0.6

Answer

Kp = 0.6

The integral term makes the loop type 1, so a step is followed without steady-state error.

9.Block diagrams

Answer: T(0) = 0.9273

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) = 3/(s + 1 + 3·1) = 3/(s + 4)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 51/(s + 4 + 51) = 51/(s + 55)

Result

T(0) = 51/55 = 0.9273

Answer

T(0) = 0.9273
10.Block diagrams

Answer: css = 1.908

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) = 8/(s + 9 + 144)
C/R = K·G/(1 + K·G) = 144/(s + 9 + 144)

Steady state

css = 2·144/153 + 0.5·8/153 = 1.908

Answer

css = 1.908

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

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