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) = 37.5/(s² + 12s + 75) and receives a unit step input. Find the percent overshoot %OS of the step response.

G(s) = 37.5/(s² + 12s + 75)
%OS = %
2.Step responseHard
k = 5000 N/mc = 42 N·s/mm = 4 kgF = 5 Nx(t)

A mass m = 4 kg hangs on a spring k = 5000 N/m with a viscous damper c = 42 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 5 N is suddenly applied. Find the damping ratio ζ.

m = 4 kgk = 5000 N/mc = 42 N·s/mF = 5 N
ζ =
3.Bode plots & marginsMedium
|H| (dB)0.11101001000−100−80−60−40−200ω (rad/s)ω = 2

Find the exact magnitude |G(jω)| of G(s) = 64/((s + 4)(s + 16)) at ω = 2 rad/s, in dB.

G(s) = 64/((s + 4)(s + 16))ω = 2 rad/s
|G(jω)| = dB
4.Bode plots & marginsHard
|H| (dB)0.010.010.10.1111010100100−80−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​

A unity-feedback loop has the open-loop transfer function L(s) = 2/(s(s + 2)). Find the phase margin PM in degrees.

L(s) = 2/(s(s + 2))
PM = °
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1(s + 1)(s + 2)(s + 3)C(s)−

A unity-feedback loop has L(s) = K/((s + 1)(s + 2)(s + 3)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/((s + 1)(s + 2)(s + 3))
Kmax =
6.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10−12−14j2j4sb​

The root locus of L(s) = K/((s + 6)(s + 12)) 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 + 6)(s + 12))
K =
7.Steady-state error & PID controlMedium
R(s)+K = 2510s + 1C(s)−

A process G(s) = 5/(10s + 1) is controlled by a proportional gain K = 2 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.

G(s) = 5/(10s + 1)K = 2
ess = %
8.Steady-state error & PID controlHard
R(s)+K = 501s(s + 5)C(s)−

A proportional controller K = 50 drives the plant G(s) = 1/(s(s + 5)) in a unity-feedback loop. Find the peak time tp of the step response.

G(s) = 1/(s(s + 5))K = 50
tp = s
9.Block diagramsMedium
R(s)+44s(s + 1)G(s)C(s)−

A unity negative feedback loop has G(s) = 44/(s(s + 1)). The reference is a ramp r(t) = 1·t. Find the steady-state error ess.

G(s) = 44/(s(s + 1))r(t) = 1·t
ess =
10.Block diagramsHard
R(s)+Ks(s + 2)(s + 7)G(s)C(s)−

The plant G(s) = K/(s(s + 2)(s + 7)) 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(s + 2)(s + 7))unity negative feedback
ω = rad/s

3Solutions

1.Step response

Answer: %OS = 4.888 %

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 = 75 → ωₙ = 8.66 rad/s
2ζωₙ = 12 → ζ = 12/(2·8.66) = 0.6928

Result

%OS = 100·e−ζπ/√(1−ζ²) = 100·e−3.018 = 4.888 %

Answer

%OS = 4.888 %
2.Step response

Answer: ζ = 0.1485

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) = √(5000/4) = 35.36 rad/s
ζ = c/(2√(km)) = 42/(2·√(5000·4)) = 0.1485

Result

ζ = 0.1485

Answer

ζ = 0.1485
3.Bode plots & margins

Answer: |G(jω)| = −1.036 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.

Multiply the magnitudes

|G(j2)| = 64/(√(2² + 4²)·√(2² + 16²)) = 64/(4.472·16.12) = 0.8875

Result

20·log₁₀(0.8875) = −1.036 dB

Answer

|G(jω)| = −1.036 dB
4.Bode plots & margins

Answer: PM = 65.53 °

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.

Gain crossover

ω⁴ + 4ω² − 4 = 0
ω² = (−4 + √(16 + 4·4))/2 = 0.8284
ωc = 0.9102 rad/s

Phase margin

∠L(jωc) = −90° − atan(0.9102/2) = −114.5°
PM = 180° + ∠L = 65.53°

Answer

PM = 65.53°
5.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 60

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 polynomial

s³ + 6s² + 11s + 6 + K = 0

Routh array

s³: 1, 11
s²: 6, 6 + K
s¹: (6·11 − 6 − K)/6
s⁰: 6 + K

Condition

The s¹ entry must stay positive.

K < 6·11 − 6 = 60

Answer

Kmax = 60

At the limit the loop oscillates at ω = √11 = 3.317 rad/s.

6.Stability: Routh–Hurwitz & root locus

Answer: K = 9

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 + 6)(s + 12) gives 2s + 18 = 0.

sb = −9

Magnitude condition

K = |−9 + 6|·|−9 + 12| = 3·3 = 9

Answer

K = 9

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

7.Steady-state error & PID control

Answer: ess = 9.091 %

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.

Position constant

Kp = 2·5 = 10

Step error

ess = 100/(1 + 10) = 9.091 %

Answer

ess = 9.091 %
8.Steady-state error & PID control

Answer: tp = 0.475 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 + 5) + 50·1 = s² + 5s + 50

ωₙ and ζ

ωₙ = √50 = 7.071 rad/s
ζ = 5/(2·7.071) = 0.3536

Peak time

ωd = 7.071·√(1 − 0.3536²) = 6.614 rad/s
tp = π/6.614 = 0.475 s

Answer

tp = 0.475 s
9.Block diagrams

Answer: ess = 0.02273

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) = 44/1 = 44

Steady-state error

ess = A/Kᵥ = 1/44 = 0.02273

Answer

ess = 0.02273
10.Block diagrams

Answer: ω = 3.742 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 + 9s² + 14s + K = 0

Routh array

s³ | 1 14
s² | 9 K
s¹ | (9·14 − K)/9
s⁰ | K

Stability limit

All entries of the first column must be positive.

Kmax = 9·14 = 126

Frequency of oscillation

9·(jω)² + 126 = 0 → ω² = 126/9 = 14
ω = 3.742 rad/s

Answer

ω = 3.742 rad/s

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