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)3?

A system has the transfer function G(s) = 267/(s² + 7s + 89) and receives a unit step input. Find the percent overshoot %OS of the step response.

G(s) = 267/(s² + 7s + 89)
%OS = %
2.Step responseHard
t (s)y(t)120 %

A second-order system without zeros must have a step response with 20 % overshoot. Find the damping ratio ζ.

%OS = 20 %
ζ =
3.Bode plots & marginsMedium
vin​C = 100 nFR = 1 kΩvout​

The RC high-pass filter shown has R = 1 kΩ, C = 100 nF. A sinusoidal input of 400 Hz is applied. Find the gain |H| = |vout/vin| in dB.

R = 1 kΩC = 100 nFf = 400 Hz
|H| = dB
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) = K/(s(s + 5)(s + 10)). Find the gain K that gives a gain margin of 15 dB.

L(s) = K/(s(s + 5)(s + 10))GM = 15 dB
K =
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+8401s(s + 4)(s + 10)C(s)−

The loop in the diagram has negative feedback and L(s) = 840/(s(s + 4)(s + 10)). How many closed-loop poles N lie in the right half-plane?

L(s) = 840/(s(s + 4)(s + 10))
N =
6.Stability: Routh–Hurwitz & root locusHard
ReImζ = 0.8−1−2−3−4−5j1

The root locus of L(s) = K/(s(s + 4)) is shown. Find the gain K for which the closed-loop poles have the damping ratio ζ = 0.8.

L(s) = K/(s(s + 4))ζ = 0.8
K =
7.Steady-state error & PID controlMedium
r(t) = 1 + 2t+5s(s + 1)(s + 4)C(s)−

The input r(t) = 1 + 2t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 5/(s(s + 1)(s + 4)). Find the steady-state error ess.

G(s) = 5/(s(s + 1)(s + 4))r(t) = 1 + 2t
ess =
8.Steady-state error & PID controlHard
R(s)+K = 81s(s + 4)C(s)−

A proportional controller K = 8 drives the plant G(s) = 1/(s(s + 4)) in a unity-feedback loop. Find the peak time tp of the step response.

G(s) = 1/(s(s + 4))K = 8
tp = s
9.Block diagramsMedium
R(s)+10K+10s + 2G(s)C(s)5H₁−−

The plant G(s) = 10/(s + 2) has an inner (minor) negative feedback loop with gain H₁ = 5. A controller gain K = 10 and an outer unity negative feedback loop close the system. Find the closed-loop pole p.

G(s) = 10/(s + 2)H₁ = 5K = 10
p =
10.Block diagramsHard
R(s)+Kcontroller+D(s)+4s + 2plantC(s)−

A proportional controller K drives the plant G(s) = 4/(s + 2) in a unity negative feedback loop. A disturbance step of size 2 enters at the plant input while R = 0. Find the smallest gain K that keeps the steady-state output at or below 0.04.

G(s) = 4/(s + 2)D = 2css ≤ 0.04
K =

3Solutions

1.Step response

Answer: %OS = 28.5 %

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 = 89 → ωₙ = 9.434 rad/s
2ζωₙ = 7 → ζ = 7/(2·9.434) = 0.371

Result

%OS = 100·e−ζπ/√(1−ζ²) = 100·e−1.255 = 28.5 %

Answer

%OS = 28.5 %
2.Step response

Answer: ζ = 0.4559

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(0.2)/√(π² + ln²(0.2)) = 1.609/√(9.87 + 2.59) = 0.4559

Answer

ζ = 0.4559
3.Bode plots & margins

Answer: |H| = −12.26 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.

Corner frequency

ωc = 1/(RC) = 1/(1000·1 × 10⁻⁷) = 10000 rad/s
fc = 1592 Hz → f/fc = 400/1592 = 0.2513

Gain

|H| = 0.2513/√(1 + 0.2513²) = 0.2437
20·log₁₀(0.2437) = −12.26 dB

Answer

|H| = −12.26 dB
4.Bode plots & margins

Answer: K = 133.4

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/5)·(ω/10) = 1 → ω180 = √(5·10) = 7.071 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 25)·√(ω² + 100)) = K/(5·10·(5 + 10)) = K/750

Required gain

750/K = 1015/20 = 5.623
K = 750/5.623 = 133.4

Answer

K = 133.4
5.Stability: Routh–Hurwitz & root locus

Answer: N = 2

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 + 14s² + 40s + 840 = 0

Routh array

s³: 1, 40
s²: 14, 840
s¹: −20
s⁰: 840

Sign changes in the first column

1 → 14 → −20 → 840
N = 2

Answer

N = 2

The gain is above the limit K = 560, so a complex pair has crossed into the right half-plane.

6.Stability: Routh–Hurwitz & root locus

Answer: K = 6.25

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.

Compare with the standard form

s² + 4s + K = s² + 2ζωₙs + ωₙ²
2·0.8·ωₙ = 4 → ωₙ = 2.5 rad/s

Gain

K = ωₙ² = 2.5² = 6.25

Answer

K = 6.25

The closed-loop poles sit where the ζ line crosses the vertical branch at s = −2.

7.Steady-state error & PID control

Answer: ess = 1.6

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) = 1.25

Step part

Kp = ∞ → estep = 0

Ramp error

ess = 2/1.25 = 1.6

Answer

ess = 1.6
8.Steady-state error & PID control

Answer: tp = 1.571 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 + 4) + 8·1 = s² + 4s + 8

ωₙ and ζ

ωₙ = √8 = 2.828 rad/s
ζ = 4/(2·2.828) = 0.7071

Peak time

ωd = 2.828·√(1 − 0.7071²) = 2 rad/s
tp = π/2 = 1.571 s

Answer

tp = 1.571 s
9.Block diagrams

Answer: p = −152

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) = 10/(s + 2 + 10·5) = 10/(s + 52)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 100/(s + 52 + 100) = 100/(s + 152)

Result

p = −152

Answer

p = −152
10.Block diagrams

Answer: K = 49.5

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).

Disturbance response

css = D·G(0)/(1 + K·G(0)) = 2·4/(2 + 4K)

Solve for K

2 + 4K = 2·4/0.04 = 200
K = (200 − 2)/4 = 49.5

Answer

K = 49.5

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