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)3ts​ = ?

A system has the transfer function G(s) = 72/(s² + 4s + 24) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 72/(s² + 4s + 24)
ts = s
2.Step responseHard
+−24 VR = 224 ΩL = 50 mHC = 1 µFvC​

A series RLC circuit with R = 224 Ω, L = 50 mH and C = 1 µF is connected to a 24 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the undamped natural frequency ωn in rad/s.

R = 224 ΩL = 50 mHC = 1 µFV = 24 V
ωn = rad/s
3.Bode plots & marginsMedium
vin​L = 5 mHC = 4.7 µFR = 20 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 20 Ω, with L = 5 mH and C = 4.7 µF. Find the quality factor Q.

R = 20 ΩL = 5 mHC = 4.7 µF
Q =
4.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 36/(s(s + 2)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 36/(s(s + 2))
ωc = rad/s
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+K = 21s(s + 1)(s + 2)C(s)−

The loop in the diagram runs with K = 2 and L(s) = K/(s(s + 1)(s + 2)). By what factor m can the gain be raised before the closed loop becomes unstable?

L(s) = K/(s(s + 1)(s + 2))K = 2
m =
6.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6j1

The root locus of L(s) = K/(s(s + 4)(s + 5)) leaves the real axis between s = 0 and s = −4. Find the break-away point sb.

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

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

G(s) = 2(s + 4)/(s(s + 4)(s + 5))r(t) = 5t
ess =
8.Steady-state error & PID controlHard
R(s)+20 + Kd​s10s(s + 4)C(s)−

A PD controller C(s) = 20 + Kd·s drives the plant G(s) = 10/(s(s + 4)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.6.

G(s) = 10/(s(s + 4))Kp = 20ζ = 0.6
Kd = s
9.Block diagramsMedium
R(s)+K(s + 3)(s + 6)G(s)C(s)−

The plant G(s) = K/((s + 3)(s + 6)) is in a unity negative feedback loop. Find the gain K for which the closed loop is critically damped (two equal real poles).

G(s) = K/((s + 3)(s + 6))unity negative feedback
K =
10.Block diagramsHard
R(s)+Kcontroller+D(s)+3s + 7plantC(s)−

A proportional controller K drives the plant G(s) = 3/(s + 7) in a unity negative feedback loop. A disturbance step of size 0.5 enters at the plant input while R = 0. Find the smallest gain K that keeps the steady-state output at or below 0.004286.

G(s) = 3/(s + 7)D = 0.5css ≤ 0.004286
K =

3Solutions

1.Step response

Answer: ts = 2 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 = 24 → ωₙ = 4.899 rad/s
2ζωₙ = 4 → ζ = 4/(2·4.899) = 0.4082

Result

ζωₙ = 2 → ts ≈ 4/2 = 2 s

Answer

ts = 2 s
2.Step response

Answer: ωn = 4472 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).

Natural frequency and damping ratio

ωₙ = 1/√(LC) = 1/√(0.05·1 × 10⁻⁶) = 4472 rad/s
ζ = (R/2)·√(C/L) = (224/2)·√(1 × 10⁻⁶/0.05) = 0.5009

Result

ωₙ = 4472 rad/s

Answer

ωn = 4472 rad/s
3.Bode plots & margins

Answer: Q = 1.631

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/√(0.005·4.7 × 10⁻⁶) = 6523 rad/s
f₀ = ω₀/(2π) = 1038 Hz

Quality factor

BW = R/L = 20/0.005 = 4000 rad/s = 636.6 Hz
Q = ω₀/BW = 6523/4000 = 1.631

Answer

Q = 1.631
4.Bode plots & margins

Answer: ωc = 5.836 rad/s

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 − 1296 = 0
ω² = (−4 + √(16 + 4·1296))/2 = 34.06
ωc = 5.836 rad/s

Answer

ωc = 5.836 rad/s
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 + 3s² + 2s + K = 0
s³: 1, 2
s²: 3, K
s¹: (3·2 − K)/3
s⁰: K

Largest stable gain

Kmax = 3·2 = 6

Factor

m = 6/2 = 3

Answer

m = 3

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

6.Stability: Routh–Hurwitz & root locus

Answer: sb = −1.472

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) = −s(s + 4)(s + 5) = −(s³ + 9s² + 20s)

Stationary points

dK/ds = −(3s² + 18s + 20) = 0
s = (−18 ± √(324 − 240))/6 = −1.472, −4.528

Pick the point on the locus

The locus lies on the real axis between 0 and −4; the other root is not on the locus.

sb = −1.472

Answer

sb = −1.472
7.Steady-state error & PID control

Answer: ess = 12.5

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) = 0.4

Ramp error

ess = 5/0.4 = 12.5

Answer

ess = 12.5
8.Steady-state error & PID control

Answer: Kd = 1.297 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) + 10(20 + Kds) = s² + (4 + 10Kd)s + 200

Natural frequency

ωₙ = √200 = 14.14 rad/s

Derivative gain

4 + 10Kd = 2·0.6·14.14 = 16.97
Kd = 1.297 s

Answer

Kd = 1.297 s

The derivative term adds damping without changing ωₙ.

9.Block diagrams

Answer: K = 2.25

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

1 + K/((s + 3)(s + 6)) = 0 → s² + 9s + 18 + K = 0

Repeated root

A quadratic s² + βs + γ has a double root when β² = 4γ.

9² = 4(18 + K) → K = 81/4 − 18 = 2.25

Answer

K = 2.25
Both closed-loop poles are then at s = −4.5.
10.Block diagrams

Answer: K = 114.3

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)) = 0.5·3/(7 + 3K)

Solve for K

7 + 3K = 0.5·3/0.004286 = 350
K = (350 − 7)/3 = 114.3

Answer

K = 114.3

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