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)0.5tp​ = ?

A system has the transfer function G(s) = 26.5/(s² + 3s + 53) and receives a unit step input. Find the peak time tp of the step response.

G(s) = 26.5/(s² + 3s + 53)
tp = s
2.Step responseHard
+−5 VR = 57 ΩL = 10 mHC = 2 µFvC​

A series RLC circuit with R = 57 Ω, L = 10 mH and C = 2 µF is connected to a 5 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the percent overshoot %OS of the capacitor voltage.

R = 57 ΩL = 10 mHC = 2 µFV = 5 V
%OS = %
3.Bode plots & marginsMedium
vin​L = 20 mHC = 0.22 µFR = 100 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 100 Ω, with L = 20 mH and C = 0.22 µF. Find the centre (resonant) frequency f0 in Hz.

R = 100 ΩL = 20 mHC = 0.22 µF
f0 = Hz
4.Bode plots & marginsHard
|H| (dB)0.10.111101010010010001000−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​

A unity-feedback loop has the open-loop transfer function L(s) = 136/(s(s + 8)). Find the phase margin PM in degrees.

L(s) = 136/(s(s + 8))
PM = °
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 10(s − 4)(s + 5)C(s)−

The plant G(s) = (s + 10)/((s − 4)(s + 5)) has an unstable pole at s = 4. A proportional gain K closes a unity-feedback loop around it. Find the smallest gain Kmin that makes the closed loop stable.

G(s) = (s + 10)/((s − 4)(s + 5))
Kmin =
6.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−7−8−9−101j1j2

The root locus of L(s) = K(s + 8)/(s(s + 3)) leaves the real axis between s = 0 and s = −3, circles round and joins the real axis again to the left of the zero. Find the break-away point sb.

L(s) = K(s + 8)/(s(s + 3))
sb =
7.Steady-state error & PID controlMedium
r(t) = 0.5·t²/2+10(s + 2)s²(s + 5)C(s)−

The parabolic input r(t) = 0.5·t²/2 is applied to a unity-feedback loop with G(s) = 10(s + 2)/(s²(s + 5)). Find the steady-state error ess.

G(s) = 10(s + 2)/(s²(s + 5))r(t) = 0.5·t²/2
ess =
8.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s1s + 4C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 1/(s + 4) in a unity-feedback loop. The closed loop must have ωₙ = 5 rad/s and ζ = 0.7. Find the proportional gain Kp.

G(s) = 1/(s + 4)ωₙ = 5 rad/sζ = 0.7
Kp =
9.Block diagramsMedium
R(s)+9K+5s + 6G(s)C(s)0.5H₁−−

The plant G(s) = 5/(s + 6) has an inner (minor) negative feedback loop with gain H₁ = 0.5. A controller gain K = 9 and an outer unity negative feedback loop close the system. Find the closed-loop pole p.

G(s) = 5/(s + 6)H₁ = 0.5K = 9
p =
10.Block diagramsHard
R(s)+Kcontroller+D(s)+10s + 2plantC(s)−

A proportional controller K drives the plant G(s) = 10/(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 0.1.

G(s) = 10/(s + 2)D = 2css ≤ 0.1
K =

3Solutions

1.Step response

Answer: tp = 0.441 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 = 53 → ωₙ = 7.28 rad/s
2ζωₙ = 3 → ζ = 3/(2·7.28) = 0.206

Result

ωd = 7.28·√(1 − 0.206²) = 7.124 rad/s → tp = π/ωd = 0.441 s

Answer

tp = 0.441 s
2.Step response

Answer: %OS = 25.07 %

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/√(0.01·2 × 10⁻⁶) = 7071 rad/s
ζ = (R/2)·√(C/L) = (57/2)·√(2 × 10⁻⁶/0.01) = 0.4031

Result

%OS = 100·e−ζπ/√(1−ζ²) = 25.07 %

Answer

%OS = 25.07 %
3.Bode plots & margins

Answer: f0 = 2399 Hz

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/√(0.02·2.2 × 10⁻⁷) = 15076 rad/s
f₀ = ω₀/(2π) = 2399 Hz

Answer

f0 = 2399 Hz
4.Bode plots & margins

Answer: PM = 37.63 °

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 + 64ω² − 18496 = 0
ω² = (−64 + √(4096 + 4·18496))/2 = 107.7
ωc = 10.38 rad/s

Phase margin

∠L(jωc) = −90° − atan(10.38/8) = −142.4°
PM = 180° + ∠L = 37.63°

Answer

PM = 37.63°
5.Stability: Routh–Hurwitz & root locus

Answer: Kmin = 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

(s − 4)(s + 5) + K(s + 10) = s² + (1 + K)s + (10K − 20)

Conditions

For s² + c₁s + c₀ all roots lie in the left half-plane exactly when c₁ > 0 and c₀ > 0.

1 + K > 0 → K > −1
10K − 20 > 0 → K > 2
Kmin = 2

Answer

Kmin = 2
6.Stability: Routh–Hurwitz & root locus

Answer: sb = −1.675

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 + 3s)/(s + 8)

Stationary points

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

(2s + 3)(s + 8) − (s² + 3s) = s² + 16s + 24 = 0
s = −8 ± √40 = −1.675, −14.32

Break-away point

sb = −1.675

Answer

sb = −1.675

Between the two break points the locus is a circle centred on the zero, with radius √40 = 6.325.

7.Steady-state error & PID control

Answer: ess = 0.125

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.

Acceleration constant

Ka = 10·2/5 = 4

Parabola error

ess = 0.5/4 = 0.125

Answer

ess = 0.125
8.Steady-state error & PID control

Answer: Kp = 3

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 + 4).

s² + (4 + 1Kp)s + 1Ki

Target polynomial

s² + 2·0.7·5s + 5² = s² + 7s + 25

Match the coefficients

4 + 1Kp = 7 → Kp = 3

Answer

Kp = 3

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

9.Block diagrams

Answer: p = −53.5

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).

Inner loop

Gᵢ = G/(1 + G·H₁) = 5/(s + 6 + 5·0.5) = 5/(s + 8.5)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 45/(s + 8.5 + 45) = 45/(s + 53.5)

Result

p = −53.5

Answer

p = −53.5
10.Block diagrams

Answer: K = 19.8

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)) = 2·10/(2 + 10K)

Solve for K

2 + 10K = 2·10/0.1 = 200
K = (200 − 2)/10 = 19.8

Answer

K = 19.8

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