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)2?

A system has the transfer function G(s) = 20/(s² + 5s + 10) and receives a unit step input. Find the percent overshoot %OS of the step response.

G(s) = 20/(s² + 5s + 10)
%OS = %
2.Step responseHard
t (s)y(t)1tp​ = 0.5 s5 %

The step response of a second-order system without zeros shows 5 % overshoot, and the first peak occurs at t = 0.5 s. Find the natural frequency ωn.

%OS = 5 %tp = 0.5 s
ωn = rad/s
3.Bode plots & marginsMedium
vin​L = 10 mHC = 0.22 µFR = 50 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 50 Ω, with L = 10 mH and C = 0.22 µF. Find the centre (resonant) frequency f0 in Hz.

R = 50 ΩL = 10 mHC = 0.22 µF
f0 = Hz
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 + 1)(s + 9)). Find the gain K that gives a gain margin of 6 dB.

L(s) = K/(s(s + 1)(s + 9))GM = 6 dB
K =
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 3)(s + 6)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 3)(s + 6)). The gain is raised until the closed loop is on the edge of stability. Find the frequency ω of the sustained oscillation.

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

For L(s) = K/((s + 8)(s² + 2s + 5)), find the angle of departure θd of the root locus from the complex pole in the upper half-plane, measured from the positive real axis.

L(s) = K/((s + 8)(s² + 2s + 5))
θd = °
7.Steady-state error & PID controlMedium
r(t) = 1t+Ks(s + 5)C(s)−

A unity-feedback loop with G(s) = K/(s(s + 5)) must follow the ramp r(t) = 1t with a steady-state error of 0.5. Find the gain K.

G(s) = K/(s(s + 5))r(t) = 1tess = 0.5
K =
8.Steady-state error & PID controlHard
r(t) = 1t+1 + 1/s4(s + 2)(s + 10)C(s)−

A PI controller C(s) = 1 + 1/s drives the plant G(s) = 4/((s + 2)(s + 10)) in a unity-feedback loop. Find the steady-state error ess for the ramp input r(t) = 1t.

C(s) = 1 + 1/sG(s) = 4/((s + 2)(s + 10))r(t) = 1t
ess =
9.Block diagramsMedium
R(s)+K(s + 7)(s + 9)G(s)C(s)−

The plant G(s) = K/((s + 7)(s + 9)) 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 + 7)(s + 9))unity negative feedback
K =
10.Block diagramsHard
R(s)+K(s + 1)(s + 7)(s + 8)G(s)C(s)−

The plant G(s) = K/((s + 1)(s + 7)(s + 8)) 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 + 1)(s + 7)(s + 8))unity negative feedback
ω = rad/s

3Solutions

1.Step response

Answer: %OS = 1.732 %

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 = 10 → ωₙ = 3.162 rad/s
2ζωₙ = 5 → ζ = 5/(2·3.162) = 0.7906

Result

%OS = 100·e−ζπ/√(1−ζ²) = 100·e−4.056 = 1.732 %

Answer

%OS = 1.732 %
2.Step response

Answer: ωn = 8.682 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).

Damping ratio

ζ = −ln(0.05)/√(π² + ln²(0.05)) = 2.996/√(9.87 + 8.974) = 0.6901

Natural frequency

ωd = π/tp = π/0.5 = 6.283 rad/s
ωₙ = ωd/√(1 − ζ²) = 6.283/0.7237 = 8.682 rad/s

Answer

ωn = 8.682 rad/s
3.Bode plots & margins

Answer: f0 = 3393 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/√(0.01·2.2 × 10⁻⁷) = 21320 rad/s
f₀ = ω₀/(2π) = 3393 Hz

Answer

f0 = 3393 Hz
4.Bode plots & margins

Answer: K = 45.11

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/1)·(ω/9) = 1 → ω180 = √(1·9) = 3 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 1)·√(ω² + 81)) = K/(1·9·(1 + 9)) = K/90

Required gain

90/K = 106/20 = 1.995
K = 90/1.995 = 45.11

Answer

K = 45.11
5.Stability: Routh–Hurwitz & root locus

Answer: ω = 4.243 rad/s

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 + 9s² + 18s + K = 0
s³: 1, 18
s²: 9, K
s¹: (9·18 − K)/9
s⁰: K

Critical gain

K = 9·18 = 162

Auxiliary equation

At the critical gain the s¹ row is zero; the s² row gives the poles on the imaginary axis.

9s² + 162 = 0 → s = ±j√18
ω = 4.243 rad/s

Answer

ω = 4.243 rad/s
6.Stability: Routh–Hurwitz & root locus

Answer: θd = 74.05 °

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.

Pole locations

s² + 2s + 5 = 0 → s = −1 ± j2
s = −8

Angles from the other poles

∠(−1 + j2 + 8) = atan2(2, 7) = 15.95°
∠(−1 + j2 − (−1 − j2)) = ∠(j4) = 90°

Angle of departure

θd = 180° − 15.95° − 90° = 74.05°

Answer

θd = 74.05°
7.Steady-state error & PID control

Answer: K = 10

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.

Required velocity constant

Kv = A/ess = 1/0.5 = 2

Gain

K = 5·Kv = 10

Answer

K = 10

A large gain makes this loop very lightly damped: ζ = 0.7906. That conflict is why compensators are used.

8.Steady-state error & PID control

Answer: ess = 5

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).

Velocity constant

s·C(s) = 1s + 1 → 1 as s → 0.

Kv = 1·4/20 = 0.2

Ramp error

ess = 1/0.2 = 5

Answer

ess = 5

The PI controller removes the step error completely; a ramp is followed with a constant lag.

9.Block diagrams

Answer: K = 1

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 + 7)(s + 9)) = 0 → s² + 16s + 63 + K = 0

Repeated root

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

16² = 4(63 + K) → K = 256/4 − 63 = 1

Answer

K = 1
Both closed-loop poles are then at s = −8.
10.Block diagrams

Answer: ω = 8.426 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 + 16s² + 71s + 56 + K = 0

Routh array

s³ | 1 71
s² | 16 56 + K
s¹ | (16·71 − 56 − K)/16
s⁰ | 56 + K

Stability limit

All entries of the first column must be positive.

Kmax = 16·71 − 56 = 1080

Frequency of oscillation

16·(jω)² + 56 + 1080 = 0 → ω² = 1136/16 = 71
ω = 8.426 rad/s

Answer

ω = 8.426 rad/s

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