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)1

A system has the transfer function G(s) = 86/(s² + 11s + 86) and receives a unit step input. Find the damping ratio ζ.

G(s) = 86/(s² + 11s + 86)
ζ =
2.Step responseHard
+−24 VR = 126 ΩL = 200 mHC = 2 µFvC​

A series RLC circuit with R = 126 Ω, L = 200 mH and C = 2 µF is connected to a 24 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the time tp of the first voltage peak, in ms.

R = 126 ΩL = 200 mHC = 2 µFV = 24 V
tp = ms
3.Bode plots & marginsMedium
|H| (dB)0.10.111101010010010001000−40−2002040ω (rad/s)φ (°)ω (rad/s)−90−450ω = 2

Find the phase ∠G(jω) of G(s) = 12(s + 9)/(s(s + 16)) at ω = 2 rad/s, in degrees.

G(s) = 12(s + 9)/(s(s + 16))ω = 2 rad/s
∠G(jω) = °
4.Bode plots & marginsHard
|H| (dB)11101010010010001000−80−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​

A unity-feedback loop has the open-loop transfer function L(s) = 740/(s(s + 10)). Find the phase margin PM in degrees.

L(s) = 740/(s(s + 10))
PM = °
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+21s(s + 4)(s + 6)C(s)+

The loop in the diagram uses positive feedback, with L(s) = 2/(s(s + 4)(s + 6)). How many closed-loop poles N lie in the right half-plane?

L(s) = 2/(s(s + 4)(s + 6))
N =
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) = 0.5t+4(s + 2)s(s + 1)(s + 4)C(s)−

A ramp r(t) = 0.5t is applied to a unity-feedback loop with G(s) = 4(s + 2)/(s(s + 1)(s + 4)). Find the steady-state error ess.

G(s) = 4(s + 2)/(s(s + 1)(s + 4))r(t) = 0.5t
ess =
8.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s5s + 1C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 5/(s + 1) in a unity-feedback loop. The closed loop must have ωₙ = 6 rad/s and ζ = 0.5. Find the proportional gain Kp.

G(s) = 5/(s + 1)ωₙ = 6 rad/sζ = 0.5
Kp =
9.Block diagramsMedium
R(s)+Ks(s + 14)G(s)C(s)−

The plant G(s) = K/(s(s + 14)) 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(s + 14))unity negative feedback
K =
10.Block diagramsHard
R(s)+56s(s + 6)G(s)C(s)1H−

The plant G(s) = 56/(s(s + 6)) is in a negative feedback loop with sensor gain H = 1. The closed-loop poles are complex. Find their imaginary part ωd (the positive value), in rad/s.

G(s) = 56/(s(s + 6))H = 1
ωd = rad/s

3Solutions

1.Step response

Answer: ζ = 0.5931

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 = 86 → ωₙ = 9.274 rad/s
2ζωₙ = 11 → ζ = 11/(2·9.274) = 0.5931

Result

ζ = 0.5931

Answer

ζ = 0.5931
2.Step response

Answer: tp = 2.028 ms

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/√(0.2·2 × 10⁻⁶) = 1581 rad/s
ζ = (R/2)·√(C/L) = (126/2)·√(2 × 10⁻⁶/0.2) = 0.1992

Result

ωd = 1549 rad/s
tp = π/ωd = 2.028 ms

Answer

tp = 2.028 ms
3.Bode plots & margins

Answer: ∠G(jω) = −84.6 °

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.

Add the angles

∠G = atan(2/9) − 90° − atan(2/16) = 12.53° − 90° − 7.125°

Result

∠G(j2) = −84.6°

Answer

∠G(jω) = −84.6°
4.Bode plots & margins

Answer: PM = 20.82 °

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 + 100ω² − 547600 = 0
ω² = (−100 + √(10000 + 4·547600))/2 = 691.7
ωc = 26.3 rad/s

Phase margin

∠L(jωc) = −90° − atan(26.3/10) = −159.2°
PM = 180° + ∠L = 20.82°

Answer

PM = 20.82°
5.Stability: Routh–Hurwitz & root locus

Answer: N = 1

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 + 10s² + 24s − 2 = 0

Routh array

s³: 1, 24
s²: 10, −2
s¹: 24.2
s⁰: −2

Sign changes in the first column

1 → 10 → 24.2 → −2
N = 1

Answer

N = 1

Positive feedback makes the constant coefficient negative: one real pole always moves into the right half-plane.

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 = 0.25

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) = 2

Ramp error

ess = 0.5/2 = 0.25

Answer

ess = 0.25
8.Steady-state error & PID control

Answer: Kp = 1

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 + 1).

s² + (1 + 5Kp)s + 5Ki

Target polynomial

s² + 2·0.5·6s + 6² = s² + 6s + 36

Match the coefficients

1 + 5Kp = 6 → Kp = 1

Answer

Kp = 1

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

9.Block diagrams

Answer: K = 49

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(s + 14)) = 0 → s² + 14s + K = 0

Repeated root

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

14² = 4K → K = 196/4 = 49

Answer

K = 49
Both closed-loop poles are then at s = −7.
10.Block diagrams

Answer: ωd = 6.856 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

s(s + 6) + 56·1 = 0 → s² + 6s + 56 = 0

Roots

s = −3 ± j·√(56 − 9) = −3 ± j6.856
ωd = 6.856 rad/s

Answer

ωd = 6.856 rad/s

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