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) = 32/(s² + s + 16) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.

G(s) = 32/(s² + s + 16)
σ =
2.Step responseHard
R(s)+Ks(s + 16)G(s)C(s)−

A unity negative feedback loop has G(s) = K/(s(s + 16)). Find the gain K that gives a step response with 30 % overshoot.

G(s) = K/(s(s + 16))%OS = 30 %
K =
3.Bode plots & marginsMedium
|H| (dB)0.10.111101010010010001000−40−20020406080ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ω = 10

Find the phase ∠G(jω) of G(s) = 17(s + 19)/(s(s + 1)) at ω = 10 rad/s, in degrees.

G(s) = 17(s + 19)/(s(s + 1))ω = 10 rad/s
∠G(jω) = °
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) = 8/(s(s + 1)(s + 6)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 8/(s(s + 1)(s + 6))
ω180 = rad/s
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+501s(s + 2)(s + 3)C(s)+

The loop in the diagram uses positive feedback, with L(s) = 50/(s(s + 2)(s + 3)). How many closed-loop poles N lie in the right half-plane?

L(s) = 50/(s(s + 2)(s + 3))
N =
6.Stability: Routh–Hurwitz & root locusHard
ReIm−5j5

The root locus of L(s) = K/((s + 4)(s + 5)(s + 6)) is sketched. Find the frequency ω at which the locus crosses the imaginary axis.

L(s) = K/((s + 4)(s + 5)(s + 6))
ω = rad/s
7.Steady-state error & PID controlMedium
r(t) = 0.5t+20s(s + 2)(s + 5)C(s)−

A ramp r(t) = 0.5t is applied to a unity-feedback loop with G(s) = 20/(s(s + 2)(s + 5)). Find the steady-state error ess.

G(s) = 20/(s(s + 2)(s + 5))r(t) = 0.5t
ess =
8.Steady-state error & PID controlHard
r(t) = 2t+2 + 5/s5(s + 4)(s + 5)C(s)−

A PI controller C(s) = 2 + 5/s drives the plant G(s) = 5/((s + 4)(s + 5)) in a unity-feedback loop. Find the steady-state error ess for the ramp input r(t) = 2t.

C(s) = 2 + 5/sG(s) = 5/((s + 4)(s + 5))r(t) = 2t
ess =
9.Block diagramsMedium
R(s)+23s + 8G(s)C(s)2H−

A plant G(s) = 23/(s + 8) is controlled in a negative feedback loop whose sensor has the gain H = 2. A step of size 2 is applied at R. Find the steady-state output css.

G(s) = 23/(s + 8)H = 2R = 2
css =
10.Block diagramsHard
R(s)+Ks(s + 4)(s + 7)G(s)C(s)−

The plant G(s) = K/(s(s + 4)(s + 7)) is in a unity negative feedback loop. Use the Routh–Hurwitz criterion to find the largest gain Kmax for which the closed loop is stable.

G(s) = K/(s(s + 4)(s + 7))unity negative feedback
Kmax =

3Solutions

1.Step response

Answer: σ = −0.5

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 = 16 → ωₙ = 4 rad/s
2ζωₙ = 1 → ζ = 1/(2·4) = 0.125

Result

σ = −ζωₙ = −1/2 = −0.5

Answer

σ = −0.5
2.Step response

Answer: K = 499.8

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).

Closed-loop transfer function

T(s) = K/(s² + 16s + K)

Damping ratio from the overshoot

ζ = −ln(0.3)/√(π² + ln²(0.3)) = 0.3579

Gain

ωₙ = 16/(2ζ) = 22.36 rad/s
K = ωₙ² = 499.8

Answer

K = 499.8
3.Bode plots & margins

Answer: ∠G(jω) = −146.5 °

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(10/19) − 90° − atan(10/1) = 27.76° − 90° − 84.29°

Result

∠G(j10) = −146.5°

Answer

∠G(jω) = −146.5°
4.Bode plots & margins

Answer: ω180 = 2.449 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.

Phase crossover

(ω/1)·(ω/6) = 1 → ω180 = √(1·6) = 2.449 rad/s

Answer

ω180 = 2.449 rad/s
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 + 5s² + 6s − 50 = 0

Routh array

s³: 1, 6
s²: 5, −50
s¹: 16
s⁰: −50

Sign changes in the first column

1 → 5 → 16 → −50
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: ω = 8.602 rad/s

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.

Critical gain

s³ + 15s² + 74s + 120 + K = 0
s¹: (15·74 − 120 − K)/15 = 0 → K = 990

Crossing frequency

15s² + 120 + 990 = 15s² + 1110 = 0 → s² = −74
ω = √74 = 8.602 rad/s

Answer

ω = 8.602 rad/s
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: ess = 1.6

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) = 2s + 5 → 5 as s → 0.

Kv = 5·5/20 = 1.25

Ramp error

ess = 2/1.25 = 1.6

Answer

ess = 1.6

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

9.Block diagrams

Answer: css = 0.8519

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).

Closed-loop transfer function

T(s) = G/(1 + G·H) = 23/(s + 8 + 46) = 23/(s + 54)

Result

css = R·T(0) = 2·23/54 = 0.8519

Answer

css = 0.8519
10.Block diagrams

Answer: Kmax = 308

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 + 11s² + 28s + K = 0

Routh array

s³ | 1 28
s² | 11 K
s¹ | (11·28 − K)/11
s⁰ | K

Stability limit

All entries of the first column must be positive.

Kmax = 11·28 = 308

Answer

Kmax = 308

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