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) = 62/(s² + 11s + 62) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.

G(s) = 62/(s² + 11s + 62)
σ =
2.Step responseHard
k = 500 N/mc = 13 N·s/mm = 1 kgF = 25 Nx(t)

A mass m = 1 kg hangs on a spring k = 500 N/m with a viscous damper c = 13 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 25 N is suddenly applied. Find the largest displacement xmax in mm.

m = 1 kgk = 500 N/mc = 13 N·s/mF = 25 N
xmax = mm
3.Bode plots & marginsMedium
vin​R = 1 kΩC = 4.7 nFvout​

The RC low-pass filter shown has R = 1 kΩ, C = 4.7 nF. A sinusoidal input of 68 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 1 kΩC = 4.7 nFf = 68 kHz
φ = °
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 3 dB.

L(s) = K/(s(s + 1)(s + 9))GM = 3 dB
K =
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+22401s(s + 4)(s + 10)C(s)−

The loop in the diagram has negative feedback and L(s) = 2240/(s(s + 4)(s + 10)). How many closed-loop poles N lie in the right half-plane?

L(s) = 2240/(s(s + 4)(s + 10))
N =
6.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−7−81j1j2

The root locus of L(s) = K(s + 7)/(s(s + 6)) leaves the real axis between s = 0 and s = −6, circles round and joins the real axis again to the left of the zero. Find the break-in point si.

L(s) = K(s + 7)/(s(s + 6))
si =
7.Steady-state error & PID controlMedium
R(s)+K = 100.510s + 1C(s)−

A process G(s) = 0.5/(10s + 1) is controlled by a proportional gain K = 10 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.

G(s) = 0.5/(10s + 1)K = 10
ess = %
8.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s10s + 1C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 10/(s + 1) in a unity-feedback loop. The closed loop must have ωₙ = 8 rad/s and ζ = 0.7. Find the integral gain Ki.

G(s) = 10/(s + 1)ωₙ = 8 rad/sζ = 0.7
Ki =
9.Block diagramsMedium
R(s)+6(s + 7)(s + 5)(s + 9)G(s)C(s)−

A unity negative feedback loop has G(s) = 6(s + 7)/((s + 5)(s + 9)). A step of size 2 is applied at the reference. Find the steady-state error ess.

G(s) = 6(s + 7)/((s + 5)(s + 9))R = 2
ess =
10.Block diagramsHard
R(s)+24s(s + 8)G(s)C(s)2H−

The plant G(s) = 24/(s(s + 8)) is in a negative feedback loop with sensor gain H = 2. The closed-loop poles are complex. Find their imaginary part ωd (the positive value), in rad/s.

G(s) = 24/(s(s + 8))H = 2
ωd = rad/s

3Solutions

1.Step response

Answer: σ = −5.5

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 = 62 → ωₙ = 7.874 rad/s
2ζωₙ = 11 → ζ = 11/(2·7.874) = 0.6985

Result

σ = −ζωₙ = −11/2 = −5.5

Answer

σ = −5.5
2.Step response

Answer: xmax = 69.25 mm

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 = √(k/m) = √(500/1) = 22.36 rad/s
ζ = c/(2√(km)) = 13/(2·√(500·1)) = 0.2907

Result

xst = F/k = 25/500 m = 50 mm
%OS = 38.5 %
xmax = 50·(1 + 0.385) = 69.25 mm

Answer

xmax = 69.25 mm
3.Bode plots & margins

Answer: φ = −63.53 °

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.

Corner frequency

ωc = 1/(RC) = 1/(1000·4.7 × 10⁻⁹) = 212766 rad/s
fc = 33863 Hz → f/fc = 68000/33863 = 2.008

Phase

φ = − atan(2.008) = −63.53°

Answer

φ = −63.53°
4.Bode plots & margins

Answer: K = 63.72

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 = 103/20 = 1.413
K = 90/1.413 = 63.72

Answer

K = 63.72
5.Stability: Routh–Hurwitz & root locus

Answer: N = 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

s³ + 14s² + 40s + 2240 = 0

Routh array

s³: 1, 40
s²: 14, 2240
s¹: −120
s⁰: 2240

Sign changes in the first column

1 → 14 → −120 → 2240
N = 2

Answer

N = 2

The gain is above the limit K = 560, so a complex pair has crossed into the right half-plane.

6.Stability: Routh–Hurwitz & root locus

Answer: si = −9.646

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 + 6s)/(s + 7)

Stationary points

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

(2s + 6)(s + 7) − (s² + 6s) = s² + 14s + 42 = 0
s = −7 ± √7 = −4.354, −9.646

Break-in point

si = −9.646

Answer

si = −9.646

Between the two break points the locus is a circle centred on the zero, with radius √7 = 2.646.

7.Steady-state error & PID control

Answer: ess = 16.67 %

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.

Position constant

Kp = 10·0.5 = 5

Step error

ess = 100/(1 + 5) = 16.67 %

Answer

ess = 16.67 %
8.Steady-state error & PID control

Answer: Ki = 6.4

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 + 10Kp)s + 10Ki

Target polynomial

s² + 2·0.7·8s + 8² = s² + 11.2s + 64

Match the coefficients

10Ki = 64 → Ki = 6.4

Answer

Ki = 6.4

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

9.Block diagrams

Answer: ess = 1.034

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).

Position error constant

Kₚ = G(0) = 6·7/(5·9) = 0.9333

Steady-state error

ess = A/(1 + Kₚ) = 2/(1 + 0.9333) = 1.034

Answer

ess = 1.034
10.Block diagrams

Answer: ωd = 5.657 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 + 8) + 24·2 = 0 → s² + 8s + 48 = 0

Roots

s = −4 ± j·√(48 − 16) = −4 ± j5.657
ωd = 5.657 rad/s

Answer

ωd = 5.657 rad/s

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