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)0.5

A system has the transfer function G(s) = 31/(s² + 8s + 62) and receives a unit step input. Find the damping ratio ζ.

G(s) = 31/(s² + 8s + 62)
ζ =
2.Step responseHard
+−10 VR = 78 ΩL = 200 mHC = 47 µFvC​

A series RLC circuit with R = 78 Ω, L = 200 mH and C = 47 µF is connected to a 10 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 = 78 ΩL = 200 mHC = 47 µFV = 10 V
%OS = %
3.Bode plots & marginsMedium
vin​L = 22 mHR = 220 Ωvout​

The RL low-pass filter shown has R = 220 Ω, L = 22 mH. A sinusoidal input of 3.2 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 220 ΩL = 22 mHf = 3.2 kHz
φ = °
4.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−270−180−900ω180​

A unity-feedback loop has the open-loop transfer function L(s) = K/(s(s + 8)(s + 10)). Find the gain K that gives a gain margin of 5 dB.

L(s) = K/(s(s + 8)(s + 10))GM = 5 dB
K =
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1(s + 2)(s + 5)(s + 6)C(s)−

A unity-feedback loop has L(s) = K/((s + 2)(s + 5)(s + 6)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/((s + 2)(s + 5)(s + 6))
Kmax =
6.Stability: Routh–Hurwitz & root locusHard
ReImζ = 0.707−2−4−6−8−10−12j2j4

The root locus of L(s) = K/(s(s + 10)) is shown. Find the gain K for which the closed-loop poles have the damping ratio ζ = 0.707.

L(s) = K/(s(s + 10))ζ = 0.707
K =
7.Steady-state error & PID controlMedium
r(t) = 5 + 3t+20(s + 2)s(s + 1)(s + 2)C(s)−

The input r(t) = 5 + 3t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 20(s + 2)/(s(s + 1)(s + 2)). Find the steady-state error ess.

G(s) = 20(s + 2)/(s(s + 1)(s + 2))r(t) = 5 + 3t
ess =
8.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s2s + 4C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 2/(s + 4) in a unity-feedback loop. The closed loop must have ωₙ = 3 rad/s and ζ = 1. Find the proportional gain Kp.

G(s) = 2/(s + 4)ωₙ = 3 rad/sζ = 1
Kp =
9.Block diagramsMedium
R(s)+7K+6s + 3G(s)C(s)4H₁−−

The plant G(s) = 6/(s + 3) has an inner (minor) negative feedback loop with gain H₁ = 4. A controller gain K = 7 and an outer unity negative feedback loop close the system. Find the closed-loop DC gain T(0).

G(s) = 6/(s + 3)H₁ = 4K = 7
T(0) =
10.Block diagramsHard
R(s)+Ks(s + 1)(s + 5)G(s)C(s)−

The plant G(s) = K/(s(s + 1)(s + 5)) 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(s + 1)(s + 5))unity negative feedback
ω = rad/s

3Solutions

1.Step response

Answer: ζ = 0.508

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 = 8 → ζ = 8/(2·7.874) = 0.508

Result

ζ = 0.508

Answer

ζ = 0.508
2.Step response

Answer: %OS = 9.603 %

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/√(0.2·4.7 × 10⁻⁵) = 326.2 rad/s
ζ = (R/2)·√(C/L) = (78/2)·√(4.7 × 10⁻⁵/0.2) = 0.5979

Result

%OS = 100·e−ζπ/√(1−ζ²) = 9.603 %

Answer

%OS = 9.603 %
3.Bode plots & margins

Answer: φ = −63.56 °

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 = R/L = 220/0.022 = 10000 rad/s
fc = 1592 Hz → f/fc = 3200/1592 = 2.011

Phase

φ = − atan(2.011) = −63.56°

Answer

φ = −63.56°
4.Bode plots & margins

Answer: K = 809.8

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/8)·(ω/10) = 1 → ω180 = √(8·10) = 8.944 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 64)·√(ω² + 100)) = K/(8·10·(8 + 10)) = K/1440

Required gain

1440/K = 105/20 = 1.778
K = 1440/1.778 = 809.8

Answer

K = 809.8
5.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 616

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 + 13s² + 52s + 60 + K = 0

Routh array

s³: 1, 52
s²: 13, 60 + K
s¹: (13·52 − 60 − K)/13
s⁰: 60 + K

Condition

The s¹ entry must stay positive.

K < 13·52 − 60 = 616

Answer

Kmax = 616

At the limit the loop oscillates at ω = √52 = 7.211 rad/s.

6.Stability: Routh–Hurwitz & root locus

Answer: K = 50.02

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.

Compare with the standard form

s² + 10s + K = s² + 2ζωₙs + ωₙ²
2·0.707·ωₙ = 10 → ωₙ = 7.072 rad/s

Gain

K = ωₙ² = 7.072² = 50.02

Answer

K = 50.02

The closed-loop poles sit where the ζ line crosses the vertical branch at s = −5.

7.Steady-state error & PID control

Answer: ess = 0.15

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) = 20

Step part

Kp = ∞ → estep = 0

Ramp error

ess = 3/20 = 0.15

Answer

ess = 0.15
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 + 4).

s² + (4 + 2Kp)s + 2Ki

Target polynomial

s² + 2·1·3s + 3² = s² + 6s + 9

Match the coefficients

4 + 2Kp = 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: T(0) = 0.6087

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) = 6/(s + 3 + 6·4) = 6/(s + 27)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 42/(s + 27 + 42) = 42/(s + 69)

Result

T(0) = 42/69 = 0.6087

Answer

T(0) = 0.6087
10.Block diagrams

Answer: ω = 2.236 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 + 6s² + 5s + K = 0

Routh array

s³ | 1 5
s² | 6 K
s¹ | (6·5 − K)/6
s⁰ | K

Stability limit

All entries of the first column must be positive.

Kmax = 6·5 = 30

Frequency of oscillation

6·(jω)² + 30 = 0 → ω² = 30/6 = 5
ω = 2.236 rad/s

Answer

ω = 2.236 rad/s

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