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)0.5ts​ = ?

A system has the transfer function G(s) = 13.5/(s² + 8s + 27) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 13.5/(s² + 8s + 27)
ts = s
2.Step responseHard
R(s)+55s(s + 2)G(s)C(s)−

A unity negative feedback loop has G(s) = 55/(s(s + 2)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 55/(s(s + 2))
ts = s
3.Bode plots & marginsMedium
vin​L = 20 mHC = 0.47 µFR = 50 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 50 Ω, with L = 20 mH and C = 0.47 µF. Find the quality factor Q.

R = 50 ΩL = 20 mHC = 0.47 µF
Q =
4.Bode plots & marginsHard
|H| (dB)0.10.111101010010010001000−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 128/(s(s + 4)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 128/(s(s + 4))
ωc = rad/s
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 7s(s + 1)(s + 2)C(s)−

A unity-feedback loop has L(s) = K(s + 7)/(s(s + 1)(s + 2)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K(s + 7)/(s(s + 1)(s + 2))
Kmax =
6.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6j1

The root locus of L(s) = K/(s(s + 3)(s + 5)) leaves the real axis between s = 0 and s = −3. Find the break-away point sb.

L(s) = K/(s(s + 3)(s + 5))
sb =
7.Steady-state error & PID controlMedium
r(t) = 10t+Ks(s + 10)C(s)−

A unity-feedback loop with G(s) = K/(s(s + 10)) must follow the ramp r(t) = 10t with a steady-state error of 0.2. Find the gain K.

G(s) = K/(s(s + 10))r(t) = 10tess = 0.2
K =
8.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s10s + 4C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 10/(s + 4) in a unity-feedback loop. The closed loop must have ωₙ = 5 rad/s and ζ = 0.5. Find the proportional gain Kp.

G(s) = 10/(s + 4)ωₙ = 5 rad/sζ = 0.5
Kp =
9.Block diagramsMedium
R(s)+15K+7s + 7G(s)C(s)5H₁−−

The plant G(s) = 7/(s + 7) has an inner (minor) negative feedback loop with gain H₁ = 5. A controller gain K = 15 and an outer unity negative feedback loop close the system. Find the closed-loop DC gain T(0).

G(s) = 7/(s + 7)H₁ = 5K = 15
T(0) =
10.Block diagramsHard
R(s)+52s(s + 6)G(s)C(s)1H−

The plant G(s) = 52/(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) = 52/(s(s + 6))H = 1
ωd = rad/s

3Solutions

1.Step response

Answer: ts = 1 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).

Compare with the standard form

The denominator must match s² + 2ζωₙs + ωₙ².

ωₙ² = 27 → ωₙ = 5.196 rad/s
2ζωₙ = 8 → ζ = 8/(2·5.196) = 0.7698

Result

ζωₙ = 4 → ts ≈ 4/4 = 1 s

Answer

ts = 1 s
2.Step response

Answer: ts = 4 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).

Closed-loop transfer function

T(s) = 55/(s² + 2s + 55): ωₙ = 7.416 rad/s, ζ = 0.1348

Settling time

ζωₙ = 2/2 = 1
ts ≈ 4/(ζωₙ) = 4 s

Answer

ts = 4 s

For this loop ζωₙ = a/2 whatever K is: raising K adds overshoot but does not shorten the settling time.

3.Bode plots & margins

Answer: Q = 4.126

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/√(0.02·4.7 × 10⁻⁷) = 10314 rad/s
f₀ = ω₀/(2π) = 1642 Hz

Quality factor

BW = R/L = 50/0.02 = 2500 rad/s = 397.9 Hz
Q = ω₀/BW = 10314/2500 = 4.126

Answer

Q = 4.126
4.Bode plots & margins

Answer: ωc = 10.97 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.

Gain crossover

ω⁴ + 16ω² − 16384 = 0
ω² = (−16 + √(256 + 4·16384))/2 = 120.2
ωc = 10.97 rad/s

Answer

ωc = 10.97 rad/s
5.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 1.5

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

s(s + 1)(s + 2) + K(s + 7) = s³ + 3s² + (2 + K)s + 7K

Routh array

s³: 1, 2 + K
s²: 3, 7K
s¹: (3·(2 + K) − 7K)/3
s⁰: 7K

Condition

6 − 4K > 0
K < 6/4 = 1.5

Answer

Kmax = 1.5

With the zero to the right of s = −3 the loop would stay stable for every K > 0.

6.Stability: Routh–Hurwitz & root locus

Answer: sb = −1.214

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 + 3)(s + 5) = −(s³ + 8s² + 15s)

Stationary points

dK/ds = −(3s² + 16s + 15) = 0
s = (−16 ± √(256 − 180))/6 = −1.214, −4.12

Pick the point on the locus

The locus lies on the real axis between 0 and −3; the other root is not on the locus.

sb = −1.214

Answer

sb = −1.214
7.Steady-state error & PID control

Answer: K = 500

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 = 10/0.2 = 50

Gain

K = 10·Kv = 500

Answer

K = 500

A large gain makes this loop very lightly damped: ζ = 0.2236. That conflict is why compensators are used.

8.Steady-state error & PID control

Answer: Kp = 0.1

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 + 10Kp)s + 10Ki

Target polynomial

s² + 2·0.5·5s + 5² = s² + 5s + 25

Match the coefficients

4 + 10Kp = 5 → Kp = 0.1

Answer

Kp = 0.1

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

9.Block diagrams

Answer: T(0) = 0.7143

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) = 7/(s + 7 + 7·5) = 7/(s + 42)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 105/(s + 42 + 105) = 105/(s + 147)

Result

T(0) = 105/147 = 0.7143

Answer

T(0) = 0.7143
10.Block diagrams

Answer: ωd = 6.557 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) + 52·1 = 0 → s² + 6s + 52 = 0

Roots

s = −3 ± j·√(52 − 9) = −3 ± j6.557
ωd = 6.557 rad/s

Answer

ωd = 6.557 rad/s

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