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)1ts​ = ?

A system has the transfer function G(s) = 74/(s² + 8s + 74) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 74/(s² + 8s + 74)
ts = s
2.Step responseHard
R(s)+Ks(s + 12)G(s)C(s)−

A unity negative feedback loop has G(s) = K/(s(s + 12)). Find the gain K that gives a step response with 10 % overshoot.

G(s) = K/(s(s + 12))%OS = 10 %
K =
3.Bode plots & marginsMedium
vin​R = 1 kΩC = 1 nFvout​

The RC low-pass filter shown has R = 1 kΩ, C = 1 nF. A sinusoidal input of 80 kHz is applied. Find the gain |H| = |vout/vin| in dB.

R = 1 kΩC = 1 nFf = 80 kHz
|H| = dB
4.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−80−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 6.5/(s(s + 1)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 6.5/(s(s + 1))
ωc = rad/s
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 6s(s + 2)(s + 3)C(s)−

A unity-feedback loop has L(s) = K(s + 6)/(s(s + 2)(s + 3)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K(s + 6)/(s(s + 2)(s + 3))
Kmax =
6.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4j2

For L(s) = K/((s + 1)(s² + 8s + 25)), find the angle of departure θd of the root locus from the complex pole in the upper half-plane, measured from the positive real axis.

L(s) = K/((s + 1)(s² + 8s + 25))
θd = °
7.Steady-state error & PID controlMedium
R(s)+K = 1950.5s + 1C(s)−

A process G(s) = 5/(0.5s + 1) is controlled by a proportional gain K = 19 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.

G(s) = 5/(0.5s + 1)K = 19
ess = %
8.Steady-state error & PID controlHard
R(s)+K = 54s(s + 5)C(s)−

A proportional controller K = 5 drives the plant G(s) = 4/(s(s + 5)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 4/(s(s + 5))K = 5
ts = s
9.Block diagramsMedium
R(s)+2K+8s + 7G(s)C(s)3H₁−−

The plant G(s) = 8/(s + 7) has an inner (minor) negative feedback loop with gain H₁ = 3. A controller gain K = 2 and an outer unity negative feedback loop close the system. Find the closed-loop pole p.

G(s) = 8/(s + 7)H₁ = 3K = 2
p =
10.Block diagramsHard
R(s)+Kcontroller+D(s)+10s + 6plantC(s)−

A proportional controller K drives the plant G(s) = 10/(s + 6) in a unity negative feedback loop. A disturbance step of size 2 enters at the plant input while R = 0. Find the smallest gain K that keeps the steady-state output at or below 0.6667.

G(s) = 10/(s + 6)D = 2css ≤ 0.6667
K =

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 = 74 → ωₙ = 8.602 rad/s
2ζωₙ = 8 → ζ = 8/(2·8.602) = 0.465

Result

ζωₙ = 4 → ts ≈ 4/4 = 1 s

Answer

ts = 1 s
2.Step response

Answer: K = 103

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 + 12s + K)

Damping ratio from the overshoot

ζ = −ln(0.1)/√(π² + ln²(0.1)) = 0.5912

Gain

ωₙ = 12/(2ζ) = 10.15 rad/s
K = ωₙ² = 103

Answer

K = 103
3.Bode plots & margins

Answer: |H| = −0.9783 dB

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/(1000·1 × 10⁻⁹) = 1000000 rad/s
fc = 159155 Hz → f/fc = 80000/159155 = 0.5027

Gain

|H| = 1/√(1 + 0.5027²) = 0.8935
20·log₁₀(0.8935) = −0.9783 dB

Answer

|H| = −0.9783 dB
4.Bode plots & margins

Answer: ωc = 2.453 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 + 1ω² − 42.25 = 0
ω² = (−1 + √(1 + 4·42.25))/2 = 6.019
ωc = 2.453 rad/s

Answer

ωc = 2.453 rad/s
5.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 30

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 + 2)(s + 3) + K(s + 6) = s³ + 5s² + (6 + K)s + 6K

Routh array

s³: 1, 6 + K
s²: 5, 6K
s¹: (5·(6 + K) − 6K)/5
s⁰: 6K

Condition

30 − 1K > 0
K < 30/1 = 30

Answer

Kmax = 30

With the zero to the right of s = −5 the loop would stay stable for every K > 0.

6.Stability: Routh–Hurwitz & root locus

Answer: θd = −45 °

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.

Pole locations

s² + 8s + 25 = 0 → s = −4 ± j3
s = −1

Angles from the other poles

∠(−4 + j3 + 1) = atan2(3, -3) = 135°
∠(−4 + j3 − (−4 − j3)) = ∠(j6) = 90°

Angle of departure

θd = 180° − 135° − 90° = −45°

Answer

θd = −45°
7.Steady-state error & PID control

Answer: ess = 1.042 %

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 = 19·5 = 95

Step error

ess = 100/(1 + 95) = 1.042 %

Answer

ess = 1.042 %
8.Steady-state error & PID control

Answer: ts = 1.6 s

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

s(s + 5) + 5·4 = s² + 5s + 20

ωₙ and ζ

ωₙ = √20 = 4.472 rad/s
ζ = 5/(2·4.472) = 0.559

Settling time

ζωₙ = 5/2 = 2.5
ts ≈ 4/2.5 = 1.6 s

Answer

ts = 1.6 s

The settling time does not depend on K here: raising K only makes the response oscillate faster.

9.Block diagrams

Answer: p = −47

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) = 8/(s + 7 + 8·3) = 8/(s + 31)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 16/(s + 31 + 16) = 16/(s + 47)

Result

p = −47

Answer

p = −47
10.Block diagrams

Answer: K = 2.4

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).

Disturbance response

css = D·G(0)/(1 + K·G(0)) = 2·10/(6 + 10K)

Solve for K

6 + 10K = 2·10/0.6667 = 30
K = (30 − 6)/10 = 2.4

Answer

K = 2.4

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