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)0.5tp​ = ?

A system has the transfer function G(s) = 16.5/(s² + 9s + 33) and receives a unit step input. Find the peak time tp of the step response.

G(s) = 16.5/(s² + 9s + 33)
tp = s
2.Step responseHard
R(s)+40s(s + 4)G(s)C(s)−

A unity negative feedback loop has G(s) = 40/(s(s + 4)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 40/(s(s + 4))
ts = s
3.Bode plots & marginsMedium
vin​R = 4.7 kΩC = 22 nFvout​

The RC low-pass filter shown has R = 4.7 kΩ, C = 22 nF. A sinusoidal input of 3.1 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 4.7 kΩC = 22 nFf = 3.1 kHz
φ = °
4.Bode plots & marginsHard
|H| (dB)11101010010010001000−80−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 145/(s(s + 10)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 145/(s(s + 10))
ωc = rad/s
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 6s(s + 1)(s + 4)C(s)−

A unity-feedback loop has L(s) = K(s + 6)/(s(s + 1)(s + 4)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K(s + 6)/(s(s + 1)(s + 4))
Kmax =
6.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−71j1j2

The root locus of L(s) = K(s + 6)/(s(s + 4)) leaves the real axis between s = 0 and s = −4, circles round and joins the real axis again to the left of the zero. Find the break-in point si.

L(s) = K(s + 6)/(s(s + 4))
si =
7.Steady-state error & PID controlMedium
R(s)+K = 915s + 1C(s)−

A process G(s) = 1/(5s + 1) is controlled by a proportional gain K = 9 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.

G(s) = 1/(5s + 1)K = 9
ess = %
8.Steady-state error & PID controlHard
R(s)+8 + Kd​s5s(s + 2)C(s)−

A PD controller C(s) = 8 + Kd·s drives the plant G(s) = 5/(s(s + 2)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.6.

G(s) = 5/(s(s + 2))Kp = 8ζ = 0.6
Kd = s
9.Block diagramsMedium
R(s)+4(s + 2)(s + 1)(s + 3)G(s)C(s)−

A unity negative feedback loop has G(s) = 4(s + 2)/((s + 1)(s + 3)). A step of size 10 is applied at the reference. Find the steady-state error ess.

G(s) = 4(s + 2)/((s + 1)(s + 3))R = 10
ess =
10.Block diagramsHard
R(s)+9s + 3G₁+10s + 6G₂C(s)2H₁−0.25H₂−

Reduce the block diagram with G₁ = 9/(s + 3), G₂ = 10/(s + 6), an inner negative feedback H₁ = 2 around G₂ and an outer negative feedback H₂ = 0.25. Find the DC gain T(0) of C(s)/R(s).

G₁ = 9/(s + 3)G₂ = 10/(s + 6)H₁ = 2H₂ = 0.25
T(0) =

3Solutions

1.Step response

Answer: tp = 0.8798 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 = 33 → ωₙ = 5.745 rad/s
2ζωₙ = 9 → ζ = 9/(2·5.745) = 0.7833

Result

ωd = 5.745·√(1 − 0.7833²) = 3.571 rad/s → tp = π/ωd = 0.8798 s

Answer

tp = 0.8798 s
2.Step response

Answer: ts = 2 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) = 40/(s² + 4s + 40): ωₙ = 6.325 rad/s, ζ = 0.3162

Settling time

ζωₙ = 4/2 = 2
ts ≈ 4/(ζωₙ) = 2 s

Answer

ts = 2 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: φ = −63.59 °

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/(4700·2.2 × 10⁻⁸) = 9671 rad/s
fc = 1539 Hz → f/fc = 3100/1539 = 2.014

Phase

φ = − atan(2.014) = −63.59°

Answer

φ = −63.59°
4.Bode plots & margins

Answer: ωc = 10.17 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 + 100ω² − 21025 = 0
ω² = (−100 + √(10000 + 4·21025))/2 = 103.4
ωc = 10.17 rad/s

Answer

ωc = 10.17 rad/s
5.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 20

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 + 4) + K(s + 6) = s³ + 5s² + (4 + K)s + 6K

Routh array

s³: 1, 4 + K
s²: 5, 6K
s¹: (5·(4 + K) − 6K)/5
s⁰: 6K

Condition

20 − 1K > 0
K < 20/1 = 20

Answer

Kmax = 20

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: si = −9.464

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 + 4s)/(s + 6)

Stationary points

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

(2s + 4)(s + 6) − (s² + 4s) = s² + 12s + 24 = 0
s = −6 ± √12 = −2.536, −9.464

Break-in point

si = −9.464

Answer

si = −9.464

Between the two break points the locus is a circle centred on the zero, with radius √12 = 3.464.

7.Steady-state error & PID control

Answer: ess = 10 %

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 = 9·1 = 9

Step error

ess = 100/(1 + 9) = 10 %

Answer

ess = 10 %
8.Steady-state error & PID control

Answer: Kd = 1.118 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 + 2) + 5(8 + Kds) = s² + (2 + 5Kd)s + 40

Natural frequency

ωₙ = √40 = 6.325 rad/s

Derivative gain

2 + 5Kd = 2·0.6·6.325 = 7.589
Kd = 1.118 s

Answer

Kd = 1.118 s

The derivative term adds damping without changing ωₙ.

9.Block diagrams

Answer: ess = 2.727

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) = 4·2/(1·3) = 2.667

Steady-state error

ess = A/(1 + Kₚ) = 10/(1 + 2.667) = 2.727

Answer

ess = 2.727
10.Block diagrams

Answer: T(0) = 0.8955

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 = G₂/(1 + G₂H₁) = 10/(s + 26)

Outer loop

T(s) = G₁Gᵢ/(1 + G₁GᵢH₂) = 90/((s + 3)(s + 26) + 22.5)

DC gain

T(0) = 90/(3·26 + 22.5) = 90/100.5 = 0.8955

Answer

T(0) = 0.8955

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