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) = 4/(s² + 3s + 4) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 4/(s² + 3s + 4)
ts = s
2.Step responseHard
R(s)+255s(s + 20)G(s)C(s)−

A unity negative feedback loop has G(s) = 255/(s(s + 20)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 255/(s(s + 20))
ts = s
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 800 Hz is applied. Find the gain |H| = |vout/vin| in dB.

R = 220 ΩL = 22 mHf = 800 Hz
|H| = dB
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 + 2)(s + 8)). Find the gain K that gives a gain margin of 6 dB.

L(s) = K/(s(s + 2)(s + 8))GM = 6 dB
K =
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 1)(s + 10)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 1)(s + 10)). The gain is raised until the closed loop is on the edge of stability. Find the frequency ω of the sustained oscillation.

L(s) = K/(s(s + 1)(s + 10))
ω = rad/s
6.Stability: Routh–Hurwitz & root locusHard
ReIm−0.5−1−1.5−2−2.50.5j0.5j1

The root locus of L(s) = K/(s(s + 1)(s + 2)) leaves the real axis between s = 0 and s = −1. Find the break-away point sb.

L(s) = K/(s(s + 1)(s + 2))
sb =
7.Steady-state error & PID controlMedium
r(t) = 10 + 5t+2(s + 4)s(s + 4)(s + 5)C(s)−

The input r(t) = 10 + 5t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 2(s + 4)/(s(s + 4)(s + 5)). Find the steady-state error ess.

G(s) = 2(s + 4)/(s(s + 4)(s + 5))r(t) = 10 + 5t
ess =
8.Steady-state error & PID controlHard
R(s)+K5s(s + 4)C(s)−

A proportional controller K drives the plant G(s) = 5/(s(s + 4)) in a unity-feedback loop. Find the gain K that gives the closed loop a damping ratio ζ = 0.4.

G(s) = 5/(s(s + 4))ζ = 0.4
K =
9.Block diagramsMedium
R(s)+Ks(s + 12)G(s)C(s)−

The plant G(s) = K/(s(s + 12)) is in a unity negative feedback loop. Find the gain K for which the closed loop is critically damped (two equal real poles).

G(s) = K/(s(s + 12))unity negative feedback
K =
10.Block diagramsHard
R(s)+K(s + 3)(s + 4)(s + 6)G(s)C(s)−

The plant G(s) = K/((s + 3)(s + 4)(s + 6)) 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 + 3)(s + 4)(s + 6))unity negative feedback
ω = rad/s

3Solutions

1.Step response

Answer: ts = 2.667 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 = 4 → ωₙ = 2 rad/s
2ζωₙ = 3 → ζ = 3/(2·2) = 0.75

Result

ζωₙ = 1.5 → ts ≈ 4/1.5 = 2.667 s

Answer

ts = 2.667 s
2.Step response

Answer: ts = 0.4 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) = 255/(s² + 20s + 255): ωₙ = 15.97 rad/s, ζ = 0.6262

Settling time

ζωₙ = 20/2 = 10
ts ≈ 4/(ζωₙ) = 0.4 s

Answer

ts = 0.4 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: |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 = R/L = 220/0.022 = 10000 rad/s
fc = 1592 Hz → f/fc = 800/1592 = 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: K = 80.19

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/2)·(ω/8) = 1 → ω180 = √(2·8) = 4 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 4)·√(ω² + 64)) = K/(2·8·(2 + 8)) = K/160

Required gain

160/K = 106/20 = 1.995
K = 160/1.995 = 80.19

Answer

K = 80.19
5.Stability: Routh–Hurwitz & root locus

Answer: ω = 3.162 rad/s

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.

Routh array

s³ + 11s² + 10s + K = 0
s³: 1, 10
s²: 11, K
s¹: (11·10 − K)/11
s⁰: K

Critical gain

K = 11·10 = 110

Auxiliary equation

At the critical gain the s¹ row is zero; the s² row gives the poles on the imaginary axis.

11s² + 110 = 0 → s = ±j√10
ω = 3.162 rad/s

Answer

ω = 3.162 rad/s
6.Stability: Routh–Hurwitz & root locus

Answer: sb = −0.4226

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 + 1)(s + 2) = −(s³ + 3s² + 2s)

Stationary points

dK/ds = −(3s² + 6s + 2) = 0
s = (−6 ± √(36 − 24))/6 = −0.4226, −1.577

Pick the point on the locus

The locus lies on the real axis between 0 and −1; the other root is not on the locus.

sb = −0.4226

Answer

sb = −0.4226
7.Steady-state error & PID control

Answer: ess = 12.5

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) = 0.4

Step part

Kp = ∞ → estep = 0

Ramp error

ess = 5/0.4 = 12.5

Answer

ess = 12.5
8.Steady-state error & PID control

Answer: K = 5

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 + 4) + 5K = s² + 4s + 5K

Natural frequency

2·0.4·ωₙ = 4 → ωₙ = 5 rad/s

Gain

5K = ωₙ² = 25 → K = 5

Answer

K = 5
9.Block diagrams

Answer: K = 36

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

1 + K/(s(s + 12)) = 0 → s² + 12s + K = 0

Repeated root

A quadratic s² + βs + γ has a double root when β² = 4γ.

12² = 4K → K = 144/4 = 36

Answer

K = 36
Both closed-loop poles are then at s = −6.
10.Block diagrams

Answer: ω = 7.348 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 + 13s² + 54s + 72 + K = 0

Routh array

s³ | 1 54
s² | 13 72 + K
s¹ | (13·54 − 72 − K)/13
s⁰ | 72 + K

Stability limit

All entries of the first column must be positive.

Kmax = 13·54 − 72 = 630

Frequency of oscillation

13·(jω)² + 72 + 630 = 0 → ω² = 702/13 = 54
ω = 7.348 rad/s

Answer

ω = 7.348 rad/s

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