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) = 25/(s² + 8s + 25) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 25/(s² + 8s + 25)
ts = s
2.Step responseHard
+−10 VR = 62 ΩL = 500 mHC = 47 µFvC​

A series RLC circuit with R = 62 Ω, L = 500 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 = 62 ΩL = 500 mHC = 47 µFV = 10 V
%OS = %
3.Bode plots & marginsMedium
|H| (dB)0.010.1110100−40−20020ω (rad/s)ω = 0.5

Find the exact magnitude |G(jω)| of G(s) = 630/((s + 9)(s + 10)) at ω = 0.5 rad/s, in dB.

G(s) = 630/((s + 9)(s + 10))ω = 0.5 rad/s
|G(jω)| = 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) = 6/(s(s + 1)(s + 4)). Find the gain margin GM in dB.

L(s) = 6/(s(s + 1)(s + 4))
GM = dB
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 10(s − 2)(s + 2)C(s)−

The plant G(s) = (s + 10)/((s − 2)(s + 2)) has an unstable pole at s = 2. A proportional gain K closes a unity-feedback loop around it. Find the smallest gain Kmin that makes the closed loop stable.

G(s) = (s + 10)/((s − 2)(s + 2))
Kmin =
6.Stability: Routh–Hurwitz & root locusHard
ReIm−0.5−1−1.5−2−2.5−3−3.5−4−4.5−50.5j0.5j1

A unity-feedback loop has L(s) = K/((s + 2)(s + 3)(s + 4)). Find the point σa where the asymptotes of the root locus cross the real axis.

L(s) = K/((s + 2)(s + 3)(s + 4))
σa =
7.Steady-state error & PID controlMedium
R(s)+K = 22.52s + 1C(s)−

A process G(s) = 2.5/(2s + 1) is controlled by a proportional gain K = 2 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.

G(s) = 2.5/(2s + 1)K = 2
ess = %
8.Steady-state error & PID controlHard
R(s)+K = 504s(s + 8)C(s)−

A proportional controller K = 50 drives the plant G(s) = 4/(s(s + 8)) in a unity-feedback loop. Find the peak time tp of the step response.

G(s) = 4/(s(s + 8))K = 50
tp = s
9.Block diagramsMedium
R(s)+15s + 7G(s)C(s)0.5H−

A plant G(s) = 15/(s + 7) is controlled in a negative feedback loop whose sensor has the gain H = 0.5. Find the time constant τ of the closed loop.

G(s) = 15/(s + 7)H = 0.5
τ = s
10.Block diagramsHard
R(s)+28s(s + 3)G(s)C(s)1H−

The plant G(s) = 28/(s(s + 3)) is in a negative feedback loop with sensor gain H = 1. The closed-loop poles are complex. Find their real part σ.

G(s) = 28/(s(s + 3))H = 1
σ =

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 = 25 → ωₙ = 5 rad/s
2ζωₙ = 8 → ζ = 8/(2·5) = 0.8

Result

ζωₙ = 4 → ts ≈ 4/4 = 1 s

Answer

ts = 1 s
2.Step response

Answer: %OS = 37.16 %

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/√(0.5·4.7 × 10⁻⁵) = 206.3 rad/s
ζ = (R/2)·√(C/L) = (62/2)·√(4.7 × 10⁻⁵/0.5) = 0.3006

Result

%OS = 100·e−ζπ/√(1−ζ²) = 37.16 %

Answer

%OS = 37.16 %
3.Bode plots & margins

Answer: |G(jω)| = 16.88 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.

Multiply the magnitudes

|G(j0.5)| = 630/(√(0.5² + 9²)·√(0.5² + 10²)) = 630/(9.014·10.01) = 6.981

Result

20·log₁₀(6.981) = 16.88 dB

Answer

|G(jω)| = 16.88 dB
4.Bode plots & margins

Answer: GM = 10.46 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.

Phase crossover

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

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 1)·√(ω² + 16)) = K/(1·4·(1 + 4)) = K/20

Gain margin

GM = 20·log₁₀(20/6) = 10.46 dB

Answer

GM = 10.46 dB
5.Stability: Routh–Hurwitz & root locus

Answer: Kmin = 0.4

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 − 2)(s + 2) + K(s + 10) = s² + (0 + K)s + (10K − 4)

Conditions

For s² + c₁s + c₀ all roots lie in the left half-plane exactly when c₁ > 0 and c₀ > 0.

0 + K > 0 → K > 0
10K − 4 > 0 → K > 0.4
Kmin = 0.4

Answer

Kmin = 0.4
6.Stability: Routh–Hurwitz & root locus

Answer: σa = −3

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.

Poles and zeros

poles: −2, −3, −4
n − m = 3 − 0 = 3

Centroid

σa = (−9)/3 = −3

Answer

σa = −3

The 3 asymptotes leave the centroid at angles of 60°, 180°, 300°.

7.Steady-state error & PID control

Answer: ess = 16.67 %

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 = 2·2.5 = 5

Step error

ess = 100/(1 + 5) = 16.67 %

Answer

ess = 16.67 %
8.Steady-state error & PID control

Answer: tp = 0.2316 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 + 8) + 50·4 = s² + 8s + 200

ωₙ and ζ

ωₙ = √200 = 14.14 rad/s
ζ = 8/(2·14.14) = 0.2828

Peak time

ωd = 14.14·√(1 − 0.2828²) = 13.56 rad/s
tp = π/13.56 = 0.2316 s

Answer

tp = 0.2316 s
9.Block diagrams

Answer: τ = 0.06897 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).

Closed-loop transfer function

T(s) = G/(1 + G·H) = 15/(s + 7 + 7.5) = 15/(s + 14.5)

Result

τ = 1/14.5 = 0.06897 s

Answer

τ = 0.06897 s
10.Block diagrams

Answer: σ = −1.5

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 + 3) + 28·1 = 0 → s² + 3s + 28 = 0

Roots

s = −1.5 ± j·√(28 − 2.25) = −1.5 ± j5.074
σ = −1.5

Answer

σ = −1.5

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