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)3ts​ = ?

A system has the transfer function G(s) = 192/(s² + 6s + 64) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 192/(s² + 6s + 64)
ts = s
2.Step responseHard
+−24 VR = 248 ΩL = 200 mHC = 4.7 µFvC​

A series RLC circuit with R = 248 Ω, L = 200 mH and C = 4.7 µF is connected to a 24 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 = 248 ΩL = 200 mHC = 4.7 µFV = 24 V
%OS = %
3.Bode plots & marginsMedium
vin​C = 2.2 nFR = 2.2 kΩvout​

The RC high-pass filter shown has R = 2.2 kΩ, C = 2.2 nF. A sinusoidal input of 6.6 kHz is applied. Find the gain |H| = |vout/vin| in dB.

R = 2.2 kΩC = 2.2 nFf = 6.6 kHz
|H| = dB
4.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−270−180−900ω180​

A unity-feedback loop has the open-loop transfer function L(s) = K/(s(s + 7)(s + 8)). Find the gain K that gives a gain margin of 3 dB.

L(s) = K/(s(s + 7)(s + 8))GM = 3 dB
K =
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+9901s(s + 5)(s + 6)C(s)−

The loop in the diagram has negative feedback and L(s) = 990/(s(s + 5)(s + 6)). How many closed-loop poles N lie in the right half-plane?

L(s) = 990/(s(s + 5)(s + 6))
N =
6.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−7−81j1j2sb​

The root locus of L(s) = K/((s + 1)(s + 7)) is shown. Find the gain K at the break-away point sb, where the two closed-loop poles meet on the real axis.

L(s) = K/((s + 1)(s + 7))
K =
7.Steady-state error & PID controlMedium
R(s)+K = 10.50.5s + 1C(s)−

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

G(s) = 0.5/(0.5s + 1)K = 1
ess = %
8.Steady-state error & PID controlHard
R(s)+25 + Kd​s4s(s + 3)C(s)−

A PD controller C(s) = 25 + Kd·s drives the plant G(s) = 4/(s(s + 3)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.9.

G(s) = 4/(s(s + 3))Kp = 25ζ = 0.9
Kd = s
9.Block diagramsMedium
R(s)+Ks(s + 5)G(s)C(s)−

The plant G(s) = K/(s(s + 5)) 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 + 5))unity negative feedback
K =
10.Block diagramsHard
R(s)+10controller+D(s)+1s + 3plantC(s)−

A proportional controller K = 10 drives the plant G(s) = 1/(s + 3) in a unity negative feedback loop. A disturbance D enters at the plant input (it is added to the controller output). With R = 0, a disturbance step of size 0.5 is applied. Find the steady-state output css.

K = 10G(s) = 1/(s + 3)D = 0.5R = 0
css =

3Solutions

1.Step response

Answer: ts = 1.333 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 = 64 → ωₙ = 8 rad/s
2ζωₙ = 6 → ζ = 6/(2·8) = 0.375

Result

ζωₙ = 3 → ts ≈ 4/3 = 1.333 s

Answer

ts = 1.333 s
2.Step response

Answer: %OS = 9.413 %

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/√(0.2·4.7 × 10⁻⁶) = 1031 rad/s
ζ = (R/2)·√(C/L) = (248/2)·√(4.7 × 10⁻⁶/0.2) = 0.6011

Result

%OS = 100·e−ζπ/√(1−ζ²) = 9.413 %

Answer

%OS = 9.413 %
3.Bode plots & margins

Answer: |H| = −14.12 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/(2200·2.2 × 10⁻⁹) = 206612 rad/s
fc = 32883 Hz → f/fc = 6600/32883 = 0.2007

Gain

|H| = 0.2007/√(1 + 0.2007²) = 0.1968
20·log₁₀(0.1968) = −14.12 dB

Answer

|H| = −14.12 dB
4.Bode plots & margins

Answer: K = 594.7

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/7)·(ω/8) = 1 → ω180 = √(7·8) = 7.483 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 49)·√(ω² + 64)) = K/(7·8·(7 + 8)) = K/840

Required gain

840/K = 103/20 = 1.413
K = 840/1.413 = 594.7

Answer

K = 594.7
5.Stability: Routh–Hurwitz & root locus

Answer: N = 2

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

s³ + 11s² + 30s + 990 = 0

Routh array

s³: 1, 30
s²: 11, 990
s¹: −60
s⁰: 990

Sign changes in the first column

1 → 11 → −60 → 990
N = 2

Answer

N = 2

The gain is above the limit K = 330, so a complex pair has crossed into the right half-plane.

6.Stability: Routh–Hurwitz & root locus

Answer: K = 9

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.

Break-away point

dK/ds = 0 for K = −(s + 1)(s + 7) gives 2s + 8 = 0.

sb = −4

Magnitude condition

K = |−4 + 1|·|−4 + 7| = 3·3 = 9

Answer

K = 9

For this gain the closed loop is critically damped (ζ = 1).

7.Steady-state error & PID control

Answer: ess = 66.67 %

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 = 1·0.5 = 0.5

Step error

ess = 100/(1 + 0.5) = 66.67 %

Answer

ess = 66.67 %
8.Steady-state error & PID control

Answer: Kd = 3.75 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 + 3) + 4(25 + Kds) = s² + (3 + 4Kd)s + 100

Natural frequency

ωₙ = √100 = 10 rad/s

Derivative gain

3 + 4Kd = 2·0.9·10 = 18
Kd = 3.75 s

Answer

Kd = 3.75 s

The derivative term adds damping without changing ωₙ.

9.Block diagrams

Answer: K = 6.25

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 + 5)) = 0 → s² + 5s + K = 0

Repeated root

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

5² = 4K → K = 25/4 = 6.25

Answer

K = 6.25
Both closed-loop poles are then at s = −2.5.
10.Block diagrams

Answer: css = 0.03846

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).

Transfer functions

C/D = G/(1 + K·G) = 1/(s + 3 + 10)

Steady state

css = 0.5·1/13 = 0.03846

Answer

css = 0.03846

A larger K reduces the effect of the disturbance, but never to zero with a proportional controller.

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