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)3

A system has the transfer function G(s) = 153/(s² + 9s + 51) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.

G(s) = 153/(s² + 9s + 51)
σ =
2.Step responseHard
k = 400 N/mc = 6 N·s/mm = 2.5 kgF = 5 Nx(t)

A mass m = 2.5 kg hangs on a spring k = 400 N/m with a viscous damper c = 6 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 5 N is suddenly applied. Find the damping ratio ζ.

m = 2.5 kgk = 400 N/mc = 6 N·s/mF = 5 N
ζ =
3.Bode plots & marginsMedium
vin​C = 1 nFR = 1 kΩvout​

The RC high-pass filter shown has R = 1 kΩ, C = 1 nF. A sinusoidal input of 480 kHz is applied. Find the gain |H| = |vout/vin| in dB.

R = 1 kΩC = 1 nFf = 480 kHz
|H| = dB
4.Bode plots & marginsHard
|H| (dB)11101010010010001000−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 3400/(s(s + 20)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 3400/(s(s + 20))
ωc = rad/s
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 9s(s + 1)(s + 4)C(s)−

A unity-feedback loop has L(s) = K(s + 9)/(s(s + 1)(s + 4)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K(s + 9)/(s(s + 1)(s + 4))
Kmax =
6.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10−12−14−16j2j4

The root locus of L(s) = K(s + 13)/(s(s + 5)) leaves the real axis between s = 0 and s = −5, circles round and joins the real axis again to the left of the zero. Find the break-away point sb.

L(s) = K(s + 13)/(s(s + 5))
sb =
7.Steady-state error & PID controlMedium
R(s)+K = 510.5s + 1C(s)−

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

G(s) = 1/(0.5s + 1)K = 5
ess = %
8.Steady-state error & PID controlHard
R(s)+K = 401s(s + 5)C(s)−

A proportional controller K = 40 drives the plant G(s) = 1/(s(s + 5)) in a unity-feedback loop. Find the percent overshoot %OS of the step response.

G(s) = 1/(s(s + 5))K = 40
%OS = %
9.Block diagramsMedium
R(s)+Ks(s + 11)G(s)C(s)−

The plant G(s) = K/(s(s + 11)) 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 + 11))unity negative feedback
K =
10.Block diagramsHard
R(s)+22controller+D(s)+6s + 7plantC(s)−

A proportional controller K = 22 drives the plant G(s) = 6/(s + 7) in a unity negative feedback loop. A disturbance D enters at the plant input (it is added to the controller output). A reference step of size 5 and a disturbance step of size 1 are applied together. Find the steady-state output css.

K = 22G(s) = 6/(s + 7)D = 1R = 5
css =

3Solutions

1.Step response

Answer: σ = −4.5

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 = 51 → ωₙ = 7.141 rad/s
2ζωₙ = 9 → ζ = 9/(2·7.141) = 0.6301

Result

σ = −ζωₙ = −9/2 = −4.5

Answer

σ = −4.5
2.Step response

Answer: ζ = 0.09487

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 = √(k/m) = √(400/2.5) = 12.65 rad/s
ζ = c/(2√(km)) = 6/(2·√(400·2.5)) = 0.09487

Result

ζ = 0.09487

Answer

ζ = 0.09487
3.Bode plots & margins

Answer: |H| = −0.453 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 = 480000/159155 = 3.016

Gain

|H| = 3.016/√(1 + 3.016²) = 0.9492
20·log₁₀(0.9492) = −0.453 dB

Answer

|H| = −0.453 dB
4.Bode plots & margins

Answer: ωc = 56.62 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 + 400ω² − 1.156 × 10⁷ = 0
ω² = (−400 + √(160000 + 4·1.156 × 10⁷))/2 = 3206
ωc = 56.62 rad/s

Answer

ωc = 56.62 rad/s
5.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 5

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 + 9) = s³ + 5s² + (4 + K)s + 9K

Routh array

s³: 1, 4 + K
s²: 5, 9K
s¹: (5·(4 + K) − 9K)/5
s⁰: 9K

Condition

20 − 4K > 0
K < 20/4 = 5

Answer

Kmax = 5

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: sb = −2.802

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 + 5s)/(s + 13)

Stationary points

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

(2s + 5)(s + 13) − (s² + 5s) = s² + 26s + 65 = 0
s = −13 ± √104 = −2.802, −23.2

Break-away point

sb = −2.802

Answer

sb = −2.802

Between the two break points the locus is a circle centred on the zero, with radius √104 = 10.2.

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 = 5·1 = 5

Step error

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

Answer

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

Answer: %OS = 25.87 %

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) + 40·1 = s² + 5s + 40

ωₙ and ζ

ωₙ = √40 = 6.325 rad/s
ζ = 5/(2·6.325) = 0.3953

Overshoot

%OS = 100·e−0.3953π/√(1 − 0.3953²) = 25.87 %

Answer

%OS = 25.87 %
9.Block diagrams

Answer: K = 30.25

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 + 11)) = 0 → s² + 11s + K = 0

Repeated root

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

11² = 4K → K = 121/4 = 30.25

Answer

K = 30.25
Both closed-loop poles are then at s = −5.5.
10.Block diagrams

Answer: css = 4.791

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) = 6/(s + 7 + 132)
C/R = K·G/(1 + K·G) = 132/(s + 7 + 132)

Steady state

css = 5·132/139 + 1·6/139 = 4.791

Answer

css = 4.791

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

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