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) = 237/(s² + 8s + 79) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.

G(s) = 237/(s² + 8s + 79)
σ =
2.Step responseHard
t (s)y(t)1tp​ = 0.8 s4.3 %

The step response of a second-order system without zeros shows 4.3 % overshoot, and the first peak occurs at t = 0.8 s. Find the natural frequency ωn.

%OS = 4.3 %tp = 0.8 s
ωn = rad/s
3.Bode plots & marginsMedium
vin​R = 220 ΩL = 10 mHvout​

The RL high-pass filter shown has R = 220 Ω, L = 10 mH. A sinusoidal input of 880 Hz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 220 ΩL = 10 mHf = 880 Hz
φ = °
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)+K1s(s + 2)(s + 10)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 2)(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 + 2)(s + 10))
ω = rad/s
6.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10−12j2

The root locus of L(s) = K(s + 10)/(s(s + 2)) leaves the real axis between s = 0 and s = −2, circles round and joins the real axis again to the left of the zero. Find the break-in point si.

L(s) = K(s + 10)/(s(s + 2))
si =
7.Steady-state error & PID controlMedium
r(t) = 10 + 3t+20s(s + 4)(s + 5)C(s)−

The input r(t) = 10 + 3t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 20/(s(s + 4)(s + 5)). Find the steady-state error ess.

G(s) = 20/(s(s + 4)(s + 5))r(t) = 10 + 3t
ess =
8.Steady-state error & PID controlHard
R(s)+K = 15s(s + 2)C(s)−

A proportional controller K = 1 drives the plant G(s) = 5/(s(s + 2)) in a unity-feedback loop. Find the percent overshoot %OS of the step response.

G(s) = 5/(s(s + 2))K = 1
%OS = %
9.Block diagramsMedium
R(s)+8K+4s + 5G(s)C(s)2H₁−−

The plant G(s) = 4/(s + 5) has an inner (minor) negative feedback loop with gain H₁ = 2. A controller gain K = 8 and an outer unity negative feedback loop close the system. Find the closed-loop DC gain T(0).

G(s) = 4/(s + 5)H₁ = 2K = 8
T(0) =
10.Block diagramsHard
R(s)+K(s + 5)(s + 6)(s + 7)G(s)C(s)−

The plant G(s) = K/((s + 5)(s + 6)(s + 7)) is in a unity negative feedback loop. Use the Routh–Hurwitz criterion to find the largest gain Kmax for which the closed loop is stable.

G(s) = K/((s + 5)(s + 6)(s + 7))unity negative feedback
Kmax =

3Solutions

1.Step response

Answer: σ = −4

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 = 79 → ωₙ = 8.888 rad/s
2ζωₙ = 8 → ζ = 8/(2·8.888) = 0.45

Result

σ = −ζωₙ = −8/2 = −4

Answer

σ = −4
2.Step response

Answer: ωn = 5.558 rad/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).

Damping ratio

ζ = −ln(0.043)/√(π² + ln²(0.043)) = 3.147/√(9.87 + 9.901) = 0.7077

Natural frequency

ωd = π/tp = π/0.8 = 3.927 rad/s
ωₙ = ωd/√(1 − ζ²) = 3.927/0.7065 = 5.558 rad/s

Answer

ωn = 5.558 rad/s
3.Bode plots & margins

Answer: φ = 75.89 °

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 = 220/0.01 = 22000 rad/s
fc = 3501 Hz → f/fc = 880/3501 = 0.2513

Phase

φ = 90° − atan(0.2513) = 75.89°

Answer

φ = 75.89°
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: ω = 4.472 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 + 12s² + 20s + K = 0
s³: 1, 20
s²: 12, K
s¹: (12·20 − K)/12
s⁰: K

Critical gain

K = 12·20 = 240

Auxiliary equation

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

12s² + 240 = 0 → s = ±j√20
ω = 4.472 rad/s

Answer

ω = 4.472 rad/s
6.Stability: Routh–Hurwitz & root locus

Answer: si = −18.94

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 + 2s)/(s + 10)

Stationary points

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

(2s + 2)(s + 10) − (s² + 2s) = s² + 20s + 20 = 0
s = −10 ± √80 = −1.056, −18.94

Break-in point

si = −18.94

Answer

si = −18.94

Between the two break points the locus is a circle centred on the zero, with radius √80 = 8.944.

7.Steady-state error & PID control

Answer: ess = 3

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) = 1

Step part

Kp = ∞ → estep = 0

Ramp error

ess = 3/1 = 3

Answer

ess = 3
8.Steady-state error & PID control

Answer: %OS = 20.79 %

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) + 1·5 = s² + 2s + 5

ωₙ and ζ

ωₙ = √5 = 2.236 rad/s
ζ = 2/(2·2.236) = 0.4472

Overshoot

%OS = 100·e−0.4472π/√(1 − 0.4472²) = 20.79 %

Answer

%OS = 20.79 %
9.Block diagrams

Answer: T(0) = 0.7111

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 = G/(1 + G·H₁) = 4/(s + 5 + 4·2) = 4/(s + 13)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 32/(s + 13 + 32) = 32/(s + 45)

Result

T(0) = 32/45 = 0.7111

Answer

T(0) = 0.7111
10.Block diagrams

Answer: Kmax = 1716

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 + 18s² + 107s + 210 + K = 0

Routh array

s³ | 1 107
s² | 18 210 + K
s¹ | (18·107 − 210 − K)/18
s⁰ | 210 + K

Stability limit

All entries of the first column must be positive.

Kmax = 18·107 − 210 = 1716

Answer

Kmax = 1716

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