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)3tp​ = ?

A system has the transfer function G(s) = 132/(s² + 6s + 44) and receives a unit step input. Find the peak time tp of the step response.

G(s) = 132/(s² + 6s + 44)
tp = s
2.Step responseHard
t (s)y(t)120 %

A second-order system without zeros must have a step response with 20 % overshoot. Find the damping ratio ζ.

%OS = 20 %
ζ =
3.Bode plots & marginsMedium
|H| (dB)0.010.11101001000−80−60−40−20020ω (rad/s)ω = 0.5

Find the exact magnitude |G(jω)| of G(s) = 512/((s + 4)(s + 16)) at ω = 0.5 rad/s, in dB.

G(s) = 512/((s + 4)(s + 16))ω = 0.5 rad/s
|G(jω)| = dB
4.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 14.5/(s(s + 5)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 14.5/(s(s + 5))
ωc = rad/s
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 9s(s + 2)(s + 3)C(s)−

A unity-feedback loop has L(s) = K(s + 9)/(s(s + 2)(s + 3)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K(s + 9)/(s(s + 2)(s + 3))
Kmax =
6.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−7−81j1j2

For L(s) = K/((s + 7)(s² + 6s + 10)), find the angle of departure θd of the root locus from the complex pole in the upper half-plane, measured from the positive real axis.

L(s) = K/((s + 7)(s² + 6s + 10))
θd = °
7.Steady-state error & PID controlMedium
r(t) = 5t+Ks(s + 2)C(s)−

A unity-feedback loop with G(s) = K/(s(s + 2)) must follow the ramp r(t) = 5t with a steady-state error of 0.02. Find the gain K.

G(s) = K/(s(s + 2))r(t) = 5tess = 0.02
K =
8.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s5s + 2C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 5/(s + 2) in a unity-feedback loop. The closed loop must have ωₙ = 6 rad/s and ζ = 1. Find the integral gain Ki.

G(s) = 5/(s + 2)ωₙ = 6 rad/sζ = 1
Ki =
9.Block diagramsMedium
R(s)+6(s + 5)(s + 4)(s + 8)G(s)C(s)−

A unity negative feedback loop has G(s) = 6(s + 5)/((s + 4)(s + 8)). A step of size 2 is applied at the reference. Find the steady-state error ess.

G(s) = 6(s + 5)/((s + 4)(s + 8))R = 2
ess =
10.Block diagramsHard
R(s)+Kcontroller+D(s)+3s + 5plantC(s)−

A proportional controller K drives the plant G(s) = 3/(s + 5) in a unity negative feedback loop. A disturbance step of size 1 enters at the plant input while R = 0. Find the smallest gain K that keeps the steady-state output at or below 0.006.

G(s) = 3/(s + 5)D = 1css ≤ 0.006
K =

3Solutions

1.Step response

Answer: tp = 0.531 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 = 44 → ωₙ = 6.633 rad/s
2ζωₙ = 6 → ζ = 6/(2·6.633) = 0.4523

Result

ωd = 6.633·√(1 − 0.4523²) = 5.916 rad/s → tp = π/ωd = 0.531 s

Answer

tp = 0.531 s
2.Step response

Answer: ζ = 0.4559

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(0.2)/√(π² + ln²(0.2)) = 1.609/√(9.87 + 2.59) = 0.4559

Answer

ζ = 0.4559
3.Bode plots & margins

Answer: |G(jω)| = 17.99 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)| = 512/(√(0.5² + 4²)·√(0.5² + 16²)) = 512/(4.031·16.01) = 7.934

Result

20·log₁₀(7.934) = 17.99 dB

Answer

|G(jω)| = 17.99 dB
4.Bode plots & margins

Answer: ωc = 2.578 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 + 25ω² − 210.3 = 0
ω² = (−25 + √(625 + 4·210.3))/2 = 6.644
ωc = 2.578 rad/s

Answer

ωc = 2.578 rad/s
5.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 7.5

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 + 2)(s + 3) + K(s + 9) = s³ + 5s² + (6 + K)s + 9K

Routh array

s³: 1, 6 + K
s²: 5, 9K
s¹: (5·(6 + K) − 9K)/5
s⁰: 9K

Condition

30 − 4K > 0
K < 30/4 = 7.5

Answer

Kmax = 7.5

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: θd = 75.96 °

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.

Pole locations

s² + 6s + 10 = 0 → s = −3 ± j1
s = −7

Angles from the other poles

∠(−3 + j1 + 7) = atan2(1, 4) = 14.04°
∠(−3 + j1 − (−3 − j1)) = ∠(j2) = 90°

Angle of departure

θd = 180° − 14.04° − 90° = 75.96°

Answer

θd = 75.96°
7.Steady-state error & PID control

Answer: K = 500

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.

Required velocity constant

Kv = A/ess = 5/0.02 = 250

Gain

K = 2·Kv = 500

Answer

K = 500

A large gain makes this loop very lightly damped: ζ = 0.04472. That conflict is why compensators are used.

8.Steady-state error & PID control

Answer: Ki = 7.2

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

1 + C(s)G(s) = 0, multiplied by s(s + 2).

s² + (2 + 5Kp)s + 5Ki

Target polynomial

s² + 2·1·6s + 6² = s² + 12s + 36

Match the coefficients

5Ki = 36 → Ki = 7.2

Answer

Ki = 7.2

The integral term makes the loop type 1, so a step is followed without steady-state error.

9.Block diagrams

Answer: ess = 1.032

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).

Position error constant

Kₚ = G(0) = 6·5/(4·8) = 0.9375

Steady-state error

ess = A/(1 + Kₚ) = 2/(1 + 0.9375) = 1.032

Answer

ess = 1.032
10.Block diagrams

Answer: K = 165

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).

Disturbance response

css = D·G(0)/(1 + K·G(0)) = 1·3/(5 + 3K)

Solve for K

5 + 3K = 1·3/0.006 = 500
K = (500 − 5)/3 = 165

Answer

K = 165

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