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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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)1

A system has the transfer function G(s) = 88/(s² + 3s + 88) and receives a unit step input. Find the natural frequency ωn in rad/s.

G(s) = 88/(s² + 3s + 88)
ωn = rad/s
2.Step responseHard
+−12 VR = 54 ΩL = 200 mHC = 100 µFvC​

A series RLC circuit with R = 54 Ω, L = 200 mH and C = 100 µF is connected to a 12 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 = 54 ΩL = 200 mHC = 100 µFV = 12 V
%OS = %
3.Bode plots & marginsMedium
|H| (dB)0.11101001000−80−60−40−20020ω (rad/s)ω = 8

Find the exact magnitude |G(jω)| of G(s) = 405/((s + 3)(s + 15)) at ω = 8 rad/s, in dB.

G(s) = 405/((s + 3)(s + 15))ω = 8 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) = 80/(s(s + 10)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 80/(s(s + 10))
ωc = rad/s
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 1(s − 1)(s + 2)C(s)−

The plant G(s) = (s + 1)/((s − 1)(s + 2)) has an unstable pole at s = 1. 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 + 1)/((s − 1)(s + 2))
Kmin =
6.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−7−8−9−101j1j2−1

For the loop L(s) = K/(s(s + 4)(s + 8)), find the gain K that puts a closed-loop pole at s = −1.

L(s) = K/(s(s + 4)(s + 8))s = −1
K =
7.Steady-state error & PID controlMedium
r(t) = 2t+Ks(s + 2)C(s)−

A unity-feedback loop with G(s) = K/(s(s + 2)) must follow the ramp r(t) = 2t with a steady-state error of 0.05. Find the gain K.

G(s) = K/(s(s + 2))r(t) = 2tess = 0.05
K =
8.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s1s + 1C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 1/(s + 1) in a unity-feedback loop. The closed loop must have ωₙ = 3 rad/s and ζ = 0.8. Find the integral gain Ki.

G(s) = 1/(s + 1)ωₙ = 3 rad/sζ = 0.8
Ki =
9.Block diagramsMedium
R(s)+6s + 8G(s)C(s)2H−

A plant G(s) = 6/(s + 8) is controlled in a negative feedback loop whose sensor has the gain H = 2. A step of size 24 is applied at R. Find the steady-state output css.

G(s) = 6/(s + 8)H = 2R = 24
css =
10.Block diagramsHard
R(s)+10controller+D(s)+7s + 1plantC(s)−

A proportional controller K = 10 drives the plant G(s) = 7/(s + 1) 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 2 are applied together. Find the steady-state output css.

K = 10G(s) = 7/(s + 1)D = 2R = 5
css =

3Solutions

1.Step response

Answer: ωn = 9.381 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).

Compare with the standard form

The denominator must match s² + 2ζωₙs + ωₙ².

ωₙ² = 88 → ωₙ = 9.381 rad/s
2ζωₙ = 3 → ζ = 3/(2·9.381) = 0.1599

Result

ωₙ = 9.381 rad/s

Answer

ωn = 9.381 rad/s
2.Step response

Answer: %OS = 9.262 %

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/√(0.2·1 × 10⁻⁴) = 223.6 rad/s
ζ = (R/2)·√(C/L) = (54/2)·√(1 × 10⁻⁴/0.2) = 0.6037

Result

%OS = 100·e−ζπ/√(1−ζ²) = 9.262 %

Answer

%OS = 9.262 %
3.Bode plots & margins

Answer: |G(jω)| = 8.907 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(j8)| = 405/(√(8² + 3²)·√(8² + 15²)) = 405/(8.544·17) = 2.788

Result

20·log₁₀(2.788) = 8.907 dB

Answer

|G(jω)| = 8.907 dB
4.Bode plots & margins

Answer: ωc = 6.659 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 + 100ω² − 6400 = 0
ω² = (−100 + √(10000 + 4·6400))/2 = 44.34
ωc = 6.659 rad/s

Answer

ωc = 6.659 rad/s
5.Stability: Routh–Hurwitz & root locus

Answer: Kmin = 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

(s − 1)(s + 2) + K(s + 1) = s² + (1 + K)s + (1K − 2)

Conditions

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

1 + K > 0 → K > −1
1K − 2 > 0 → K > 2
Kmin = 2

Answer

Kmin = 2
6.Stability: Routh–Hurwitz & root locus

Answer: K = 21

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.

Magnitude condition

K = |s|·|s + 4|·|s + 8|

Evaluate at the point

K = 1·3·7 = 21

Answer

K = 21

Check: s(s + 4)(s + 8) + K is zero at s = −1.

7.Steady-state error & PID control

Answer: K = 80

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 = 2/0.05 = 40

Gain

K = 2·Kv = 80

Answer

K = 80

A large gain makes this loop very lightly damped: ζ = 0.1118. That conflict is why compensators are used.

8.Steady-state error & PID control

Answer: Ki = 9

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 + 1).

s² + (1 + 1Kp)s + 1Ki

Target polynomial

s² + 2·0.8·3s + 3² = s² + 4.8s + 9

Match the coefficients

1Ki = 9 → Ki = 9

Answer

Ki = 9

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

9.Block diagrams

Answer: css = 7.2

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) = 6/(s + 8 + 12) = 6/(s + 20)

Result

css = R·T(0) = 24·6/20 = 7.2

Answer

css = 7.2
10.Block diagrams

Answer: css = 5.127

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) = 7/(s + 1 + 70)
C/R = K·G/(1 + K·G) = 70/(s + 1 + 70)

Steady state

css = 5·70/71 + 2·7/71 = 5.127

Answer

css = 5.127

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

ExamProblems · examproblems.com/courses/control-engineering-2