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) = 17/(s² + 3s + 17) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.

G(s) = 17/(s² + 3s + 17)
σ =
2.Step responseHard
+−5 VR = 224 ΩL = 50 mHC = 1 µFvC​

A series RLC circuit with R = 224 Ω, L = 50 mH and C = 1 µF is connected to a 5 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the peak capacitor voltage vC,max.

R = 224 ΩL = 50 mHC = 1 µFV = 5 V
vC,max = V
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 phase shift φ of vout relative to vin, in degrees.

R = 1 kΩC = 1 nFf = 480 kHz
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) = K/(s(s + 4)(s + 7)). Find the gain K that gives a gain margin of 12 dB.

L(s) = K/(s(s + 4)(s + 7))GM = 12 dB
K =
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 1)(s + 2)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 1)(s + 2)). Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/(s(s + 1)(s + 2))
Kmax =
6.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10−12j2sb​

The root locus of L(s) = K/((s + 2)(s + 10)) 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 + 2)(s + 10))
K =
7.Steady-state error & PID controlMedium
r(t) = 0.5t+5(s + 4)s(s + 1)C(s)−

A ramp r(t) = 0.5t is applied to a unity-feedback loop with G(s) = 5(s + 4)/(s(s + 1)). Find the steady-state error ess.

G(s) = 5(s + 4)/(s(s + 1))r(t) = 0.5t
ess =
8.Steady-state error & PID controlHard
R(s)+K = 504s(s + 5)C(s)−

A proportional controller K = 50 drives the plant G(s) = 4/(s(s + 5)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 4/(s(s + 5))K = 50
ts = s
9.Block diagramsMedium
R(s)+9(s + 5)(s + 4)(s + 6)G(s)C(s)−

A unity negative feedback loop has G(s) = 9(s + 5)/((s + 4)(s + 6)). A step of size 10 is applied at the reference. Find the steady-state error ess.

G(s) = 9(s + 5)/((s + 4)(s + 6))R = 10
ess =
10.Block diagramsHard
R(s)+Ks(s + 2)(s + 10)G(s)C(s)−

The plant G(s) = K/(s(s + 2)(s + 10)) is in a unity negative feedback loop. At the largest stable gain the loop oscillates. Find the frequency of oscillation ω in rad/s.

G(s) = K/(s(s + 2)(s + 10))unity negative feedback
ω = rad/s

3Solutions

1.Step response

Answer: σ = −1.5

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 = 17 → ωₙ = 4.123 rad/s
2ζωₙ = 3 → ζ = 3/(2·4.123) = 0.3638

Result

σ = −ζωₙ = −3/2 = −1.5

Answer

σ = −1.5
2.Step response

Answer: vC,max = 5.812 V

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/√(0.05·1 × 10⁻⁶) = 4472 rad/s
ζ = (R/2)·√(C/L) = (224/2)·√(1 × 10⁻⁶/0.05) = 0.5009

Result

%OS = 16.23 %
vC,max = 5·(1 + 0.1623) = 5.812 V

Answer

vC,max = 5.812 V
3.Bode plots & margins

Answer: φ = 18.34 °

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

Phase

φ = 90° − atan(3.016) = 18.34°

Answer

φ = 18.34°
4.Bode plots & margins

Answer: K = 77.37

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/4)·(ω/7) = 1 → ω180 = √(4·7) = 5.292 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 16)·√(ω² + 49)) = K/(4·7·(4 + 7)) = K/308

Required gain

308/K = 1012/20 = 3.981
K = 308/3.981 = 77.37

Answer

K = 77.37
5.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 6

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 equation

s³ + 3s² + 2s + K = 0

Routh array

s³: 1, 2
s²: 3, K
s¹: (3·2 − K)/3
s⁰: K

Conditions

Every entry of the first column must be positive.

K > 0
3·2 − K > 0 → K < 6

Answer

Kmax = 6

At K = 6 the closed loop oscillates at ω = √2 = 1.414 rad/s.

6.Stability: Routh–Hurwitz & root locus

Answer: K = 16

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 + 2)(s + 10) gives 2s + 12 = 0.

sb = −6

Magnitude condition

K = |−6 + 2|·|−6 + 10| = 4·4 = 16

Answer

K = 16

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

7.Steady-state error & PID control

Answer: ess = 0.025

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) = 20

Ramp error

ess = 0.5/20 = 0.025

Answer

ess = 0.025
8.Steady-state error & PID control

Answer: ts = 1.6 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 + 5) + 50·4 = s² + 5s + 200

ωₙ and ζ

ωₙ = √200 = 14.14 rad/s
ζ = 5/(2·14.14) = 0.1768

Settling time

ζωₙ = 5/2 = 2.5
ts ≈ 4/2.5 = 1.6 s

Answer

ts = 1.6 s

The settling time does not depend on K here: raising K only makes the response oscillate faster.

9.Block diagrams

Answer: ess = 3.478

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) = 9·5/(4·6) = 1.875

Steady-state error

ess = A/(1 + Kₚ) = 10/(1 + 1.875) = 3.478

Answer

ess = 3.478
10.Block diagrams

Answer: ω = 4.472 rad/s

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 + 12s² + 20s + K = 0

Routh array

s³ | 1 20
s² | 12 K
s¹ | (12·20 − K)/12
s⁰ | K

Stability limit

All entries of the first column must be positive.

Kmax = 12·20 = 240

Frequency of oscillation

12·(jω)² + 240 = 0 → ω² = 240/12 = 20
ω = 4.472 rad/s

Answer

ω = 4.472 rad/s

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