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)1tp​ = ?

A system has the transfer function G(s) = 43/(s² + 10s + 43) and receives a unit step input. Find the peak time tp of the step response.

G(s) = 43/(s² + 10s + 43)
tp = s
2.Step responseHard
R(s)+Ks(s + 10)G(s)C(s)−

A unity negative feedback loop has G(s) = K/(s(s + 10)). Find the gain K that gives a step response with 5 % overshoot.

G(s) = K/(s(s + 10))%OS = 5 %
K =
3.Bode plots & marginsMedium
vin​C = 1 nFR = 4.7 kΩvout​

The RC high-pass filter shown has R = 4.7 kΩ, C = 1 nF. A sinusoidal input of 6.8 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 4.7 kΩC = 1 nFf = 6.8 kHz
φ = °
4.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−100−80−60−40−20020ω (rad/s)φ (°)ω (rad/s)−270−180−900ω180​

A unity-feedback loop has the open-loop transfer function L(s) = K/(s(s + 2)(s + 3)). Find the gain K that gives a gain margin of 15 dB.

L(s) = K/(s(s + 2)(s + 3))GM = 15 dB
K =
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 4)(s + 8)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 4)(s + 8)). Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/(s(s + 4)(s + 8))
Kmax =
6.Stability: Routh–Hurwitz & root locusHard
ReImζ = 0.4−1−2j1j2

The root locus of L(s) = K/(s(s + 2)) is shown. Find the gain K for which the closed-loop poles have the damping ratio ζ = 0.4.

L(s) = K/(s(s + 2))ζ = 0.4
K =
7.Steady-state error & PID controlMedium
R(s)+K = 1210s + 1C(s)−

A process G(s) = 2/(10s + 1) is controlled by a proportional gain K = 1 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.

G(s) = 2/(10s + 1)K = 1
ess = %
8.Steady-state error & PID controlHard
R(s)+K4s(s + 6)C(s)−

A proportional controller K drives the plant G(s) = 4/(s(s + 6)) in a unity-feedback loop. Find the gain K that gives the closed loop a damping ratio ζ = 0.5.

G(s) = 4/(s(s + 6))ζ = 0.5
K =
9.Block diagramsMedium
R(s)+10K+4s + 7G(s)C(s)3H₁−−

The plant G(s) = 4/(s + 7) has an inner (minor) negative feedback loop with gain H₁ = 3. A controller gain K = 10 and an outer unity negative feedback loop close the system. Find the closed-loop DC gain T(0).

G(s) = 4/(s + 7)H₁ = 3K = 10
T(0) =
10.Block diagramsHard
R(s)+78s(s + 10)G(s)C(s)2H−

The plant G(s) = 78/(s(s + 10)) is in a negative feedback loop with sensor gain H = 2. The closed-loop poles are complex. Find their real part σ.

G(s) = 78/(s(s + 10))H = 2
σ =

3Solutions

1.Step response

Answer: tp = 0.7405 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 = 43 → ωₙ = 6.557 rad/s
2ζωₙ = 10 → ζ = 10/(2·6.557) = 0.7625

Result

ωd = 6.557·√(1 − 0.7625²) = 4.243 rad/s → tp = π/ωd = 0.7405 s

Answer

tp = 0.7405 s
2.Step response

Answer: K = 52.49

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).

Closed-loop transfer function

T(s) = K/(s² + 10s + K)

Damping ratio from the overshoot

ζ = −ln(0.05)/√(π² + ln²(0.05)) = 0.6901

Gain

ωₙ = 10/(2ζ) = 7.245 rad/s
K = ωₙ² = 52.49

Answer

K = 52.49
3.Bode plots & margins

Answer: φ = 78.65 °

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/(4700·1 × 10⁻⁹) = 212766 rad/s
fc = 33863 Hz → f/fc = 6800/33863 = 0.2008

Phase

φ = 90° − atan(0.2008) = 78.65°

Answer

φ = 78.65°
4.Bode plots & margins

Answer: K = 5.335

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/2)·(ω/3) = 1 → ω180 = √(2·3) = 2.449 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 4)·√(ω² + 9)) = K/(2·3·(2 + 3)) = K/30

Required gain

30/K = 1015/20 = 5.623
K = 30/5.623 = 5.335

Answer

K = 5.335
5.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 384

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 + 12s² + 32s + K = 0

Routh array

s³: 1, 32
s²: 12, K
s¹: (12·32 − K)/12
s⁰: K

Conditions

Every entry of the first column must be positive.

K > 0
12·32 − K > 0 → K < 384

Answer

Kmax = 384

At K = 384 the closed loop oscillates at ω = √32 = 5.657 rad/s.

6.Stability: Routh–Hurwitz & root locus

Answer: K = 6.25

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.

Compare with the standard form

s² + 2s + K = s² + 2ζωₙs + ωₙ²
2·0.4·ωₙ = 2 → ωₙ = 2.5 rad/s

Gain

K = ωₙ² = 2.5² = 6.25

Answer

K = 6.25

The closed-loop poles sit where the ζ line crosses the vertical branch at s = −1.

7.Steady-state error & PID control

Answer: ess = 33.33 %

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 = 1·2 = 2

Step error

ess = 100/(1 + 2) = 33.33 %

Answer

ess = 33.33 %
8.Steady-state error & PID control

Answer: K = 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

s(s + 6) + 4K = s² + 6s + 4K

Natural frequency

2·0.5·ωₙ = 6 → ωₙ = 6 rad/s

Gain

4K = ωₙ² = 36 → K = 9

Answer

K = 9
9.Block diagrams

Answer: T(0) = 0.678

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 + 7 + 4·3) = 4/(s + 19)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 40/(s + 19 + 40) = 40/(s + 59)

Result

T(0) = 40/59 = 0.678

Answer

T(0) = 0.678
10.Block diagrams

Answer: σ = −5

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

s(s + 10) + 78·2 = 0 → s² + 10s + 156 = 0

Roots

s = −5 ± j·√(156 − 25) = −5 ± j11.45
σ = −5

Answer

σ = −5

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