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) = 58/(s² + 11s + 58) and receives a unit step input. Find the peak time tp of the step response.

G(s) = 58/(s² + 11s + 58)
tp = s
2.Step responseHard
R(s)+Ks(s + 6)G(s)C(s)−

A unity negative feedback loop has G(s) = K/(s(s + 6)). Find the gain K that gives a step response with 25 % overshoot.

G(s) = K/(s(s + 6))%OS = 25 %
K =
3.Bode plots & marginsMedium
vin​C = 10 nFR = 3.3 kΩvout​

The RC high-pass filter shown has R = 3.3 kΩ, C = 10 nF. A sinusoidal input of 960 Hz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 3.3 kΩC = 10 nFf = 960 Hz
φ = °
4.Bode plots & marginsHard
|H| (dB)11101010010010001000−80−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 700/(s(s + 20)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 700/(s(s + 20))
ωc = rad/s
5.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 8)(s + 10)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 8)(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 + 8)(s + 10))
ω = rad/s
6.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−71j1j2

A unity-feedback loop has L(s) = K(s + 6)/((s + 5)(s + 1)(s + 4)). Find the point σa where the asymptotes of the root locus cross the real axis.

L(s) = K(s + 6)/((s + 5)(s + 1)(s + 4))
σa =
7.Steady-state error & PID controlMedium
R(s)+K = 520.5s + 1C(s)−

A process G(s) = 2/(0.5s + 1) is controlled by a proportional gain K = 5 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.

G(s) = 2/(0.5s + 1)K = 5
ess = %
8.Steady-state error & PID controlHard
r(t) = 2t+5 + 2/s5(s + 2)(s + 5)C(s)−

A PI controller C(s) = 5 + 2/s drives the plant G(s) = 5/((s + 2)(s + 5)) in a unity-feedback loop. Find the steady-state error ess for the ramp input r(t) = 2t.

C(s) = 5 + 2/sG(s) = 5/((s + 2)(s + 5))r(t) = 2t
ess =
9.Block diagramsMedium
R(s)+Ks(s + 13)G(s)C(s)−

The plant G(s) = K/(s(s + 13)) is in a unity negative feedback loop. Find the gain K for which the closed loop is critically damped (two equal real poles).

G(s) = K/(s(s + 13))unity negative feedback
K =
10.Block diagramsHard
R(s)+17s + 1G₁+1s + 6G₂C(s)4H₁−0.1H₂−

Reduce the block diagram with G₁ = 17/(s + 1), G₂ = 1/(s + 6), an inner negative feedback H₁ = 4 around G₂ and an outer negative feedback H₂ = 0.1. Find the DC gain T(0) of C(s)/R(s).

G₁ = 17/(s + 1)G₂ = 1/(s + 6)H₁ = 4H₂ = 0.1
T(0) =

3Solutions

1.Step response

Answer: tp = 0.5964 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 = 58 → ωₙ = 7.616 rad/s
2ζωₙ = 11 → ζ = 11/(2·7.616) = 0.7222

Result

ωd = 7.616·√(1 − 0.7222²) = 5.268 rad/s → tp = π/ωd = 0.5964 s

Answer

tp = 0.5964 s
2.Step response

Answer: K = 55.22

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 + 6s + K)

Damping ratio from the overshoot

ζ = −ln(0.25)/√(π² + ln²(0.25)) = 0.4037

Gain

ωₙ = 6/(2ζ) = 7.431 rad/s
K = ωₙ² = 55.22

Answer

K = 55.22
3.Bode plots & margins

Answer: φ = 78.74 °

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/(3300·1 × 10⁻⁸) = 30303 rad/s
fc = 4823 Hz → f/fc = 960/4823 = 0.1991

Phase

φ = 90° − atan(0.1991) = 78.74°

Answer

φ = 78.74°
4.Bode plots & margins

Answer: ωc = 22.98 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 + 400ω² − 490000 = 0
ω² = (−400 + √(160000 + 4·490000))/2 = 528
ωc = 22.98 rad/s

Answer

ωc = 22.98 rad/s
5.Stability: Routh–Hurwitz & root locus

Answer: ω = 8.944 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 + 18s² + 80s + K = 0
s³: 1, 80
s²: 18, K
s¹: (18·80 − K)/18
s⁰: K

Critical gain

K = 18·80 = 1440

Auxiliary equation

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

18s² + 1440 = 0 → s = ±j√80
ω = 8.944 rad/s

Answer

ω = 8.944 rad/s
6.Stability: Routh–Hurwitz & root locus

Answer: σa = −2

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.

Poles and zeros

poles: −5, −1, −4
zeros: −6
n − m = 3 − 1 = 2

Centroid

σa = (−10 − (−6))/2 = −2

Answer

σa = −2

The 2 asymptotes leave the centroid at angles of 90°, 270°.

7.Steady-state error & PID control

Answer: ess = 9.091 %

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 = 5·2 = 10

Step error

ess = 100/(1 + 10) = 9.091 %

Answer

ess = 9.091 %
8.Steady-state error & PID control

Answer: ess = 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).

Velocity constant

s·C(s) = 5s + 2 → 2 as s → 0.

Kv = 2·5/10 = 1

Ramp error

ess = 2/1 = 2

Answer

ess = 2

The PI controller removes the step error completely; a ramp is followed with a constant lag.

9.Block diagrams

Answer: K = 42.25

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

1 + K/(s(s + 13)) = 0 → s² + 13s + K = 0

Repeated root

A quadratic s² + βs + γ has a double root when β² = 4γ.

13² = 4K → K = 169/4 = 42.25

Answer

K = 42.25
Both closed-loop poles are then at s = −6.5.
10.Block diagrams

Answer: T(0) = 1.453

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 = G₂/(1 + G₂H₁) = 1/(s + 10)

Outer loop

T(s) = G₁Gᵢ/(1 + G₁GᵢH₂) = 17/((s + 1)(s + 10) + 1.7)

DC gain

T(0) = 17/(1·10 + 1.7) = 17/11.7 = 1.453

Answer

T(0) = 1.453

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