Control Engineering 1: cram pack

Formula sheets, the key method steps for every topic and 12 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.Laplace transformsMedium
t (s)f(t)t = 0.4f(0.4) = ?

Find f(t) for F(s) = 60/(s(s + 3)(s + 4)) by partial fractions, and evaluate f(0.4).

F(s) = 60/(s(s + 3)(s + 4))t = 0.4 s
f(0.4) =
2.Laplace transformsHard
t (s)y(t)t = 0.25y(0.25) = ?

A system obeys y″ + 8y′ + 15y = 15·u(t), with y(0) = 1 and y′(0) = 0. At t = 0 a step input u(t) = 2 is applied. Use the Laplace transform to find y(0.25).

y″ + 8y′ + 15y = 15·u(t)y(0) = 1y′(0) = 0u(t) = 2t = 0.25 s
y(0.25) =
3.Block diagramsMedium
R(s)+38s(s + 1)G(s)C(s)−

A unity negative feedback loop has G(s) = 38/(s(s + 1)). The reference is a ramp r(t) = 5·t. Find the steady-state error ess.

G(s) = 38/(s(s + 1))r(t) = 5·t
ess =
4.Block diagramsHard
R(s)+15controller+D(s)+9s + 3plantC(s)−

A proportional controller K = 15 drives the plant G(s) = 9/(s + 3) 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 1 and a disturbance step of size 2 are applied together. Find the steady-state output css.

K = 15G(s) = 9/(s + 3)D = 2R = 1
css =
5.Step responseMedium
t (s)y(t)0.5

A system has the transfer function G(s) = 25.5/(s² + 11s + 51) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.

G(s) = 25.5/(s² + 11s + 51)
σ =
6.Step responseHard
+−5 VR = 600 ΩL = 500 mHC = 2 µFvC​

A series RLC circuit with R = 600 Ω, L = 500 mH and C = 2 µF is connected to a 5 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the undamped natural frequency ωn in rad/s.

R = 600 ΩL = 500 mHC = 2 µFV = 5 V
ωn = rad/s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 10(s − 1)(s + 5)C(s)−

The plant G(s) = (s + 10)/((s − 1)(s + 5)) 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 + 10)/((s − 1)(s + 5))
Kmin =
8.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 =
9.Steady-state error & PID controlMedium
R(s)+K(s + 5)(s + 10)C(s)−

A unity-feedback loop has G(s) = K/((s + 5)(s + 10)). Find the gain K that gives a steady-state error of 4 % for a step input.

G(s) = K/((s + 5)(s + 10))ess = 4 %
K =
10.Steady-state error & PID controlHard
R(s)+10 + Kd​s2s(s + 1)C(s)−

A PD controller C(s) = 10 + Kd·s drives the plant G(s) = 2/(s(s + 1)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.9.

G(s) = 2/(s(s + 1))Kp = 10ζ = 0.9
Kd = s
11.Bode plots & marginsMedium
vin​L = 1 mHR = 22 Ωvout​

The RL low-pass filter shown has R = 22 Ω, L = 1 mH. A sinusoidal input of 11 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 22 ΩL = 1 mHf = 11 kHz
φ = °
12.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) = 10/(s(s + 1)(s + 4)). Find the gain margin GM in dB.

L(s) = 10/(s(s + 1)(s + 4))
GM = dB

3Solutions

1.Laplace transforms

Answer: f(0.4) = 2.005

Laplace transform: F(s) = ∫₀∞ f(t)·e−st dt turns derivatives into multiplications by s: ℒ{f′} = sF(s) − f(0), ℒ{f″} = s²F(s) − s·f(0) − f′(0). Standard pairs: 1 ↔ 1/s, e−at ↔ 1/(s + a), t·e−at ↔ 1/(s + a)², sin ωt ↔ ω/(s² + ω²), cos ωt ↔ s/(s² + ω²), and a shift s → s + σ multiplies the time function by e−σt. To invert a rational F(s), split it into partial fractions: the residue of a simple pole p is A = [(s − p)·F(s)] at s = p. Final value theorem: f(∞) = lims→0 s·F(s), valid only when all poles of s·F(s) lie in the left half-plane. Initial value theorem: f(0⁺) = lims→∞ s·F(s).

Partial fractions

F(s) = A/s + B/(s + 3) + C/(s + 4)
A = 60/(3·4) = 5
B = 60/((−3)·1) = −20
C = 60/((−4)·(−1)) = 15

Inverse transform

f(t) = 5 − 20·e−3t + 15·e−4t

Evaluate

f(0.4) = 5 − 20·e−3·0.4 + 15·e−4·0.4 = 2.005

Answer

f(0.4) = 2.005
2.Laplace transforms

Answer: y(0.25) = 1.249

Laplace transform: F(s) = ∫₀∞ f(t)·e−st dt turns derivatives into multiplications by s: ℒ{f′} = sF(s) − f(0), ℒ{f″} = s²F(s) − s·f(0) − f′(0). Standard pairs: 1 ↔ 1/s, e−at ↔ 1/(s + a), t·e−at ↔ 1/(s + a)², sin ωt ↔ ω/(s² + ω²), cos ωt ↔ s/(s² + ω²), and a shift s → s + σ multiplies the time function by e−σt. To invert a rational F(s), split it into partial fractions: the residue of a simple pole p is A = [(s − p)·F(s)] at s = p. Final value theorem: f(∞) = lims→0 s·F(s), valid only when all poles of s·F(s) lie in the left half-plane. Initial value theorem: f(0⁺) = lims→∞ s·F(s).

Transform the equation

(s² + 8s + 15)·Y(s) = 1s + 8 + 30/s
Y(s) = (s² + 8s + 30)/(s(s + 3)(s + 5))

Partial fractions

A = 30/(3·5) = 2
B = 15/((−3)·2) = −2.5
C = 15/((−5)·(−2)) = 1.5

Result

y(t) = 2 − 2.5·e−3t + 1.5·e−5t
y(0.25) = 1.249

Answer

y(0.25) = 1.249

Check: y(0) = A + B + C = 1.

3.Block diagrams

Answer: ess = 0.1316

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).

Velocity error constant

Kᵥ = lims→0 s·G(s) = 38/1 = 38

Steady-state error

ess = A/Kᵥ = 5/38 = 0.1316

Answer

ess = 0.1316
4.Block diagrams

Answer: css = 1.109

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) = 9/(s + 3 + 135)
C/R = K·G/(1 + K·G) = 135/(s + 3 + 135)

Steady state

css = 1·135/138 + 2·9/138 = 1.109

Answer

css = 1.109

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

5.Step response

Answer: σ = −5.5

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 = 51 → ωₙ = 7.141 rad/s
2ζωₙ = 11 → ζ = 11/(2·7.141) = 0.7702

Result

σ = −ζωₙ = −11/2 = −5.5

Answer

σ = −5.5
6.Step response

Answer: ωn = 1000 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).

Natural frequency and damping ratio

ωₙ = 1/√(LC) = 1/√(0.5·2 × 10⁻⁶) = 1000 rad/s
ζ = (R/2)·√(C/L) = (600/2)·√(2 × 10⁻⁶/0.5) = 0.6

Result

ωₙ = 1000 rad/s

Answer

ωn = 1000 rad/s
7.Stability: Routh–Hurwitz & root locus

Answer: Kmin = 0.5

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 − 1)(s + 5) + K(s + 10) = s² + (4 + K)s + (10K − 5)

Conditions

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

4 + K > 0 → K > −4
10K − 5 > 0 → K > 0.5
Kmin = 0.5

Answer

Kmin = 0.5
8.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).

9.Steady-state error & PID control

Answer: K = 1200

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 error constant

1/(1 + Kp) = 0.04 → Kp = 24

Gain

K/50 = 24 → K = 1200

Answer

K = 1200

Check the closed-loop stability as well: for this second-order loop any K > 0 is stable.

10.Steady-state error & PID control

Answer: Kd = 3.525 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 + 1) + 2(10 + Kds) = s² + (1 + 2Kd)s + 20

Natural frequency

ωₙ = √20 = 4.472 rad/s

Derivative gain

1 + 2Kd = 2·0.9·4.472 = 8.05
Kd = 3.525 s

Answer

Kd = 3.525 s

The derivative term adds damping without changing ωₙ.

11.Bode plots & margins

Answer: φ = −72.34 °

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 = R/L = 22/0.001 = 22000 rad/s
fc = 3501 Hz → f/fc = 11000/3501 = 3.142

Phase

φ = − atan(3.142) = −72.34°

Answer

φ = −72.34°
12.Bode plots & margins

Answer: GM = 6.021 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.

Phase crossover

(ω/1)·(ω/4) = 1 → ω180 = √(1·4) = 2 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 1)·√(ω² + 16)) = K/(1·4·(1 + 4)) = K/20

Gain margin

GM = 20·log₁₀(20/10) = 6.021 dB

Answer

GM = 6.021 dB

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