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
ReIm−1−2−3−4−5−6−7−812

Find f(t) for F(s) = (4s + 28)/(s² + 2s + 5) by completing the square, and evaluate f(0.4).

F(s) = (4s + 28)/(s² + 2s + 5)t = 0.4 s
f(0.4) =
2.Laplace transformsHard
t (s)y(t)t = 0.2y(0.2) = ?

A system obeys y″ + 7y′ + 10y = 40·u(t), with y(0) = 2 and y′(0) = 0. At t = 0 a step input u(t) = 5 is applied. Use the Laplace transform to find y(0.2).

y″ + 7y′ + 10y = 40·u(t)y(0) = 2y′(0) = 0u(t) = 5t = 0.2 s
y(0.2) =
3.Block diagramsMedium
R(s)+8s + 4G(s)C(s)0.25H−

A plant G(s) = 8/(s + 4) is controlled in a negative feedback loop whose sensor has the gain H = 0.25. Find the time constant τ of the closed loop.

G(s) = 8/(s + 4)H = 0.25
τ = s
4.Block diagramsHard
R(s)+4s + 3G₁+10s + 5G₂C(s)1H₁−0.2H₂−

Reduce the block diagram with G₁ = 4/(s + 3), G₂ = 10/(s + 5), an inner negative feedback H₁ = 1 around G₂ and an outer negative feedback H₂ = 0.2. Find the DC gain T(0) of C(s)/R(s).

G₁ = 4/(s + 3)G₂ = 10/(s + 5)H₁ = 1H₂ = 0.2
T(0) =
5.Step responseMedium
t (s)y(t)0.5

A system has the transfer function G(s) = 14.5/(s² + 4s + 29) and receives a unit step input. Find the damping ratio ζ.

G(s) = 14.5/(s² + 4s + 29)
ζ =
6.Step responseHard
+−24 VR = 40 ΩL = 20 mHC = 2 µFvC​

A series RLC circuit with R = 40 Ω, L = 20 mH and C = 2 µF is connected to a 24 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the time tp of the first voltage peak, in ms.

R = 40 ΩL = 20 mHC = 2 µFV = 24 V
tp = ms
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+9601s(s + 4)(s + 6)C(s)−

The loop in the diagram has negative feedback and L(s) = 960/(s(s + 4)(s + 6)). How many closed-loop poles N lie in the right half-plane?

L(s) = 960/(s(s + 4)(s + 6))
N =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6j1−1.5

For the loop L(s) = K/(s(s + 3)(s + 5)), find the gain K that puts a closed-loop pole at s = −1.5.

L(s) = K/(s(s + 3)(s + 5))s = −1.5
K =
9.Steady-state error & PID controlMedium
R(s)+K(s + 2)(s + 4)C(s)−

A unity-feedback loop has G(s) = K/((s + 2)(s + 4)). Find the gain K that gives a steady-state error of 2 % for a step input.

G(s) = K/((s + 2)(s + 4))ess = 2 %
K =
10.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s1s + 4C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 1/(s + 4) in a unity-feedback loop. The closed loop must have ωₙ = 3 rad/s and ζ = 1. Find the proportional gain Kp.

G(s) = 1/(s + 4)ωₙ = 3 rad/sζ = 1
Kp =
11.Bode plots & marginsMedium
vin​L = 1 mHR = 47 Ωvout​

The RL low-pass filter shown has R = 47 Ω, L = 1 mH. A sinusoidal input of 75 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 47 ΩL = 1 mHf = 75 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) = 328/(s(s + 5)(s + 8)). Find the gain margin GM in dB.

L(s) = 328/(s(s + 5)(s + 8))
GM = dB

3Solutions

1.Laplace transforms

Answer: f(0.4) = 7.638

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).

Complete the square

s² + 2s + 5 = (s + 1)² + 4 = (s + 1)² + 2²

Match the numerator

F(s) = 4·(s + 1)/((s + 1)² + 2²) + (24/2)·2/((s + 1)² + 2²)

Inverse transform

f(t) = e−t·(4·cos 2t + 12·sin 2t)
f(0.4) = e−0.4·(4·cos(2·0.4) + 12·sin(2·0.4)) = 7.638

Answer

f(0.4) = 7.638
2.Laplace transforms

Answer: y(0.2) = 4.305

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 + 7s + 10)·Y(s) = 2s + 14 + 200/s
Y(s) = (2s² + 14s + 200)/(s(s + 2)(s + 5))

Partial fractions

A = 200/(2·5) = 20
B = 180/((−2)·3) = −30
C = 180/((−5)·(−3)) = 12

Result

y(t) = 20 − 30·e−2t + 12·e−5t
y(0.2) = 4.305

Answer

y(0.2) = 4.305

Check: y(0) = A + B + C = 2.

3.Block diagrams

Answer: τ = 0.1667 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).

Closed-loop transfer function

T(s) = G/(1 + G·H) = 8/(s + 4 + 2) = 8/(s + 6)

Result

τ = 1/6 = 0.1667 s

Answer

τ = 0.1667 s
4.Block diagrams

Answer: T(0) = 0.7547

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) = 10/(s + 15)

Outer loop

T(s) = G₁Gᵢ/(1 + G₁GᵢH₂) = 40/((s + 3)(s + 15) + 8)

DC gain

T(0) = 40/(3·15 + 8) = 40/53 = 0.7547

Answer

T(0) = 0.7547
5.Step response

Answer: ζ = 0.3714

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 = 29 → ωₙ = 5.385 rad/s
2ζωₙ = 4 → ζ = 4/(2·5.385) = 0.3714

Result

ζ = 0.3714

Answer

ζ = 0.3714
6.Step response

Answer: tp = 0.6413 ms

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/√(0.02·2 × 10⁻⁶) = 5000 rad/s
ζ = (R/2)·√(C/L) = (40/2)·√(2 × 10⁻⁶/0.02) = 0.2

Result

ωd = 4899 rad/s
tp = π/ωd = 0.6413 ms

Answer

tp = 0.6413 ms
7.Stability: Routh–Hurwitz & root locus

Answer: N = 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

s³ + 10s² + 24s + 960 = 0

Routh array

s³: 1, 24
s²: 10, 960
s¹: −72
s⁰: 960

Sign changes in the first column

1 → 10 → −72 → 960
N = 2

Answer

N = 2

The gain is above the limit K = 240, so a complex pair has crossed into the right half-plane.

8.Stability: Routh–Hurwitz & root locus

Answer: K = 7.875

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 + 3|·|s + 5|

Evaluate at the point

K = 1.5·1.5·3.5 = 7.875

Answer

K = 7.875

Check: s(s + 3)(s + 5) + K is zero at s = −1.5.

9.Steady-state error & PID control

Answer: K = 392

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) = 0.02 → Kp = 49

Gain

K/8 = 49 → K = 392

Answer

K = 392

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

10.Steady-state error & PID control

Answer: Kp = 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).

Characteristic polynomial

1 + C(s)G(s) = 0, multiplied by s(s + 4).

s² + (4 + 1Kp)s + 1Ki

Target polynomial

s² + 2·1·3s + 3² = s² + 6s + 9

Match the coefficients

4 + 1Kp = 6 → Kp = 2

Answer

Kp = 2

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

11.Bode plots & margins

Answer: φ = −84.3 °

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 = 47/0.001 = 47000 rad/s
fc = 7480 Hz → f/fc = 75000/7480 = 10.03

Phase

φ = − atan(10.03) = −84.3°

Answer

φ = −84.3°
12.Bode plots & margins

Answer: GM = 4.003 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/5)·(ω/8) = 1 → ω180 = √(5·8) = 6.325 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 25)·√(ω² + 64)) = K/(5·8·(5 + 8)) = K/520

Gain margin

GM = 20·log₁₀(520/328) = 4.003 dB

Answer

GM = 4.003 dB

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