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 = 0.5f(0.5) = ?

Find f(t) for F(s) = 35/(s(s + 2)(s + 4)) by partial fractions, and evaluate f(0.5).

F(s) = 35/(s(s + 2)(s + 4))t = 0.5 s
f(0.5) =
2.Laplace transformsHard
t (s)y(t)t = 0.2y(0.2) = ?

Solve y″ + 10y′ + 24y = 0 with y(0) = 9 and y′(0) = −14 using the Laplace transform, and find y(0.2).

y″ + 10y′ + 24y = 0y(0) = 9y′(0) = −14t = 0.2 s
y(0.2) =
3.Block diagramsMedium
R(s)+4(s + 1)(s + 4)(s + 9)G(s)C(s)−

A unity negative feedback loop has G(s) = 4(s + 1)/((s + 4)(s + 9)). A step of size 10 is applied at the reference. Find the steady-state error ess.

G(s) = 4(s + 1)/((s + 4)(s + 9))R = 10
ess =
4.Block diagramsHard
R(s)+K(s + 5)(s + 6)(s + 7)G(s)C(s)−

The plant G(s) = K/((s + 5)(s + 6)(s + 7)) is in a unity negative feedback loop. Use the Routh–Hurwitz criterion to find the largest gain Kmax for which the closed loop is stable.

G(s) = K/((s + 5)(s + 6)(s + 7))unity negative feedback
Kmax =
5.Step responseMedium
t (s)y(t)1

A system has the transfer function G(s) = 67/(s² + 2s + 67) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.

G(s) = 67/(s² + 2s + 67)
σ =
6.Step responseHard
+−12 VR = 141 ΩL = 20 mHC = 1 µFvC​

A series RLC circuit with R = 141 Ω, L = 20 mH and C = 1 µF is connected to a 12 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 = 141 ΩL = 20 mHC = 1 µFV = 12 V
ωn = rad/s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 5(s − 2)(s + 5)C(s)−

The plant G(s) = (s + 5)/((s − 2)(s + 5)) has an unstable pole at s = 2. 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 + 5)/((s − 2)(s + 5))
Kmin =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10−12j2

The root locus of L(s) = K(s + 11)/(s(s + 7)) leaves the real axis between s = 0 and s = −7, circles round and joins the real axis again to the left of the zero. Find the break-away point sb.

L(s) = K(s + 11)/(s(s + 7))
sb =
9.Steady-state error & PID controlMedium
r(t) = 10 + 2t+20(s + 2)s(s + 1)(s + 2)C(s)−

The input r(t) = 10 + 2t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 20(s + 2)/(s(s + 1)(s + 2)). Find the steady-state error ess.

G(s) = 20(s + 2)/(s(s + 1)(s + 2))r(t) = 10 + 2t
ess =
10.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s1s + 1C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 1/(s + 1) in a unity-feedback loop. The closed loop must have ωₙ = 10 rad/s and ζ = 0.5. Find the proportional gain Kp.

G(s) = 1/(s + 1)ωₙ = 10 rad/sζ = 0.5
Kp =
11.Bode plots & marginsMedium
vin​L = 100 mHC = 0.47 µFR = 10 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 10 Ω, with L = 100 mH and C = 0.47 µF. Find the quality factor Q.

R = 10 ΩL = 100 mHC = 0.47 µF
Q =
12.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​

A unity-feedback loop has the open-loop transfer function L(s) = 11.2/(s(s + 4)). Find the phase margin PM in degrees.

L(s) = 11.2/(s(s + 4))
PM = °

3Solutions

1.Laplace transforms

Answer: f(0.5) = 1.748

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 + 2) + C/(s + 4)
A = 35/(2·4) = 4.375
B = 35/((−2)·2) = −8.75
C = 35/((−4)·(−2)) = 4.375

Inverse transform

f(t) = 4.375 − 8.75·e−2t + 4.375·e−4t

Evaluate

f(0.5) = 4.375 − 8.75·e−2·0.5 + 4.375·e−4·0.5 = 1.748

Answer

f(0.5) = 1.748
2.Laplace transforms

Answer: y(0.2) = 5.673

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

s²Y − 9s − (−14) + 10(sY − 9) + 24Y = 0
Y(s) = (9s + 76)/((s + 4)(s + 6))

Partial fractions

A = 40/2 = 20
B = 22/(−2) = −11

Result

y(t) = 20·e−4t − 11·e−6t
y(0.2) = 5.673

Answer

y(0.2) = 5.673

Check: y(0) = A + B = 9.

3.Block diagrams

Answer: ess = 9

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) = 4·1/(4·9) = 0.1111

Steady-state error

ess = A/(1 + Kₚ) = 10/(1 + 0.1111) = 9

Answer

ess = 9
4.Block diagrams

Answer: Kmax = 1716

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 + 18s² + 107s + 210 + K = 0

Routh array

s³ | 1 107
s² | 18 210 + K
s¹ | (18·107 − 210 − K)/18
s⁰ | 210 + K

Stability limit

All entries of the first column must be positive.

Kmax = 18·107 − 210 = 1716

Answer

Kmax = 1716
5.Step response

Answer: σ = −1

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 = 67 → ωₙ = 8.185 rad/s
2ζωₙ = 2 → ζ = 2/(2·8.185) = 0.1222

Result

σ = −ζωₙ = −2/2 = −1

Answer

σ = −1
6.Step response

Answer: ωn = 7071 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/√(0.02·1 × 10⁻⁶) = 7071 rad/s
ζ = (R/2)·√(C/L) = (141/2)·√(1 × 10⁻⁶/0.02) = 0.4985

Result

ωₙ = 7071 rad/s

Answer

ωn = 7071 rad/s
7.Stability: Routh–Hurwitz & root locus

Answer: Kmin = 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

(s − 2)(s + 5) + K(s + 5) = s² + (3 + K)s + (5K − 10)

Conditions

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

3 + K > 0 → K > −3
5K − 10 > 0 → K > 2
Kmin = 2

Answer

Kmin = 2
8.Stability: Routh–Hurwitz & root locus

Answer: sb = −4.367

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.

Gain along the real axis

K(s) = −(s² + 7s)/(s + 11)

Stationary points

Differentiate with the quotient rule and set the numerator to zero.

(2s + 7)(s + 11) − (s² + 7s) = s² + 22s + 77 = 0
s = −11 ± √44 = −4.367, −17.63

Break-away point

sb = −4.367

Answer

sb = −4.367

Between the two break points the locus is a circle centred on the zero, with radius √44 = 6.633.

9.Steady-state error & PID control

Answer: ess = 0.1

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

Step part

Kp = ∞ → estep = 0

Ramp error

ess = 2/20 = 0.1

Answer

ess = 0.1
10.Steady-state error & PID control

Answer: Kp = 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

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

s² + (1 + 1Kp)s + 1Ki

Target polynomial

s² + 2·0.5·10s + 10² = s² + 10s + 100

Match the coefficients

1 + 1Kp = 10 → Kp = 9

Answer

Kp = 9

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

11.Bode plots & margins

Answer: Q = 46.13

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.

Resonant frequency

ω₀ = 1/√(LC) = 1/√(0.1·4.7 × 10⁻⁷) = 4613 rad/s
f₀ = ω₀/(2π) = 734.1 Hz

Quality factor

BW = R/L = 10/0.1 = 100 rad/s = 15.92 Hz
Q = ω₀/BW = 4613/100 = 46.13

Answer

Q = 46.13
12.Bode plots & margins

Answer: PM = 59.03 °

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 + 16ω² − 125.4 = 0
ω² = (−16 + √(256 + 4·125.4))/2 = 5.764
ωc = 2.401 rad/s

Phase margin

∠L(jωc) = −90° − atan(2.401/4) = −121°
PM = 180° + ∠L = 59.03°

Answer

PM = 59.03°

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