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−1−2−3−4−5−6−7−8−91

F(s) = (4s + 3)/(s² + 10s + 16). Its inverse transform has the form f(t) = A·e−2t + B·e−8t. Find the coefficient A by partial fractions.

F(s) = (4s + 3)/(s² + 10s + 16)
A =
2.Laplace transformsHard
t (s)y(t)t = 0.25y(0.25) = ?

Solve y″ + 7y′ + 12y = 0 with y(0) = 4 and y′(0) = −10 using the Laplace transform, and find y(0.25).

y″ + 7y′ + 12y = 0y(0) = 4y′(0) = −10t = 0.25 s
y(0.25) =
3.Block diagramsMedium
R(s)+13K+4s + 4G(s)C(s)3H₁−−

The plant G(s) = 4/(s + 4) has an inner (minor) negative feedback loop with gain H₁ = 3. A controller gain K = 13 and an outer unity negative feedback loop close the system. Find the closed-loop pole p.

G(s) = 4/(s + 4)H₁ = 3K = 13
p =
4.Block diagramsHard
R(s)+23controller+D(s)+10s + 6plantC(s)−

A proportional controller K = 23 drives the plant G(s) = 10/(s + 6) in a unity negative feedback loop. A disturbance D enters at the plant input (it is added to the controller output). With R = 0, a disturbance step of size 0.5 is applied. Find the steady-state output css.

K = 23G(s) = 10/(s + 6)D = 0.5R = 0
css =
5.Step responseMedium
t (s)y(t)3tp​ = ?

A system has the transfer function G(s) = 288/(s² + 5s + 96) and receives a unit step input. Find the peak time tp of the step response.

G(s) = 288/(s² + 5s + 96)
tp = s
6.Step responseHard
k = 500 N/mc = 28 N·s/mm = 2.5 kgF = 5 Nx(t)

A mass m = 2.5 kg hangs on a spring k = 500 N/m with a viscous damper c = 28 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 5 N is suddenly applied. Find the time tp at which the displacement is largest.

m = 2.5 kgk = 500 N/mc = 28 N·s/mF = 5 N
tp = s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 2(s − 4)(s + 1)C(s)−

The plant G(s) = (s + 2)/((s − 4)(s + 1)) has an unstable pole at s = 4. 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 + 2)/((s − 4)(s + 1))
Kmin =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4j2

For L(s) = K/((s + 3)(s² + 8s + 25)), find the angle of departure θd of the root locus from the complex pole in the upper half-plane, measured from the positive real axis.

L(s) = K/((s + 3)(s² + 8s + 25))
θd = °
9.Steady-state error & PID controlMedium
r(t) = 5t+Ks(s + 10)C(s)−

A unity-feedback loop with G(s) = K/(s(s + 10)) must follow the ramp r(t) = 5t with a steady-state error of 0.02. Find the gain K.

G(s) = K/(s(s + 10))r(t) = 5tess = 0.02
K =
10.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s10s + 4C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 10/(s + 4) in a unity-feedback loop. The closed loop must have ωₙ = 10 rad/s and ζ = 0.5. Find the integral gain Ki.

G(s) = 10/(s + 4)ωₙ = 10 rad/sζ = 0.5
Ki =
11.Bode plots & marginsMedium
vin​R = 47 kΩC = 47 nFvout​

The RC low-pass filter shown has R = 47 kΩ, C = 47 nF. A sinusoidal input of 720 Hz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 47 kΩC = 47 nFf = 720 Hz
φ = °
12.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−270−180−900ω180​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 1020/(s(s + 8)(s + 10)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 1020/(s(s + 8)(s + 10))
ω180 = rad/s

3Solutions

1.Laplace transforms

Answer: A = −0.8333

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).

Factor the denominator

s² + 10s + 16 = (s + 2)(s + 8)
F(s) = A/(s + 2) + B/(s + 8)

Cover-up rule

Multiply by (s + 2) and set s = −2: the B-term vanishes.

A = (4s + 3)/(s + 8) |s = −2 = −5/6 = −0.8333

Answer

A = −0.8333
(B = 4.833, so f(t) = −0.8333·e−2t + 4.833·e−8t)
2.Laplace transforms

Answer: y(0.25) = 2.098

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 − 4s − (−10) + 7(sY − 4) + 12Y = 0
Y(s) = (4s + 18)/((s + 3)(s + 4))

Partial fractions

A = 6/1 = 6
B = 2/(−1) = −2

Result

y(t) = 6·e−3t − 2·e−4t
y(0.25) = 2.098

Answer

y(0.25) = 2.098

Check: y(0) = A + B = 4.

3.Block diagrams

Answer: p = −68

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 + 4 + 4·3) = 4/(s + 16)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 52/(s + 16 + 52) = 52/(s + 68)

Result

p = −68

Answer

p = −68
4.Block diagrams

Answer: css = 0.02119

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) = 10/(s + 6 + 230)

Steady state

css = 0.5·10/236 = 0.02119

Answer

css = 0.02119

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

5.Step response

Answer: tp = 0.3316 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 = 96 → ωₙ = 9.798 rad/s
2ζωₙ = 5 → ζ = 5/(2·9.798) = 0.2552

Result

ωd = 9.798·√(1 − 0.2552²) = 9.474 rad/s → tp = π/ωd = 0.3316 s

Answer

tp = 0.3316 s
6.Step response

Answer: tp = 0.2419 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 = √(k/m) = √(500/2.5) = 14.14 rad/s
ζ = c/(2√(km)) = 28/(2·√(500·2.5)) = 0.396

Result

ωd = 12.99 rad/s
tp = π/ωd = 0.2419 s

Answer

tp = 0.2419 s
7.Stability: Routh–Hurwitz & root locus

Answer: Kmin = 3

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 − 4)(s + 1) + K(s + 2) = s² + (-3 + K)s + (2K − 4)

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
2K − 4 > 0 → K > 2
Kmin = 3

Answer

Kmin = 3
8.Stability: Routh–Hurwitz & root locus

Answer: θd = −18.43 °

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.

Pole locations

s² + 8s + 25 = 0 → s = −4 ± j3
s = −3

Angles from the other poles

∠(−4 + j3 + 3) = atan2(3, -1) = 108.4°
∠(−4 + j3 − (−4 − j3)) = ∠(j6) = 90°

Angle of departure

θd = 180° − 108.4° − 90° = −18.43°

Answer

θd = −18.43°
9.Steady-state error & PID control

Answer: K = 2500

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 velocity constant

Kv = A/ess = 5/0.02 = 250

Gain

K = 10·Kv = 2500

Answer

K = 2500

A large gain makes this loop very lightly damped: ζ = 0.1. That conflict is why compensators are used.

10.Steady-state error & PID control

Answer: Ki = 10

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 + 10Kp)s + 10Ki

Target polynomial

s² + 2·0.5·10s + 10² = s² + 10s + 100

Match the coefficients

10Ki = 100 → Ki = 10

Answer

Ki = 10

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

11.Bode plots & margins

Answer: φ = −84.29 °

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/(47000·4.7 × 10⁻⁸) = 452.7 rad/s
fc = 72.05 Hz → f/fc = 720/72.05 = 9.993

Phase

φ = − atan(9.993) = −84.29°

Answer

φ = −84.29°
12.Bode plots & margins

Answer: ω180 = 8.944 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.

Phase crossover

(ω/8)·(ω/10) = 1 → ω180 = √(8·10) = 8.944 rad/s

Answer

ω180 = 8.944 rad/s

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