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 = 1f(1) = ?

Find f(t) for F(s) = 55/(s(s + 1)(s + 4)) by partial fractions, and evaluate f(1).

F(s) = 55/(s(s + 1)(s + 4))t = 1 s
f(1) =
2.Laplace transformsHard
t (s)y(t)t = 0.4y(0.4) = ?

A system obeys y″ + 3y′ + 2y = 8·u(t), with y(0) = 1 and y′(0) = 0. At t = 0 a step input u(t) = 5 is applied. Use the Laplace transform to find y(0.4).

y″ + 3y′ + 2y = 8·u(t)y(0) = 1y′(0) = 0u(t) = 5t = 0.4 s
y(0.4) =
3.Block diagramsMedium
R(s)+2K+7s + 8G(s)C(s)4H₁−−

The plant G(s) = 7/(s + 8) has an inner (minor) negative feedback loop with gain H₁ = 4. A controller gain K = 2 and an outer unity negative feedback loop close the system. Find the closed-loop pole p.

G(s) = 7/(s + 8)H₁ = 4K = 2
p =
4.Block diagramsHard
R(s)+20s + 1G₁+8s + 7G₂C(s)2H₁−0.5H₂−

Reduce the block diagram with G₁ = 20/(s + 1), G₂ = 8/(s + 7), an inner negative feedback H₁ = 2 around G₂ and an outer negative feedback H₂ = 0.5. Find the DC gain T(0) of C(s)/R(s).

G₁ = 20/(s + 1)G₂ = 8/(s + 7)H₁ = 2H₂ = 0.5
T(0) =
5.Step responseMedium
t (s)y(t)1?

A system has the transfer function G(s) = 97/(s² + 10s + 97) and receives a unit step input. Find the percent overshoot %OS of the step response.

G(s) = 97/(s² + 10s + 97)
%OS = %
6.Step responseHard
R(s)+200s(s + 6)G(s)C(s)−

A unity negative feedback loop has G(s) = 200/(s(s + 6)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 200/(s(s + 6))
ts = s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 4)(s + 8)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 4)(s + 8)). Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/(s(s + 4)(s + 8))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8j2

For L(s) = K/((s + 7)(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 + 7)(s² + 8s + 25))
θd = °
9.Steady-state error & PID controlMedium
r(t) = 1 + 5t+20(s + 2)s(s + 1)(s + 5)C(s)−

The input r(t) = 1 + 5t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 20(s + 2)/(s(s + 1)(s + 5)). Find the steady-state error ess.

G(s) = 20(s + 2)/(s(s + 1)(s + 5))r(t) = 1 + 5t
ess =
10.Steady-state error & PID controlHard
R(s)+K = 401s(s + 2)C(s)−

A proportional controller K = 40 drives the plant G(s) = 1/(s(s + 2)) in a unity-feedback loop. Find the percent overshoot %OS of the step response.

G(s) = 1/(s(s + 2))K = 40
%OS = %
11.Bode plots & marginsMedium
vin​C = 1 nFR = 47 kΩvout​

The RC high-pass filter shown has R = 47 kΩ, C = 1 nF. A sinusoidal input of 1.7 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 47 kΩC = 1 nFf = 1.7 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) = 20/(s(s + 1)(s + 10)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 20/(s(s + 1)(s + 10))
ω180 = rad/s

3Solutions

1.Laplace transforms

Answer: f(1) = 7.089

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 + 1) + C/(s + 4)
A = 55/(1·4) = 13.75
B = 55/((−1)·3) = −18.33
C = 55/((−4)·(−3)) = 4.583

Inverse transform

f(t) = 13.75 − 18.33·e−t + 4.583·e−4t

Evaluate

f(1) = 13.75 − 18.33·e−1 + 4.583·e−4·1 = 7.089

Answer

f(1) = 7.089
2.Laplace transforms

Answer: y(0.4) = 3.065

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 + 3s + 2)·Y(s) = 1s + 3 + 40/s
Y(s) = (s² + 3s + 40)/(s(s + 1)(s + 2))

Partial fractions

A = 40/(1·2) = 20
B = 38/((−1)·1) = −38
C = 38/((−2)·(−1)) = 19

Result

y(t) = 20 − 38·e−t + 19·e−2t
y(0.4) = 3.065

Answer

y(0.4) = 3.065

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

3.Block diagrams

Answer: p = −50

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) = 7/(s + 8 + 7·4) = 7/(s + 36)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 14/(s + 36 + 14) = 14/(s + 50)

Result

p = −50

Answer

p = −50
4.Block diagrams

Answer: T(0) = 1.553

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) = 8/(s + 23)

Outer loop

T(s) = G₁Gᵢ/(1 + G₁GᵢH₂) = 160/((s + 1)(s + 23) + 80)

DC gain

T(0) = 160/(1·23 + 80) = 160/103 = 1.553

Answer

T(0) = 1.553
5.Step response

Answer: %OS = 15.7 %

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 = 97 → ωₙ = 9.849 rad/s
2ζωₙ = 10 → ζ = 10/(2·9.849) = 0.5077

Result

%OS = 100·e−ζπ/√(1−ζ²) = 100·e−1.851 = 15.7 %

Answer

%OS = 15.7 %
6.Step response

Answer: ts = 1.333 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).

Closed-loop transfer function

T(s) = 200/(s² + 6s + 200): ωₙ = 14.14 rad/s, ζ = 0.2121

Settling time

ζωₙ = 6/2 = 3
ts ≈ 4/(ζωₙ) = 1.333 s

Answer

ts = 1.333 s

For this loop ζωₙ = a/2 whatever K is: raising K adds overshoot but does not shorten the settling time.

7.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 384

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 equation

s³ + 12s² + 32s + K = 0

Routh array

s³: 1, 32
s²: 12, K
s¹: (12·32 − K)/12
s⁰: K

Conditions

Every entry of the first column must be positive.

K > 0
12·32 − K > 0 → K < 384

Answer

Kmax = 384

At K = 384 the closed loop oscillates at ω = √32 = 5.657 rad/s.

8.Stability: Routh–Hurwitz & root locus

Answer: θd = 45 °

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 = −7

Angles from the other poles

∠(−4 + j3 + 7) = atan2(3, 3) = 45°
∠(−4 + j3 − (−4 − j3)) = ∠(j6) = 90°

Angle of departure

θd = 180° − 45° − 90° = 45°

Answer

θd = 45°
9.Steady-state error & PID control

Answer: ess = 0.625

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) = 8

Step part

Kp = ∞ → estep = 0

Ramp error

ess = 5/8 = 0.625

Answer

ess = 0.625
10.Steady-state error & PID control

Answer: %OS = 60.47 %

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 + 2) + 40·1 = s² + 2s + 40

ωₙ and ζ

ωₙ = √40 = 6.325 rad/s
ζ = 2/(2·6.325) = 0.1581

Overshoot

%OS = 100·e−0.1581π/√(1 − 0.1581²) = 60.47 %

Answer

%OS = 60.47 %
11.Bode plots & margins

Answer: φ = 63.34 °

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 = 1/(RC) = 1/(47000·1 × 10⁻⁹) = 21277 rad/s
fc = 3386 Hz → f/fc = 1700/3386 = 0.502

Phase

φ = 90° − atan(0.502) = 63.34°

Answer

φ = 63.34°
12.Bode plots & margins

Answer: ω180 = 3.162 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/1)·(ω/10) = 1 → ω180 = √(1·10) = 3.162 rad/s

Answer

ω180 = 3.162 rad/s

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