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−22

Find f(t) for F(s) = 33/(s² + 6s + 18) by completing the square, and evaluate f(0.25).

F(s) = 33/(s² + 6s + 18)t = 0.25 s
f(0.25) =
2.Laplace transformsHard
t (s)y(t)t = 0.2y(0.2) = ?

A system obeys y″ + 7y′ + 12y = 60·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 5 for t ≥ 0. Use the Laplace transform to find the output y(0.2).

y″ + 7y′ + 12y = 60·u(t)u(t) = 5y(0) = 0, y′(0) = 0t = 0.2 s
y(0.2) =
3.Block diagramsMedium
R(s)+14K+7s + 8G(s)C(s)5H₁−−

The plant G(s) = 7/(s + 8) has an inner (minor) negative feedback loop with gain H₁ = 5. A controller gain K = 14 and an outer unity negative feedback loop close the system. Find the closed-loop DC gain T(0).

G(s) = 7/(s + 8)H₁ = 5K = 14
T(0) =
4.Block diagramsHard
R(s)+Ks(s + 6)(s + 10)G(s)C(s)−

The plant G(s) = K/(s(s + 6)(s + 10)) 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(s + 6)(s + 10))unity negative feedback
Kmax =
5.Step responseMedium
t (s)y(t)3tp​ = ?

A system has the transfer function G(s) = 75/(s² + 2s + 25) and receives a unit step input. Find the peak time tp of the step response.

G(s) = 75/(s² + 2s + 25)
tp = s
6.Step responseHard
R(s)+Ks(s + 2)G(s)C(s)−

A unity negative feedback loop has G(s) = K/(s(s + 2)). Find the gain K that gives a step response with 20 % overshoot.

G(s) = K/(s(s + 2))%OS = 20 %
K =
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 8s(s + 2)(s + 3)C(s)−

A unity-feedback loop has L(s) = K(s + 8)/(s(s + 2)(s + 3)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K(s + 8)/(s(s + 2)(s + 3))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−0.5−1−1.5−2−2.5−3−3.5−40.5j0.5j1

The root locus of L(s) = K/(s(s + 1)(s + 3)) leaves the real axis between s = 0 and s = −1. Find the break-away point sb.

L(s) = K/(s(s + 1)(s + 3))
sb =
9.Steady-state error & PID controlMedium
R(s)+K = 9410s + 1C(s)−

A process G(s) = 4/(10s + 1) is controlled by a proportional gain K = 9 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.

G(s) = 4/(10s + 1)K = 9
ess = %
10.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s10s + 2C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 10/(s + 2) in a unity-feedback loop. The closed loop must have ωₙ = 3 rad/s and ζ = 0.8. Find the integral gain Ki.

G(s) = 10/(s + 2)ωₙ = 3 rad/sζ = 0.8
Ki =
11.Bode plots & marginsMedium
vin​L = 2 mHC = 0.22 µFR = 50 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 50 Ω, with L = 2 mH and C = 0.22 µF. Find the centre (resonant) frequency f0 in Hz.

R = 50 ΩL = 2 mHC = 0.22 µF
f0 = Hz
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) = 376/(s(s + 7)(s + 10)). Find the gain margin GM in dB.

L(s) = 376/(s(s + 7)(s + 10))
GM = dB

3Solutions

1.Laplace transforms

Answer: f(0.25) = 3.542

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 + 6s + 18 = (s + 3)² + 9 = (s + 3)² + 3²

Match the numerator

F(s) = (33/3)·3/((s + 3)² + 3²)

Inverse transform

f(t) = e−3t·(11·sin 3t)
f(0.25) = e−3·0.25·(11·sin(3·0.25)) = 3.542

Answer

f(0.25) = 3.542
2.Laplace transforms

Answer: y(0.2) = 3.819

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

Y(s) = 300/(s(s + 3)(s + 4))

Partial fractions

A = 300/(3·4) = 25
B = 300/((−3)·1) = −100
C = 300/((−4)·(−1)) = 75

Output

y(t) = 25 − 100·e−3t + 75·e−4t
y(0.2) = 3.819

Answer

y(0.2) = 3.819
3.Block diagrams

Answer: T(0) = 0.695

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 + 7·5) = 7/(s + 43)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 98/(s + 43 + 98) = 98/(s + 141)

Result

T(0) = 98/141 = 0.695

Answer

T(0) = 0.695
4.Block diagrams

Answer: Kmax = 960

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 + 16s² + 60s + K = 0

Routh array

s³ | 1 60
s² | 16 K
s¹ | (16·60 − K)/16
s⁰ | K

Stability limit

All entries of the first column must be positive.

Kmax = 16·60 = 960

Answer

Kmax = 960
5.Step response

Answer: tp = 0.6413 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 = 25 → ωₙ = 5 rad/s
2ζωₙ = 2 → ζ = 2/(2·5) = 0.2

Result

ωd = 5·√(1 − 0.2²) = 4.899 rad/s → tp = π/ωd = 0.6413 s

Answer

tp = 0.6413 s
6.Step response

Answer: K = 4.81

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) = K/(s² + 2s + K)

Damping ratio from the overshoot

ζ = −ln(0.2)/√(π² + ln²(0.2)) = 0.4559

Gain

ωₙ = 2/(2ζ) = 2.193 rad/s
K = ωₙ² = 4.81

Answer

K = 4.81
7.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 10

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(s + 2)(s + 3) + K(s + 8) = s³ + 5s² + (6 + K)s + 8K

Routh array

s³: 1, 6 + K
s²: 5, 8K
s¹: (5·(6 + K) − 8K)/5
s⁰: 8K

Condition

30 − 3K > 0
K < 30/3 = 10

Answer

Kmax = 10

With the zero to the right of s = −5 the loop would stay stable for every K > 0.

8.Stability: Routh–Hurwitz & root locus

Answer: sb = −0.4514

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) = −s(s + 1)(s + 3) = −(s³ + 4s² + 3s)

Stationary points

dK/ds = −(3s² + 8s + 3) = 0
s = (−8 ± √(64 − 36))/6 = −0.4514, −2.215

Pick the point on the locus

The locus lies on the real axis between 0 and −1; the other root is not on the locus.

sb = −0.4514

Answer

sb = −0.4514
9.Steady-state error & PID control

Answer: ess = 2.703 %

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.

Position constant

Kp = 9·4 = 36

Step error

ess = 100/(1 + 36) = 2.703 %

Answer

ess = 2.703 %
10.Steady-state error & PID control

Answer: Ki = 0.9

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 + 2).

s² + (2 + 10Kp)s + 10Ki

Target polynomial

s² + 2·0.8·3s + 3² = s² + 4.8s + 9

Match the coefficients

10Ki = 9 → Ki = 0.9

Answer

Ki = 0.9

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

11.Bode plots & margins

Answer: f0 = 7587 Hz

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/√(0.002·2.2 × 10⁻⁷) = 47673 rad/s
f₀ = ω₀/(2π) = 7587 Hz

Answer

f0 = 7587 Hz
12.Bode plots & margins

Answer: GM = 10.01 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/7)·(ω/10) = 1 → ω180 = √(7·10) = 8.367 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 49)·√(ω² + 100)) = K/(7·10·(7 + 10)) = K/1190

Gain margin

GM = 20·log₁₀(1190/376) = 10.01 dB

Answer

GM = 10.01 dB

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