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 = 0.75f(0.75) = ?

Find f(t) for F(s) = 10/(s(s + 1)(s + 3)) by partial fractions, and evaluate f(0.75).

F(s) = 10/(s(s + 1)(s + 3))t = 0.75 s
f(0.75) =
2.Laplace transformsHard
ReIm−1−212

A system is described by the differential equation y″ + 2y′ + 5y = 3u′ + 6u, with input u(t) and output y(t). Find its transfer function G(s) = Y(s)/U(s) (zero initial conditions) and evaluate the impulse response g(t) at t = 0.2 s. Find g(0.2).

y″ + 2y′ + 5y = 3u′ + 6ut = 0.2 s
g(0.2) =
3.Block diagramsMedium
R(s)+6K+9s + 4G(s)C(s)1H₁−−

The plant G(s) = 9/(s + 4) has an inner (minor) negative feedback loop with gain H₁ = 1. A controller gain K = 6 and an outer unity negative feedback loop close the system. Find the closed-loop DC gain T(0).

G(s) = 9/(s + 4)H₁ = 1K = 6
T(0) =
4.Block diagramsHard
R(s)+30s(s + 10)G(s)C(s)2H−

The plant G(s) = 30/(s(s + 10)) is in a negative feedback loop with sensor gain H = 2. The closed-loop poles are complex. Find their real part σ.

G(s) = 30/(s(s + 10))H = 2
σ =
5.Step responseMedium
t (s)y(t)1?

A system has the transfer function G(s) = 98/(s² + 16s + 98) and receives a unit step input. Find the percent overshoot %OS of the step response.

G(s) = 98/(s² + 16s + 98)
%OS = %
6.Step responseHard
+−10 VR = 85 ΩL = 20 mHC = 1 µFvC​

A series RLC circuit with R = 85 Ω, L = 20 mH and C = 1 µF is connected to a 10 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the percent overshoot %OS of the capacitor voltage.

R = 85 ΩL = 20 mHC = 1 µFV = 10 V
%OS = %
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 3)(s + 4)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 3)(s + 4)). The gain is raised until the closed loop is on the edge of stability. Find the frequency ω of the sustained oscillation.

L(s) = K/(s(s + 3)(s + 4))
ω = rad/s
8.Stability: Routh–Hurwitz & root locusHard
ReImζ = 0.707−0.5−1−1.5−2−2.50.5j0.5j1

The root locus of L(s) = K/(s(s + 2)) is shown. Find the gain K for which the closed-loop poles have the damping ratio ζ = 0.707.

L(s) = K/(s(s + 2))ζ = 0.707
K =
9.Steady-state error & PID controlMedium
r(t) = 10t+Ks(s + 5)C(s)−

A unity-feedback loop with G(s) = K/(s(s + 5)) must follow the ramp r(t) = 10t with a steady-state error of 0.5. Find the gain K.

G(s) = K/(s(s + 5))r(t) = 10tess = 0.5
K =
10.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s2s + 1C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 2/(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) = 2/(s + 1)ωₙ = 10 rad/sζ = 0.5
Kp =
11.Bode plots & marginsMedium
|H| (dB)0.10.111101010010010001000−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ω = 5

Find the phase ∠G(jω) of G(s) = 7(s + 19)/(s(s + 3)) at ω = 5 rad/s, in degrees.

G(s) = 7(s + 19)/(s(s + 3))ω = 5 rad/s
∠G(jω) = °
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) = K/(s(s + 2)(s + 9)). Find the gain K that gives a gain margin of 4 dB.

L(s) = K/(s(s + 2)(s + 9))GM = 4 dB
K =

3Solutions

1.Laplace transforms

Answer: f(0.75) = 1.147

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 + 3)
A = 10/(1·3) = 3.333
B = 10/((−1)·2) = −5
C = 10/((−3)·(−2)) = 1.667

Inverse transform

f(t) = 3.333 − 5·e−t + 1.667·e−3t

Evaluate

f(0.75) = 3.333 − 5·e−0.75 + 1.667·e−3·0.75 = 1.147

Answer

f(0.75) = 1.147
2.Laplace transforms

Answer: g(0.2) = 2.741

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).

Transfer function

G(s) = Y(s)/U(s) = (3s + 6)/(s² + 2s + 5)

Partial fractions

s² + 2s + 5 = (s + 1)² + 2²
G(s) = 3·(s + 1)/((s + 1)² + 2²) + (3/2)·2/((s + 1)² + 2²)

Impulse response

g(t) = e−t·(3·cos 2t + 1.5·sin 2t)
g(0.2) = 2.741

Answer

g(0.2) = 2.741
3.Block diagrams

Answer: T(0) = 0.806

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) = 9/(s + 4 + 9·1) = 9/(s + 13)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 54/(s + 13 + 54) = 54/(s + 67)

Result

T(0) = 54/67 = 0.806

Answer

T(0) = 0.806
4.Block diagrams

Answer: σ = −5

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

s(s + 10) + 30·2 = 0 → s² + 10s + 60 = 0

Roots

s = −5 ± j·√(60 − 25) = −5 ± j5.916
σ = −5

Answer

σ = −5
5.Step response

Answer: %OS = 1.343 %

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 = 98 → ωₙ = 9.899 rad/s
2ζωₙ = 16 → ζ = 16/(2·9.899) = 0.8081

Result

%OS = 100·e−ζπ/√(1−ζ²) = 100·e−4.31 = 1.343 %

Answer

%OS = 1.343 %
6.Step response

Answer: %OS = 37.16 %

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) = (85/2)·√(1 × 10⁻⁶/0.02) = 0.3005

Result

%OS = 100·e−ζπ/√(1−ζ²) = 37.16 %

Answer

%OS = 37.16 %
7.Stability: Routh–Hurwitz & root locus

Answer: ω = 3.464 rad/s

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.

Routh array

s³ + 7s² + 12s + K = 0
s³: 1, 12
s²: 7, K
s¹: (7·12 − K)/7
s⁰: K

Critical gain

K = 7·12 = 84

Auxiliary equation

At the critical gain the s¹ row is zero; the s² row gives the poles on the imaginary axis.

7s² + 84 = 0 → s = ±j√12
ω = 3.464 rad/s

Answer

ω = 3.464 rad/s
8.Stability: Routh–Hurwitz & root locus

Answer: K = 2.001

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.

Compare with the standard form

s² + 2s + K = s² + 2ζωₙs + ωₙ²
2·0.707·ωₙ = 2 → ωₙ = 1.414 rad/s

Gain

K = ωₙ² = 1.414² = 2.001

Answer

K = 2.001

The closed-loop poles sit where the ζ line crosses the vertical branch at s = −1.

9.Steady-state error & PID control

Answer: K = 100

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 = 10/0.5 = 20

Gain

K = 5·Kv = 100

Answer

K = 100

A large gain makes this loop very lightly damped: ζ = 0.25. That conflict is why compensators are used.

10.Steady-state error & PID control

Answer: Kp = 4.5

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 + 2Kp)s + 2Ki

Target polynomial

s² + 2·0.5·10s + 10² = s² + 10s + 100

Match the coefficients

1 + 2Kp = 10 → Kp = 4.5

Answer

Kp = 4.5

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

11.Bode plots & margins

Answer: ∠G(jω) = −134.3 °

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.

Add the angles

∠G = atan(5/19) − 90° − atan(5/3) = 14.74° − 90° − 59.04°

Result

∠G(j5) = −134.3°

Answer

∠G(jω) = −134.3°
12.Bode plots & margins

Answer: K = 124.9

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/2)·(ω/9) = 1 → ω180 = √(2·9) = 4.243 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 4)·√(ω² + 81)) = K/(2·9·(2 + 9)) = K/198

Required gain

198/K = 104/20 = 1.585
K = 198/1.585 = 124.9

Answer

K = 124.9

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