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) = (6s + 35)/(s² + 4s + 4) and evaluate f(1).

F(s) = (6s + 35)/(s² + 4s + 4)t = 1 s
f(1) =
2.Laplace transformsHard
ReIm−0.5−1−1.5−2−2.5−3−3.5−4−4.50.51

A system is described by the differential equation y″ + 7y′ + 12y = 4u′ + 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″ + 7y′ + 12y = 4u′ + 6ut = 0.2 s
g(0.2) =
3.Block diagramsMedium
R(s)+K(s + 3)(s + 6)G(s)C(s)−

The plant G(s) = K/((s + 3)(s + 6)) is in a unity negative feedback loop. Find the gain K for which the closed loop is critically damped (two equal real poles).

G(s) = K/((s + 3)(s + 6))unity negative feedback
K =
4.Block diagramsHard
R(s)+44s(s + 12)G(s)C(s)2H−

The plant G(s) = 44/(s(s + 12)) is in a negative feedback loop with sensor gain H = 2. The closed-loop poles are complex. Find their real part σ.

G(s) = 44/(s(s + 12))H = 2
σ =
5.Step responseMedium
t (s)y(t)0.5

A system has the transfer function G(s) = 45.5/(s² + 15s + 91) and receives a unit step input. Find the damping ratio ζ.

G(s) = 45.5/(s² + 15s + 91)
ζ =
6.Step responseHard
+−10 VR = 85 ΩL = 10 mHC = 2 µFvC​

A series RLC circuit with R = 85 Ω, L = 10 mH and C = 2 µF is connected to a 10 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the time tp of the first voltage peak, in ms.

R = 85 ΩL = 10 mHC = 2 µFV = 10 V
tp = ms
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1(s + 1)(s + 3)(s + 6)C(s)−

A unity-feedback loop has L(s) = K/((s + 1)(s + 3)(s + 6)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/((s + 1)(s + 3)(s + 6))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10−12j2

A unity-feedback loop has L(s) = K/(s(s + 2)(s + 10)(s + 3)). Find the point σa where the asymptotes of the root locus cross the real axis.

L(s) = K/(s(s + 2)(s + 10)(s + 3))
σa =
9.Steady-state error & PID controlMedium
r(t) = 2 + 5t+5(s + 4)s(s + 5)(s + 10)C(s)−

The input r(t) = 2 + 5t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 5(s + 4)/(s(s + 5)(s + 10)). Find the steady-state error ess.

G(s) = 5(s + 4)/(s(s + 5)(s + 10))r(t) = 2 + 5t
ess =
10.Steady-state error & PID controlHard
R(s)+K = 44s(s + 2)C(s)−

A proportional controller K = 4 drives the plant G(s) = 4/(s(s + 2)) in a unity-feedback loop. Find the percent overshoot %OS of the step response.

G(s) = 4/(s(s + 2))K = 4
%OS = %
11.Bode plots & marginsMedium
|H| (dB)0.1110100−40−2002040ω (rad/s)ω = 1

Find the exact magnitude |G(jω)| of G(s) = 255/((s + 3)(s + 5)) at ω = 1 rad/s, in dB.

G(s) = 255/((s + 3)(s + 5))ω = 1 rad/s
|G(jω)| = dB
12.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−40−200204060ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 104/(s(s + 4)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 104/(s(s + 4))
ωc = rad/s

3Solutions

1.Laplace transforms

Answer: f(1) = 3.925

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 + 4s + 4 = (s + 2)²

Partial fractions

6s + 35 = 6(s + 2) + 23
F(s) = 6/(s + 2) + 23/(s + 2)²

Inverse transform

f(t) = 6·e−2t + 23·t·e−2t
f(1) = (6 + 23·1)·e−2·1 = 3.925

Answer

f(1) = 3.925
2.Laplace transforms

Answer: g(0.2) = 1.2

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) = (4s + 6)/(s² + 7s + 12)

Partial fractions

s² + 7s + 12 = (s + 3)(s + 4)
A = −6, B = 10

Impulse response

g(t) = −6·e−3t + 10·e−4t
g(0.2) = 1.2

Answer

g(0.2) = 1.2
3.Block diagrams

Answer: K = 2.25

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

1 + K/((s + 3)(s + 6)) = 0 → s² + 9s + 18 + K = 0

Repeated root

A quadratic s² + βs + γ has a double root when β² = 4γ.

9² = 4(18 + K) → K = 81/4 − 18 = 2.25

Answer

K = 2.25
Both closed-loop poles are then at s = −4.5.
4.Block diagrams

Answer: σ = −6

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 + 12) + 44·2 = 0 → s² + 12s + 88 = 0

Roots

s = −6 ± j·√(88 − 36) = −6 ± j7.211
σ = −6

Answer

σ = −6
5.Step response

Answer: ζ = 0.7862

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 = 91 → ωₙ = 9.539 rad/s
2ζωₙ = 15 → ζ = 15/(2·9.539) = 0.7862

Result

ζ = 0.7862

Answer

ζ = 0.7862
6.Step response

Answer: tp = 0.5559 ms

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/√(0.01·2 × 10⁻⁶) = 7071 rad/s
ζ = (R/2)·√(C/L) = (85/2)·√(2 × 10⁻⁶/0.01) = 0.601

Result

ωd = 5651 rad/s
tp = π/ωd = 0.5559 ms

Answer

tp = 0.5559 ms
7.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 252

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

s³ + 10s² + 27s + 18 + K = 0

Routh array

s³: 1, 27
s²: 10, 18 + K
s¹: (10·27 − 18 − K)/10
s⁰: 18 + K

Condition

The s¹ entry must stay positive.

K < 10·27 − 18 = 252

Answer

Kmax = 252

At the limit the loop oscillates at ω = √27 = 5.196 rad/s.

8.Stability: Routh–Hurwitz & root locus

Answer: σa = −3.75

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.

Poles and zeros

poles: 0, −2, −10, −3
n − m = 4 − 0 = 4

Centroid

σa = (−15)/4 = −3.75

Answer

σa = −3.75

The 4 asymptotes leave the centroid at angles of 45°, 135°, 225°, 315°.

9.Steady-state error & PID control

Answer: ess = 12.5

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) = 0.4

Step part

Kp = ∞ → estep = 0

Ramp error

ess = 5/0.4 = 12.5

Answer

ess = 12.5
10.Steady-state error & PID control

Answer: %OS = 44.43 %

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) + 4·4 = s² + 2s + 16

ωₙ and ζ

ωₙ = √16 = 4 rad/s
ζ = 2/(2·4) = 0.25

Overshoot

%OS = 100·e−0.25π/√(1 − 0.25²) = 44.43 %

Answer

%OS = 44.43 %
11.Bode plots & margins

Answer: |G(jω)| = 23.98 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.

Multiply the magnitudes

|G(j1)| = 255/(√(1² + 3²)·√(1² + 5²)) = 255/(3.162·5.099) = 15.81

Result

20·log₁₀(15.81) = 23.98 dB

Answer

|G(jω)| = 23.98 dB
12.Bode plots & margins

Answer: ωc = 9.814 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.

Gain crossover

ω⁴ + 16ω² − 10816 = 0
ω² = (−16 + √(256 + 4·10816))/2 = 96.31
ωc = 9.814 rad/s

Answer

ωc = 9.814 rad/s

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