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−5−10−155

Find f(t) for F(s) = (s + 17)/(s² + 8s + 41) by completing the square, and evaluate f(0.4).

F(s) = (s + 17)/(s² + 8s + 41)t = 0.4 s
f(0.4) =
2.Laplace transformsHard
t (s)y(t)t = 0.3y(0.3) = ?

Solve y″ + 9y′ + 18y = 0 with y(0) = 3 and y′(0) = 9 using the Laplace transform, and find y(0.3).

y″ + 9y′ + 18y = 0y(0) = 3y′(0) = 9t = 0.3 s
y(0.3) =
3.Block diagramsMedium
R(s)+K(s + 3)(s + 12)G(s)C(s)−

The plant G(s) = K/((s + 3)(s + 12)) 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 + 12))unity negative feedback
K =
4.Block diagramsHard
R(s)+48s(s + 8)G(s)C(s)2H−

The plant G(s) = 48/(s(s + 8)) is in a negative feedback loop with sensor gain H = 2. The closed-loop poles are complex. Find their real part σ.

G(s) = 48/(s(s + 8))H = 2
σ =
5.Step responseMedium
t (s)y(t)1tp​ = ?

A system has the transfer function G(s) = 83/(s² + 5s + 83) and receives a unit step input. Find the peak time tp of the step response.

G(s) = 83/(s² + 5s + 83)
tp = s
6.Step responseHard
k = 200 N/mc = 25 N·s/mm = 5 kgF = 5 Nx(t)

A mass m = 5 kg hangs on a spring k = 200 N/m with a viscous damper c = 25 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 = 5 kgk = 200 N/mc = 25 N·s/mF = 5 N
tp = s
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−5j5

The root locus of L(s) = K/((s + 1)(s + 5)(s + 6)) is sketched. Find the frequency ω at which the locus crosses the imaginary axis.

L(s) = K/((s + 1)(s + 5)(s + 6))
ω = rad/s
9.Steady-state error & PID controlMedium
r(t) = 2 + 1t+50s(s + 1)(s + 5)C(s)−

The input r(t) = 2 + 1t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 50/(s(s + 1)(s + 5)). Find the steady-state error ess.

G(s) = 50/(s(s + 1)(s + 5))r(t) = 2 + 1t
ess =
10.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s4s + 2C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 4/(s + 2) in a unity-feedback loop. The closed loop must have ωₙ = 5 rad/s and ζ = 0.7. Find the integral gain Ki.

G(s) = 4/(s + 2)ωₙ = 5 rad/sζ = 0.7
Ki =
11.Bode plots & marginsMedium
vin​C = 1 nFR = 10 kΩvout​

The RC high-pass filter shown has R = 10 kΩ, C = 1 nF. A sinusoidal input of 64 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 10 kΩC = 1 nFf = 64 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) = 8/(s(s + 1)(s + 4)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 8/(s(s + 1)(s + 4))
ω180 = rad/s

3Solutions

1.Laplace transforms

Answer: f(0.4) = 0.3933

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 + 8s + 41 = (s + 4)² + 25 = (s + 4)² + 5²

Match the numerator

F(s) = 1·(s + 4)/((s + 4)² + 5²) + (13/5)·5/((s + 4)² + 5²)

Inverse transform

f(t) = e−4t·(cos 5t + 2.6·sin 5t)
f(0.4) = e−4·0.4·(cos(5·0.4) + 2.6·sin(5·0.4)) = 0.3933

Answer

f(0.4) = 0.3933
2.Laplace transforms

Answer: y(0.3) = 2.667

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 − 3s − 9 + 9(sY − 3) + 18Y = 0
Y(s) = (3s + 36)/((s + 3)(s + 6))

Partial fractions

A = 27/3 = 9
B = 18/(−3) = −6

Result

y(t) = 9·e−3t − 6·e−6t
y(0.3) = 2.667

Answer

y(0.3) = 2.667

Check: y(0) = A + B = 3.

3.Block diagrams

Answer: K = 20.25

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 + 12)) = 0 → s² + 15s + 36 + K = 0

Repeated root

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

15² = 4(36 + K) → K = 225/4 − 36 = 20.25

Answer

K = 20.25
Both closed-loop poles are then at s = −7.5.
4.Block diagrams

Answer: σ = −4

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 + 8) + 48·2 = 0 → s² + 8s + 96 = 0

Roots

s = −4 ± j·√(96 − 16) = −4 ± j8.944
σ = −4

Answer

σ = −4
5.Step response

Answer: tp = 0.3586 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 = 83 → ωₙ = 9.11 rad/s
2ζωₙ = 5 → ζ = 5/(2·9.11) = 0.2744

Result

ωd = 9.11·√(1 − 0.2744²) = 8.761 rad/s → tp = π/ωd = 0.3586 s

Answer

tp = 0.3586 s
6.Step response

Answer: tp = 0.5408 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) = √(200/5) = 6.325 rad/s
ζ = c/(2√(km)) = 25/(2·√(200·5)) = 0.3953

Result

ωd = 5.809 rad/s
tp = π/ωd = 0.5408 s

Answer

tp = 0.5408 s
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: ω = 6.403 rad/s

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.

Critical gain

s³ + 12s² + 41s + 30 + K = 0
s¹: (12·41 − 30 − K)/12 = 0 → K = 462

Crossing frequency

12s² + 30 + 462 = 12s² + 492 = 0 → s² = −41
ω = √41 = 6.403 rad/s

Answer

ω = 6.403 rad/s
9.Steady-state error & PID control

Answer: ess = 0.1

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) = 10

Step part

Kp = ∞ → estep = 0

Ramp error

ess = 1/10 = 0.1

Answer

ess = 0.1
10.Steady-state error & PID control

Answer: Ki = 6.25

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 + 4Kp)s + 4Ki

Target polynomial

s² + 2·0.7·5s + 5² = s² + 7s + 25

Match the coefficients

4Ki = 25 → Ki = 6.25

Answer

Ki = 6.25

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

11.Bode plots & margins

Answer: φ = 13.97 °

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/(10000·1 × 10⁻⁹) = 100000 rad/s
fc = 15915 Hz → f/fc = 64000/15915 = 4.021

Phase

φ = 90° − atan(4.021) = 13.97°

Answer

φ = 13.97°
12.Bode plots & margins

Answer: ω180 = 2 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/4) = 1 → ω180 = √(1·4) = 2 rad/s

Answer

ω180 = 2 rad/s

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