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−224

Find f(t) for F(s) = (3s + 6)/(s² + 2s + 17) by completing the square, and evaluate f(0.3).

F(s) = (3s + 6)/(s² + 2s + 17)t = 0.3 s
f(0.3) =
2.Laplace transformsHard
ReIm−1−2−3−4−5−6−7−81

A system is described by the differential equation y″ + 6y′ + 8y = u′ + 7u, 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 = 0.25 s. Find g(0.25).

y″ + 6y′ + 8y = u′ + 7ut = 0.25 s
g(0.25) =
3.Block diagramsMedium
R(s)+16K+1s + 2G(s)C(s)4H₁−−

The plant G(s) = 1/(s + 2) has an inner (minor) negative feedback loop with gain H₁ = 4. A controller gain K = 16 and an outer unity negative feedback loop close the system. Find the closed-loop DC gain T(0).

G(s) = 1/(s + 2)H₁ = 4K = 16
T(0) =
4.Block diagramsHard
R(s)+23controller+D(s)+6s + 1plantC(s)−

A proportional controller K = 23 drives the plant G(s) = 6/(s + 1) in a unity negative feedback loop. A disturbance D enters at the plant input (it is added to the controller output). A reference step of size 2 and a disturbance step of size 0.5 are applied together. Find the steady-state output css.

K = 23G(s) = 6/(s + 1)D = 0.5R = 2
css =
5.Step responseMedium
t (s)y(t)1ts​ = ?

A system has the transfer function G(s) = 87/(s² + 6s + 87) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 87/(s² + 6s + 87)
ts = s
6.Step responseHard
k = 200 N/mc = 14 N·s/mm = 1 kgF = 10 Nx(t)

A mass m = 1 kg hangs on a spring k = 200 N/m with a viscous damper c = 14 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 10 N is suddenly applied. Find the damping ratio ζ.

m = 1 kgk = 200 N/mc = 14 N·s/mF = 10 N
ζ =
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 2)(s + 4)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 2)(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 + 2)(s + 4))
ω = rad/s
8.Stability: Routh–Hurwitz & root locusHard
ReIm−0.5−1−1.5−2−2.5−3−3.5−4−4.5−50.5j0.5j1−0.5

For the loop L(s) = K/(s(s + 2)(s + 4)), find the gain K that puts a closed-loop pole at s = −0.5.

L(s) = K/(s(s + 2)(s + 4))s = −0.5
K =
9.Steady-state error & PID controlMedium
r(t) = 2t+20(s + 4)s(s + 5)C(s)−

A ramp r(t) = 2t is applied to a unity-feedback loop with G(s) = 20(s + 4)/(s(s + 5)). Find the steady-state error ess.

G(s) = 20(s + 4)/(s(s + 5))r(t) = 2t
ess =
10.Steady-state error & PID controlHard
R(s)+K = 25s(s + 4)C(s)−

A proportional controller K = 2 drives the plant G(s) = 5/(s(s + 4)) in a unity-feedback loop. Find the percent overshoot %OS of the step response.

G(s) = 5/(s(s + 4))K = 2
%OS = %
11.Bode plots & marginsMedium
vin​L = 1 mHC = 2.2 µFR = 2 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 2 Ω, with L = 1 mH and C = 2.2 µF. Find the −3 dB bandwidth BW in Hz.

R = 2 ΩL = 1 mHC = 2.2 µF
BW = 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) = 38/(s(s + 3)(s + 5)). Find the gain margin GM in dB.

L(s) = 38/(s(s + 3)(s + 5))
GM = dB

3Solutions

1.Laplace transforms

Answer: f(0.3) = 1.323

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 + 2s + 17 = (s + 1)² + 16 = (s + 1)² + 4²

Match the numerator

F(s) = 3·(s + 1)/((s + 1)² + 4²) + (3/4)·4/((s + 1)² + 4²)

Inverse transform

f(t) = e−t·(3·cos 4t + 0.75·sin 4t)
f(0.3) = e−0.3·(3·cos(4·0.3) + 0.75·sin(4·0.3)) = 1.323

Answer

f(0.3) = 1.323
2.Laplace transforms

Answer: g(0.25) = 0.9645

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) = (s + 7)/(s² + 6s + 8)

Partial fractions

s² + 6s + 8 = (s + 2)(s + 4)
A = 2.5, B = −1.5

Impulse response

g(t) = 2.5·e−2t − 1.5·e−4t
g(0.25) = 0.9645

Answer

g(0.25) = 0.9645
3.Block diagrams

Answer: T(0) = 0.7273

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) = 1/(s + 2 + 1·4) = 1/(s + 6)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 16/(s + 6 + 16) = 16/(s + 22)

Result

T(0) = 16/22 = 0.7273

Answer

T(0) = 0.7273
4.Block diagrams

Answer: css = 2.007

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).

Transfer functions

C/D = G/(1 + K·G) = 6/(s + 1 + 138)
C/R = K·G/(1 + K·G) = 138/(s + 1 + 138)

Steady state

css = 2·138/139 + 0.5·6/139 = 2.007

Answer

css = 2.007

A larger K reduces the effect of the disturbance, but never to zero with a proportional controller.

5.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).

Compare with the standard form

The denominator must match s² + 2ζωₙs + ωₙ².

ωₙ² = 87 → ωₙ = 9.327 rad/s
2ζωₙ = 6 → ζ = 6/(2·9.327) = 0.3216

Result

ζωₙ = 3 → ts ≈ 4/3 = 1.333 s

Answer

ts = 1.333 s
6.Step response

Answer: ζ = 0.495

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) = √(200/1) = 14.14 rad/s
ζ = c/(2√(km)) = 14/(2·√(200·1)) = 0.495

Result

ζ = 0.495

Answer

ζ = 0.495
7.Stability: Routh–Hurwitz & root locus

Answer: ω = 2.828 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 + 6s² + 8s + K = 0
s³: 1, 8
s²: 6, K
s¹: (6·8 − K)/6
s⁰: K

Critical gain

K = 6·8 = 48

Auxiliary equation

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

6s² + 48 = 0 → s = ±j√8
ω = 2.828 rad/s

Answer

ω = 2.828 rad/s
8.Stability: Routh–Hurwitz & root locus

Answer: K = 2.625

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.

Magnitude condition

K = |s|·|s + 2|·|s + 4|

Evaluate at the point

K = 0.5·1.5·3.5 = 2.625

Answer

K = 2.625

Check: s(s + 2)(s + 4) + K is zero at s = −0.5.

9.Steady-state error & PID control

Answer: ess = 0.125

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) = 16

Ramp error

ess = 2/16 = 0.125

Answer

ess = 0.125
10.Steady-state error & PID control

Answer: %OS = 7.691 %

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 + 4) + 2·5 = s² + 4s + 10

ωₙ and ζ

ωₙ = √10 = 3.162 rad/s
ζ = 4/(2·3.162) = 0.6325

Overshoot

%OS = 100·e−0.6325π/√(1 − 0.6325²) = 7.691 %

Answer

%OS = 7.691 %
11.Bode plots & margins

Answer: BW = 318.3 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/√(0.001·2.2 × 10⁻⁶) = 21320 rad/s
f₀ = ω₀/(2π) = 3393 Hz

Bandwidth

BW = R/L = 2/0.001 = 2000 rad/s = 318.3 Hz

Answer

BW = 318.3 Hz
12.Bode plots & margins

Answer: GM = 9.988 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/3)·(ω/5) = 1 → ω180 = √(3·5) = 3.873 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 9)·√(ω² + 25)) = K/(3·5·(3 + 5)) = K/120

Gain margin

GM = 20·log₁₀(120/38) = 9.988 dB

Answer

GM = 9.988 dB

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