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 = 0.2f(0.2) = ?

Find f(t) for F(s) = (s + 7)/(s² + 10s + 25) and evaluate f(0.2).

F(s) = (s + 7)/(s² + 10s + 25)t = 0.2 s
f(0.2) =
2.Laplace transformsHard
ReIm−1−2−3−4−5−6−7−8−91

A system is described by the differential equation y″ + 5y′ + 6y = 2u′ + 16u, 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″ + 5y′ + 6y = 2u′ + 16ut = 0.25 s
g(0.25) =
3.Block diagramsMedium
R(s)+3(s + 6)(s + 3)(s + 7)G(s)C(s)−

A unity negative feedback loop has G(s) = 3(s + 6)/((s + 3)(s + 7)). A step of size 5 is applied at the reference. Find the steady-state error ess.

G(s) = 3(s + 6)/((s + 3)(s + 7))R = 5
ess =
4.Block diagramsHard
R(s)+50s(s + 2)G(s)C(s)1H−

The plant G(s) = 50/(s(s + 2)) is in a negative feedback loop with sensor gain H = 1. The closed-loop poles are complex. Find their real part σ.

G(s) = 50/(s(s + 2))H = 1
σ =
5.Step responseMedium
t (s)y(t)1

A system has the transfer function G(s) = 92/(s² + 11s + 92) and receives a unit step input. Find the natural frequency ωn in rad/s.

G(s) = 92/(s² + 11s + 92)
ωn = rad/s
6.Step responseHard
R(s)+Ks(s + 6)G(s)C(s)−

A unity negative feedback loop has G(s) = K/(s(s + 6)). Find the gain K that gives a step response with 30 % overshoot.

G(s) = K/(s(s + 6))%OS = 30 %
K =
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 4(s − 1)(s + 2)C(s)−

The plant G(s) = (s + 4)/((s − 1)(s + 2)) has an unstable pole at s = 1. A proportional gain K closes a unity-feedback loop around it. Find the smallest gain Kmin that makes the closed loop stable.

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

A unity-feedback loop has L(s) = K/((s + 6)(s + 3)(s + 10)(s + 5)). Find the point σa where the asymptotes of the root locus cross the real axis.

L(s) = K/((s + 6)(s + 3)(s + 10)(s + 5))
σa =
9.Steady-state error & PID controlMedium
R(s)+K(s + 2)(s + 5)C(s)−

A unity-feedback loop has G(s) = K/((s + 2)(s + 5)). Find the gain K that gives a steady-state error of 4 % for a step input.

G(s) = K/((s + 2)(s + 5))ess = 4 %
K =
10.Steady-state error & PID controlHard
R(s)+K = 510s(s + 4)C(s)−

A proportional controller K = 5 drives the plant G(s) = 10/(s(s + 4)) in a unity-feedback loop. Find the peak time tp of the step response.

G(s) = 10/(s(s + 4))K = 5
tp = s
11.Bode plots & marginsMedium
|H| (dB)0.010.11101001000−80−60−40−20020ω (rad/s)ω = 0.5

Find the exact magnitude |G(jω)| of G(s) = 960/((s + 10)(s + 12)) at ω = 0.5 rad/s, in dB.

G(s) = 960/((s + 10)(s + 12))ω = 0.5 rad/s
|G(jω)| = dB
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) = 254/(s(s + 7)(s + 9)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 254/(s(s + 7)(s + 9))
ω180 = rad/s

3Solutions

1.Laplace transforms

Answer: f(0.2) = 0.515

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 + 10s + 25 = (s + 5)²

Partial fractions

s + 7 = 1(s + 5) + 2
F(s) = 1/(s + 5) + 2/(s + 5)²

Inverse transform

f(t) = e−5t + 2·t·e−5t
f(0.2) = (1 + 2·0.2)·e−5·0.2 = 0.515

Answer

f(0.2) = 0.515
2.Laplace transforms

Answer: g(0.25) = 2.555

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) = (2s + 16)/(s² + 5s + 6)

Partial fractions

s² + 5s + 6 = (s + 2)(s + 3)
A = 12, B = −10

Impulse response

g(t) = 12·e−2t − 10·e−3t
g(0.25) = 2.555

Answer

g(0.25) = 2.555
3.Block diagrams

Answer: ess = 2.692

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).

Position error constant

Kₚ = G(0) = 3·6/(3·7) = 0.8571

Steady-state error

ess = A/(1 + Kₚ) = 5/(1 + 0.8571) = 2.692

Answer

ess = 2.692
4.Block diagrams

Answer: σ = −1

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 + 2) + 50·1 = 0 → s² + 2s + 50 = 0

Roots

s = −1 ± j·√(50 − 1) = −1 ± j7
σ = −1

Answer

σ = −1
5.Step response

Answer: ωn = 9.592 rad/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 = 92 → ωₙ = 9.592 rad/s
2ζωₙ = 11 → ζ = 11/(2·9.592) = 0.5734

Result

ωₙ = 9.592 rad/s

Answer

ωn = 9.592 rad/s
6.Step response

Answer: K = 70.28

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 + 6s + K)

Damping ratio from the overshoot

ζ = −ln(0.3)/√(π² + ln²(0.3)) = 0.3579

Gain

ωₙ = 6/(2ζ) = 8.383 rad/s
K = ωₙ² = 70.28

Answer

K = 70.28
7.Stability: Routh–Hurwitz & root locus

Answer: Kmin = 0.5

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 − 1)(s + 2) + K(s + 4) = s² + (1 + K)s + (4K − 2)

Conditions

For s² + c₁s + c₀ all roots lie in the left half-plane exactly when c₁ > 0 and c₀ > 0.

1 + K > 0 → K > −1
4K − 2 > 0 → K > 0.5
Kmin = 0.5

Answer

Kmin = 0.5
8.Stability: Routh–Hurwitz & root locus

Answer: σa = −6

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: −6, −3, −10, −5
n − m = 4 − 0 = 4

Centroid

σa = (−24)/4 = −6

Answer

σa = −6

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

9.Steady-state error & PID control

Answer: K = 240

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 error constant

1/(1 + Kp) = 0.04 → Kp = 24

Gain

K/10 = 24 → K = 240

Answer

K = 240

Check the closed-loop stability as well: for this second-order loop any K > 0 is stable.

10.Steady-state error & PID control

Answer: tp = 0.4632 s

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) + 5·10 = s² + 4s + 50

ωₙ and ζ

ωₙ = √50 = 7.071 rad/s
ζ = 4/(2·7.071) = 0.2828

Peak time

ωd = 7.071·√(1 − 0.2828²) = 6.782 rad/s
tp = π/6.782 = 0.4632 s

Answer

tp = 0.4632 s
11.Bode plots & margins

Answer: |G(jω)| = 18.04 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(j0.5)| = 960/(√(0.5² + 10²)·√(0.5² + 12²)) = 960/(10.01·12.01) = 7.983

Result

20·log₁₀(7.983) = 18.04 dB

Answer

|G(jω)| = 18.04 dB
12.Bode plots & margins

Answer: ω180 = 7.937 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/7)·(ω/9) = 1 → ω180 = √(7·9) = 7.937 rad/s

Answer

ω180 = 7.937 rad/s

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