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 = 0.1f(0.1) = ?

Find f(t) for F(s) = (s + 15)/(s² + 10s + 25) and evaluate f(0.1).

F(s) = (s + 15)/(s² + 10s + 25)t = 0.1 s
f(0.1) =
2.Laplace transformsHard
t (s)y(t)t = 0.2y(0.2) = ?

Solve y″ + 11y′ + 30y = 0 with y(0) = 9 and y′(0) = −1 using the Laplace transform, and find y(0.2).

y″ + 11y′ + 30y = 0y(0) = 9y′(0) = −1t = 0.2 s
y(0.2) =
3.Block diagramsMedium
R(s)+11(s + 1)(s + 3)(s + 7)G(s)C(s)−

A unity negative feedback loop has G(s) = 11(s + 1)/((s + 3)(s + 7)). A step of size 5 is applied at the reference. Find the steady-state error ess.

G(s) = 11(s + 1)/((s + 3)(s + 7))R = 5
ess =
4.Block diagramsHard
R(s)+11s + 5G₁+6s + 7G₂C(s)4H₁−0.5H₂−

Reduce the block diagram with G₁ = 11/(s + 5), G₂ = 6/(s + 7), an inner negative feedback H₁ = 4 around G₂ and an outer negative feedback H₂ = 0.5. Find the DC gain T(0) of C(s)/R(s).

G₁ = 11/(s + 5)G₂ = 6/(s + 7)H₁ = 4H₂ = 0.5
T(0) =
5.Step responseMedium
t (s)y(t)2tp​ = ?

A system has the transfer function G(s) = 42/(s² + 3s + 21) and receives a unit step input. Find the peak time tp of the step response.

G(s) = 42/(s² + 3s + 21)
tp = s
6.Step responseHard
+−10 VR = 90 ΩL = 500 mHC = 22 µFvC​

A series RLC circuit with R = 90 Ω, L = 500 mH and C = 22 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 = 90 ΩL = 500 mHC = 22 µFV = 10 V
%OS = %
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 9s(s + 3)(s + 4)C(s)−

A unity-feedback loop has L(s) = K(s + 9)/(s(s + 3)(s + 4)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K(s + 9)/(s(s + 3)(s + 4))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−0.5−1−1.5−2−2.5−3−3.5−4−4.5−50.5j0.5j1

The root locus of L(s) = K/(s(s + 3)(s + 4)) leaves the real axis between s = 0 and s = −3. Find the break-away point sb.

L(s) = K/(s(s + 3)(s + 4))
sb =
9.Steady-state error & PID controlMedium
R(s)+K(s + 4)(s + 10)C(s)−

A unity-feedback loop has G(s) = K/((s + 4)(s + 10)). Find the gain K that gives a steady-state error of 5 % for a step input.

G(s) = K/((s + 4)(s + 10))ess = 5 %
K =
10.Steady-state error & PID controlHard
R(s)+K = 251s(s + 6)C(s)−

A proportional controller K = 25 drives the plant G(s) = 1/(s(s + 6)) in a unity-feedback loop. Find the peak time tp of the step response.

G(s) = 1/(s(s + 6))K = 25
tp = s
11.Bode plots & marginsMedium
|H| (dB)0.11101001000−80−60−40−20020ω (rad/s)ω = 4

Find the exact magnitude |G(jω)| of G(s) = 440/((s + 11)(s + 20)) at ω = 4 rad/s, in dB.

G(s) = 440/((s + 11)(s + 20))ω = 4 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) = 6/(s(s + 1)(s + 5)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 6/(s(s + 1)(s + 5))
ω180 = rad/s

3Solutions

1.Laplace transforms

Answer: f(0.1) = 1.213

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 + 15 = 1(s + 5) + 10
F(s) = 1/(s + 5) + 10/(s + 5)²

Inverse transform

f(t) = e−5t + 10·t·e−5t
f(0.1) = (1 + 10·0.1)·e−5·0.1 = 1.213

Answer

f(0.1) = 1.213
2.Laplace transforms

Answer: y(0.2) = 6.245

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 − 9s − (−1) + 11(sY − 9) + 30Y = 0
Y(s) = (9s + 98)/((s + 5)(s + 6))

Partial fractions

A = 53/1 = 53
B = 44/(−1) = −44

Result

y(t) = 53·e−5t − 44·e−6t
y(0.2) = 6.245

Answer

y(0.2) = 6.245

Check: y(0) = A + B = 9.

3.Block diagrams

Answer: ess = 3.281

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) = 11·1/(3·7) = 0.5238

Steady-state error

ess = A/(1 + Kₚ) = 5/(1 + 0.5238) = 3.281

Answer

ess = 3.281
4.Block diagrams

Answer: T(0) = 0.3511

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 = G₂/(1 + G₂H₁) = 6/(s + 31)

Outer loop

T(s) = G₁Gᵢ/(1 + G₁GᵢH₂) = 66/((s + 5)(s + 31) + 33)

DC gain

T(0) = 66/(5·31 + 33) = 66/188 = 0.3511

Answer

T(0) = 0.3511
5.Step response

Answer: tp = 0.7255 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 = 21 → ωₙ = 4.583 rad/s
2ζωₙ = 3 → ζ = 3/(2·4.583) = 0.3273

Result

ωd = 4.583·√(1 − 0.3273²) = 4.33 rad/s → tp = π/ωd = 0.7255 s

Answer

tp = 0.7255 s
6.Step response

Answer: %OS = 37.44 %

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/√(0.5·2.2 × 10⁻⁵) = 301.5 rad/s
ζ = (R/2)·√(C/L) = (90/2)·√(2.2 × 10⁻⁵/0.5) = 0.2985

Result

%OS = 100·e−ζπ/√(1−ζ²) = 37.44 %

Answer

%OS = 37.44 %
7.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 42

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(s + 3)(s + 4) + K(s + 9) = s³ + 7s² + (12 + K)s + 9K

Routh array

s³: 1, 12 + K
s²: 7, 9K
s¹: (7·(12 + K) − 9K)/7
s⁰: 9K

Condition

84 − 2K > 0
K < 84/2 = 42

Answer

Kmax = 42

With the zero to the right of s = −7 the loop would stay stable for every K > 0.

8.Stability: Routh–Hurwitz & root locus

Answer: sb = −1.131

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.

Gain along the real axis

K(s) = −s(s + 3)(s + 4) = −(s³ + 7s² + 12s)

Stationary points

dK/ds = −(3s² + 14s + 12) = 0
s = (−14 ± √(196 − 144))/6 = −1.131, −3.535

Pick the point on the locus

The locus lies on the real axis between 0 and −3; the other root is not on the locus.

sb = −1.131

Answer

sb = −1.131
9.Steady-state error & PID control

Answer: K = 760

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) = 0.05 → Kp = 19

Gain

K/40 = 19 → K = 760

Answer

K = 760

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 = 0.7854 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 + 6) + 25·1 = s² + 6s + 25

ωₙ and ζ

ωₙ = √25 = 5 rad/s
ζ = 6/(2·5) = 0.6

Peak time

ωd = 5·√(1 − 0.6²) = 4 rad/s
tp = π/4 = 0.7854 s

Answer

tp = 0.7854 s
11.Bode plots & margins

Answer: |G(jω)| = 5.311 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(j4)| = 440/(√(4² + 11²)·√(4² + 20²)) = 440/(11.7·20.4) = 1.843

Result

20·log₁₀(1.843) = 5.311 dB

Answer

|G(jω)| = 5.311 dB
12.Bode plots & margins

Answer: ω180 = 2.236 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/5) = 1 → ω180 = √(1·5) = 2.236 rad/s

Answer

ω180 = 2.236 rad/s

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