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−2−4−6−8

F(s) = (4s + 34)/(s² + 12s + 32). Its inverse transform has the form f(t) = A·e−4t + B·e−8t. Find the coefficient A by partial fractions.

F(s) = (4s + 34)/(s² + 12s + 32)
A =
2.Laplace transformsHard
t (s)y(t)t = 0.1y(0.1) = ?

A system obeys y″ + 8y′ + 15y = 15·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 10 for t ≥ 0. Use the Laplace transform to find the output y(0.1).

y″ + 8y′ + 15y = 15·u(t)u(t) = 10y(0) = 0, y′(0) = 0t = 0.1 s
y(0.1) =
3.Block diagramsMedium
R(s)+K(s + 1)(s + 11)G(s)C(s)−

The plant G(s) = K/((s + 1)(s + 11)) 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 + 1)(s + 11))unity negative feedback
K =
4.Block diagramsHard
R(s)+9s + 1G₁+7s + 8G₂C(s)1H₁−1H₂−

Reduce the block diagram with G₁ = 9/(s + 1), G₂ = 7/(s + 8), an inner negative feedback H₁ = 1 around G₂ and an outer negative feedback H₂ = 1. Find the DC gain T(0) of C(s)/R(s).

G₁ = 9/(s + 1)G₂ = 7/(s + 8)H₁ = 1H₂ = 1
T(0) =
5.Step responseMedium
t (s)y(t)1?

A system has the transfer function G(s) = 12/(s² + 4s + 12) and receives a unit step input. Find the percent overshoot %OS of the step response.

G(s) = 12/(s² + 4s + 12)
%OS = %
6.Step responseHard
t (s)y(t)1tp​ = 0.25 s50 %

The step response of a second-order system without zeros shows 50 % overshoot, and the first peak occurs at t = 0.25 s. Find the natural frequency ωn.

%OS = 50 %tp = 0.25 s
ωn = rad/s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 9s(s + 1)(s + 3)C(s)−

A unity-feedback loop has L(s) = K(s + 9)/(s(s + 1)(s + 3)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K(s + 9)/(s(s + 1)(s + 3))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10−12−14−162j2j4

The root locus of L(s) = K(s + 14)/(s(s + 8)) leaves the real axis between s = 0 and s = −8, circles round and joins the real axis again to the left of the zero. Find the break-in point si.

L(s) = K(s + 14)/(s(s + 8))
si =
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 25 % for a step input.

G(s) = K/((s + 2)(s + 5))ess = 25 %
K =
10.Steady-state error & PID controlHard
R(s)+K = 810s(s + 4)C(s)−

A proportional controller K = 8 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 = 8
tp = s
11.Bode plots & marginsMedium
|H| (dB)0.1110100−40−2002040ω (rad/s)ω = 3

Find the exact magnitude |G(jω)| of G(s) = 120/((s + 2)(s + 3)) at ω = 3 rad/s, in dB.

G(s) = 120/((s + 2)(s + 3))ω = 3 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) = K/(s(s + 1)(s + 2)). Find the gain K that gives a gain margin of 8 dB.

L(s) = K/(s(s + 1)(s + 2))GM = 8 dB
K =

3Solutions

1.Laplace transforms

Answer: A = 4.5

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 + 12s + 32 = (s + 4)(s + 8)
F(s) = A/(s + 4) + B/(s + 8)

Cover-up rule

Multiply by (s + 4) and set s = −4: the B-term vanishes.

A = (4s + 34)/(s + 8) |s = −4 = 18/4 = 4.5

Answer

A = 4.5
(B = −0.5, so f(t) = 4.5·e−4t − 0.5·e−8t)
2.Laplace transforms

Answer: y(0.1) = 0.5775

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

Y(s) = 150/(s(s + 3)(s + 5))

Partial fractions

A = 150/(3·5) = 10
B = 150/((−3)·2) = −25
C = 150/((−5)·(−2)) = 15

Output

y(t) = 10 − 25·e−3t + 15·e−5t
y(0.1) = 0.5775

Answer

y(0.1) = 0.5775
3.Block diagrams

Answer: K = 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 + 1)(s + 11)) = 0 → s² + 12s + 11 + K = 0

Repeated root

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

12² = 4(11 + K) → K = 144/4 − 11 = 25

Answer

K = 25
Both closed-loop poles are then at s = −6.
4.Block diagrams

Answer: T(0) = 0.8077

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) = 7/(s + 15)

Outer loop

T(s) = G₁Gᵢ/(1 + G₁GᵢH₂) = 63/((s + 1)(s + 15) + 63)

DC gain

T(0) = 63/(1·15 + 63) = 63/78 = 0.8077

Answer

T(0) = 0.8077
5.Step response

Answer: %OS = 10.85 %

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 = 12 → ωₙ = 3.464 rad/s
2ζωₙ = 4 → ζ = 4/(2·3.464) = 0.5774

Result

%OS = 100·e−ζπ/√(1−ζ²) = 100·e−2.221 = 10.85 %

Answer

%OS = 10.85 %
6.Step response

Answer: ωn = 12.87 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).

Damping ratio

ζ = −ln(0.5)/√(π² + ln²(0.5)) = 0.6931/√(9.87 + 0.4805) = 0.2155

Natural frequency

ωd = π/tp = π/0.25 = 12.57 rad/s
ωₙ = ωd/√(1 − ζ²) = 12.57/0.9765 = 12.87 rad/s

Answer

ωn = 12.87 rad/s
7.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 2.4

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 + 1)(s + 3) + K(s + 9) = s³ + 4s² + (3 + K)s + 9K

Routh array

s³: 1, 3 + K
s²: 4, 9K
s¹: (4·(3 + K) − 9K)/4
s⁰: 9K

Condition

12 − 5K > 0
K < 12/5 = 2.4

Answer

Kmax = 2.4

With the zero to the right of s = −4 the loop would stay stable for every K > 0.

8.Stability: Routh–Hurwitz & root locus

Answer: si = −23.17

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) = −(s² + 8s)/(s + 14)

Stationary points

Differentiate with the quotient rule and set the numerator to zero.

(2s + 8)(s + 14) − (s² + 8s) = s² + 28s + 112 = 0
s = −14 ± √84 = −4.835, −23.17

Break-in point

si = −23.17

Answer

si = −23.17

Between the two break points the locus is a circle centred on the zero, with radius √84 = 9.165.

9.Steady-state error & PID control

Answer: K = 30

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) = 0.25 → Kp = 3

Gain

K/10 = 3 → K = 30

Answer

K = 30

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 = 0.3604 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) + 8·10 = s² + 4s + 80

ωₙ and ζ

ωₙ = √80 = 8.944 rad/s
ζ = 4/(2·8.944) = 0.2236

Peak time

ωd = 8.944·√(1 − 0.2236²) = 8.718 rad/s
tp = π/8.718 = 0.3604 s

Answer

tp = 0.3604 s
11.Bode plots & margins

Answer: |G(jω)| = 17.89 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(j3)| = 120/(√(3² + 2²)·√(3² + 3²)) = 120/(3.606·4.243) = 7.845

Result

20·log₁₀(7.845) = 17.89 dB

Answer

|G(jω)| = 17.89 dB
12.Bode plots & margins

Answer: K = 2.389

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/2) = 1 → ω180 = √(1·2) = 1.414 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 1)·√(ω² + 4)) = K/(1·2·(1 + 2)) = K/6

Required gain

6/K = 108/20 = 2.512
K = 6/2.512 = 2.389

Answer

K = 2.389

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