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−22

Find f(t) for F(s) = (s + 2)/(s² + 2s + 10) by completing the square, and evaluate f(1.2).

F(s) = (s + 2)/(s² + 2s + 10)t = 1.2 s
f(1.2) =
2.Laplace transformsHard
t (s)y(t)t = 0.25y(0.25) = ?

A system obeys y″ + 7y′ + 6y = 6·u(t), with y(0) = 5 and y′(0) = 0. At t = 0 a step input u(t) = 10 is applied. Use the Laplace transform to find y(0.25).

y″ + 7y′ + 6y = 6·u(t)y(0) = 5y′(0) = 0u(t) = 10t = 0.25 s
y(0.25) =
3.Block diagramsMedium
R(s)+10(s + 7)(s + 1)(s + 6)G(s)C(s)−

A unity negative feedback loop has G(s) = 10(s + 7)/((s + 1)(s + 6)). A step of size 5 is applied at the reference. Find the steady-state error ess.

G(s) = 10(s + 7)/((s + 1)(s + 6))R = 5
ess =
4.Block diagramsHard
R(s)+20s + 2G₁+5s + 8G₂C(s)4H₁−0.5H₂−

Reduce the block diagram with G₁ = 20/(s + 2), G₂ = 5/(s + 8), 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 = 20/(s + 2)G₂ = 5/(s + 8)H₁ = 4H₂ = 0.5
T(0) =
5.Step responseMedium
t (s)y(t)1ts​ = ?

A system has the transfer function G(s) = 72/(s² + 6s + 72) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 72/(s² + 6s + 72)
ts = s
6.Step responseHard
t (s)y(t)1tp​ = 1.5 s5 %

The step response of a second-order system without zeros shows 5 % overshoot, and the first peak occurs at t = 1.5 s. Find the natural frequency ωn.

%OS = 5 %tp = 1.5 s
ωn = rad/s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1(s + 3)(s + 4)(s + 5)C(s)−

A unity-feedback loop has L(s) = K/((s + 3)(s + 4)(s + 5)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/((s + 3)(s + 4)(s + 5))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−71j1j2

A unity-feedback loop has L(s) = K/(s(s + 6)(s + 4)). Find the point σa where the asymptotes of the root locus cross the real axis.

L(s) = K/(s(s + 6)(s + 4))
σa =
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 2 % for a step input.

G(s) = K/((s + 4)(s + 10))ess = 2 %
K =
10.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s4s + 5C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 4/(s + 5) in a unity-feedback loop. The closed loop must have ωₙ = 8 rad/s and ζ = 1. Find the proportional gain Kp.

G(s) = 4/(s + 5)ωₙ = 8 rad/sζ = 1
Kp =
11.Bode plots & marginsMedium
vin​C = 100 nFR = 3.3 kΩvout​

The RC high-pass filter shown has R = 3.3 kΩ, C = 100 nF. A sinusoidal input of 240 Hz is applied. Find the gain |H| = |vout/vin| in dB.

R = 3.3 kΩC = 100 nFf = 240 Hz
|H| = dB
12.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​

A unity-feedback loop has the open-loop transfer function L(s) = 22/(s(s + 10)). Find the phase margin PM in degrees.

L(s) = 22/(s(s + 10))
PM = °

3Solutions

1.Laplace transforms

Answer: f(1.2) = −0.3145

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 + 10 = (s + 1)² + 9 = (s + 1)² + 3²

Match the numerator

F(s) = 1·(s + 1)/((s + 1)² + 3²) + (1/3)·3/((s + 1)² + 3²)

Inverse transform

f(t) = e−t·(cos 3t + 0.3333·sin 3t)
f(1.2) = e−1.2·(cos(3·1.2) + 0.3333·sin(3·1.2)) = −0.3145

Answer

f(1.2) = −0.3145
2.Laplace transforms

Answer: y(0.25) = 5.55

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

(s² + 7s + 6)·Y(s) = 5s + 35 + 60/s
Y(s) = (5s² + 35s + 60)/(s(s + 1)(s + 6))

Partial fractions

A = 60/(1·6) = 10
B = 30/((−1)·5) = −6
C = 30/((−6)·(−5)) = 1

Result

y(t) = 10 − 6·e−t + e−6t
y(0.25) = 5.55

Answer

y(0.25) = 5.55

Check: y(0) = A + B + C = 5.

3.Block diagrams

Answer: ess = 0.3947

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) = 10·7/(1·6) = 11.67

Steady-state error

ess = A/(1 + Kₚ) = 5/(1 + 11.67) = 0.3947

Answer

ess = 0.3947
4.Block diagrams

Answer: T(0) = 0.9434

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) = 5/(s + 28)

Outer loop

T(s) = G₁Gᵢ/(1 + G₁GᵢH₂) = 100/((s + 2)(s + 28) + 50)

DC gain

T(0) = 100/(2·28 + 50) = 100/106 = 0.9434

Answer

T(0) = 0.9434
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 = 72 → ωₙ = 8.485 rad/s
2ζωₙ = 6 → ζ = 6/(2·8.485) = 0.3536

Result

ζωₙ = 3 → ts ≈ 4/3 = 1.333 s

Answer

ts = 1.333 s
6.Step response

Answer: ωn = 2.894 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(0.05)/√(π² + ln²(0.05)) = 2.996/√(9.87 + 8.974) = 0.6901

Natural frequency

ωd = π/tp = π/1.5 = 2.094 rad/s
ωₙ = ωd/√(1 − ζ²) = 2.094/0.7237 = 2.894 rad/s

Answer

ωn = 2.894 rad/s
7.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 504

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 + 12s² + 47s + 60 + K = 0

Routh array

s³: 1, 47
s²: 12, 60 + K
s¹: (12·47 − 60 − K)/12
s⁰: 60 + K

Condition

The s¹ entry must stay positive.

K < 12·47 − 60 = 504

Answer

Kmax = 504

At the limit the loop oscillates at ω = √47 = 6.856 rad/s.

8.Stability: Routh–Hurwitz & root locus

Answer: σa = −3.333

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: 0, −6, −4
n − m = 3 − 0 = 3

Centroid

σa = (−10)/3 = −3.333

Answer

σa = −3.333

The 3 asymptotes leave the centroid at angles of 60°, 180°, 300°.

9.Steady-state error & PID control

Answer: K = 1960

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) = 0.02 → Kp = 49

Gain

K/40 = 49 → K = 1960

Answer

K = 1960

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

10.Steady-state error & PID control

Answer: Kp = 2.75

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 + 5).

s² + (5 + 4Kp)s + 4Ki

Target polynomial

s² + 2·1·8s + 8² = s² + 16s + 64

Match the coefficients

5 + 4Kp = 16 → Kp = 2.75

Answer

Kp = 2.75

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

11.Bode plots & margins

Answer: |H| = −7.023 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.

Corner frequency

ωc = 1/(RC) = 1/(3300·1 × 10⁻⁷) = 3030 rad/s
fc = 482.3 Hz → f/fc = 240/482.3 = 0.4976

Gain

|H| = 0.4976/√(1 + 0.4976²) = 0.4455
20·log₁₀(0.4455) = −7.023 dB

Answer

|H| = −7.023 dB
12.Bode plots & margins

Answer: PM = 77.86 °

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.

Gain crossover

ω⁴ + 100ω² − 484 = 0
ω² = (−100 + √(10000 + 4·484))/2 = 4.626
ωc = 2.151 rad/s

Phase margin

∠L(jωc) = −90° − atan(2.151/10) = −102.1°
PM = 180° + ∠L = 77.86°

Answer

PM = 77.86°

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