Control Engineering 1: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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−2−4−6−8−102

Find f(t) for F(s) = (4s + 39)/(s² + 4s + 8) by completing the square, and evaluate f(0.6).

F(s) = (4s + 39)/(s² + 4s + 8)t = 0.6 s
f(0.6) =
2.Laplace transformsHard
t (s)y(t)t = 0.25y(0.25) = ?

A system obeys y″ + 7y′ + 10y = 10·u(t), with y(0) = 1 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′ + 10y = 10·u(t)y(0) = 1y′(0) = 0u(t) = 10t = 0.25 s
y(0.25) =
3.Block diagramsMedium
R(s)+48s(s + 2)G(s)C(s)−

A unity negative feedback loop has G(s) = 48/(s(s + 2)). The reference is a ramp r(t) = 1·t. Find the steady-state error ess.

G(s) = 48/(s(s + 2))r(t) = 1·t
ess =
4.Block diagramsHard
R(s)+11s + 4G₁+3s + 2G₂C(s)2H₁−1H₂−

Reduce the block diagram with G₁ = 11/(s + 4), G₂ = 3/(s + 2), an inner negative feedback H₁ = 2 around G₂ and an outer negative feedback H₂ = 1. Find the DC gain T(0) of C(s)/R(s).

G₁ = 11/(s + 4)G₂ = 3/(s + 2)H₁ = 2H₂ = 1
T(0) =
5.Step responseMedium
t (s)y(t)2

A system has the transfer function G(s) = 138/(s² + 5s + 69) and receives a unit step input. Find the natural frequency ωn in rad/s.

G(s) = 138/(s² + 5s + 69)
ωn = rad/s
6.Step responseHard
R(s)+Ks(s + 8)G(s)C(s)−

A unity negative feedback loop has G(s) = K/(s(s + 8)). Find the gain K that gives a step response with 25 % overshoot.

G(s) = K/(s(s + 8))%OS = 25 %
K =
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K = 41s(s + 1)(s + 3)C(s)−

The loop in the diagram runs with K = 4 and L(s) = K/(s(s + 1)(s + 3)). By what factor m can the gain be raised before the closed loop becomes unstable?

L(s) = K/(s(s + 1)(s + 3))K = 4
m =
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 + 1)(s + 4)C(s)−

A unity-feedback loop has G(s) = K/((s + 1)(s + 4)). Find the gain K that gives a steady-state error of 10 % for a step input.

G(s) = K/((s + 1)(s + 4))ess = 10 %
K =
10.Steady-state error & PID controlHard
R(s)+K = 165s(s + 8)C(s)−

A proportional controller K = 16 drives the plant G(s) = 5/(s(s + 8)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 5/(s(s + 8))K = 16
ts = s
11.Bode plots & marginsMedium
vin​C = 47 nFR = 2.2 kΩvout​

The RC high-pass filter shown has R = 2.2 kΩ, C = 47 nF. A sinusoidal input of 310 Hz is applied. Find the gain |H| = |vout/vin| in dB.

R = 2.2 kΩC = 47 nFf = 310 Hz
|H| = dB
12.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 33/(s(s + 2)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 33/(s(s + 2))
ωc = rad/s

3Solutions

1.Laplace transforms

Answer: f(0.6) = 4.788

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 + 4s + 8 = (s + 2)² + 4 = (s + 2)² + 2²

Match the numerator

F(s) = 4·(s + 2)/((s + 2)² + 2²) + (31/2)·2/((s + 2)² + 2²)

Inverse transform

f(t) = e−2t·(4·cos 2t + 15.5·sin 2t)
f(0.6) = e−2·0.6·(4·cos(2·0.6) + 15.5·sin(2·0.6)) = 4.788

Answer

f(0.6) = 4.788
2.Laplace transforms

Answer: y(0.25) = 2.621

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 + 10)·Y(s) = 1s + 7 + 100/s
Y(s) = (s² + 7s + 100)/(s(s + 2)(s + 5))

Partial fractions

A = 100/(2·5) = 10
B = 90/((−2)·3) = −15
C = 90/((−5)·(−3)) = 6

Result

y(t) = 10 − 15·e−2t + 6·e−5t
y(0.25) = 2.621

Answer

y(0.25) = 2.621

Check: y(0) = A + B + C = 1.

3.Block diagrams

Answer: ess = 0.04167

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).

Velocity error constant

Kᵥ = lims→0 s·G(s) = 48/2 = 24

Steady-state error

ess = A/Kᵥ = 1/24 = 0.04167

Answer

ess = 0.04167
4.Block diagrams

Answer: T(0) = 0.5077

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) = 3/(s + 8)

Outer loop

T(s) = G₁Gᵢ/(1 + G₁GᵢH₂) = 33/((s + 4)(s + 8) + 33)

DC gain

T(0) = 33/(4·8 + 33) = 33/65 = 0.5077

Answer

T(0) = 0.5077
5.Step response

Answer: ωn = 8.307 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 = 69 → ωₙ = 8.307 rad/s
2ζωₙ = 5 → ζ = 5/(2·8.307) = 0.301

Result

ωₙ = 8.307 rad/s

Answer

ωn = 8.307 rad/s
6.Step response

Answer: K = 98.17

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 + 8s + K)

Damping ratio from the overshoot

ζ = −ln(0.25)/√(π² + ln²(0.25)) = 0.4037

Gain

ωₙ = 8/(2ζ) = 9.908 rad/s
K = ωₙ² = 98.17

Answer

K = 98.17
7.Stability: Routh–Hurwitz & root locus

Answer: m = 3

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 + 4s² + 3s + K = 0
s³: 1, 3
s²: 4, K
s¹: (4·3 − K)/4
s⁰: K

Largest stable gain

Kmax = 4·3 = 12

Factor

m = 12/4 = 3

Answer

m = 3

This factor is the gain margin as a ratio: 9.542 dB.

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 = 36

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) = 0.1 → Kp = 9

Gain

K/4 = 9 → K = 36

Answer

K = 36

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

10.Steady-state error & PID control

Answer: ts = 1 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 + 8) + 16·5 = s² + 8s + 80

ωₙ and ζ

ωₙ = √80 = 8.944 rad/s
ζ = 8/(2·8.944) = 0.4472

Settling time

ζωₙ = 8/2 = 4
ts ≈ 4/4 = 1 s

Answer

ts = 1 s

The settling time does not depend on K here: raising K only makes the response oscillate faster.

11.Bode plots & margins

Answer: |H| = −14.09 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/(2200·4.7 × 10⁻⁸) = 9671 rad/s
fc = 1539 Hz → f/fc = 310/1539 = 0.2014

Gain

|H| = 0.2014/√(1 + 0.2014²) = 0.1974
20·log₁₀(0.1974) = −14.09 dB

Answer

|H| = −14.09 dB
12.Bode plots & margins

Answer: ωc = 5.573 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.

Gain crossover

ω⁴ + 4ω² − 1089 = 0
ω² = (−4 + √(16 + 4·1089))/2 = 31.06
ωc = 5.573 rad/s

Answer

ωc = 5.573 rad/s

ExamProblems · examproblems.com/courses/control-engineering-1