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 = 0.5f(0.5) = ?

Find f(t) for F(s) = 45/(s(s + 3)(s + 5)) by partial fractions, and evaluate f(0.5).

F(s) = 45/(s(s + 3)(s + 5))t = 0.5 s
f(0.5) =
2.Laplace transformsHard
t (s)y(t)t = 0.1y(0.1) = ?

Solve y″ + 7y′ + 12y = 0 with y(0) = 7 and y′(0) = 15 using the Laplace transform, and find y(0.1).

y″ + 7y′ + 12y = 0y(0) = 7y′(0) = 15t = 0.1 s
y(0.1) =
3.Block diagramsMedium
R(s)+18K+4s + 1G(s)C(s)0.5H₁−−

The plant G(s) = 4/(s + 1) has an inner (minor) negative feedback loop with gain H₁ = 0.5. A controller gain K = 18 and an outer unity negative feedback loop close the system. Find the closed-loop DC gain T(0).

G(s) = 4/(s + 1)H₁ = 0.5K = 18
T(0) =
4.Block diagramsHard
R(s)+Ks(s + 4)(s + 5)G(s)C(s)−

The plant G(s) = K/(s(s + 4)(s + 5)) is in a unity negative feedback loop. At the largest stable gain the loop oscillates. Find the frequency of oscillation ω in rad/s.

G(s) = K/(s(s + 4)(s + 5))unity negative feedback
ω = rad/s
5.Step responseMedium
t (s)y(t)1

A system has the transfer function G(s) = 44/(s² + 8s + 44) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.

G(s) = 44/(s² + 8s + 44)
σ =
6.Step responseHard
R(s)+45s(s + 5)G(s)C(s)−

A unity negative feedback loop has G(s) = 45/(s(s + 5)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 45/(s(s + 5))
ts = s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+4951s(s + 5)(s + 6)C(s)−

The loop in the diagram has negative feedback and L(s) = 495/(s(s + 5)(s + 6)). How many closed-loop poles N lie in the right half-plane?

L(s) = 495/(s(s + 5)(s + 6))
N =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6j2j4

The root locus of L(s) = K/((s + 1)(s + 3)(s + 5)) is sketched. Find the frequency ω at which the locus crosses the imaginary axis.

L(s) = K/((s + 1)(s + 3)(s + 5))
ω = rad/s
9.Steady-state error & PID controlMedium
r(t) = 5 + 5t+2s(s + 4)C(s)−

The input r(t) = 5 + 5t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 2/(s(s + 4)). Find the steady-state error ess.

G(s) = 2/(s(s + 4))r(t) = 5 + 5t
ess =
10.Steady-state error & PID controlHard
R(s)+K = 1010s(s + 6)C(s)−

A proportional controller K = 10 drives the plant G(s) = 10/(s(s + 6)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 10/(s(s + 6))K = 10
ts = s
11.Bode plots & marginsMedium
vin​C = 2.2 nFR = 2.2 kΩvout​

The RC high-pass filter shown has R = 2.2 kΩ, C = 2.2 nF. A sinusoidal input of 66 kHz is applied. Find the gain |H| = |vout/vin| in dB.

R = 2.2 kΩC = 2.2 nFf = 66 kHz
|H| = dB
12.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−270−180−900ω180​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 908/(s(s + 8)(s + 10)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 908/(s(s + 8)(s + 10))
ω180 = rad/s

3Solutions

1.Laplace transforms

Answer: f(0.5) = 1.696

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).

Partial fractions

F(s) = A/s + B/(s + 3) + C/(s + 5)
A = 45/(3·5) = 3
B = 45/((−3)·2) = −7.5
C = 45/((−5)·(−2)) = 4.5

Inverse transform

f(t) = 3 − 7.5·e−3t + 4.5·e−5t

Evaluate

f(0.5) = 3 − 7.5·e−3·0.5 + 4.5·e−5·0.5 = 1.696

Answer

f(0.5) = 1.696
2.Laplace transforms

Answer: y(0.1) = 7.724

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 − 7s − 15 + 7(sY − 7) + 12Y = 0
Y(s) = (7s + 64)/((s + 3)(s + 4))

Partial fractions

A = 43/1 = 43
B = 36/(−1) = −36

Result

y(t) = 43·e−3t − 36·e−4t
y(0.1) = 7.724

Answer

y(0.1) = 7.724

Check: y(0) = A + B = 7.

3.Block diagrams

Answer: T(0) = 0.96

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 = G/(1 + G·H₁) = 4/(s + 1 + 4·0.5) = 4/(s + 3)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 72/(s + 3 + 72) = 72/(s + 75)

Result

T(0) = 72/75 = 0.96

Answer

T(0) = 0.96
4.Block diagrams

Answer: ω = 4.472 rad/s

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

s³ + 9s² + 20s + K = 0

Routh array

s³ | 1 20
s² | 9 K
s¹ | (9·20 − K)/9
s⁰ | K

Stability limit

All entries of the first column must be positive.

Kmax = 9·20 = 180

Frequency of oscillation

9·(jω)² + 180 = 0 → ω² = 180/9 = 20
ω = 4.472 rad/s

Answer

ω = 4.472 rad/s
5.Step response

Answer: σ = −4

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 = 44 → ωₙ = 6.633 rad/s
2ζωₙ = 8 → ζ = 8/(2·6.633) = 0.603

Result

σ = −ζωₙ = −8/2 = −4

Answer

σ = −4
6.Step response

Answer: ts = 1.6 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).

Closed-loop transfer function

T(s) = 45/(s² + 5s + 45): ωₙ = 6.708 rad/s, ζ = 0.3727

Settling time

ζωₙ = 5/2 = 2.5
ts ≈ 4/(ζωₙ) = 1.6 s

Answer

ts = 1.6 s

For this loop ζωₙ = a/2 whatever K is: raising K adds overshoot but does not shorten the settling time.

7.Stability: Routh–Hurwitz & root locus

Answer: N = 2

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 + 11s² + 30s + 495 = 0

Routh array

s³: 1, 30
s²: 11, 495
s¹: −15
s⁰: 495

Sign changes in the first column

1 → 11 → −15 → 495
N = 2

Answer

N = 2

The gain is above the limit K = 330, so a complex pair has crossed into the right half-plane.

8.Stability: Routh–Hurwitz & root locus

Answer: ω = 4.796 rad/s

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.

Critical gain

s³ + 9s² + 23s + 15 + K = 0
s¹: (9·23 − 15 − K)/9 = 0 → K = 192

Crossing frequency

9s² + 15 + 192 = 9s² + 207 = 0 → s² = −23
ω = √23 = 4.796 rad/s

Answer

ω = 4.796 rad/s
9.Steady-state error & PID control

Answer: ess = 10

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.

Velocity constant

Kv = lim s·G(s) = 0.5

Step part

Kp = ∞ → estep = 0

Ramp error

ess = 5/0.5 = 10

Answer

ess = 10
10.Steady-state error & PID control

Answer: ts = 1.333 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) + 10·10 = s² + 6s + 100

ωₙ and ζ

ωₙ = √100 = 10 rad/s
ζ = 6/(2·10) = 0.3

Settling time

ζωₙ = 6/2 = 3
ts ≈ 4/3 = 1.333 s

Answer

ts = 1.333 s

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

11.Bode plots & margins

Answer: |H| = −0.963 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/(2200·2.2 × 10⁻⁹) = 206612 rad/s
fc = 32883 Hz → f/fc = 66000/32883 = 2.007

Gain

|H| = 2.007/√(1 + 2.007²) = 0.8951
20·log₁₀(0.8951) = −0.963 dB

Answer

|H| = −0.963 dB
12.Bode plots & margins

Answer: ω180 = 8.944 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/8)·(ω/10) = 1 → ω180 = √(8·10) = 8.944 rad/s

Answer

ω180 = 8.944 rad/s

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