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) = 30/(s(s + 3)(s + 6)) by partial fractions, and evaluate f(0.1).

F(s) = 30/(s(s + 3)(s + 6))t = 0.1 s
f(0.1) =
2.Laplace transformsHard
t (s)y(t)t = 1y(1) = ?

A system obeys y″ + 4y′ + 3y = 9·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 1 for t ≥ 0. Use the Laplace transform to find the output y(1).

y″ + 4y′ + 3y = 9·u(t)u(t) = 1y(0) = 0, y′(0) = 0t = 1 s
y(1) =
3.Block diagramsMedium
R(s)+22s(s + 3)G(s)C(s)−

A unity negative feedback loop has G(s) = 22/(s(s + 3)). The reference is a ramp r(t) = 5·t. Find the steady-state error ess.

G(s) = 22/(s(s + 3))r(t) = 5·t
ess =
4.Block diagramsHard
R(s)+64s(s + 8)G(s)C(s)2H−

The plant G(s) = 64/(s(s + 8)) is in a negative feedback loop with sensor gain H = 2. The closed-loop poles are complex. Find their real part σ.

G(s) = 64/(s(s + 8))H = 2
σ =
5.Step responseMedium
t (s)y(t)1?

A system has the transfer function G(s) = 44/(s² + 9s + 44) and receives a unit step input. Find the percent overshoot %OS of the step response.

G(s) = 44/(s² + 9s + 44)
%OS = %
6.Step responseHard
t (s)y(t)125 %

A second-order system without zeros must have a step response with 25 % overshoot. Find the damping ratio ζ.

%OS = 25 %
ζ =
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1(s + 1)(s + 5)(s + 6)C(s)−

A unity-feedback loop has L(s) = K/((s + 1)(s + 5)(s + 6)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/((s + 1)(s + 5)(s + 6))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−7−8−9−101j1j2

The root locus of L(s) = K/(s(s + 6)(s + 8)) leaves the real axis between s = 0 and s = −6. Find the break-away point sb.

L(s) = K/(s(s + 6)(s + 8))
sb =
9.Steady-state error & PID controlMedium
R(s)+K(s + 2)(s + 4)C(s)−

A unity-feedback loop has G(s) = K/((s + 2)(s + 4)). Find the gain K that gives a steady-state error of 20 % for a step input.

G(s) = K/((s + 2)(s + 4))ess = 20 %
K =
10.Steady-state error & PID controlHard
R(s)+K = 104s(s + 5)C(s)−

A proportional controller K = 10 drives the plant G(s) = 4/(s(s + 5)) in a unity-feedback loop. Find the peak time tp of the step response.

G(s) = 4/(s(s + 5))K = 10
tp = s
11.Bode plots & marginsMedium
vin​C = 10 nFR = 10 kΩvout​

The RC high-pass filter shown has R = 10 kΩ, C = 10 nF. A sinusoidal input of 320 Hz is applied. Find the gain |H| = |vout/vin| in dB.

R = 10 kΩC = 10 nFf = 320 Hz
|H| = 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 + 7)(s + 8)). Find the gain K that gives a gain margin of 12 dB.

L(s) = K/(s(s + 7)(s + 8))GM = 12 dB
K =

3Solutions

1.Laplace transforms

Answer: f(0.1) = 0.112

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 + 6)
A = 30/(3·6) = 1.667
B = 30/((−3)·3) = −3.333
C = 30/((−6)·(−3)) = 1.667

Inverse transform

f(t) = 1.667 − 3.333·e−3t + 1.667·e−6t

Evaluate

f(0.1) = 1.667 − 3.333·e−3·0.1 + 1.667·e−6·0.1 = 0.112

Answer

f(0.1) = 0.112
2.Laplace transforms

Answer: y(1) = 1.419

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) = 9/(s(s + 1)(s + 3))

Partial fractions

A = 9/(1·3) = 3
B = 9/((−1)·2) = −4.5
C = 9/((−3)·(−2)) = 1.5

Output

y(t) = 3 − 4.5·e−t + 1.5·e−3t
y(1) = 1.419

Answer

y(1) = 1.419
3.Block diagrams

Answer: ess = 0.6818

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) = 22/3 = 7.333

Steady-state error

ess = A/Kᵥ = 5/7.333 = 0.6818

Answer

ess = 0.6818
4.Block diagrams

Answer: σ = −4

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

s(s + 8) + 64·2 = 0 → s² + 8s + 128 = 0

Roots

s = −4 ± j·√(128 − 16) = −4 ± j10.58
σ = −4

Answer

σ = −4
5.Step response

Answer: %OS = 5.497 %

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 = 9 → ζ = 9/(2·6.633) = 0.6784

Result

%OS = 100·e−ζπ/√(1−ζ²) = 100·e−2.901 = 5.497 %

Answer

%OS = 5.497 %
6.Step response

Answer: ζ = 0.4037

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(0.25)/√(π² + ln²(0.25)) = 1.386/√(9.87 + 1.922) = 0.4037

Answer

ζ = 0.4037
7.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 462

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 + 41s + 30 + K = 0

Routh array

s³: 1, 41
s²: 12, 30 + K
s¹: (12·41 − 30 − K)/12
s⁰: 30 + K

Condition

The s¹ entry must stay positive.

K < 12·41 − 30 = 462

Answer

Kmax = 462

At the limit the loop oscillates at ω = √41 = 6.403 rad/s.

8.Stability: Routh–Hurwitz & root locus

Answer: sb = −2.263

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 + 6)(s + 8) = −(s³ + 14s² + 48s)

Stationary points

dK/ds = −(3s² + 28s + 48) = 0
s = (−28 ± √(784 − 576))/6 = −2.263, −7.07

Pick the point on the locus

The locus lies on the real axis between 0 and −6; the other root is not on the locus.

sb = −2.263

Answer

sb = −2.263
9.Steady-state error & PID control

Answer: K = 32

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) = 0.2 → Kp = 4

Gain

K/8 = 4 → K = 32

Answer

K = 32

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 = 0.5408 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 + 5) + 10·4 = s² + 5s + 40

ωₙ and ζ

ωₙ = √40 = 6.325 rad/s
ζ = 5/(2·6.325) = 0.3953

Peak time

ωd = 6.325·√(1 − 0.3953²) = 5.809 rad/s
tp = π/5.809 = 0.5408 s

Answer

tp = 0.5408 s
11.Bode plots & margins

Answer: |H| = −14.11 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/(10000·1 × 10⁻⁸) = 10000 rad/s
fc = 1592 Hz → f/fc = 320/1592 = 0.2011

Gain

|H| = 0.2011/√(1 + 0.2011²) = 0.1971
20·log₁₀(0.1971) = −14.11 dB

Answer

|H| = −14.11 dB
12.Bode plots & margins

Answer: K = 211

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/7)·(ω/8) = 1 → ω180 = √(7·8) = 7.483 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 49)·√(ω² + 64)) = K/(7·8·(7 + 8)) = K/840

Required gain

840/K = 1012/20 = 3.981
K = 840/3.981 = 211

Answer

K = 211

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