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 = 0.75f(0.75) = ?

Find f(t) for F(s) = (5s + 26)/(s² + 4s + 4) and evaluate f(0.75).

F(s) = (5s + 26)/(s² + 4s + 4)t = 0.75 s
f(0.75) =
2.Laplace transformsHard
t (s)y(t)t = 0.1y(0.1) = ?

A system obeys y″ + 9y′ + 20y = 10·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(0.1).

y″ + 9y′ + 20y = 10·u(t)u(t) = 1y(0) = 0, y′(0) = 0t = 0.1 s
y(0.1) =
3.Block diagramsMedium
R(s)+40s + 2G(s)C(s)2H−

A plant G(s) = 40/(s + 2) is controlled in a negative feedback loop whose sensor has the gain H = 2. Find the time constant τ of the closed loop.

G(s) = 40/(s + 2)H = 2
τ = s
4.Block diagramsHard
R(s)+Kcontroller+D(s)+1s + 2plantC(s)−

A proportional controller K drives the plant G(s) = 1/(s + 2) in a unity negative feedback loop. A disturbance step of size 1 enters at the plant input while R = 0. Find the smallest gain K that keeps the steady-state output at or below 0.005.

G(s) = 1/(s + 2)D = 1css ≤ 0.005
K =
5.Step responseMedium
t (s)y(t)3

A system has the transfer function G(s) = 99/(s² + 7s + 33) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.

G(s) = 99/(s² + 7s + 33)
σ =
6.Step responseHard
R(s)+285s(s + 4)G(s)C(s)−

A unity negative feedback loop has G(s) = 285/(s(s + 4)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 285/(s(s + 4))
ts = s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K = 601s(s + 4)(s + 6)C(s)−

The loop in the diagram runs with K = 60 and L(s) = K/(s(s + 4)(s + 6)). By what factor m can the gain be raised before the closed loop becomes unstable?

L(s) = K/(s(s + 4)(s + 6))K = 60
m =
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 + 8)/(s(s + 2)) leaves the real axis between s = 0 and s = −2, circles round and joins the real axis again to the left of the zero. Find the break-away point sb.

L(s) = K(s + 8)/(s(s + 2))
sb =
9.Steady-state error & PID controlMedium
r(t) = 1 + 3t+4(s + 2)s(s + 1)(s + 2)C(s)−

The input r(t) = 1 + 3t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 4(s + 2)/(s(s + 1)(s + 2)). Find the steady-state error ess.

G(s) = 4(s + 2)/(s(s + 1)(s + 2))r(t) = 1 + 3t
ess =
10.Steady-state error & PID controlHard
R(s)+K = 504s(s + 5)C(s)−

A proportional controller K = 50 drives the plant G(s) = 4/(s(s + 5)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 4/(s(s + 5))K = 50
ts = s
11.Bode plots & marginsMedium
|H| (dB)0.11101001000−60−40−2002040ω (rad/s)ω = 10

Find the exact magnitude |G(jω)| of G(s) = 1872/((s + 9)(s + 13)) at ω = 10 rad/s, in dB.

G(s) = 1872/((s + 9)(s + 13))ω = 10 rad/s
|G(jω)| = dB
12.Bode plots & marginsHard
|H| (dB)0.10.111101010010010001000−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​

A unity-feedback loop has the open-loop transfer function L(s) = 216/(s(s + 4)). Find the phase margin PM in degrees.

L(s) = 216/(s(s + 4))
PM = °

3Solutions

1.Laplace transforms

Answer: f(0.75) = 3.793

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 + 4s + 4 = (s + 2)²

Partial fractions

5s + 26 = 5(s + 2) + 16
F(s) = 5/(s + 2) + 16/(s + 2)²

Inverse transform

f(t) = 5·e−2t + 16·t·e−2t
f(0.75) = (5 + 16·0.75)·e−2·0.75 = 3.793

Answer

f(0.75) = 3.793
2.Laplace transforms

Answer: y(0.1) = 0.03726

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) = 10/(s(s + 4)(s + 5))

Partial fractions

A = 10/(4·5) = 0.5
B = 10/((−4)·1) = −2.5
C = 10/((−5)·(−1)) = 2

Output

y(t) = 0.5 − 2.5·e−4t + 2·e−5t
y(0.1) = 0.03726

Answer

y(0.1) = 0.03726
3.Block diagrams

Answer: τ = 0.0122 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).

Closed-loop transfer function

T(s) = G/(1 + G·H) = 40/(s + 2 + 80) = 40/(s + 82)

Result

τ = 1/82 = 0.0122 s

Answer

τ = 0.0122 s
4.Block diagrams

Answer: K = 198

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).

Disturbance response

css = D·G(0)/(1 + K·G(0)) = 1·1/(2 + 1K)

Solve for K

2 + 1K = 1·1/0.005 = 200
K = (200 − 2)/1 = 198

Answer

K = 198
5.Step response

Answer: σ = −3.5

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 = 33 → ωₙ = 5.745 rad/s
2ζωₙ = 7 → ζ = 7/(2·5.745) = 0.6093

Result

σ = −ζωₙ = −7/2 = −3.5

Answer

σ = −3.5
6.Step response

Answer: ts = 2 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) = 285/(s² + 4s + 285): ωₙ = 16.88 rad/s, ζ = 0.1185

Settling time

ζωₙ = 4/2 = 2
ts ≈ 4/(ζωₙ) = 2 s

Answer

ts = 2 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: m = 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.

Routh array

s³ + 10s² + 24s + K = 0
s³: 1, 24
s²: 10, K
s¹: (10·24 − K)/10
s⁰: K

Largest stable gain

Kmax = 10·24 = 240

Factor

m = 240/60 = 4

Answer

m = 4

This factor is the gain margin as a ratio: 12.04 dB.

8.Stability: Routh–Hurwitz & root locus

Answer: sb = −1.072

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 + 2s)/(s + 8)

Stationary points

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

(2s + 2)(s + 8) − (s² + 2s) = s² + 16s + 16 = 0
s = −8 ± √48 = −1.072, −14.93

Break-away point

sb = −1.072

Answer

sb = −1.072

Between the two break points the locus is a circle centred on the zero, with radius √48 = 6.928.

9.Steady-state error & PID control

Answer: ess = 0.75

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) = 4

Step part

Kp = ∞ → estep = 0

Ramp error

ess = 3/4 = 0.75

Answer

ess = 0.75
10.Steady-state error & PID control

Answer: ts = 1.6 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) + 50·4 = s² + 5s + 200

ωₙ and ζ

ωₙ = √200 = 14.14 rad/s
ζ = 5/(2·14.14) = 0.1768

Settling time

ζωₙ = 5/2 = 2.5
ts ≈ 4/2.5 = 1.6 s

Answer

ts = 1.6 s

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

11.Bode plots & margins

Answer: |G(jω)| = 18.57 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(j10)| = 1872/(√(10² + 9²)·√(10² + 13²)) = 1872/(13.45·16.4) = 8.484

Result

20·log₁₀(8.484) = 18.57 dB

Answer

|G(jω)| = 18.57 dB
12.Bode plots & margins

Answer: PM = 15.5 °

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 + 16ω² − 46656 = 0
ω² = (−16 + √(256 + 4·46656))/2 = 208.1
ωc = 14.43 rad/s

Phase margin

∠L(jωc) = −90° − atan(14.43/4) = −164.5°
PM = 180° + ∠L = 15.5°

Answer

PM = 15.5°

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