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−5−10−15−20−25−30

Find f(t) for F(s) = (s + 30)/(s² + 8s + 17) by completing the square, and evaluate f(0.4).

F(s) = (s + 30)/(s² + 8s + 17)t = 0.4 s
f(0.4) =
2.Laplace transformsHard
t (s)y(t)t = 0.2y(0.2) = ?

Solve y″ + 4y′ + 3y = 0 with y(0) = −1 and y′(0) = −1 using the Laplace transform, and find y(0.2).

y″ + 4y′ + 3y = 0y(0) = −1y′(0) = −1t = 0.2 s
y(0.2) =
3.Block diagramsMedium
R(s)+Ks(s + 12)G(s)C(s)−

The plant G(s) = K/(s(s + 12)) is in a unity negative feedback loop. Find the gain K for which the closed loop is critically damped (two equal real poles).

G(s) = K/(s(s + 12))unity negative feedback
K =
4.Block diagramsHard
R(s)+23controller+D(s)+8s + 8plantC(s)−

A proportional controller K = 23 drives the plant G(s) = 8/(s + 8) in a unity negative feedback loop. A disturbance D enters at the plant input (it is added to the controller output). A reference step of size 10 and a disturbance step of size 1 are applied together. Find the steady-state output css.

K = 23G(s) = 8/(s + 8)D = 1R = 10
css =
5.Step responseMedium
t (s)y(t)1tp​ = ?

A system has the transfer function G(s) = 91/(s² + 13s + 91) and receives a unit step input. Find the peak time tp of the step response.

G(s) = 91/(s² + 13s + 91)
tp = s
6.Step responseHard
k = 5000 N/mc = 15 N·s/mm = 0.5 kgF = 100 Nx(t)

A mass m = 0.5 kg hangs on a spring k = 5000 N/m with a viscous damper c = 15 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 100 N is suddenly applied. Find the largest displacement xmax in mm.

m = 0.5 kgk = 5000 N/mc = 15 N·s/mF = 100 N
xmax = mm
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 2)(s + 8)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 2)(s + 8)). The gain is raised until the closed loop is on the edge of stability. Find the frequency ω of the sustained oscillation.

L(s) = K/(s(s + 2)(s + 8))
ω = rad/s
8.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−7−81j1j2

The root locus of L(s) = K(s + 7)/(s(s + 3)) leaves the real axis between s = 0 and s = −3, circles round and joins the real axis again to the left of the zero. Find the break-in point si.

L(s) = K(s + 7)/(s(s + 3))
si =
9.Steady-state error & PID controlMedium
R(s)+K = 510.5s + 1C(s)−

A process G(s) = 1/(0.5s + 1) is controlled by a proportional gain K = 5 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.

G(s) = 1/(0.5s + 1)K = 5
ess = %
10.Steady-state error & PID controlHard
R(s)+K10s(s + 10)C(s)−

A proportional controller K drives the plant G(s) = 10/(s(s + 10)) in a unity-feedback loop. Find the gain K that gives the closed loop a damping ratio ζ = 0.707.

G(s) = 10/(s(s + 10))ζ = 0.707
K =
11.Bode plots & marginsMedium
vin​C = 4.7 nFR = 22 kΩvout​

The RC high-pass filter shown has R = 22 kΩ, C = 4.7 nF. A sinusoidal input of 3.1 kHz is applied. Find the gain |H| = |vout/vin| in dB.

R = 22 kΩC = 4.7 nFf = 3.1 kHz
|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 + 8)(s + 9)). Find the gain K that gives a gain margin of 10 dB.

L(s) = K/(s(s + 8)(s + 9))GM = 10 dB
K =

3Solutions

1.Laplace transforms

Answer: f(0.4) = 2.23

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 + 8s + 17 = (s + 4)² + 1 = (s + 4)² + 1²

Match the numerator

F(s) = 1·(s + 4)/((s + 4)² + 1²) + (26/1)·1/((s + 4)² + 1²)

Inverse transform

f(t) = e−4t·(cos 1t + 26·sin 1t)
f(0.4) = e−4·0.4·(cos(1·0.4) + 26·sin(1·0.4)) = 2.23

Answer

f(0.4) = 2.23
2.Laplace transforms

Answer: y(0.2) = −1.089

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 − (−1)s − (−1) + 4(sY − (−1)) + 3Y = 0
Y(s) = (−s − 5)/((s + 1)(s + 3))

Partial fractions

A = −4/2 = −2
B = −2/(−2) = 1

Result

y(t) = −2·e−t + e−3t
y(0.2) = −1.089

Answer

y(0.2) = −1.089

Check: y(0) = A + B = −1.

3.Block diagrams

Answer: K = 36

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

1 + K/(s(s + 12)) = 0 → s² + 12s + K = 0

Repeated root

A quadratic s² + βs + γ has a double root when β² = 4γ.

12² = 4K → K = 144/4 = 36

Answer

K = 36
Both closed-loop poles are then at s = −6.
4.Block diagrams

Answer: css = 9.625

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).

Transfer functions

C/D = G/(1 + K·G) = 8/(s + 8 + 184)
C/R = K·G/(1 + K·G) = 184/(s + 8 + 184)

Steady state

css = 10·184/192 + 1·8/192 = 9.625

Answer

css = 9.625

A larger K reduces the effect of the disturbance, but never to zero with a proportional controller.

5.Step response

Answer: tp = 0.4499 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 = 91 → ωₙ = 9.539 rad/s
2ζωₙ = 13 → ζ = 13/(2·9.539) = 0.6814

Result

ωd = 9.539·√(1 − 0.6814²) = 6.982 rad/s → tp = π/ωd = 0.4499 s

Answer

tp = 0.4499 s
6.Step response

Answer: xmax = 32.42 mm

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).

Natural frequency and damping ratio

ωₙ = √(k/m) = √(5000/0.5) = 100 rad/s
ζ = c/(2√(km)) = 15/(2·√(5000·0.5)) = 0.15

Result

xst = F/k = 100/5000 m = 20 mm
%OS = 62.09 %
xmax = 20·(1 + 0.6209) = 32.42 mm

Answer

xmax = 32.42 mm
7.Stability: Routh–Hurwitz & root locus

Answer: ω = 4 rad/s

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 + 16s + K = 0
s³: 1, 16
s²: 10, K
s¹: (10·16 − K)/10
s⁰: K

Critical gain

K = 10·16 = 160

Auxiliary equation

At the critical gain the s¹ row is zero; the s² row gives the poles on the imaginary axis.

10s² + 160 = 0 → s = ±j√16
ω = 4 rad/s

Answer

ω = 4 rad/s
8.Stability: Routh–Hurwitz & root locus

Answer: si = −12.29

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 + 3s)/(s + 7)

Stationary points

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

(2s + 3)(s + 7) − (s² + 3s) = s² + 14s + 21 = 0
s = −7 ± √28 = −1.708, −12.29

Break-in point

si = −12.29

Answer

si = −12.29

Between the two break points the locus is a circle centred on the zero, with radius √28 = 5.292.

9.Steady-state error & PID control

Answer: ess = 16.67 %

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.

Position constant

Kp = 5·1 = 5

Step error

ess = 100/(1 + 5) = 16.67 %

Answer

ess = 16.67 %
10.Steady-state error & PID control

Answer: K = 5.002

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 + 10) + 10K = s² + 10s + 10K

Natural frequency

2·0.707·ωₙ = 10 → ωₙ = 7.072 rad/s

Gain

10K = ωₙ² = 50.02 → K = 5.002

Answer

K = 5.002
11.Bode plots & margins

Answer: |H| = −0.957 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/(22000·4.7 × 10⁻⁹) = 9671 rad/s
fc = 1539 Hz → f/fc = 3100/1539 = 2.014

Gain

|H| = 2.014/√(1 + 2.014²) = 0.8957
20·log₁₀(0.8957) = −0.957 dB

Answer

|H| = −0.957 dB
12.Bode plots & margins

Answer: K = 387.1

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/9) = 1 → ω180 = √(8·9) = 8.485 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 64)·√(ω² + 81)) = K/(8·9·(8 + 9)) = K/1224

Required gain

1224/K = 1010/20 = 3.162
K = 1224/3.162 = 387.1

Answer

K = 387.1

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