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−2−424

Find f(t) for F(s) = (5s + 14)/(s² + 8s + 32) by completing the square, and evaluate f(0.4).

F(s) = (5s + 14)/(s² + 8s + 32)t = 0.4 s
f(0.4) =
2.Laplace transformsHard
t (s)y(t)t = 0.25y(0.25) = ?

Solve y″ + 7y′ + 6y = 0 with y(0) = −3 and y′(0) = 6 using the Laplace transform, and find y(0.25).

y″ + 7y′ + 6y = 0y(0) = −3y′(0) = 6t = 0.25 s
y(0.25) =
3.Block diagramsMedium
R(s)+9(s + 4)(s + 1)(s + 3)G(s)C(s)−

A unity negative feedback loop has G(s) = 9(s + 4)/((s + 1)(s + 3)). A step of size 5 is applied at the reference. Find the steady-state error ess.

G(s) = 9(s + 4)/((s + 1)(s + 3))R = 5
ess =
4.Block diagramsHard
R(s)+K(s + 1)(s + 4)(s + 8)G(s)C(s)−

The plant G(s) = K/((s + 1)(s + 4)(s + 8)) is in a unity negative feedback loop. Use the Routh–Hurwitz criterion to find the largest gain Kmax for which the closed loop is stable.

G(s) = K/((s + 1)(s + 4)(s + 8))unity negative feedback
Kmax =
5.Step responseMedium
t (s)y(t)1tp​ = ?

A system has the transfer function G(s) = 99/(s² + 7s + 99) and receives a unit step input. Find the peak time tp of the step response.

G(s) = 99/(s² + 7s + 99)
tp = s
6.Step responseHard
R(s)+Ks(s + 20)G(s)C(s)−

A unity negative feedback loop has G(s) = K/(s(s + 20)). Find the gain K that gives a step response with 5 % overshoot.

G(s) = K/(s(s + 20))%OS = 5 %
K =
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+201s(s + 5)(s + 8)C(s)+

The loop in the diagram uses positive feedback, with L(s) = 20/(s(s + 5)(s + 8)). How many closed-loop poles N lie in the right half-plane?

L(s) = 20/(s(s + 5)(s + 8))
N =
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 + 4)/(s(s + 1)) leaves the real axis between s = 0 and s = −1, circles round and joins the real axis again to the left of the zero. Find the break-away point sb.

L(s) = K(s + 4)/(s(s + 1))
sb =
9.Steady-state error & PID controlMedium
R(s)+K = 555s + 1C(s)−

A process G(s) = 5/(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) = 5/(5s + 1)K = 5
ess = %
10.Steady-state error & PID controlHard
R(s)+K1s(s + 4)C(s)−

A proportional controller K drives the plant G(s) = 1/(s(s + 4)) in a unity-feedback loop. Find the gain K that gives the closed loop a damping ratio ζ = 0.707.

G(s) = 1/(s(s + 4))ζ = 0.707
K =
11.Bode plots & marginsMedium
|H| (dB)0.11101001000−80−60−40−20020ω (rad/s)ω = 15

Find the exact magnitude |G(jω)| of G(s) = 720/((s + 5)(s + 18)) at ω = 15 rad/s, in dB.

G(s) = 720/((s + 5)(s + 18))ω = 15 rad/s
|G(jω)| = 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) = 30/(s(s + 1)(s + 6)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 30/(s(s + 1)(s + 6))
ω180 = rad/s

3Solutions

1.Laplace transforms

Answer: f(0.4) = −0.3322

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 + 32 = (s + 4)² + 16 = (s + 4)² + 4²

Match the numerator

F(s) = 5·(s + 4)/((s + 4)² + 4²) + (−6/4)·4/((s + 4)² + 4²)

Inverse transform

f(t) = e−4t·(5·cos 4t − 1.5·sin 4t)
f(0.4) = e−4·0.4·(5·cos(4·0.4) − 1.5·sin(4·0.4)) = −0.3322

Answer

f(0.4) = −0.3322
2.Laplace transforms

Answer: y(0.25) = −2.003

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 − (−3)s − 6 + 7(sY − (−3)) + 6Y = 0
Y(s) = (−3s − 15)/((s + 1)(s + 6))

Partial fractions

A = −12/5 = −2.4
B = 3/(−5) = −0.6

Result

y(t) = −2.4·e−t − 0.6·e−6t
y(0.25) = −2.003

Answer

y(0.25) = −2.003

Check: y(0) = A + B = −3.

3.Block diagrams

Answer: ess = 0.3846

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).

Position error constant

Kₚ = G(0) = 9·4/(1·3) = 12

Steady-state error

ess = A/(1 + Kₚ) = 5/(1 + 12) = 0.3846

Answer

ess = 0.3846
4.Block diagrams

Answer: Kmax = 540

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 + 13s² + 44s + 32 + K = 0

Routh array

s³ | 1 44
s² | 13 32 + K
s¹ | (13·44 − 32 − K)/13
s⁰ | 32 + K

Stability limit

All entries of the first column must be positive.

Kmax = 13·44 − 32 = 540

Answer

Kmax = 540
5.Step response

Answer: tp = 0.3373 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 = 99 → ωₙ = 9.95 rad/s
2ζωₙ = 7 → ζ = 7/(2·9.95) = 0.3518

Result

ωd = 9.95·√(1 − 0.3518²) = 9.314 rad/s → tp = π/ωd = 0.3373 s

Answer

tp = 0.3373 s
6.Step response

Answer: K = 210

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 + 20s + K)

Damping ratio from the overshoot

ζ = −ln(0.05)/√(π² + ln²(0.05)) = 0.6901

Gain

ωₙ = 20/(2ζ) = 14.49 rad/s
K = ωₙ² = 210

Answer

K = 210
7.Stability: Routh–Hurwitz & root locus

Answer: N = 1

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 + 13s² + 40s − 20 = 0

Routh array

s³: 1, 40
s²: 13, −20
s¹: 41.54
s⁰: −20

Sign changes in the first column

1 → 13 → 41.54 → −20
N = 1

Answer

N = 1

Positive feedback makes the constant coefficient negative: one real pole always moves into the right half-plane.

8.Stability: Routh–Hurwitz & root locus

Answer: sb = −0.5359

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 + 1s)/(s + 4)

Stationary points

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

(2s + 1)(s + 4) − (s² + 1s) = s² + 8s + 4 = 0
s = −4 ± √12 = −0.5359, −7.464

Break-away point

sb = −0.5359

Answer

sb = −0.5359

Between the two break points the locus is a circle centred on the zero, with radius √12 = 3.464.

9.Steady-state error & PID control

Answer: ess = 3.846 %

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 = 5·5 = 25

Step error

ess = 100/(1 + 25) = 3.846 %

Answer

ess = 3.846 %
10.Steady-state error & PID control

Answer: K = 8.002

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 + 4) + 1K = s² + 4s + 1K

Natural frequency

2·0.707·ωₙ = 4 → ωₙ = 2.829 rad/s

Gain

1K = ωₙ² = 8.002 → K = 8.002

Answer

K = 8.002
11.Bode plots & margins

Answer: |G(jω)| = 5.772 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(j15)| = 720/(√(15² + 5²)·√(15² + 18²)) = 720/(15.81·23.43) = 1.943

Result

20·log₁₀(1.943) = 5.772 dB

Answer

|G(jω)| = 5.772 dB
12.Bode plots & margins

Answer: ω180 = 2.449 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/1)·(ω/6) = 1 → ω180 = √(1·6) = 2.449 rad/s

Answer

ω180 = 2.449 rad/s

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