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) = 25/(s(s + 1)(s + 6)) by partial fractions, and evaluate f(0.75).

F(s) = 25/(s(s + 1)(s + 6))t = 0.75 s
f(0.75) =
2.Laplace transformsHard
t (s)y(t)t = 0.25y(0.25) = ?

Solve y″ + 5y′ + 4y = 0 with y(0) = −8 and y′(0) = −3 using the Laplace transform, and find y(0.25).

y″ + 5y′ + 4y = 0y(0) = −8y′(0) = −3t = 0.25 s
y(0.25) =
3.Block diagramsMedium
R(s)+20K+8s + 6G(s)C(s)0.5H₁−−

The plant G(s) = 8/(s + 6) has an inner (minor) negative feedback loop with gain H₁ = 0.5. A controller gain K = 20 and an outer unity negative feedback loop close the system. Find the closed-loop pole p.

G(s) = 8/(s + 6)H₁ = 0.5K = 20
p =
4.Block diagramsHard
R(s)+K(s + 5)(s + 6)(s + 7)G(s)C(s)−

The plant G(s) = K/((s + 5)(s + 6)(s + 7)) 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 + 5)(s + 6)(s + 7))unity negative feedback
Kmax =
5.Step responseMedium
t (s)y(t)1?

A system has the transfer function G(s) = 40/(s² + 10s + 40) and receives a unit step input. Find the percent overshoot %OS of the step response.

G(s) = 40/(s² + 10s + 40)
%OS = %
6.Step responseHard
t (s)y(t)1tp​ = 0.8 s10 %

The step response of a second-order system without zeros shows 10 % overshoot, and the first peak occurs at t = 0.8 s. Find the natural frequency ωn.

%OS = 10 %tp = 0.8 s
ωn = rad/s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 9s(s + 1)(s + 2)C(s)−

A unity-feedback loop has L(s) = K(s + 9)/(s(s + 1)(s + 2)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K(s + 9)/(s(s + 1)(s + 2))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−7−81j1j2sb​

The root locus of L(s) = K/((s + 3)(s + 7)) is shown. Find the gain K at the break-away point sb, where the two closed-loop poles meet on the real axis.

L(s) = K/((s + 3)(s + 7))
K =
9.Steady-state error & PID controlMedium
R(s)+K(s + 4)(s + 10)C(s)−

A unity-feedback loop has G(s) = K/((s + 4)(s + 10)). Find the gain K that gives a steady-state error of 20 % for a step input.

G(s) = K/((s + 4)(s + 10))ess = 20 %
K =
10.Steady-state error & PID controlHard
R(s)+K = 2010s(s + 5)C(s)−

A proportional controller K = 20 drives the plant G(s) = 10/(s(s + 5)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 10/(s(s + 5))K = 20
ts = s
11.Bode plots & marginsMedium
vin​C = 100 nFR = 47 kΩvout​

The RC high-pass filter shown has R = 47 kΩ, C = 100 nF. A sinusoidal input of 6.8 Hz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 47 kΩC = 100 nFf = 6.8 Hz
φ = °
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) = 36/(s(s + 2)(s + 5)). Find the gain margin GM in dB.

L(s) = 36/(s(s + 2)(s + 5))
GM = dB

3Solutions

1.Laplace transforms

Answer: f(0.75) = 1.814

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 + 1) + C/(s + 6)
A = 25/(1·6) = 4.167
B = 25/((−1)·5) = −5
C = 25/((−6)·(−5)) = 0.8333

Inverse transform

f(t) = 4.167 − 5·e−t + 0.8333·e−6t

Evaluate

f(0.75) = 4.167 − 5·e−0.75 + 0.8333·e−6·0.75 = 1.814

Answer

f(0.75) = 1.814
2.Laplace transforms

Answer: y(0.25) = −7.737

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 − (−8)s − (−3) + 5(sY − (−8)) + 4Y = 0
Y(s) = (−8s − 43)/((s + 1)(s + 4))

Partial fractions

A = −35/3 = −11.67
B = −11/(−3) = 3.667

Result

y(t) = −11.67·e−t + 3.667·e−4t
y(0.25) = −7.737

Answer

y(0.25) = −7.737

Check: y(0) = A + B = −8.

3.Block diagrams

Answer: p = −170

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) = 8/(s + 6 + 8·0.5) = 8/(s + 10)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 160/(s + 10 + 160) = 160/(s + 170)

Result

p = −170

Answer

p = −170
4.Block diagrams

Answer: Kmax = 1716

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 + 18s² + 107s + 210 + K = 0

Routh array

s³ | 1 107
s² | 18 210 + K
s¹ | (18·107 − 210 − K)/18
s⁰ | 210 + K

Stability limit

All entries of the first column must be positive.

Kmax = 18·107 − 210 = 1716

Answer

Kmax = 1716
5.Step response

Answer: %OS = 1.732 %

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 = 40 → ωₙ = 6.325 rad/s
2ζωₙ = 10 → ζ = 10/(2·6.325) = 0.7906

Result

%OS = 100·e−ζπ/√(1−ζ²) = 100·e−4.056 = 1.732 %

Answer

%OS = 1.732 %
6.Step response

Answer: ωn = 4.869 rad/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).

Damping ratio

ζ = −ln(0.1)/√(π² + ln²(0.1)) = 2.303/√(9.87 + 5.302) = 0.5912

Natural frequency

ωd = π/tp = π/0.8 = 3.927 rad/s
ωₙ = ωd/√(1 − ζ²) = 3.927/0.8066 = 4.869 rad/s

Answer

ωn = 4.869 rad/s
7.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 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

s(s + 1)(s + 2) + K(s + 9) = s³ + 3s² + (2 + K)s + 9K

Routh array

s³: 1, 2 + K
s²: 3, 9K
s¹: (3·(2 + K) − 9K)/3
s⁰: 9K

Condition

6 − 6K > 0
K < 6/6 = 1

Answer

Kmax = 1

With the zero to the right of s = −3 the loop would stay stable for every K > 0.

8.Stability: Routh–Hurwitz & root locus

Answer: K = 4

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.

Break-away point

dK/ds = 0 for K = −(s + 3)(s + 7) gives 2s + 10 = 0.

sb = −5

Magnitude condition

K = |−5 + 3|·|−5 + 7| = 2·2 = 4

Answer

K = 4

For this gain the closed loop is critically damped (ζ = 1).

9.Steady-state error & PID control

Answer: K = 160

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/40 = 4 → K = 160

Answer

K = 160

Check the closed-loop stability as well: for this second-order loop any K > 0 is stable.

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) + 20·10 = 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: φ = 78.65 °

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/(47000·1 × 10⁻⁷) = 212.8 rad/s
fc = 33.86 Hz → f/fc = 6.8/33.86 = 0.2008

Phase

φ = 90° − atan(0.2008) = 78.65°

Answer

φ = 78.65°
12.Bode plots & margins

Answer: GM = 5.776 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.

Phase crossover

(ω/2)·(ω/5) = 1 → ω180 = √(2·5) = 3.162 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 4)·√(ω² + 25)) = K/(2·5·(2 + 5)) = K/70

Gain margin

GM = 20·log₁₀(70/36) = 5.776 dB

Answer

GM = 5.776 dB

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