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
ReIm5

Find f(t) for F(s) = 37/(s² + 8s + 41) by completing the square, and evaluate f(0.25).

F(s) = 37/(s² + 8s + 41)t = 0.25 s
f(0.25) =
2.Laplace transformsHard
t (s)y(t)t = 0.25y(0.25) = ?

Solve y″ + 7y′ + 12y = 0 with y(0) = 9 and y′(0) = 0 using the Laplace transform, and find y(0.25).

y″ + 7y′ + 12y = 0y(0) = 9y′(0) = 0t = 0.25 s
y(0.25) =
3.Block diagramsMedium
R(s)+30s(s + 10)G(s)C(s)−

A unity negative feedback loop has G(s) = 30/(s(s + 10)). The reference is a ramp r(t) = 10·t. Find the steady-state error ess.

G(s) = 30/(s(s + 10))r(t) = 10·t
ess =
4.Block diagramsHard
R(s)+24s(s + 9)G(s)C(s)2H−

The plant G(s) = 24/(s(s + 9)) is in a negative feedback loop with sensor gain H = 2. The closed-loop poles are complex. Find their imaginary part ωd (the positive value), in rad/s.

G(s) = 24/(s(s + 9))H = 2
ωd = rad/s
5.Step responseMedium
t (s)y(t)0.5

A system has the transfer function G(s) = 8.5/(s² + 4s + 17) and receives a unit step input. Find the damping ratio ζ.

G(s) = 8.5/(s² + 4s + 17)
ζ =
6.Step responseHard
R(s)+160s(s + 12)G(s)C(s)−

A unity negative feedback loop has G(s) = 160/(s(s + 12)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 160/(s(s + 12))
ts = s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 5(s − 4)(s + 5)C(s)−

The plant G(s) = (s + 5)/((s − 4)(s + 5)) has an unstable pole at s = 4. A proportional gain K closes a unity-feedback loop around it. Find the smallest gain Kmin that makes the closed loop stable.

G(s) = (s + 5)/((s − 4)(s + 5))
Kmin =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6j2j4

The root locus of L(s) = K/((s + 1)(s + 2)(s + 5)) is sketched. Find the frequency ω at which the locus crosses the imaginary axis.

L(s) = K/((s + 1)(s + 2)(s + 5))
ω = rad/s
9.Steady-state error & PID controlMedium
r(t) = 5t+Ks(s + 1)C(s)−

A unity-feedback loop with G(s) = K/(s(s + 1)) must follow the ramp r(t) = 5t with a steady-state error of 0.2. Find the gain K.

G(s) = K/(s(s + 1))r(t) = 5tess = 0.2
K =
10.Steady-state error & PID controlHard
R(s)+K = 510s(s + 2)C(s)−

A proportional controller K = 5 drives the plant G(s) = 10/(s(s + 2)) in a unity-feedback loop. Find the percent overshoot %OS of the step response.

G(s) = 10/(s(s + 2))K = 5
%OS = %
11.Bode plots & marginsMedium
vin​L = 2.2 mHR = 22 Ωvout​

The RL low-pass filter shown has R = 22 Ω, L = 2.2 mH. A sinusoidal input of 8 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 22 ΩL = 2.2 mHf = 8 kHz
φ = °
12.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−100−80−60−40−20020ω (rad/s)φ (°)ω (rad/s)−270−180−900ω180​

A unity-feedback loop has the open-loop transfer function L(s) = K/(s(s + 3)(s + 4)). Find the gain K that gives a gain margin of 20 dB.

L(s) = K/(s(s + 3)(s + 4))GM = 20 dB
K =

3Solutions

1.Laplace transforms

Answer: f(0.25) = 2.583

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 + 41 = (s + 4)² + 25 = (s + 4)² + 5²

Match the numerator

F(s) = (37/5)·5/((s + 4)² + 5²)

Inverse transform

f(t) = e−4t·(7.4·sin 5t)
f(0.25) = e−4·0.25·(7.4·sin(5·0.25)) = 2.583

Answer

f(0.25) = 2.583
2.Laplace transforms

Answer: y(0.25) = 7.072

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 − 9s − 0 + 7(sY − 9) + 12Y = 0
Y(s) = (9s + 63)/((s + 3)(s + 4))

Partial fractions

A = 36/1 = 36
B = 27/(−1) = −27

Result

y(t) = 36·e−3t − 27·e−4t
y(0.25) = 7.072

Answer

y(0.25) = 7.072

Check: y(0) = A + B = 9.

3.Block diagrams

Answer: ess = 3.333

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) = 30/10 = 3

Steady-state error

ess = A/Kᵥ = 10/3 = 3.333

Answer

ess = 3.333
4.Block diagrams

Answer: ωd = 5.268 rad/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).

Characteristic equation

s(s + 9) + 24·2 = 0 → s² + 9s + 48 = 0

Roots

s = −4.5 ± j·√(48 − 20.25) = −4.5 ± j5.268
ωd = 5.268 rad/s

Answer

ωd = 5.268 rad/s
5.Step response

Answer: ζ = 0.4851

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 = 17 → ωₙ = 4.123 rad/s
2ζωₙ = 4 → ζ = 4/(2·4.123) = 0.4851

Result

ζ = 0.4851

Answer

ζ = 0.4851
6.Step response

Answer: ts = 0.6667 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) = 160/(s² + 12s + 160): ωₙ = 12.65 rad/s, ζ = 0.4743

Settling time

ζωₙ = 12/2 = 6
ts ≈ 4/(ζωₙ) = 0.6667 s

Answer

ts = 0.6667 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: Kmin = 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.

Characteristic polynomial

(s − 4)(s + 5) + K(s + 5) = s² + (1 + K)s + (5K − 20)

Conditions

For s² + c₁s + c₀ all roots lie in the left half-plane exactly when c₁ > 0 and c₀ > 0.

1 + K > 0 → K > −1
5K − 20 > 0 → K > 4
Kmin = 4

Answer

Kmin = 4
8.Stability: Routh–Hurwitz & root locus

Answer: ω = 4.123 rad/s

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.

Critical gain

s³ + 8s² + 17s + 10 + K = 0
s¹: (8·17 − 10 − K)/8 = 0 → K = 126

Crossing frequency

8s² + 10 + 126 = 8s² + 136 = 0 → s² = −17
ω = √17 = 4.123 rad/s

Answer

ω = 4.123 rad/s
9.Steady-state error & PID control

Answer: K = 25

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 velocity constant

Kv = A/ess = 5/0.2 = 25

Gain

K = 1·Kv = 25

Answer

K = 25

A large gain makes this loop very lightly damped: ζ = 0.1. That conflict is why compensators are used.

10.Steady-state error & PID control

Answer: %OS = 63.84 %

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 + 2) + 5·10 = s² + 2s + 50

ωₙ and ζ

ωₙ = √50 = 7.071 rad/s
ζ = 2/(2·7.071) = 0.1414

Overshoot

%OS = 100·e−0.1414π/√(1 − 0.1414²) = 63.84 %

Answer

%OS = 63.84 %
11.Bode plots & margins

Answer: φ = −78.75 °

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 = R/L = 22/0.0022 = 10000 rad/s
fc = 1592 Hz → f/fc = 8000/1592 = 5.027

Phase

φ = − atan(5.027) = −78.75°

Answer

φ = −78.75°
12.Bode plots & margins

Answer: K = 8.4

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/3)·(ω/4) = 1 → ω180 = √(3·4) = 3.464 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 9)·√(ω² + 16)) = K/(3·4·(3 + 4)) = K/84

Required gain

84/K = 1020/20 = 10
K = 84/10 = 8.4

Answer

K = 8.4

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