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 = 1.5f(1.5) = ?

Find f(t) for F(s) = 50/(s(s + 1)(s + 5)) by partial fractions, and evaluate f(1.5).

F(s) = 50/(s(s + 1)(s + 5))t = 1.5 s
f(1.5) =
2.Laplace transformsHard
t (s)y(t)t = 0.5y(0.5) = ?

Solve y″ + 6y′ + 5y = 0 with y(0) = −5 and y′(0) = −9 using the Laplace transform, and find y(0.5).

y″ + 6y′ + 5y = 0y(0) = −5y′(0) = −9t = 0.5 s
y(0.5) =
3.Block diagramsMedium
R(s)+4s(s + 1)G(s)C(s)−

A unity negative feedback loop has G(s) = 4/(s(s + 1)). The reference is a ramp r(t) = 10·t. Find the steady-state error ess.

G(s) = 4/(s(s + 1))r(t) = 10·t
ess =
4.Block diagramsHard
R(s)+22controller+D(s)+1s + 9plantC(s)−

A proportional controller K = 22 drives the plant G(s) = 1/(s + 9) 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 5 and a disturbance step of size 5 are applied together. Find the steady-state output css.

K = 22G(s) = 1/(s + 9)D = 5R = 5
css =
5.Step responseMedium
t (s)y(t)0.5

A system has the transfer function G(s) = 12/(s² + 5s + 24) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.

G(s) = 12/(s² + 5s + 24)
σ =
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)+Ks + 4(s − 1)(s + 3)C(s)−

The plant G(s) = (s + 4)/((s − 1)(s + 3)) has an unstable pole at s = 1. 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 + 4)/((s − 1)(s + 3))
Kmin =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−0.5−1−1.5−2−2.5−3−3.5−4−4.5−50.5j0.5j1−1.5

For the loop L(s) = K/(s(s + 2)(s + 4)), find the gain K that puts a closed-loop pole at s = −1.5.

L(s) = K/(s(s + 2)(s + 4))s = −1.5
K =
9.Steady-state error & PID controlMedium
r(t) = 1·t²/2+4(s + 1)s²(s + 2)C(s)−

The parabolic input r(t) = 1·t²/2 is applied to a unity-feedback loop with G(s) = 4(s + 1)/(s²(s + 2)). Find the steady-state error ess.

G(s) = 4(s + 1)/(s²(s + 2))r(t) = 1·t²/2
ess =
10.Steady-state error & PID controlHard
R(s)+K4s(s + 6)C(s)−

A proportional controller K drives the plant G(s) = 4/(s(s + 6)) in a unity-feedback loop. Find the gain K that gives the closed loop a damping ratio ζ = 0.8.

G(s) = 4/(s(s + 6))ζ = 0.8
K =
11.Bode plots & marginsMedium
|H| (dB)0.10.111101010010010001000−40−20020406080ω (rad/s)φ (°)ω (rad/s)−135−90−450ω = 5

Find the phase ∠G(jω) of G(s) = 40(s + 17)/(s(s + 4)) at ω = 5 rad/s, in degrees.

G(s) = 40(s + 17)/(s(s + 4))ω = 5 rad/s
∠G(jω) = °
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) = 78/(s(s + 4)(s + 7)). Find the gain margin GM in dB.

L(s) = 78/(s(s + 4)(s + 7))
GM = dB

3Solutions

1.Laplace transforms

Answer: f(1.5) = 7.212

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 + 5)
A = 50/(1·5) = 10
B = 50/((−1)·4) = −12.5
C = 50/((−5)·(−4)) = 2.5

Inverse transform

f(t) = 10 − 12.5·e−t + 2.5·e−5t

Evaluate

f(1.5) = 10 − 12.5·e−1.5 + 2.5·e−5·1.5 = 7.212

Answer

f(1.5) = 7.212
2.Laplace transforms

Answer: y(0.5) = −4.868

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 − (−5)s − (−9) + 6(sY − (−5)) + 5Y = 0
Y(s) = (−5s − 39)/((s + 1)(s + 5))

Partial fractions

A = −34/4 = −8.5
B = −14/(−4) = 3.5

Result

y(t) = −8.5·e−t + 3.5·e−5t
y(0.5) = −4.868

Answer

y(0.5) = −4.868

Check: y(0) = A + B = −5.

3.Block diagrams

Answer: ess = 2.5

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) = 4/1 = 4

Steady-state error

ess = A/Kᵥ = 10/4 = 2.5

Answer

ess = 2.5
4.Block diagrams

Answer: css = 3.71

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) = 1/(s + 9 + 22)
C/R = K·G/(1 + K·G) = 22/(s + 9 + 22)

Steady state

css = 5·22/31 + 5·1/31 = 3.71

Answer

css = 3.71

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

5.Step response

Answer: σ = −2.5

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 = 24 → ωₙ = 4.899 rad/s
2ζωₙ = 5 → ζ = 5/(2·4.899) = 0.5103

Result

σ = −ζωₙ = −5/2 = −2.5

Answer

σ = −2.5
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: Kmin = 0.75

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 − 1)(s + 3) + K(s + 4) = s² + (2 + K)s + (4K − 3)

Conditions

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

2 + K > 0 → K > −2
4K − 3 > 0 → K > 0.75
Kmin = 0.75

Answer

Kmin = 0.75
8.Stability: Routh–Hurwitz & root locus

Answer: K = 1.875

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.

Magnitude condition

K = |s|·|s + 2|·|s + 4|

Evaluate at the point

K = 1.5·0.5·2.5 = 1.875

Answer

K = 1.875

Check: s(s + 2)(s + 4) + K is zero at s = −1.5.

9.Steady-state error & PID control

Answer: ess = 0.5

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.

Acceleration constant

Ka = 4·1/2 = 2

Parabola error

ess = 1/2 = 0.5

Answer

ess = 0.5
10.Steady-state error & PID control

Answer: K = 3.516

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 + 6) + 4K = s² + 6s + 4K

Natural frequency

2·0.8·ωₙ = 6 → ωₙ = 3.75 rad/s

Gain

4K = ωₙ² = 14.06 → K = 3.516

Answer

K = 3.516
11.Bode plots & margins

Answer: ∠G(jω) = −125 °

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.

Add the angles

∠G = atan(5/17) − 90° − atan(5/4) = 16.39° − 90° − 51.34°

Result

∠G(j5) = −125°

Answer

∠G(jω) = −125°
12.Bode plots & margins

Answer: GM = 11.93 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/4)·(ω/7) = 1 → ω180 = √(4·7) = 5.292 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 16)·√(ω² + 49)) = K/(4·7·(4 + 7)) = K/308

Gain margin

GM = 20·log₁₀(308/78) = 11.93 dB

Answer

GM = 11.93 dB

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