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 = 0.25f(0.25) = ?

Find f(t) for F(s) = (5s + 32)/(s² + 10s + 25) and evaluate f(0.25).

F(s) = (5s + 32)/(s² + 10s + 25)t = 0.25 s
f(0.25) =
2.Laplace transformsHard
ReIm−1−2−3−4−5−6−7−81

A system is described by the differential equation y″ + 4y′ + 3y = u′ + 7u, with input u(t) and output y(t). Find its transfer function G(s) = Y(s)/U(s) (zero initial conditions) and evaluate the impulse response g(t) at t = 0.75 s. Find g(0.75).

y″ + 4y′ + 3y = u′ + 7ut = 0.75 s
g(0.75) =
3.Block diagramsMedium
R(s)+11(s + 1)(s + 7)(s + 9)G(s)C(s)−

A unity negative feedback loop has G(s) = 11(s + 1)/((s + 7)(s + 9)). A step of size 2 is applied at the reference. Find the steady-state error ess.

G(s) = 11(s + 1)/((s + 7)(s + 9))R = 2
ess =
4.Block diagramsHard
R(s)+54s(s + 4)G(s)C(s)1H−

The plant G(s) = 54/(s(s + 4)) is in a negative feedback loop with sensor gain H = 1. The closed-loop poles are complex. Find their real part σ.

G(s) = 54/(s(s + 4))H = 1
σ =
5.Step responseMedium
t (s)y(t)0.5?

A system has the transfer function G(s) = 3/(s² + 4s + 6) and receives a unit step input. Find the percent overshoot %OS of the step response.

G(s) = 3/(s² + 4s + 6)
%OS = %
6.Step responseHard
t (s)y(t)1tp​ = 0.5 s5 %

The step response of a second-order system without zeros shows 5 % overshoot, and the first peak occurs at t = 0.5 s. Find the natural frequency ωn.

%OS = 5 %tp = 0.5 s
ωn = rad/s
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)). Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/(s(s + 2)(s + 8))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−7−8−9−101j1j2−3

For the loop L(s) = K/(s(s + 5)(s + 8)), find the gain K that puts a closed-loop pole at s = −3.

L(s) = K/(s(s + 5)(s + 8))s = −3
K =
9.Steady-state error & PID controlMedium
r(t) = 3t+2s(s + 2)(s + 10)C(s)−

A ramp r(t) = 3t is applied to a unity-feedback loop with G(s) = 2/(s(s + 2)(s + 10)). Find the steady-state error ess.

G(s) = 2/(s(s + 2)(s + 10))r(t) = 3t
ess =
10.Steady-state error & PID controlHard
r(t) = 1t+1 + 10/s4(s + 1)(s + 4)C(s)−

A PI controller C(s) = 1 + 10/s drives the plant G(s) = 4/((s + 1)(s + 4)) in a unity-feedback loop. Find the steady-state error ess for the ramp input r(t) = 1t.

C(s) = 1 + 10/sG(s) = 4/((s + 1)(s + 4))r(t) = 1t
ess =
11.Bode plots & marginsMedium
vin​R = 4.7 kΩC = 10 nFvout​

The RC low-pass filter shown has R = 4.7 kΩ, C = 10 nF. A sinusoidal input of 10 kHz is applied. Find the gain |H| = |vout/vin| in dB.

R = 4.7 kΩC = 10 nFf = 10 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 + 3)(s + 8)). Find the gain K that gives a gain margin of 4 dB.

L(s) = K/(s(s + 3)(s + 8))GM = 4 dB
K =

3Solutions

1.Laplace transforms

Answer: f(0.25) = 1.934

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).

Factor the denominator

s² + 10s + 25 = (s + 5)²

Partial fractions

5s + 32 = 5(s + 5) + 7
F(s) = 5/(s + 5) + 7/(s + 5)²

Inverse transform

f(t) = 5·e−5t + 7·t·e−5t
f(0.25) = (5 + 7·0.25)·e−5·0.25 = 1.934

Answer

f(0.25) = 1.934
2.Laplace transforms

Answer: g(0.75) = 1.206

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).

Transfer function

G(s) = Y(s)/U(s) = (s + 7)/(s² + 4s + 3)

Partial fractions

s² + 4s + 3 = (s + 1)(s + 3)
A = 3, B = −2

Impulse response

g(t) = 3·e−t − 2·e−3t
g(0.75) = 1.206

Answer

g(0.75) = 1.206
3.Block diagrams

Answer: ess = 1.703

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) = 11·1/(7·9) = 0.1746

Steady-state error

ess = A/(1 + Kₚ) = 2/(1 + 0.1746) = 1.703

Answer

ess = 1.703
4.Block diagrams

Answer: σ = −2

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 + 4) + 54·1 = 0 → s² + 4s + 54 = 0

Roots

s = −2 ± j·√(54 − 4) = −2 ± j7.071
σ = −2

Answer

σ = −2
5.Step response

Answer: %OS = 1.176 %

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 = 6 → ωₙ = 2.449 rad/s
2ζωₙ = 4 → ζ = 4/(2·2.449) = 0.8165

Result

%OS = 100·e−ζπ/√(1−ζ²) = 100·e−4.443 = 1.176 %

Answer

%OS = 1.176 %
6.Step response

Answer: ωn = 8.682 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(0.05)/√(π² + ln²(0.05)) = 2.996/√(9.87 + 8.974) = 0.6901

Natural frequency

ωd = π/tp = π/0.5 = 6.283 rad/s
ωₙ = ωd/√(1 − ζ²) = 6.283/0.7237 = 8.682 rad/s

Answer

ωn = 8.682 rad/s
7.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 160

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 equation

s³ + 10s² + 16s + K = 0

Routh array

s³: 1, 16
s²: 10, K
s¹: (10·16 − K)/10
s⁰: K

Conditions

Every entry of the first column must be positive.

K > 0
10·16 − K > 0 → K < 160

Answer

Kmax = 160

At K = 160 the closed loop oscillates at ω = √16 = 4 rad/s.

8.Stability: Routh–Hurwitz & root locus

Answer: K = 30

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 + 5|·|s + 8|

Evaluate at the point

K = 3·2·5 = 30

Answer

K = 30

Check: s(s + 5)(s + 8) + K is zero at s = −3.

9.Steady-state error & PID control

Answer: ess = 30

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.

Velocity constant

Kv = lim s·G(s) = 0.1

Ramp error

ess = 3/0.1 = 30

Answer

ess = 30
10.Steady-state error & PID control

Answer: ess = 0.1

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).

Velocity constant

s·C(s) = 1s + 10 → 10 as s → 0.

Kv = 10·4/4 = 10

Ramp error

ess = 1/10 = 0.1

Answer

ess = 0.1

The PI controller removes the step error completely; a ramp is followed with a constant lag.

11.Bode plots & margins

Answer: |H| = −9.877 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/(4700·1 × 10⁻⁸) = 21277 rad/s
fc = 3386 Hz → f/fc = 10000/3386 = 2.953

Gain

|H| = 1/√(1 + 2.953²) = 0.3207
20·log₁₀(0.3207) = −9.877 dB

Answer

|H| = −9.877 dB
12.Bode plots & margins

Answer: K = 166.6

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/8) = 1 → ω180 = √(3·8) = 4.899 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 9)·√(ω² + 64)) = K/(3·8·(3 + 8)) = K/264

Required gain

264/K = 104/20 = 1.585
K = 264/1.585 = 166.6

Answer

K = 166.6

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