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 = 0.1f(0.1) = ?

Find f(t) for F(s) = (6s + 2)/(s² + 8s + 16) and evaluate f(0.1).

F(s) = (6s + 2)/(s² + 8s + 16)t = 0.1 s
f(0.1) =
2.Laplace transformsHard
t (s)y(t)t = 0.5y(0.5) = ?

A system obeys y″ + 3y′ + 2y = 8·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 1 for t ≥ 0. Use the Laplace transform to find the output y(0.5).

y″ + 3y′ + 2y = 8·u(t)u(t) = 1y(0) = 0, y′(0) = 0t = 0.5 s
y(0.5) =
3.Block diagramsMedium
R(s)+6s + 6G(s)C(s)0.2H−

A plant G(s) = 6/(s + 6) is controlled in a negative feedback loop whose sensor has the gain H = 0.2. A step of size 12 is applied at R. Find the steady-state output css.

G(s) = 6/(s + 6)H = 0.2R = 12
css =
4.Block diagramsHard
R(s)+17controller+D(s)+7s + 7plantC(s)−

A proportional controller K = 17 drives the plant G(s) = 7/(s + 7) in a unity negative feedback loop. A disturbance D enters at the plant input (it is added to the controller output). With R = 0, a disturbance step of size 10 is applied. Find the steady-state output css.

K = 17G(s) = 7/(s + 7)D = 10R = 0
css =
5.Step responseMedium
t (s)y(t)2ts​ = ?

A system has the transfer function G(s) = 178/(s² + 6s + 89) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 178/(s² + 6s + 89)
ts = s
6.Step responseHard
t (s)y(t)1tp​ = 0.8 s25 %

The step response of a second-order system without zeros shows 25 % overshoot, and the first peak occurs at t = 0.8 s. Find the natural frequency ωn.

%OS = 25 %tp = 0.8 s
ωn = rad/s
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 5(s − 3)(s + 3)C(s)−

The plant G(s) = (s + 5)/((s − 3)(s + 3)) has an unstable pole at s = 3. 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 − 3)(s + 3))
Kmin =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−0.5−1−1.5−2−2.5−3−3.5−40.5j0.5j1

The root locus of L(s) = K/(s(s + 2)(s + 3)) leaves the real axis between s = 0 and s = −2. Find the break-away point sb.

L(s) = K/(s(s + 2)(s + 3))
sb =
9.Steady-state error & PID controlMedium
R(s)+K = 141s + 1C(s)−

A process G(s) = 4/(1s + 1) is controlled by a proportional gain K = 1 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.

G(s) = 4/(1s + 1)K = 1
ess = %
10.Steady-state error & PID controlHard
R(s)+25 + Kd​s2s(s + 1)C(s)−

A PD controller C(s) = 25 + Kd·s drives the plant G(s) = 2/(s(s + 1)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.6.

G(s) = 2/(s(s + 1))Kp = 25ζ = 0.6
Kd = s
11.Bode plots & marginsMedium
vin​R = 3.3 kΩC = 2.2 nFvout​

The RC low-pass filter shown has R = 3.3 kΩ, C = 2.2 nF. A sinusoidal input of 11 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 3.3 kΩC = 2.2 nFf = 11 kHz
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) = 238/(s(s + 5)(s + 10)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 238/(s(s + 5)(s + 10))
ω180 = rad/s

3Solutions

1.Laplace transforms

Answer: f(0.1) = 2.547

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 + 8s + 16 = (s + 4)²

Partial fractions

6s + 2 = 6(s + 4) + −22
F(s) = 6/(s + 4) − 22/(s + 4)²

Inverse transform

f(t) = 6·e−4t − 22·t·e−4t
f(0.1) = (6 − 22·0.1)·e−4·0.1 = 2.547

Answer

f(0.1) = 2.547
2.Laplace transforms

Answer: y(0.5) = 0.6193

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

Y(s) = 8/(s(s + 1)(s + 2))

Partial fractions

A = 8/(1·2) = 4
B = 8/((−1)·1) = −8
C = 8/((−2)·(−1)) = 4

Output

y(t) = 4 − 8·e−t + 4·e−2t
y(0.5) = 0.6193

Answer

y(0.5) = 0.6193
3.Block diagrams

Answer: css = 10

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).

Closed-loop transfer function

T(s) = G/(1 + G·H) = 6/(s + 6 + 1.2) = 6/(s + 7.2)

Result

css = R·T(0) = 12·6/7.2 = 10

Answer

css = 10
4.Block diagrams

Answer: css = 0.5556

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) = 7/(s + 7 + 119)

Steady state

css = 10·7/126 = 0.5556

Answer

css = 0.5556

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

5.Step response

Answer: ts = 1.333 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 = 89 → ωₙ = 9.434 rad/s
2ζωₙ = 6 → ζ = 6/(2·9.434) = 0.318

Result

ζωₙ = 3 → ts ≈ 4/3 = 1.333 s

Answer

ts = 1.333 s
6.Step response

Answer: ωn = 4.292 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(0.25)/√(π² + ln²(0.25)) = 1.386/√(9.87 + 1.922) = 0.4037

Natural frequency

ωd = π/tp = π/0.8 = 3.927 rad/s
ωₙ = ωd/√(1 − ζ²) = 3.927/0.9149 = 4.292 rad/s

Answer

ωn = 4.292 rad/s
7.Stability: Routh–Hurwitz & root locus

Answer: Kmin = 1.8

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 − 3)(s + 3) + K(s + 5) = s² + (0 + K)s + (5K − 9)

Conditions

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

0 + K > 0 → K > 0
5K − 9 > 0 → K > 1.8
Kmin = 1.8

Answer

Kmin = 1.8
8.Stability: Routh–Hurwitz & root locus

Answer: sb = −0.7847

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) = −s(s + 2)(s + 3) = −(s³ + 5s² + 6s)

Stationary points

dK/ds = −(3s² + 10s + 6) = 0
s = (−10 ± √(100 − 72))/6 = −0.7847, −2.549

Pick the point on the locus

The locus lies on the real axis between 0 and −2; the other root is not on the locus.

sb = −0.7847

Answer

sb = −0.7847
9.Steady-state error & PID control

Answer: ess = 20 %

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 = 1·4 = 4

Step error

ess = 100/(1 + 4) = 20 %

Answer

ess = 20 %
10.Steady-state error & PID control

Answer: Kd = 3.743 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 + 1) + 2(25 + Kds) = s² + (1 + 2Kd)s + 50

Natural frequency

ωₙ = √50 = 7.071 rad/s

Derivative gain

1 + 2Kd = 2·0.6·7.071 = 8.485
Kd = 3.743 s

Answer

Kd = 3.743 s

The derivative term adds damping without changing ωₙ.

11.Bode plots & margins

Answer: φ = −26.65 °

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/(3300·2.2 × 10⁻⁹) = 137741 rad/s
fc = 21922 Hz → f/fc = 11000/21922 = 0.5018

Phase

φ = − atan(0.5018) = −26.65°

Answer

φ = −26.65°
12.Bode plots & margins

Answer: ω180 = 7.071 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/5)·(ω/10) = 1 → ω180 = √(5·10) = 7.071 rad/s

Answer

ω180 = 7.071 rad/s

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