Control Engineering 1: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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) = (3s + 35)/(s² + 4s + 4) and evaluate f(0.75).

F(s) = (3s + 35)/(s² + 4s + 4)t = 0.75 s
f(0.75) =
2.Laplace transformsHard
t (s)y(t)t = 0.4y(0.4) = ?

A system obeys y″ + 5y′ + 6y = 6·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 5 for t ≥ 0. Use the Laplace transform to find the output y(0.4).

y″ + 5y′ + 6y = 6·u(t)u(t) = 5y(0) = 0, y′(0) = 0t = 0.4 s
y(0.4) =
3.Block diagramsMedium
R(s)+K(s + 2)(s + 10)G(s)C(s)−

The plant G(s) = K/((s + 2)(s + 10)) is in a unity negative feedback loop. Find the gain K for which the closed loop is critically damped (two equal real poles).

G(s) = K/((s + 2)(s + 10))unity negative feedback
K =
4.Block diagramsHard
R(s)+52s(s + 8)G(s)C(s)2H−

The plant G(s) = 52/(s(s + 8)) is in a negative feedback loop with sensor gain H = 2. The closed-loop poles are complex. Find their real part σ.

G(s) = 52/(s(s + 8))H = 2
σ =
5.Step responseMedium
t (s)y(t)1ts​ = ?

A system has the transfer function G(s) = 55/(s² + 12s + 55) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 55/(s² + 12s + 55)
ts = s
6.Step responseHard
k = 2000 N/mc = 27 N·s/mm = 1 kgF = 25 Nx(t)

A mass m = 1 kg hangs on a spring k = 2000 N/m with a viscous damper c = 27 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 25 N is suddenly applied. Find the percent overshoot %OS of the displacement.

m = 1 kgk = 2000 N/mc = 27 N·s/mF = 25 N
%OS = %
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1(s + 1)(s + 3)(s + 4)C(s)−

A unity-feedback loop has L(s) = K/((s + 1)(s + 3)(s + 4)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/((s + 1)(s + 3)(s + 4))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10−12−14−16j2j4

The root locus of L(s) = K(s + 13)/(s(s + 6)) leaves the real axis between s = 0 and s = −6, circles round and joins the real axis again to the left of the zero. Find the break-in point si.

L(s) = K(s + 13)/(s(s + 6))
si =
9.Steady-state error & PID controlMedium
r(t) = 5 + 1t+5s(s + 1)(s + 4)C(s)−

The input r(t) = 5 + 1t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 5/(s(s + 1)(s + 4)). Find the steady-state error ess.

G(s) = 5/(s(s + 1)(s + 4))r(t) = 5 + 1t
ess =
10.Steady-state error & PID controlHard
R(s)+8 + Kd​s5s(s + 3)C(s)−

A PD controller C(s) = 8 + Kd·s drives the plant G(s) = 5/(s(s + 3)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.8.

G(s) = 5/(s(s + 3))Kp = 8ζ = 0.8
Kd = s
11.Bode plots & marginsMedium
vin​C = 2.2 nFR = 22 kΩvout​

The RC high-pass filter shown has R = 22 kΩ, C = 2.2 nF. A sinusoidal input of 820 Hz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 22 kΩC = 2.2 nFf = 820 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) = 266/(s(s + 5)(s + 7)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 266/(s(s + 5)(s + 7))
ω180 = rad/s

3Solutions

1.Laplace transforms

Answer: f(0.75) = 5.522

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 + 4s + 4 = (s + 2)²

Partial fractions

3s + 35 = 3(s + 2) + 29
F(s) = 3/(s + 2) + 29/(s + 2)²

Inverse transform

f(t) = 3·e−2t + 29·t·e−2t
f(0.75) = (3 + 29·0.75)·e−2·0.75 = 5.522

Answer

f(0.75) = 5.522
2.Laplace transforms

Answer: y(0.4) = 1.272

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) = 30/(s(s + 2)(s + 3))

Partial fractions

A = 30/(2·3) = 5
B = 30/((−2)·1) = −15
C = 30/((−3)·(−1)) = 10

Output

y(t) = 5 − 15·e−2t + 10·e−3t
y(0.4) = 1.272

Answer

y(0.4) = 1.272
3.Block diagrams

Answer: K = 16

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

1 + K/((s + 2)(s + 10)) = 0 → s² + 12s + 20 + K = 0

Repeated root

A quadratic s² + βs + γ has a double root when β² = 4γ.

12² = 4(20 + K) → K = 144/4 − 20 = 16

Answer

K = 16
Both closed-loop poles are then at s = −6.
4.Block diagrams

Answer: σ = −4

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 + 8) + 52·2 = 0 → s² + 8s + 104 = 0

Roots

s = −4 ± j·√(104 − 16) = −4 ± j9.381
σ = −4

Answer

σ = −4
5.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).

Compare with the standard form

The denominator must match s² + 2ζωₙs + ωₙ².

ωₙ² = 55 → ωₙ = 7.416 rad/s
2ζωₙ = 12 → ζ = 12/(2·7.416) = 0.809

Result

ζωₙ = 6 → ts ≈ 4/6 = 0.6667 s

Answer

ts = 0.6667 s
6.Step response

Answer: %OS = 36.98 %

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).

Natural frequency and damping ratio

ωₙ = √(k/m) = √(2000/1) = 44.72 rad/s
ζ = c/(2√(km)) = 27/(2·√(2000·1)) = 0.3019

Result

%OS = 100·e−ζπ/√(1−ζ²) = 36.98 %

Answer

%OS = 36.98 %
7.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 140

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

s³ + 8s² + 19s + 12 + K = 0

Routh array

s³: 1, 19
s²: 8, 12 + K
s¹: (8·19 − 12 − K)/8
s⁰: 12 + K

Condition

The s¹ entry must stay positive.

K < 8·19 − 12 = 140

Answer

Kmax = 140

At the limit the loop oscillates at ω = √19 = 4.359 rad/s.

8.Stability: Routh–Hurwitz & root locus

Answer: si = −22.54

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) = −(s² + 6s)/(s + 13)

Stationary points

Differentiate with the quotient rule and set the numerator to zero.

(2s + 6)(s + 13) − (s² + 6s) = s² + 26s + 78 = 0
s = −13 ± √91 = −3.461, −22.54

Break-in point

si = −22.54

Answer

si = −22.54

Between the two break points the locus is a circle centred on the zero, with radius √91 = 9.539.

9.Steady-state error & PID control

Answer: ess = 0.8

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) = 1.25

Step part

Kp = ∞ → estep = 0

Ramp error

ess = 1/1.25 = 0.8

Answer

ess = 0.8
10.Steady-state error & PID control

Answer: Kd = 1.424 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 + 3) + 5(8 + Kds) = s² + (3 + 5Kd)s + 40

Natural frequency

ωₙ = √40 = 6.325 rad/s

Derivative gain

3 + 5Kd = 2·0.8·6.325 = 10.12
Kd = 1.424 s

Answer

Kd = 1.424 s

The derivative term adds damping without changing ωₙ.

11.Bode plots & margins

Answer: φ = 76 °

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/(22000·2.2 × 10⁻⁹) = 20661 rad/s
fc = 3288 Hz → f/fc = 820/3288 = 0.2494

Phase

φ = 90° − atan(0.2494) = 76°

Answer

φ = 76°
12.Bode plots & margins

Answer: ω180 = 5.916 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/7) = 1 → ω180 = √(5·7) = 5.916 rad/s

Answer

ω180 = 5.916 rad/s

ExamProblems · examproblems.com/courses/control-engineering-1