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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
Write the closed-loop characteristic polynomial: the denominator of 1 + K·G(s) = 0, multiplied out.
Put the coefficients alternately into the first two rows of the Routh array and build the remaining rows.
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.
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
Count the poles of the open loop C(s)·G(s) at s = 0: that is the system type.
Work out the error constant that matches the input: Kp for a step, Kv for a ramp, Ka for a parabola.
Apply ess = A/(1 + Kp), A/Kv or A/Ka, or solve it for the gain you need.
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
Find f(t) for F(s) = (3s + 27)/(s² + 2s + 37) by completing the square, and evaluate f(0.25).
F(s) = (3s + 27)/(s² + 2s + 37)t = 0.25 s
f(0.25) =
2.Laplace transformsHard
A system obeys y″ + 8y′ + 12y = 12·u(t), with y(0) = 2 and y′(0) = 0. At t = 0 a step input u(t) = 1 is applied. Use the Laplace transform to find y(0.6).
The plant G(s) = K/(s(s + 5)) 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(s + 5))unity negative feedback
K =
4.Block diagramsHard
Reduce the block diagram with G₁ = 3/(s + 1), G₂ = 5/(s + 1), an inner negative feedback H₁ = 0.5 around G₂ and an outer negative feedback H₂ = 0.1. Find the DC gain T(0) of C(s)/R(s).
G₁ = 3/(s + 1)G₂ = 5/(s + 1)H₁ = 0.5H₂ = 0.1
T(0) =
5.Step responseMedium
A system has the transfer function G(s) = 71/(s² + 10s + 71) and receives a unit step input. Find the peak time tp of the step response.
G(s) = 71/(s² + 10s + 71)
tp = s
6.Step responseHard
A mass m = 0.5 kg hangs on a spring k = 400 N/m with a viscous damper c = 8 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 50 N is suddenly applied. Find the percent overshoot %OS of the displacement.
m = 0.5 kgk = 400 N/mc = 8 N·s/mF = 50 N
%OS = %
7.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram runs with K = 16 and L(s) = K/(s(s + 2)(s + 6)). By what factor m can the gain be raised before the closed loop becomes unstable?
L(s) = K/(s(s + 2)(s + 6))K = 16
m =
8.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K/((s + 2)(s + 4)(s + 6)) is sketched. Find the frequency ω at which the locus crosses the imaginary axis.
L(s) = K/((s + 2)(s + 4)(s + 6))
ω = rad/s
9.Steady-state error & PID controlMedium
A unity-feedback loop has G(s) = K/((s + 4)(s + 5)). Find the gain K that gives a steady-state error of 2 % for a step input.
G(s) = K/((s + 4)(s + 5))ess = 2 %
K =
10.Steady-state error & PID controlHard
A PD controller C(s) = 20 + Kd·s drives the plant G(s) = 2/(s(s + 3)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.5.
G(s) = 2/(s(s + 3))Kp = 20ζ = 0.5
Kd = s
11.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 27(s + 18)/(s(s + 14)) at ω = 8 rad/s, in degrees.
G(s) = 27(s + 18)/(s(s + 14))ω = 8 rad/s
∠G(jω) = °
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 234/(s(s + 5)(s + 6)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.
L(s) = 234/(s(s + 5)(s + 6))
ω180 = rad/s
3Solutions
1.Laplace transforms
Answer: f(0.25) = 3.273
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).
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
(s² + 8s + 12)·Y(s) = 2s + 16 + 12/s
Y(s) = (2s² + 16s + 12)/(s(s + 2)(s + 6))
Partial fractions
A = 12/(2·6) = 1
B = −12/((−2)·4) = 1.5
C = −12/((−6)·(−4)) = −0.5
Result
y(t) = 1 + 1.5·e−2t − 0.5·e−6t
y(0.6) = 1.438
Answer
y(0.6) = 1.438
Check: y(0) = A + B + C = 2.
3.Block diagrams
Answer: K = 6.25
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(s + 5)) = 0 → s² + 5s + K = 0
Repeated root
A quadratic s² + βs + γ has a double root when β² = 4γ.
5² = 4K → K = 25/4 = 6.25
Answer
K = 6.25
Both closed-loop poles are then at s = −2.5.
4.Block diagrams
Answer: T(0) = 3
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).
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).
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) = √(400/0.5) = 28.28 rad/s
ζ = c/(2√(km)) = 8/(2·√(400·0.5)) = 0.2828
Result
%OS = 100·e−ζπ/√(1−ζ²) = 39.6 %
Answer
%OS = 39.6 %
7.Stability: Routh–Hurwitz & root locus
Answer: m = 6
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.
Routh array
s³ + 8s² + 12s + K = 0
s³: 1, 12
s²: 8, K
s¹: (8·12 − K)/8
s⁰: K
Largest stable gain
Kmax = 8·12 = 96
Factor
m = 96/16 = 6
Answer
m = 6
This factor is the gain margin as a ratio: 15.56 dB.
8.Stability: Routh–Hurwitz & root locus
Answer: ω = 6.633 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 + 12s² + 44s + 48 + K = 0
s¹: (12·44 − 48 − K)/12 = 0 → K = 480
Crossing frequency
12s² + 48 + 480 = 12s² + 528 = 0 → s² = −44
ω = √44 = 6.633 rad/s
Answer
ω = 6.633 rad/s
9.Steady-state error & PID control
Answer: K = 980
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 error constant
1/(1 + Kp) = 0.02 → Kp = 49
Gain
K/20 = 49 → K = 980
Answer
K = 980
Check the closed-loop stability as well: for this second-order loop any K > 0 is stable.
10.Steady-state error & PID control
Answer: Kd = 1.662 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) + 2(20 + Kds) = s² + (3 + 2Kd)s + 40
Natural frequency
ωₙ = √40 = 6.325 rad/s
Derivative gain
3 + 2Kd = 2·0.5·6.325 = 6.325
Kd = 1.662 s
Answer
Kd = 1.662 s
The derivative term adds damping without changing ωₙ.
11.Bode plots & margins
Answer: ∠G(jω) = −95.78 °
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.
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.