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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.Step responseMedium
A system has the transfer function G(s) = 166/(s² + 3s + 83) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 166/(s² + 3s + 83)
ts = s
2.Step responseHard
A mass m = 0.5 kg hangs on a spring k = 500 N/m with a viscous damper c = 9 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 20 N is suddenly applied. Find the largest displacement xmax in mm.
m = 0.5 kgk = 500 N/mc = 9 N·s/mF = 20 N
xmax = mm
3.Bode plots & marginsMedium
The RL high-pass filter shown has R = 1000 Ω, L = 22 mH. A sinusoidal input of 36 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.
R = 1000 ΩL = 22 mHf = 36 kHz
φ = °
4.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = K/(s(s + 3)(s + 6)). Find the gain K that gives a gain margin of 3 dB.
L(s) = K/(s(s + 3)(s + 6))GM = 3 dB
K =
5.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 2)(s + 10)). Find the largest gain Kmax for which the closed loop is stable.
L(s) = K/(s(s + 2)(s + 10))
Kmax =
6.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K/(s(s + 1)(s + 8)) leaves the real axis between s = 0 and s = −1. Find the break-away point sb.
L(s) = K/(s(s + 1)(s + 8))
sb =
7.Steady-state error & PID controlMedium
The input r(t) = 5 + 5t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 50/(s(s + 1)(s + 2)). Find the steady-state error ess.
G(s) = 50/(s(s + 1)(s + 2))r(t) = 5 + 5t
ess =
8.Steady-state error & PID controlHard
A proportional controller K = 40 drives the plant G(s) = 10/(s(s + 8)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 10/(s(s + 8))K = 40
ts = s
9.Block diagramsMedium
The plant G(s) = K/(s(s + 3)) 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 + 3))unity negative feedback
K =
10.Block diagramsHard
Reduce the block diagram with G₁ = 17/(s + 4), G₂ = 10/(s + 2), an inner negative feedback H₁ = 0.5 around G₂ and an outer negative feedback H₂ = 0.25. Find the DC gain T(0) of C(s)/R(s).
G₁ = 17/(s + 4)G₂ = 10/(s + 2)H₁ = 0.5H₂ = 0.25
T(0) =
3Solutions
1.Step response
Answer: ts = 2.667 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 = 83 → ωₙ = 9.11 rad/s
2ζωₙ = 3 → ζ = 3/(2·9.11) = 0.1646
Result
ζωₙ = 1.5 → ts ≈ 4/1.5 = 2.667 s
Answer
ts = 2.667 s
2.Step response
Answer: xmax = 55.74 mm
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) = √(500/0.5) = 31.62 rad/s
ζ = c/(2√(km)) = 9/(2·√(500·0.5)) = 0.2846
Result
xst = F/k = 20/500 m = 40 mm
%OS = 39.35 %
xmax = 40·(1 + 0.3935) = 55.74 mm
Answer
xmax = 55.74 mm
3.Bode plots & margins
Answer: φ = 11.36 °
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 = R/L = 1000/0.022 = 45455 rad/s
fc = 7234 Hz → f/fc = 36000/7234 = 4.976
Phase
φ = 90° − atan(4.976) = 11.36°
Answer
φ = 11.36°
4.Bode plots & margins
Answer: K = 114.7
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.
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 + 12s² + 20s + K = 0
Routh array
s³: 1, 20
s²: 12, K
s¹: (12·20 − K)/12
s⁰: K
Conditions
Every entry of the first column must be positive.
K > 0
12·20 − K > 0 → K < 240
Answer
Kmax = 240
At K = 240 the closed loop oscillates at ω = √20 = 4.472 rad/s.
6.Stability: Routh–Hurwitz & root locus
Answer: sb = −0.4834
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 + 1)(s + 8) = −(s³ + 9s² + 8s)
Stationary points
dK/ds = −(3s² + 18s + 8) = 0
s = (−18 ± √(324 − 96))/6 = −0.4834, −5.517
Pick the point on the locus
The locus lies on the real axis between 0 and −1; the other root is not on the locus.
sb = −0.4834
Answer
sb = −0.4834
7.Steady-state error & PID control
Answer: ess = 0.2
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) = 25
Step part
Kp = ∞ → estep = 0
Ramp error
ess = 5/25 = 0.2
Answer
ess = 0.2
8.Steady-state error & PID control
Answer: ts = 1 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 + 8) + 40·10 = s² + 8s + 400
ωₙ and ζ
ωₙ = √400 = 20 rad/s
ζ = 8/(2·20) = 0.2
Settling time
ζωₙ = 8/2 = 4
ts ≈ 4/4 = 1 s
Answer
ts = 1 s
The settling time does not depend on K here: raising K only makes the response oscillate faster.
9.Block diagrams
Answer: K = 2.25
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 + 3)) = 0 → s² + 3s + K = 0
Repeated root
A quadratic s² + βs + γ has a double root when β² = 4γ.
3² = 4K → K = 9/4 = 2.25
Answer
K = 2.25
Both closed-loop poles are then at s = −1.5.
10.Block diagrams
Answer: T(0) = 2.411
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).