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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) = 198/(s² + 13s + 99) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.
G(s) = 198/(s² + 13s + 99)
σ =
2.Step responseHard
A mass m = 20 kg hangs on a spring k = 2000 N/m with a viscous damper c = 80 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 = 20 kgk = 2000 N/mc = 80 N·s/mF = 20 N
xmax = mm
3.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 29(s + 3)/(s(s + 16)) at ω = 15 rad/s, in degrees.
G(s) = 29(s + 3)/(s(s + 16))ω = 15 rad/s
∠G(jω) = °
4.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 450/(s(s + 10)). Find the gain crossover frequency ωc, where |L(jω)| = 1.
L(s) = 450/(s(s + 10))
ωc = rad/s
5.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 1)(s + 6)). The gain is raised until the closed loop is on the edge of stability. Find the frequency ω of the sustained oscillation.
L(s) = K/(s(s + 1)(s + 6))
ω = rad/s
6.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K(s + 10)/(s(s + 5)) leaves the real axis between s = 0 and s = −5, circles round and joins the real axis again to the left of the zero. Find the break-in point si.
L(s) = K(s + 10)/(s(s + 5))
si =
7.Steady-state error & PID controlMedium
A unity-feedback loop has G(s) = K/((s + 1)(s + 10)). Find the gain K that gives a steady-state error of 5 % for a step input.
G(s) = K/((s + 1)(s + 10))ess = 5 %
K =
8.Steady-state error & PID controlHard
A proportional controller K = 40 drives the plant G(s) = 2/(s(s + 10)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 2/(s(s + 10))K = 40
ts = s
9.Block diagramsMedium
The plant G(s) = K/((s + 1)(s + 2)) 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 + 1)(s + 2))unity negative feedback
K =
10.Block diagramsHard
The plant G(s) = 70/(s(s + 9)) is in a negative feedback loop with sensor gain H = 2. The closed-loop poles are complex. Find their imaginary part ωd (the positive value), in rad/s.
G(s) = 70/(s(s + 9))H = 2
ωd = rad/s
3Solutions
1.Step response
Answer: σ = −6.5
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 = 99 → ωₙ = 9.95 rad/s
2ζωₙ = 13 → ζ = 13/(2·9.95) = 0.6533
Result
σ = −ζωₙ = −13/2 = −6.5
Answer
σ = −6.5
2.Step response
Answer: xmax = 15.27 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) = √(2000/20) = 10 rad/s
ζ = c/(2√(km)) = 80/(2·√(2000·20)) = 0.2
Result
xst = F/k = 20/2000 m = 10 mm
%OS = 52.66 %
xmax = 10·(1 + 0.5266) = 15.27 mm
Answer
xmax = 15.27 mm
3.Bode plots & margins
Answer: ∠G(jω) = −54.46 °
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.
Gain crossover
ω⁴ + 100ω² − 202500 = 0
ω² = (−100 + √(10000 + 4·202500))/2 = 402.8
ωc = 20.07 rad/s
Answer
ωc = 20.07 rad/s
5.Stability: Routh–Hurwitz & root locus
Answer: ω = 2.449 rad/s
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 + 7s² + 6s + K = 0
s³: 1, 6
s²: 7, K
s¹: (7·6 − K)/7
s⁰: K
Critical gain
K = 7·6 = 42
Auxiliary equation
At the critical gain the s¹ row is zero; the s² row gives the poles on the imaginary axis.
7s² + 42 = 0 → s = ±j√6
ω = 2.449 rad/s
Answer
ω = 2.449 rad/s
6.Stability: Routh–Hurwitz & root locus
Answer: si = −17.07
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 + 5s)/(s + 10)
Stationary points
Differentiate with the quotient rule and set the numerator to zero.
(2s + 5)(s + 10) − (s² + 5s) = s² + 20s + 50 = 0
s = −10 ± √50 = −2.929, −17.07
Break-in point
si = −17.07
Answer
si = −17.07
Between the two break points the locus is a circle centred on the zero, with radius √50 = 7.071.
7.Steady-state error & PID control
Answer: K = 190
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) = 0.05 → Kp = 19
Gain
K/10 = 19 → K = 190
Answer
K = 190
Check the closed-loop stability as well: for this second-order loop any K > 0 is stable.
8.Steady-state error & PID control
Answer: ts = 0.8 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 + 10) + 40·2 = s² + 10s + 80
ωₙ and ζ
ωₙ = √80 = 8.944 rad/s
ζ = 10/(2·8.944) = 0.559
Settling time
ζωₙ = 10/2 = 5
ts ≈ 4/5 = 0.8 s
Answer
ts = 0.8 s
The settling time does not depend on K here: raising K only makes the response oscillate faster.
9.Block diagrams
Answer: K = 0.25
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 + 1)(s + 2)) = 0 → s² + 3s + 2 + K = 0
Repeated root
A quadratic s² + βs + γ has a double root when β² = 4γ.
3² = 4(2 + K) → K = 9/4 − 2 = 0.25
Answer
K = 0.25
Both closed-loop poles are then at s = −1.5.
10.Block diagrams
Answer: ωd = 10.94 rad/s
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).