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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) = 98/(s² + 9s + 98) and receives a unit step input. Find the peak time tp of the step response.
G(s) = 98/(s² + 9s + 98)
tp = s
2.Step responseHard
A mass m = 10 kg hangs on a spring k = 5000 N/m with a viscous damper c = 268 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 10 N is suddenly applied. Find the largest displacement xmax in mm.
m = 10 kgk = 5000 N/mc = 268 N·s/mF = 10 N
xmax = mm
3.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 50(s + 2)/(s(s + 10)) at ω = 10 rad/s, in degrees.
G(s) = 50(s + 2)/(s(s + 10))ω = 10 rad/s
∠G(jω) = °
4.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 118/(s(s + 4)(s + 9)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.
L(s) = 118/(s(s + 4)(s + 9))
ω180 = rad/s
5.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 6)(s + 8)). Find the largest gain Kmax for which the closed loop is stable.
L(s) = K/(s(s + 6)(s + 8))
Kmax =
6.Stability: Routh–Hurwitz & root locusHard
For the loop L(s) = K/(s(s + 4)(s + 10)), find the gain K that puts a closed-loop pole at s = −3.
L(s) = K/(s(s + 4)(s + 10))s = −3
K =
7.Steady-state error & PID controlMedium
The input r(t) = 2 + 1t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 2/(s(s + 1)(s + 10)). Find the steady-state error ess.
G(s) = 2/(s(s + 1)(s + 10))r(t) = 2 + 1t
ess =
8.Steady-state error & PID controlHard
A proportional controller K = 2 drives the plant G(s) = 10/(s(s + 5)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 10/(s(s + 5))K = 2
ts = s
9.Block diagramsMedium
A unity negative feedback loop has G(s) = 16/(s(s + 9)). The reference is a ramp r(t) = 1·t. Find the steady-state error ess.
G(s) = 16/(s(s + 9))r(t) = 1·t
ess =
10.Block diagramsHard
The plant G(s) = K/(s(s + 2)(s + 9)) is in a unity negative feedback loop. Use the Routh–Hurwitz criterion to find the largest gain Kmax for which the closed loop is stable.
G(s) = K/(s(s + 2)(s + 9))unity negative feedback
Kmax =
3Solutions
1.Step response
Answer: tp = 0.3563 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).
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) = √(5000/10) = 22.36 rad/s
ζ = c/(2√(km)) = 268/(2·√(5000·10)) = 0.5993
Result
xst = F/k = 10/5000 m = 2 mm
%OS = 9.521 %
xmax = 2·(1 + 0.09521) = 2.19 mm
Answer
xmax = 2.19 mm
3.Bode plots & margins
Answer: ∠G(jω) = −56.31 °
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.
Phase crossover
(ω/4)·(ω/9) = 1 → ω180 = √(4·9) = 6 rad/s
Answer
ω180 = 6 rad/s
5.Stability: Routh–Hurwitz & root locus
Answer: Kmax = 672
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 + 14s² + 48s + K = 0
Routh array
s³: 1, 48
s²: 14, K
s¹: (14·48 − K)/14
s⁰: K
Conditions
Every entry of the first column must be positive.
K > 0
14·48 − K > 0 → K < 672
Answer
Kmax = 672
At K = 672 the closed loop oscillates at ω = √48 = 6.928 rad/s.
6.Stability: Routh–Hurwitz & root locus
Answer: K = 21
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.
Magnitude condition
K = |s|·|s + 4|·|s + 10|
Evaluate at the point
K = 3·1·7 = 21
Answer
K = 21
Check: s(s + 4)(s + 10) + K is zero at s = −3.
7.Steady-state error & PID control
Answer: ess = 5
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) = 0.2
Step part
Kp = ∞ → estep = 0
Ramp error
ess = 1/0.2 = 5
Answer
ess = 5
8.Steady-state error & PID control
Answer: ts = 1.6 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 + 5) + 2·10 = s² + 5s + 20
ωₙ and ζ
ωₙ = √20 = 4.472 rad/s
ζ = 5/(2·4.472) = 0.559
Settling time
ζωₙ = 5/2 = 2.5
ts ≈ 4/2.5 = 1.6 s
Answer
ts = 1.6 s
The settling time does not depend on K here: raising K only makes the response oscillate faster.
9.Block diagrams
Answer: ess = 0.5625
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).
Velocity error constant
Kᵥ = lims→0 s·G(s) = 16/9 = 1.778
Steady-state error
ess = A/Kᵥ = 1/1.778 = 0.5625
Answer
ess = 0.5625
10.Block diagrams
Answer: Kmax = 198
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).