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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) = 172/(s² + 8s + 86) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 172/(s² + 8s + 86)
ts = s
2.Step responseHard
A series RLC circuit with R = 18 Ω, L = 10 mH and C = 4.7 µF is connected to a 5 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the undamped natural frequency ωn in rad/s.
R = 18 ΩL = 10 mHC = 4.7 µFV = 5 V
ωn = rad/s
3.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 33(s + 11)/(s(s + 10)) at ω = 5 rad/s, in degrees.
G(s) = 33(s + 11)/(s(s + 10))ω = 5 rad/s
∠G(jω) = °
4.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 596/(s(s + 7)(s + 10)). Find the gain margin GM in dB.
L(s) = 596/(s(s + 7)(s + 10))
GM = dB
5.Stability: Routh–Hurwitz & root locusMedium
A unity-feedback loop has L(s) = K(s + 9)/(s(s + 1)(s + 2)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.
L(s) = K(s + 9)/(s(s + 1)(s + 2))
Kmax =
6.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K(s + 7)/(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 + 7)/(s(s + 6))
si =
7.Steady-state error & PID controlMedium
A unity-feedback loop with G(s) = K/(s(s + 5)) must follow the ramp r(t) = 1t with a steady-state error of 0.5. Find the gain K.
G(s) = K/(s(s + 5))r(t) = 1tess = 0.5
K =
8.Steady-state error & PID controlHard
A proportional controller K = 50 drives the plant G(s) = 1/(s(s + 4)) in a unity-feedback loop. Find the peak time tp of the step response.
G(s) = 1/(s(s + 4))K = 50
tp = s
9.Block diagramsMedium
A plant G(s) = 20/(s + 9) is controlled in a negative feedback loop whose sensor has the gain H = 0.1. Find the time constant τ of the closed loop.
G(s) = 20/(s + 9)H = 0.1
τ = s
10.Block diagramsHard
The plant G(s) = K/(s(s + 4)(s + 10)) 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.
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 = 86 → ωₙ = 9.274 rad/s
2ζωₙ = 8 → ζ = 8/(2·9.274) = 0.4313
Result
ζωₙ = 4 → ts ≈ 4/4 = 1 s
Answer
ts = 1 s
2.Step response
Answer: ωn = 4613 rad/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).
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.
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.
With the zero to the right of s = −3 the loop would stay stable for every K > 0.
6.Stability: Routh–Hurwitz & root locus
Answer: si = −9.646
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 + 7)
Stationary points
Differentiate with the quotient rule and set the numerator to zero.
(2s + 6)(s + 7) − (s² + 6s) = s² + 14s + 42 = 0
s = −7 ± √7 = −4.354, −9.646
Break-in point
si = −9.646
Answer
si = −9.646
Between the two break points the locus is a circle centred on the zero, with radius √7 = 2.646.
7.Steady-state error & PID control
Answer: K = 10
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 velocity constant
Kv = A/ess = 1/0.5 = 2
Gain
K = 5·Kv = 10
Answer
K = 10
A large gain makes this loop very lightly damped: ζ = 0.7906. That conflict is why compensators are used.
8.Steady-state error & PID control
Answer: tp = 0.4632 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 + 4) + 50·1 = s² + 4s + 50
ωₙ and ζ
ωₙ = √50 = 7.071 rad/s
ζ = 4/(2·7.071) = 0.2828
Peak time
ωd = 7.071·√(1 − 0.2828²) = 6.782 rad/s
tp = π/6.782 = 0.4632 s
Answer
tp = 0.4632 s
9.Block diagrams
Answer: τ = 0.09091 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).
Closed-loop transfer function
T(s) = G/(1 + G·H) = 20/(s + 9 + 2) = 20/(s + 11)
Result
τ = 1/11 = 0.09091 s
Answer
τ = 0.09091 s
10.Block diagrams
Answer: Kmax = 560
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).