In the print window, choose “Save as PDF” to keep a copy on your device.
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) = 132/(s² + 12s + 66) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.
G(s) = 132/(s² + 12s + 66)
σ =
2.Step responseHard
A unity negative feedback loop has G(s) = 250/(s(s + 12)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).
G(s) = 250/(s(s + 12))
ts = s
3.Bode plots & marginsMedium
In the series RLC band-pass filter shown the output is taken across R = 2 Ω, with L = 5 mH and C = 0.47 µF. Find the centre (resonant) frequency f0 in Hz.
R = 2 ΩL = 5 mHC = 0.47 µF
f0 = Hz
4.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 98/(s(s + 4)(s + 7)). Find the gain margin GM in dB.
L(s) = 98/(s(s + 4)(s + 7))
GM = dB
5.Stability: Routh–Hurwitz & root locusMedium
A unity-feedback loop has L(s) = K/((s + 2)(s + 3)(s + 5)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.
L(s) = K/((s + 2)(s + 3)(s + 5))
Kmax =
6.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K(s + 6)/(s(s + 4)) leaves the real axis between s = 0 and s = −4, circles round and joins the real axis again to the left of the zero. Find the break-away point sb.
L(s) = K(s + 6)/(s(s + 4))
sb =
7.Steady-state error & PID controlMedium
A unity-feedback loop with G(s) = K/(s(s + 1)) must follow the ramp r(t) = 10t with a steady-state error of 0.05. Find the gain K.
G(s) = K/(s(s + 1))r(t) = 10tess = 0.05
K =
8.Steady-state error & PID controlHard
A PD controller C(s) = 4 + Kd·s drives the plant G(s) = 10/(s(s + 1)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.8.
G(s) = 10/(s(s + 1))Kp = 4ζ = 0.8
Kd = s
9.Block diagramsMedium
The plant G(s) = 10/(s + 4) has an inner (minor) negative feedback loop with gain H₁ = 0.5. A controller gain K = 13 and an outer unity negative feedback loop close the system. Find the closed-loop pole p.
G(s) = 10/(s + 4)H₁ = 0.5K = 13
p =
10.Block diagramsHard
The plant G(s) = 66/(s(s + 8)) 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) = 66/(s(s + 8))H = 2
ωd = rad/s
3Solutions
1.Step response
Answer: σ = −6
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 = 66 → ωₙ = 8.124 rad/s
2ζωₙ = 12 → ζ = 12/(2·8.124) = 0.7385
Result
σ = −ζωₙ = −12/2 = −6
Answer
σ = −6
2.Step response
Answer: ts = 0.6667 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).
For this loop ζωₙ = a/2 whatever K is: raising K adds overshoot but does not shorten the settling time.
3.Bode plots & margins
Answer: f0 = 3283 Hz
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.
Characteristic polynomial
s³ + 10s² + 31s + 30 + K = 0
Routh array
s³: 1, 31
s²: 10, 30 + K
s¹: (10·31 − 30 − K)/10
s⁰: 30 + K
Condition
The s¹ entry must stay positive.
K < 10·31 − 30 = 280
Answer
Kmax = 280
At the limit the loop oscillates at ω = √31 = 5.568 rad/s.
6.Stability: Routh–Hurwitz & root locus
Answer: sb = −2.536
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 + 4s)/(s + 6)
Stationary points
Differentiate with the quotient rule and set the numerator to zero.
(2s + 4)(s + 6) − (s² + 4s) = s² + 12s + 24 = 0
s = −6 ± √12 = −2.536, −9.464
Break-away point
sb = −2.536
Answer
sb = −2.536
Between the two break points the locus is a circle centred on the zero, with radius √12 = 3.464.
7.Steady-state error & PID control
Answer: K = 200
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 = 10/0.05 = 200
Gain
K = 1·Kv = 200
Answer
K = 200
A large gain makes this loop very lightly damped: ζ = 0.03536. That conflict is why compensators are used.
8.Steady-state error & PID control
Answer: Kd = 0.9119 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 + 1) + 10(4 + Kds) = s² + (1 + 10Kd)s + 40
Natural frequency
ωₙ = √40 = 6.325 rad/s
Derivative gain
1 + 10Kd = 2·0.8·6.325 = 10.12
Kd = 0.9119 s
Answer
Kd = 0.9119 s
The derivative term adds damping without changing ωₙ.
9.Block diagrams
Answer: p = −139
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).
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).