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) = 41/(s² + 10s + 41) and receives a unit step input. Find the peak time tp of the step response.
G(s) = 41/(s² + 10s + 41)
tp = s
2.Step responseHard
A series RLC circuit with R = 28 Ω, L = 200 mH and C = 10 µF is connected to a 10 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the peak capacitor voltage vC,max.
R = 28 ΩL = 200 mHC = 10 µFV = 10 V
vC,max = V
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 = 1 µF. Find the quality factor Q.
R = 2 ΩL = 5 mHC = 1 µF
Q =
4.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = K/(s(s + 3)(s + 9)). Find the gain K that gives a gain margin of 5 dB.
L(s) = K/(s(s + 3)(s + 9))GM = 5 dB
K =
5.Stability: Routh–Hurwitz & root locusMedium
A unity-feedback loop has L(s) = K/((s + 2)(s + 4)(s + 6)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.
L(s) = K/((s + 2)(s + 4)(s + 6))
Kmax =
6.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K/(s(s + 2)(s + 4)) leaves the real axis between s = 0 and s = −2. Find the break-away point sb.
L(s) = K/(s(s + 2)(s + 4))
sb =
7.Steady-state error & PID controlMedium
A unity-feedback loop with G(s) = K/(s(s + 5)) must follow the ramp r(t) = 2t with a steady-state error of 0.05. Find the gain K.
G(s) = K/(s(s + 5))r(t) = 2tess = 0.05
K =
8.Steady-state error & PID controlHard
A PD controller C(s) = 25 + 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 = 0.707.
G(s) = 10/(s(s + 1))Kp = 25ζ = 0.707
Kd = s
9.Block diagramsMedium
A unity negative feedback loop has G(s) = 2(s + 4)/((s + 1)(s + 8)). A step of size 5 is applied at the reference. Find the steady-state error ess.
G(s) = 2(s + 4)/((s + 1)(s + 8))R = 5
ess =
10.Block diagramsHard
Reduce the block diagram with G₁ = 2/(s + 1), G₂ = 1/(s + 7), an inner negative feedback H₁ = 1 around G₂ and an outer negative feedback H₂ = 1. Find the DC gain T(0) of C(s)/R(s).
G₁ = 2/(s + 1)G₂ = 1/(s + 7)H₁ = 1H₂ = 1
T(0) =
3Solutions
1.Step response
Answer: tp = 0.7854 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).
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.
Resonant frequency
ω₀ = 1/√(LC) = 1/√(0.005·1 × 10⁻⁶) = 14142 rad/s
f₀ = ω₀/(2π) = 2251 Hz
Quality factor
BW = R/L = 2/0.005 = 400 rad/s = 63.66 Hz
Q = ω₀/BW = 14142/400 = 35.36
Answer
Q = 35.36
4.Bode plots & margins
Answer: K = 182.2
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 + 12s² + 44s + 48 + K = 0
Routh array
s³: 1, 44
s²: 12, 48 + K
s¹: (12·44 − 48 − K)/12
s⁰: 48 + K
Condition
The s¹ entry must stay positive.
K < 12·44 − 48 = 480
Answer
Kmax = 480
At the limit the loop oscillates at ω = √44 = 6.633 rad/s.
6.Stability: Routh–Hurwitz & root locus
Answer: sb = −0.8453
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 + 2)(s + 4) = −(s³ + 6s² + 8s)
Stationary points
dK/ds = −(3s² + 12s + 8) = 0
s = (−12 ± √(144 − 96))/6 = −0.8453, −3.155
Pick the point on the locus
The locus lies on the real axis between 0 and −2; the other root is not on the locus.
sb = −0.8453
Answer
sb = −0.8453
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 = 2/0.05 = 40
Gain
K = 5·Kv = 200
Answer
K = 200
A large gain makes this loop very lightly damped: ζ = 0.1768. That conflict is why compensators are used.
8.Steady-state error & PID control
Answer: Kd = 2.136 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(25 + Kds) = s² + (1 + 10Kd)s + 250
Natural frequency
ωₙ = √250 = 15.81 rad/s
Derivative gain
1 + 10Kd = 2·0.707·15.81 = 22.36
Kd = 2.136 s
Answer
Kd = 2.136 s
The derivative term adds damping without changing ωₙ.
9.Block diagrams
Answer: ess = 2.5
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).
Position error constant
Kₚ = G(0) = 2·4/(1·8) = 1
Steady-state error
ess = A/(1 + Kₚ) = 5/(1 + 1) = 2.5
Answer
ess = 2.5
10.Block diagrams
Answer: T(0) = 0.2
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).