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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) = 54/(s² + 8s + 54) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.
G(s) = 54/(s² + 8s + 54)
σ =
2.Step responseHard
A series RLC circuit with R = 100 Ω, L = 100 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 = 100 ΩL = 100 mHC = 10 µFV = 10 V
vC,max = V
3.Bode plots & marginsMedium
The RC low-pass filter shown has R = 10 kΩ, C = 22 nF. A sinusoidal input of 360 Hz is applied. Find the phase shift φ of vout relative to vin, in degrees.
R = 10 kΩC = 22 nFf = 360 Hz
φ = °
4.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 92/(s(s + 8)). Find the gain crossover frequency ωc, where |L(jω)| = 1.
L(s) = 92/(s(s + 8))
ωc = rad/s
5.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram runs with K = 14.4 and L(s) = K/(s(s + 1)(s + 8)). By what factor m can the gain be raised before the closed loop becomes unstable?
L(s) = K/(s(s + 1)(s + 8))K = 14.4
m =
6.Stability: Routh–Hurwitz & root locusHard
For L(s) = K/((s + 5)(s² + 8s + 32)), find the angle of departure θd of the root locus from the complex pole in the upper half-plane, measured from the positive real axis.
L(s) = K/((s + 5)(s² + 8s + 32))
θd = °
7.Steady-state error & PID controlMedium
The input r(t) = 1 + 3t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 20/(s(s + 10)). Find the steady-state error ess.
G(s) = 20/(s(s + 10))r(t) = 1 + 3t
ess =
8.Steady-state error & PID controlHard
A PI controller C(s) = 1 + 5/s drives the plant G(s) = 2/((s + 5)(s + 10)) in a unity-feedback loop. Find the steady-state error ess for the ramp input r(t) = 2t.
C(s) = 1 + 5/sG(s) = 2/((s + 5)(s + 10))r(t) = 2t
ess =
9.Block diagramsMedium
The plant G(s) = K/(s(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(s + 2))unity negative feedback
K =
10.Block diagramsHard
The plant G(s) = K/(s(s + 1)(s + 8)) is in a unity negative feedback loop. At the largest stable gain the loop oscillates. Find the frequency of oscillation ω in rad/s.
G(s) = K/(s(s + 1)(s + 8))unity negative feedback
ω = rad/s
3Solutions
1.Step response
Answer: σ = −4
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 = 54 → ωₙ = 7.348 rad/s
2ζωₙ = 8 → ζ = 8/(2·7.348) = 0.5443
Result
σ = −ζωₙ = −8/2 = −4
Answer
σ = −4
2.Step response
Answer: vC,max = 11.63 V
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 = 1/√(LC) = 1/√(0.1·1 × 10⁻⁵) = 1000 rad/s
ζ = (R/2)·√(C/L) = (100/2)·√(1 × 10⁻⁵/0.1) = 0.5
Result
%OS = 16.3 %
vC,max = 10·(1 + 0.163) = 11.63 V
Answer
vC,max = 11.63 V
3.Bode plots & margins
Answer: φ = −26.46 °
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.
Corner frequency
ωc = 1/(RC) = 1/(10000·2.2 × 10⁻⁸) = 4545 rad/s
fc = 723.4 Hz → f/fc = 360/723.4 = 0.4976
Phase
φ = − atan(0.4976) = −26.46°
Answer
φ = −26.46°
4.Bode plots & margins
Answer: ωc = 8.087 rad/s
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 + 64ω² − 8464 = 0
ω² = (−64 + √(4096 + 4·8464))/2 = 65.41
ωc = 8.087 rad/s
Answer
ωc = 8.087 rad/s
5.Stability: Routh–Hurwitz & root locus
Answer: m = 5
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 + 9s² + 8s + K = 0
s³: 1, 8
s²: 9, K
s¹: (9·8 − K)/9
s⁰: K
Largest stable gain
Kmax = 9·8 = 72
Factor
m = 72/14.4 = 5
Answer
m = 5
This factor is the gain margin as a ratio: 13.98 dB.
6.Stability: Routh–Hurwitz & root locus
Answer: θd = 14.04 °
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.
Pole locations
s² + 8s + 32 = 0 → s = −4 ± j4
s = −5
Angles from the other poles
∠(−4 + j4 + 5) = atan2(4, 1) = 75.96°
∠(−4 + j4 − (−4 − j4)) = ∠(j8) = 90°
Angle of departure
θd = 180° − 75.96° − 90° = 14.04°
Answer
θd = 14.04°
7.Steady-state error & PID control
Answer: ess = 1.5
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) = 2
Step part
Kp = ∞ → estep = 0
Ramp error
ess = 3/2 = 1.5
Answer
ess = 1.5
8.Steady-state error & PID control
Answer: ess = 10
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).
Velocity constant
s·C(s) = 1s + 5 → 5 as s → 0.
Kv = 5·2/50 = 0.2
Ramp error
ess = 2/0.2 = 10
Answer
ess = 10
The PI controller removes the step error completely; a ramp is followed with a constant lag.
9.Block diagrams
Answer: K = 1
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(s + 2)) = 0 → s² + 2s + K = 0
Repeated root
A quadratic s² + βs + γ has a double root when β² = 4γ.
2² = 4K → K = 4/4 = 1
Answer
K = 1
Both closed-loop poles are then at s = −1.
10.Block diagrams
Answer: ω = 2.828 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).