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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) = 182/(s² + 4s + 91) and receives a unit step input. Find the percent overshoot %OS of the step response.
G(s) = 182/(s² + 4s + 91)
%OS = %
2.Step responseHard
A unity negative feedback loop has G(s) = K/(s(s + 5)). Find the gain K that gives a step response with 15 % overshoot.
G(s) = K/(s(s + 5))%OS = 15 %
K =
3.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 20(s + 15)/(s(s + 6)) at ω = 1 rad/s, in degrees.
G(s) = 20(s + 15)/(s(s + 6))ω = 1 rad/s
∠G(jω) = °
4.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 1210/(s(s + 9)(s + 10)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.
L(s) = 1210/(s(s + 9)(s + 10))
ω180 = rad/s
5.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram runs with K = 66 and L(s) = K/(s(s + 5)(s + 6)). By what factor m can the gain be raised before the closed loop becomes unstable?
L(s) = K/(s(s + 5)(s + 6))K = 66
m =
6.Stability: Routh–Hurwitz & root locusHard
For L(s) = K/((s + 6)(s² + 6s + 10)), 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 + 6)(s² + 6s + 10))
θd = °
7.Steady-state error & PID controlMedium
The parabolic input r(t) = 2·t²/2 is applied to a unity-feedback loop with G(s) = 4(s + 5)/(s²(s + 4)). Find the steady-state error ess.
G(s) = 4(s + 5)/(s²(s + 4))r(t) = 2·t²/2
ess =
8.Steady-state error & PID controlHard
A PD controller C(s) = 25 + Kd·s drives the plant G(s) = 10/(s(s + 3)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.9.
G(s) = 10/(s(s + 3))Kp = 25ζ = 0.9
Kd = s
9.Block diagramsMedium
A plant G(s) = 15/(s + 1) is controlled in a negative feedback loop whose sensor has the gain H = 0.2. Find the time constant τ of the closed loop.
G(s) = 15/(s + 1)H = 0.2
τ = s
10.Block diagramsHard
A proportional controller K drives the plant G(s) = 1/(s + 8) in a unity negative feedback loop. A disturbance step of size 10 enters at the plant input while R = 0. Find the smallest gain K that keeps the steady-state output at or below 0.0625.
G(s) = 1/(s + 8)D = 10css ≤ 0.0625
K =
3Solutions
1.Step response
Answer: %OS = 50.99 %
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 = 91 → ωₙ = 9.539 rad/s
2ζωₙ = 4 → ζ = 4/(2·9.539) = 0.2097
Result
%OS = 100·e−ζπ/√(1−ζ²) = 100·e−0.6736 = 50.99 %
Answer
%OS = 50.99 %
2.Step response
Answer: K = 23.39
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).
Closed-loop transfer function
T(s) = K/(s² + 5s + K)
Damping ratio from the overshoot
ζ = −ln(0.15)/√(π² + ln²(0.15)) = 0.5169
Gain
ωₙ = 5/(2ζ) = 4.836 rad/s
K = ωₙ² = 23.39
Answer
K = 23.39
3.Bode plots & margins
Answer: ∠G(jω) = −95.65 °
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/9)·(ω/10) = 1 → ω180 = √(9·10) = 9.487 rad/s
Answer
ω180 = 9.487 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 + 11s² + 30s + K = 0
s³: 1, 30
s²: 11, K
s¹: (11·30 − K)/11
s⁰: K
Largest stable gain
Kmax = 11·30 = 330
Factor
m = 330/66 = 5
Answer
m = 5
This factor is the gain margin as a ratio: 13.98 dB.
6.Stability: Routh–Hurwitz & root locus
Answer: θd = 71.57 °
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 + 6s + 10 = 0 → s = −3 ± j1
s = −6
Angles from the other poles
∠(−3 + j1 + 6) = atan2(1, 3) = 18.43°
∠(−3 + j1 − (−3 − j1)) = ∠(j2) = 90°
Angle of departure
θd = 180° − 18.43° − 90° = 71.57°
Answer
θd = 71.57°
7.Steady-state error & PID control
Answer: ess = 0.4
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.
Acceleration constant
Ka = 4·5/4 = 5
Parabola error
ess = 2/5 = 0.4
Answer
ess = 0.4
8.Steady-state error & PID control
Answer: Kd = 2.546 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 + 3) + 10(25 + Kds) = s² + (3 + 10Kd)s + 250
Natural frequency
ωₙ = √250 = 15.81 rad/s
Derivative gain
3 + 10Kd = 2·0.9·15.81 = 28.46
Kd = 2.546 s
Answer
Kd = 2.546 s
The derivative term adds damping without changing ωₙ.
9.Block diagrams
Answer: τ = 0.25 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) = 15/(s + 1 + 3) = 15/(s + 4)
Result
τ = 1/4 = 0.25 s
Answer
τ = 0.25 s
10.Block diagrams
Answer: K = 152
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).