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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) = 70/(s² + 13s + 70) and receives a unit step input. Find the damping ratio ζ.
G(s) = 70/(s² + 13s + 70)
ζ =
2.Step responseHard
The step response of a second-order system without zeros shows 15 % overshoot, and the first peak occurs at t = 0.2 s. Find the natural frequency ωn.
%OS = 15 %tp = 0.2 s
ωn = rad/s
3.Bode plots & marginsMedium
The RL high-pass filter shown has R = 22 Ω, L = 47 mH. A sinusoidal input of 19 Hz is applied. Find the gain |H| = |vout/vin| in dB.
R = 22 ΩL = 47 mHf = 19 Hz
|H| = dB
4.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 60/(s(s + 10)). Find the phase margin PM in degrees.
L(s) = 60/(s(s + 10))
PM = °
5.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram has negative feedback and L(s) = 440/(s(s + 1)(s + 10)). How many closed-loop poles N lie in the right half-plane?
L(s) = 440/(s(s + 1)(s + 10))
N =
6.Stability: Routh–Hurwitz & root locusHard
For the loop L(s) = K/(s(s + 5)(s + 10)), find the gain K that puts a closed-loop pole at s = −3.
L(s) = K/(s(s + 5)(s + 10))s = −3
K =
7.Steady-state error & PID controlMedium
The input r(t) = 10 + 5t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 2(s + 3)/(s(s + 2)(s + 5)). Find the steady-state error ess.
G(s) = 2(s + 3)/(s(s + 2)(s + 5))r(t) = 10 + 5t
ess =
8.Steady-state error & PID controlHard
A proportional controller K = 50 drives the plant G(s) = 10/(s(s + 8)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 10/(s(s + 8))K = 50
ts = s
9.Block diagramsMedium
A unity negative feedback loop has G(s) = 16/(s(s + 9)). The reference is a ramp r(t) = 5·t. Find the steady-state error ess.
G(s) = 16/(s(s + 9))r(t) = 5·t
ess =
10.Block diagramsHard
A proportional controller K = 15 drives the plant G(s) = 1/(s + 8) in a unity negative feedback loop. A disturbance D enters at the plant input (it is added to the controller output). A reference step of size 2 and a disturbance step of size 5 are applied together. Find the steady-state output css.
K = 15G(s) = 1/(s + 8)D = 5R = 2
css =
3Solutions
1.Step response
Answer: ζ = 0.7769
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 = 70 → ωₙ = 8.367 rad/s
2ζωₙ = 13 → ζ = 13/(2·8.367) = 0.7769
Result
ζ = 0.7769
Answer
ζ = 0.7769
2.Step response
Answer: ωn = 18.35 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.
Corner frequency
ωc = R/L = 22/0.047 = 468.1 rad/s
fc = 74.5 Hz → f/fc = 19/74.5 = 0.255
Gain
|H| = 0.255/√(1 + 0.255²) = 0.2471
20·log₁₀(0.2471) = −12.14 dB
Answer
|H| = −12.14 dB
4.Bode plots & margins
Answer: PM = 62.07 °
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 + 100ω² − 3600 = 0
ω² = (−100 + √(10000 + 4·3600))/2 = 28.1
ωc = 5.301 rad/s
Phase margin
∠L(jωc) = −90° − atan(5.301/10) = −117.9°
PM = 180° + ∠L = 62.07°
Answer
PM = 62.07°
5.Stability: Routh–Hurwitz & root locus
Answer: N = 2
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 + 11s² + 10s + 440 = 0
Routh array
s³: 1, 10
s²: 11, 440
s¹: −30
s⁰: 440
Sign changes in the first column
1 → 11 → −30 → 440
N = 2
Answer
N = 2
The gain is above the limit K = 110, so a complex pair has crossed into the right half-plane.
6.Stability: Routh–Hurwitz & root locus
Answer: K = 42
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.
Magnitude condition
K = |s|·|s + 5|·|s + 10|
Evaluate at the point
K = 3·2·7 = 42
Answer
K = 42
Check: s(s + 5)(s + 10) + K is zero at s = −3.
7.Steady-state error & PID control
Answer: ess = 8.333
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) = 0.6
Step part
Kp = ∞ → estep = 0
Ramp error
ess = 5/0.6 = 8.333
Answer
ess = 8.333
8.Steady-state error & PID control
Answer: ts = 1 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 + 8) + 50·10 = s² + 8s + 500
ωₙ and ζ
ωₙ = √500 = 22.36 rad/s
ζ = 8/(2·22.36) = 0.1789
Settling time
ζωₙ = 8/2 = 4
ts ≈ 4/4 = 1 s
Answer
ts = 1 s
The settling time does not depend on K here: raising K only makes the response oscillate faster.
9.Block diagrams
Answer: ess = 2.813
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).
Velocity error constant
Kᵥ = lims→0 s·G(s) = 16/9 = 1.778
Steady-state error
ess = A/Kᵥ = 5/1.778 = 2.813
Answer
ess = 2.813
10.Block diagrams
Answer: css = 1.522
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).
Transfer functions
C/D = G/(1 + K·G) = 1/(s + 8 + 15)
C/R = K·G/(1 + K·G) = 15/(s + 8 + 15)
Steady state
css = 2·15/23 + 5·1/23 = 1.522
Answer
css = 1.522
A larger K reduces the effect of the disturbance, but never to zero with a proportional controller.