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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.Laplace transformsMedium
Find f(t) for F(s) = (3s + 7)/(s² + 2s + 1) and evaluate f(0.8).
F(s) = (3s + 7)/(s² + 2s + 1)t = 0.8 s
f(0.8) =
2.Laplace transformsHard
A system is described by the differential equation y″ + 2y′ + 2y = u′ + 17u, with input u(t) and output y(t). Find its transfer function G(s) = Y(s)/U(s) (zero initial conditions) and evaluate the impulse response g(t) at t = 0.4 s. Find g(0.4).
y″ + 2y′ + 2y = u′ + 17ut = 0.4 s
g(0.4) =
3.Block diagramsMedium
The plant G(s) = 1/(s + 5) has an inner (minor) negative feedback loop with gain H₁ = 0.5. A controller gain K = 14 and an outer unity negative feedback loop close the system. Find the closed-loop pole p.
G(s) = 1/(s + 5)H₁ = 0.5K = 14
p =
4.Block diagramsHard
The plant G(s) = K/((s + 3)(s + 4)(s + 7)) is in a unity negative feedback loop. At the largest stable gain the loop oscillates. Find the frequency of oscillation ω in rad/s.
A system has the transfer function G(s) = 26/(s² + s + 13) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.
G(s) = 26/(s² + s + 13)
σ =
6.Step responseHard
The step response of a second-order system without zeros shows 30 % overshoot, and the first peak occurs at t = 0.8 s. Find the natural frequency ωn.
%OS = 30 %tp = 0.8 s
ωn = rad/s
7.Stability: Routh–Hurwitz & root locusMedium
A unity-feedback loop has L(s) = K/((s + 1)(s + 2)(s + 3)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.
L(s) = K/((s + 1)(s + 2)(s + 3))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
For the loop L(s) = K/(s(s + 3)(s + 5)), find the gain K that puts a closed-loop pole at s = −1.5.
L(s) = K/(s(s + 3)(s + 5))s = −1.5
K =
9.Steady-state error & PID controlMedium
A ramp r(t) = 2t is applied to a unity-feedback loop with G(s) = 50/(s(s + 4)). Find the steady-state error ess.
G(s) = 50/(s(s + 4))r(t) = 2t
ess =
10.Steady-state error & PID controlHard
A proportional controller K = 40 drives the plant G(s) = 5/(s(s + 5)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 5/(s(s + 5))K = 40
ts = s
11.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 7(s + 6)/(s(s + 4)) at ω = 20 rad/s, in degrees.
G(s) = 7(s + 6)/(s(s + 4))ω = 20 rad/s
∠G(jω) = °
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 16/(s(s + 2)(s + 4)). Find the gain margin GM in dB.
L(s) = 16/(s(s + 2)(s + 4))
GM = dB
3Solutions
1.Laplace transforms
Answer: f(0.8) = 2.786
Laplace transform: F(s) = ∫₀∞ f(t)·e−st dt turns derivatives into multiplications by s: ℒ{f′} = sF(s) − f(0), ℒ{f″} = s²F(s) − s·f(0) − f′(0). Standard pairs: 1 ↔ 1/s, e−at ↔ 1/(s + a), t·e−at ↔ 1/(s + a)², sin ωt ↔ ω/(s² + ω²), cos ωt ↔ s/(s² + ω²), and a shift s → s + σ multiplies the time function by e−σt. To invert a rational F(s), split it into partial fractions: the residue of a simple pole p is A = [(s − p)·F(s)] at s = p. Final value theorem: f(∞) = lims→0 s·F(s), valid only when all poles of s·F(s) lie in the left half-plane. Initial value theorem: f(0⁺) = lims→∞ s·F(s).
Factor the denominator
s² + 2s + 1 = (s + 1)²
Partial fractions
3s + 7 = 3(s + 1) + 4
F(s) = 3/(s + 1) + 4/(s + 1)²
Inverse transform
f(t) = 3·e−t + 4·t·e−t
f(0.8) = (3 + 4·0.8)·e−0.8 = 2.786
Answer
f(0.8) = 2.786
2.Laplace transforms
Answer: g(0.4) = 4.794
Laplace transform: F(s) = ∫₀∞ f(t)·e−st dt turns derivatives into multiplications by s: ℒ{f′} = sF(s) − f(0), ℒ{f″} = s²F(s) − s·f(0) − f′(0). Standard pairs: 1 ↔ 1/s, e−at ↔ 1/(s + a), t·e−at ↔ 1/(s + a)², sin ωt ↔ ω/(s² + ω²), cos ωt ↔ s/(s² + ω²), and a shift s → s + σ multiplies the time function by e−σt. To invert a rational F(s), split it into partial fractions: the residue of a simple pole p is A = [(s − p)·F(s)] at s = p. Final value theorem: f(∞) = lims→0 s·F(s), valid only when all poles of s·F(s) lie in the left half-plane. Initial value theorem: f(0⁺) = lims→∞ s·F(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).
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
s³ + 14s² + 61s + 84 + K = 0
Routh array
s³ | 1 61
s² | 14 84 + K
s¹ | (14·61 − 84 − K)/14
s⁰ | 84 + K
Stability limit
All entries of the first column must be positive.
Kmax = 14·61 − 84 = 770
Frequency of oscillation
14·(jω)² + 84 + 770 = 0 → ω² = 854/14 = 61
ω = 7.81 rad/s
Answer
ω = 7.81 rad/s
5.Step response
Answer: σ = −0.5
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 = 13 → ωₙ = 3.606 rad/s
2ζωₙ = 1 → ζ = 1/(2·3.606) = 0.1387
Result
σ = −ζωₙ = −1/2 = −0.5
Answer
σ = −0.5
6.Step response
Answer: ωn = 4.205 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).
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 + 6s² + 11s + 6 + K = 0
Routh array
s³: 1, 11
s²: 6, 6 + K
s¹: (6·11 − 6 − K)/6
s⁰: 6 + K
Condition
The s¹ entry must stay positive.
K < 6·11 − 6 = 60
Answer
Kmax = 60
At the limit the loop oscillates at ω = √11 = 3.317 rad/s.
8.Stability: Routh–Hurwitz & root locus
Answer: K = 7.875
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 + 3|·|s + 5|
Evaluate at the point
K = 1.5·1.5·3.5 = 7.875
Answer
K = 7.875
Check: s(s + 3)(s + 5) + K is zero at s = −1.5.
9.Steady-state error & PID control
Answer: ess = 0.16
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) = 12.5
Ramp error
ess = 2/12.5 = 0.16
Answer
ess = 0.16
10.Steady-state error & PID control
Answer: ts = 1.6 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 + 5) + 40·5 = s² + 5s + 200
ωₙ and ζ
ωₙ = √200 = 14.14 rad/s
ζ = 5/(2·14.14) = 0.1768
Settling time
ζωₙ = 5/2 = 2.5
ts ≈ 4/2.5 = 1.6 s
Answer
ts = 1.6 s
The settling time does not depend on K here: raising K only makes the response oscillate faster.
11.Bode plots & margins
Answer: ∠G(jω) = −95.39 °
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.