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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) = 50/(s(s + 2)(s + 6)) by partial fractions, and evaluate f(0.25).
F(s) = 50/(s(s + 2)(s + 6))t = 0.25 s
f(0.25) =
2.Laplace transformsHard
Solve y″ + 9y′ + 18y = 0 with y(0) = 8 and y′(0) = 2 using the Laplace transform, and find y(0.25).
y″ + 9y′ + 18y = 0y(0) = 8y′(0) = 2t = 0.25 s
y(0.25) =
3.Block diagramsMedium
A unity negative feedback loop has G(s) = 38/(s(s + 3)). The reference is a ramp r(t) = 10·t. Find the steady-state error ess.
G(s) = 38/(s(s + 3))r(t) = 10·t
ess =
4.Block diagramsHard
A proportional controller K = 8 drives the plant G(s) = 10/(s + 2) 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 2 are applied together. Find the steady-state output css.
K = 8G(s) = 10/(s + 2)D = 2R = 2
css =
5.Step responseMedium
A system has the transfer function G(s) = 68/(s² + 4s + 68) and receives a unit step input. Find the percent overshoot %OS of the step response.
G(s) = 68/(s² + 4s + 68)
%OS = %
6.Step responseHard
A unity negative feedback loop has G(s) = 60/(s(s + 6)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).
G(s) = 60/(s(s + 6))
ts = s
7.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram runs with K = 32.4 and L(s) = K/(s(s + 3)(s + 6)). By what factor m can the gain be raised before the closed loop becomes unstable?
L(s) = K/(s(s + 3)(s + 6))K = 32.4
m =
8.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K(s + 10)/(s(s + 4)) leaves the real axis between s = 0 and s = −4, circles round and joins the real axis again to the left of the zero. Find the break-away point sb.
L(s) = K(s + 10)/(s(s + 4))
sb =
9.Steady-state error & PID controlMedium
A unity-feedback loop with G(s) = K/(s(s + 5)) must follow the ramp r(t) = 10t with a steady-state error of 0.05. Find the gain K.
G(s) = K/(s(s + 5))r(t) = 10tess = 0.05
K =
10.Steady-state error & PID controlHard
A proportional controller K drives the plant G(s) = 5/(s(s + 5)) in a unity-feedback loop. Find the gain K that gives the closed loop a damping ratio ζ = 0.6.
G(s) = 5/(s(s + 5))ζ = 0.6
K =
11.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 20(s + 4)/(s(s + 17)) at ω = 2 rad/s, in degrees.
G(s) = 20(s + 4)/(s(s + 17))ω = 2 rad/s
∠G(jω) = °
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 148/(s(s + 3)(s + 8)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.
L(s) = 148/(s(s + 3)(s + 8))
ω180 = rad/s
3Solutions
1.Laplace transforms
Answer: f(0.25) = 0.8407
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).
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).
Transform the equation
s²Y − 8s − 2 + 9(sY − 8) + 18Y = 0
Y(s) = (8s + 74)/((s + 3)(s + 6))
Partial fractions
A = 50/3 = 16.67
B = 26/(−3) = −8.667
Result
y(t) = 16.67·e−3t − 8.667·e−6t
y(0.25) = 5.939
Answer
y(0.25) = 5.939
Check: y(0) = A + B = 8.
3.Block diagrams
Answer: ess = 0.7895
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) = 38/3 = 12.67
Steady-state error
ess = A/Kᵥ = 10/12.67 = 0.7895
Answer
ess = 0.7895
4.Block diagrams
Answer: css = 2.195
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) = 10/(s + 2 + 80)
C/R = K·G/(1 + K·G) = 80/(s + 2 + 80)
Steady state
css = 2·80/82 + 2·10/82 = 2.195
Answer
css = 2.195
A larger K reduces the effect of the disturbance, but never to zero with a proportional controller.
5.Step response
Answer: %OS = 45.59 %
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 = 68 → ωₙ = 8.246 rad/s
2ζωₙ = 4 → ζ = 4/(2·8.246) = 0.2425
Result
%OS = 100·e−ζπ/√(1−ζ²) = 100·e−0.7854 = 45.59 %
Answer
%OS = 45.59 %
6.Step response
Answer: ts = 1.333 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).
For this loop ζωₙ = a/2 whatever K is: raising K adds overshoot but does not shorten the settling time.
7.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 + 18s + K = 0
s³: 1, 18
s²: 9, K
s¹: (9·18 − K)/9
s⁰: K
Largest stable gain
Kmax = 9·18 = 162
Factor
m = 162/32.4 = 5
Answer
m = 5
This factor is the gain margin as a ratio: 13.98 dB.
8.Stability: Routh–Hurwitz & root locus
Answer: sb = −2.254
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) = −(s² + 4s)/(s + 10)
Stationary points
Differentiate with the quotient rule and set the numerator to zero.
(2s + 4)(s + 10) − (s² + 4s) = s² + 20s + 40 = 0
s = −10 ± √60 = −2.254, −17.75
Break-away point
sb = −2.254
Answer
sb = −2.254
Between the two break points the locus is a circle centred on the zero, with radius √60 = 7.746.
9.Steady-state error & PID control
Answer: K = 1000
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 = 10/0.05 = 200
Gain
K = 5·Kv = 1000
Answer
K = 1000
A large gain makes this loop very lightly damped: ζ = 0.07906. That conflict is why compensators are used.
10.Steady-state error & PID control
Answer: K = 3.472
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) + 5K = s² + 5s + 5K
Natural frequency
2·0.6·ωₙ = 5 → ωₙ = 4.167 rad/s
Gain
5K = ωₙ² = 17.36 → K = 3.472
Answer
K = 3.472
11.Bode plots & margins
Answer: ∠G(jω) = −70.14 °
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.