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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) = (s + 10)/(s² + 4s + 4) and evaluate f(1).
F(s) = (s + 10)/(s² + 4s + 4)t = 1 s
f(1) =
2.Laplace transformsHard
A system obeys y″ + 8y′ + 12y = 12·u(t), with y(0) = 2 and y′(0) = 0. At t = 0 a step input u(t) = 10 is applied. Use the Laplace transform to find y(0.25).
The plant G(s) = K/(s(s + 16)) 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 + 16))unity negative feedback
K =
4.Block diagramsHard
A proportional controller K = 17 drives the plant G(s) = 5/(s + 5) 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 0.5 are applied together. Find the steady-state output css.
K = 17G(s) = 5/(s + 5)D = 0.5R = 2
css =
5.Step responseMedium
A system has the transfer function G(s) = 104/(s² + 3s + 52) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.
G(s) = 104/(s² + 3s + 52)
σ =
6.Step responseHard
A mass m = 10 kg hangs on a spring k = 1000 N/m with a viscous damper c = 100 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 50 N is suddenly applied. Find the largest displacement xmax in mm.
m = 10 kgk = 1000 N/mc = 100 N·s/mF = 50 N
xmax = mm
7.Stability: Routh–Hurwitz & root locusMedium
The plant G(s) = (s + 4)/((s − 1)(s + 1)) has an unstable pole at s = 1. A proportional gain K closes a unity-feedback loop around it. Find the smallest gain Kmin that makes the closed loop stable.
G(s) = (s + 4)/((s − 1)(s + 1))
Kmin =
8.Stability: Routh–Hurwitz & root locusHard
For L(s) = K/((s + 1)(s² + 6s + 18)), 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 + 1)(s² + 6s + 18))
θd = °
9.Steady-state error & PID controlMedium
A process G(s) = 4/(1s + 1) is controlled by a proportional gain K = 10 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.
G(s) = 4/(1s + 1)K = 10
ess = %
10.Steady-state error & PID controlHard
A PD controller C(s) = 10 + 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 = 0.707.
G(s) = 10/(s(s + 3))Kp = 10ζ = 0.707
Kd = s
11.Bode plots & marginsMedium
The RC low-pass filter shown has R = 3.3 kΩ, C = 47 nF. A sinusoidal input of 10 kHz is applied. Find the gain |H| = |vout/vin| in dB.
R = 3.3 kΩC = 47 nFf = 10 kHz
|H| = dB
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 12/(s(s + 2)(s + 4)). Find the gain margin GM in dB.
L(s) = 12/(s(s + 2)(s + 4))
GM = dB
3Solutions
1.Laplace transforms
Answer: f(1) = 1.218
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 + 4s + 4 = (s + 2)²
Partial fractions
s + 10 = 1(s + 2) + 8
F(s) = 1/(s + 2) + 8/(s + 2)²
Inverse transform
f(t) = e−2t + 8·t·e−2t
f(1) = (1 + 8·1)·e−2·1 = 1.218
Answer
f(1) = 1.218
2.Laplace transforms
Answer: y(0.25) = 3.614
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
(s² + 8s + 12)·Y(s) = 2s + 16 + 120/s
Y(s) = (2s² + 16s + 120)/(s(s + 2)(s + 6))
Partial fractions
A = 120/(2·6) = 10
B = 96/((−2)·4) = −12
C = 96/((−6)·(−4)) = 4
Result
y(t) = 10 − 12·e−2t + 4·e−6t
y(0.25) = 3.614
Answer
y(0.25) = 3.614
Check: y(0) = A + B + C = 2.
3.Block diagrams
Answer: K = 64
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 + 16)) = 0 → s² + 16s + K = 0
Repeated root
A quadratic s² + βs + γ has a double root when β² = 4γ.
16² = 4K → K = 256/4 = 64
Answer
K = 64
Both closed-loop poles are then at s = −8.
4.Block diagrams
Answer: css = 1.917
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) = 5/(s + 5 + 85)
C/R = K·G/(1 + K·G) = 85/(s + 5 + 85)
Steady state
css = 2·85/90 + 0.5·5/90 = 1.917
Answer
css = 1.917
A larger K reduces the effect of the disturbance, but never to zero with a proportional controller.
5.Step response
Answer: σ = −1.5
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 = 52 → ωₙ = 7.211 rad/s
2ζωₙ = 3 → ζ = 3/(2·7.211) = 0.208
Result
σ = −ζωₙ = −3/2 = −1.5
Answer
σ = −1.5
6.Step response
Answer: xmax = 58.15 mm
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 = √(k/m) = √(1000/10) = 10 rad/s
ζ = c/(2√(km)) = 100/(2·√(1000·10)) = 0.5
Result
xst = F/k = 50/1000 m = 50 mm
%OS = 16.3 %
xmax = 50·(1 + 0.163) = 58.15 mm
Answer
xmax = 58.15 mm
7.Stability: Routh–Hurwitz & root locus
Answer: Kmin = 0.25
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.
For s² + c₁s + c₀ all roots lie in the left half-plane exactly when c₁ > 0 and c₀ > 0.
0 + K > 0 → K > 0
4K − 1 > 0 → K > 0.25
Kmin = 0.25
Answer
Kmin = 0.25
8.Stability: Routh–Hurwitz & root locus
Answer: θd = −33.69 °
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 + 18 = 0 → s = −3 ± j3
s = −1
Angles from the other poles
∠(−3 + j3 + 1) = atan2(3, -2) = 123.7°
∠(−3 + j3 − (−3 − j3)) = ∠(j6) = 90°
Angle of departure
θd = 180° − 123.7° − 90° = −33.69°
Answer
θd = −33.69°
9.Steady-state error & PID control
Answer: ess = 2.439 %
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.
Position constant
Kp = 10·4 = 40
Step error
ess = 100/(1 + 40) = 2.439 %
Answer
ess = 2.439 %
10.Steady-state error & PID control
Answer: Kd = 1.114 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(10 + Kds) = s² + (3 + 10Kd)s + 100
Natural frequency
ωₙ = √100 = 10 rad/s
Derivative gain
3 + 10Kd = 2·0.707·10 = 14.14
Kd = 1.114 s
Answer
Kd = 1.114 s
The derivative term adds damping without changing ωₙ.
11.Bode plots & margins
Answer: |H| = −19.82 dB
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/(3300·4.7 × 10⁻⁸) = 6447 rad/s
fc = 1026 Hz → f/fc = 10000/1026 = 9.745
Gain
|H| = 1/√(1 + 9.745²) = 0.1021
20·log₁₀(0.1021) = −19.82 dB
Answer
|H| = −19.82 dB
12.Bode plots & margins
Answer: GM = 12.04 dB
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.