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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
F(s) = 15/(s² + 9s + 18). Its inverse transform has the form f(t) = A·e−3t + B·e−6t. Find the coefficient A by partial fractions.
F(s) = 15/(s² + 9s + 18)
A =
2.Laplace transformsHard
A system obeys y″ + 6y′ + 5y = 20·u(t), with y(0) = 5 and y′(0) = 0. At t = 0 a step input u(t) = 1 is applied. Use the Laplace transform to find y(0.2).
A unity negative feedback loop has G(s) = 16/(s(s + 9)). The reference is a ramp r(t) = 1·t. Find the steady-state error ess.
G(s) = 16/(s(s + 9))r(t) = 1·t
ess =
4.Block diagramsHard
A proportional controller K = 18 drives the plant G(s) = 1/(s + 4) in a unity negative feedback loop. A disturbance D enters at the plant input (it is added to the controller output). With R = 0, a disturbance step of size 2 is applied. Find the steady-state output css.
K = 18G(s) = 1/(s + 4)D = 2R = 0
css =
5.Step responseMedium
A system has the transfer function G(s) = 21/(s² + 10s + 42) and receives a unit step input. Find the peak time tp of the step response.
G(s) = 21/(s² + 10s + 42)
tp = s
6.Step responseHard
A mass m = 2 kg hangs on a spring k = 500 N/m with a viscous damper c = 38 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 25 N is suddenly applied. Find the percent overshoot %OS of the displacement.
m = 2 kgk = 500 N/mc = 38 N·s/mF = 25 N
%OS = %
7.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram runs with K = 165 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 = 165
m =
8.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K/((s + 1)(s + 3)(s + 6)) is sketched. Find the frequency ω at which the locus crosses the imaginary axis.
L(s) = K/((s + 1)(s + 3)(s + 6))
ω = rad/s
9.Steady-state error & PID controlMedium
The parabolic input r(t) = 5·t²/2 is applied to a unity-feedback loop with G(s) = 10(s + 1)/(s²(s + 4)). Find the steady-state error ess.
G(s) = 10(s + 1)/(s²(s + 4))r(t) = 5·t²/2
ess =
10.Steady-state error & PID controlHard
A proportional controller K = 4 drives the plant G(s) = 4/(s(s + 4)) in a unity-feedback loop. Find the percent overshoot %OS of the step response.
G(s) = 4/(s(s + 4))K = 4
%OS = %
11.Bode plots & marginsMedium
In the series RLC band-pass filter shown the output is taken across R = 5 Ω, with L = 2 mH and C = 0.47 µF. Find the −3 dB bandwidth BW in Hz.
R = 5 ΩL = 2 mHC = 0.47 µF
BW = Hz
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 258/(s(s + 3)(s + 9)). Find the gain margin GM in dB.
L(s) = 258/(s(s + 3)(s + 9))
GM = dB
3Solutions
1.Laplace transforms
Answer: A = 5
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 + 9s + 18 = (s + 3)(s + 6)
F(s) = A/(s + 3) + B/(s + 6)
Cover-up rule
Multiply by (s + 3) and set s = −3: the B-term vanishes.
A = 15/(s + 6) |s = −3 = 15/3 = 5
Answer
A = 5
(B = −5, so f(t) = 5·e−3t − 5·e−6t)
2.Laplace transforms
Answer: y(0.2) = 4.931
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 + 6s + 5)·Y(s) = 5s + 30 + 20/s
Y(s) = (5s² + 30s + 20)/(s(s + 1)(s + 5))
Partial fractions
A = 20/(1·5) = 4
B = −5/((−1)·4) = 1.25
C = −5/((−5)·(−4)) = −0.25
Result
y(t) = 4 + 1.25·e−t − 0.25·e−5t
y(0.2) = 4.931
Answer
y(0.2) = 4.931
Check: y(0) = A + B + C = 5.
3.Block diagrams
Answer: ess = 0.5625
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 = 1/1.778 = 0.5625
Answer
ess = 0.5625
4.Block diagrams
Answer: css = 0.09091
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 + 4 + 18)
Steady state
css = 2·1/22 = 0.09091
Answer
css = 0.09091
A larger K reduces the effect of the disturbance, but never to zero with a proportional controller.
5.Step response
Answer: tp = 0.7619 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).
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) = √(500/2) = 15.81 rad/s
ζ = c/(2√(km)) = 38/(2·√(500·2)) = 0.6008
Result
%OS = 100·e−ζπ/√(1−ζ²) = 9.43 %
Answer
%OS = 9.43 %
7.Stability: Routh–Hurwitz & root locus
Answer: m = 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.
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 = 330/165 = 2
Answer
m = 2
This factor is the gain margin as a ratio: 6.021 dB.
8.Stability: Routh–Hurwitz & root locus
Answer: ω = 5.196 rad/s
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.
Critical gain
s³ + 10s² + 27s + 18 + K = 0
s¹: (10·27 − 18 − K)/10 = 0 → K = 252
Crossing frequency
10s² + 18 + 252 = 10s² + 270 = 0 → s² = −27
ω = √27 = 5.196 rad/s
Answer
ω = 5.196 rad/s
9.Steady-state error & PID control
Answer: ess = 2
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 = 10·1/4 = 2.5
Parabola error
ess = 5/2.5 = 2
Answer
ess = 2
10.Steady-state error & PID control
Answer: %OS = 16.3 %
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 + 4) + 4·4 = s² + 4s + 16
ωₙ and ζ
ωₙ = √16 = 4 rad/s
ζ = 4/(2·4) = 0.5
Overshoot
%OS = 100·e−0.5π/√(1 − 0.5²) = 16.3 %
Answer
%OS = 16.3 %
11.Bode plots & margins
Answer: BW = 397.9 Hz
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.