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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) = 26/(s² + 2s + 5) by completing the square, and evaluate f(0.2).
F(s) = 26/(s² + 2s + 5)t = 0.2 s
f(0.2) =
2.Laplace transformsHard
A system obeys y″ + 6y′ + 5y = 5·u(t), with y(0) = 1 and y′(0) = 0. At t = 0 a step input u(t) = 4 is applied. Use the Laplace transform to find y(0.3).
The plant G(s) = 4/(s + 6) has an inner (minor) negative feedback loop with gain H₁ = 3. A controller gain K = 13 and an outer unity negative feedback loop close the system. Find the closed-loop pole p.
G(s) = 4/(s + 6)H₁ = 3K = 13
p =
4.Block diagramsHard
A proportional controller K = 5 drives the plant G(s) = 5/(s + 7) 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 10 is applied. Find the steady-state output css.
K = 5G(s) = 5/(s + 7)D = 10R = 0
css =
5.Step responseMedium
A system has the transfer function G(s) = 291/(s² + 10s + 97) and receives a unit step input. Find the peak time tp of the step response.
G(s) = 291/(s² + 10s + 97)
tp = s
6.Step responseHard
A series RLC circuit with R = 283 Ω, L = 500 mH and C = 1 µF is connected to a 5 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the peak capacitor voltage vC,max.
R = 283 ΩL = 500 mHC = 1 µFV = 5 V
vC,max = V
7.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram runs with K = 60 and L(s) = K/(s(s + 4)(s + 5)). By what factor m can the gain be raised before the closed loop becomes unstable?
L(s) = K/(s(s + 4)(s + 5))K = 60
m =
8.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K(s + 5)/(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 + 5)/(s(s + 4))
sb =
9.Steady-state error & PID controlMedium
The parabolic input r(t) = 4·t²/2 is applied to a unity-feedback loop with G(s) = 2(s + 1)/(s²(s + 10)). Find the steady-state error ess.
G(s) = 2(s + 1)/(s²(s + 10))r(t) = 4·t²/2
ess =
10.Steady-state error & PID controlHard
A PI controller C(s) = 5 + 10/s drives the plant G(s) = 4/((s + 1)(s + 10)) in a unity-feedback loop. Find the steady-state error ess for the ramp input r(t) = 2t.
In the series RLC band-pass filter shown the output is taken across R = 1 Ω, with L = 1 mH and C = 1 µF. Find the centre (resonant) frequency f0 in Hz.
R = 1 ΩL = 1 mHC = 1 µF
f0 = Hz
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 70/(s(s + 3)(s + 10)). Find the gain margin GM in dB.
L(s) = 70/(s(s + 3)(s + 10))
GM = dB
3Solutions
1.Laplace transforms
Answer: f(0.2) = 4.145
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).
Complete the square
s² + 2s + 5 = (s + 1)² + 4 = (s + 1)² + 2²
Match the numerator
F(s) = (26/2)·2/((s + 1)² + 2²)
Inverse transform
f(t) = e−t·(13·sin 2t)
f(0.2) = e−0.2·(13·sin(2·0.2)) = 4.145
Answer
f(0.2) = 4.145
2.Laplace transforms
Answer: y(0.3) = 1.389
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) = 1s + 6 + 20/s
Y(s) = (s² + 6s + 20)/(s(s + 1)(s + 5))
Partial fractions
A = 20/(1·5) = 4
B = 15/((−1)·4) = −3.75
C = 15/((−5)·(−4)) = 0.75
Result
y(t) = 4 − 3.75·e−t + 0.75·e−5t
y(0.3) = 1.389
Answer
y(0.3) = 1.389
Check: y(0) = A + B + C = 1.
3.Block diagrams
Answer: p = −70
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).
Transfer functions
C/D = G/(1 + K·G) = 5/(s + 7 + 25)
Steady state
css = 10·5/32 = 1.563
Answer
css = 1.563
A larger K reduces the effect of the disturbance, but never to zero with a proportional controller.
5.Step response
Answer: tp = 0.3702 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).
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 + 20s + K = 0
s³: 1, 20
s²: 9, K
s¹: (9·20 − K)/9
s⁰: K
Largest stable gain
Kmax = 9·20 = 180
Factor
m = 180/60 = 3
Answer
m = 3
This factor is the gain margin as a ratio: 9.542 dB.
8.Stability: Routh–Hurwitz & root locus
Answer: sb = −2.764
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 + 5)
Stationary points
Differentiate with the quotient rule and set the numerator to zero.
(2s + 4)(s + 5) − (s² + 4s) = s² + 10s + 20 = 0
s = −5 ± √5 = −2.764, −7.236
Break-away point
sb = −2.764
Answer
sb = −2.764
Between the two break points the locus is a circle centred on the zero, with radius √5 = 2.236.
9.Steady-state error & PID control
Answer: ess = 20
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 = 2·1/10 = 0.2
Parabola error
ess = 4/0.2 = 20
Answer
ess = 20
10.Steady-state error & PID control
Answer: ess = 0.5
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).
Velocity constant
s·C(s) = 5s + 10 → 10 as s → 0.
Kv = 10·4/10 = 4
Ramp error
ess = 2/4 = 0.5
Answer
ess = 0.5
The PI controller removes the step error completely; a ramp is followed with a constant lag.
11.Bode plots & margins
Answer: f0 = 5033 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.
Resonant frequency
ω₀ = 1/√(LC) = 1/√(0.001·1 × 10⁻⁶) = 31623 rad/s
f₀ = ω₀/(2π) = 5033 Hz
Answer
f0 = 5033 Hz
12.Bode plots & margins
Answer: GM = 14.92 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.