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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) = (5s + 26)/(s² + 13s + 40). Its inverse transform has the form f(t) = A·e−5t + B·e−8t. Find the coefficient A by partial fractions.
F(s) = (5s + 26)/(s² + 13s + 40)
A =
2.Laplace transformsHard
A system obeys y″ + 5y′ + 4y = 12·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(0.3).
A unity negative feedback loop has G(s) = 6(s + 3)/((s + 1)(s + 9)). A step of size 10 is applied at the reference. Find the steady-state error ess.
G(s) = 6(s + 3)/((s + 1)(s + 9))R = 10
ess =
4.Block diagramsHard
Reduce the block diagram with G₁ = 8/(s + 3), G₂ = 9/(s + 3), an inner negative feedback H₁ = 2 around G₂ and an outer negative feedback H₂ = 0.5. Find the DC gain T(0) of C(s)/R(s).
G₁ = 8/(s + 3)G₂ = 9/(s + 3)H₁ = 2H₂ = 0.5
T(0) =
5.Step responseMedium
A system has the transfer function G(s) = 154/(s² + 6s + 77) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 154/(s² + 6s + 77)
ts = s
6.Step responseHard
A series RLC circuit with R = 18 Ω, L = 50 mH and C = 100 µF is connected to a 10 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the peak capacitor voltage vC,max.
R = 18 ΩL = 50 mHC = 100 µFV = 10 V
vC,max = V
7.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram uses positive feedback, with L(s) = 10/(s(s + 3)(s + 8)). How many closed-loop poles N lie in the right half-plane?
L(s) = 10/(s(s + 3)(s + 8))
N =
8.Stability: Routh–Hurwitz & root locusHard
For the loop L(s) = K/(s(s + 2)(s + 6)), find the gain K that puts a closed-loop pole at s = −0.5.
L(s) = K/(s(s + 2)(s + 6))s = −0.5
K =
9.Steady-state error & PID controlMedium
A process G(s) = 2.5/(10s + 1) is controlled by a proportional gain K = 9 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.
G(s) = 2.5/(10s + 1)K = 9
ess = %
10.Steady-state error & PID controlHard
A proportional controller K = 25 drives the plant G(s) = 2/(s(s + 8)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 2/(s(s + 8))K = 25
ts = s
11.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 50(s + 5)/(s(s + 3)) at ω = 2 rad/s, in degrees.
G(s) = 50(s + 5)/(s(s + 3))ω = 2 rad/s
∠G(jω) = °
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 190/(s(s + 4)(s + 6)). Find the gain margin GM in dB.
L(s) = 190/(s(s + 4)(s + 6))
GM = dB
3Solutions
1.Laplace transforms
Answer: A = 0.3333
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 + 13s + 40 = (s + 5)(s + 8)
F(s) = A/(s + 5) + B/(s + 8)
Cover-up rule
Multiply by (s + 5) and set s = −5: the B-term vanishes.
A = (5s + 26)/(s + 8) |s = −5 = 1/3 = 0.3333
Answer
A = 0.3333
(B = 4.667, so f(t) = 0.3333·e−5t + 4.667·e−8t)
2.Laplace transforms
Answer: y(0.3) = 4.775
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 + 5s + 4)·Y(s) = 5s + 25 + 12/s
Y(s) = (5s² + 25s + 12)/(s(s + 1)(s + 4))
Partial fractions
A = 12/(1·4) = 3
B = −8/((−1)·3) = 2.667
C = −8/((−4)·(−3)) = −0.6667
Result
y(t) = 3 + 2.667·e−t − 0.6667·e−4t
y(0.3) = 4.775
Answer
y(0.3) = 4.775
Check: y(0) = A + B + C = 5.
3.Block diagrams
Answer: ess = 3.333
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).
Position error constant
Kₚ = G(0) = 6·3/(1·9) = 2
Steady-state error
ess = A/(1 + Kₚ) = 10/(1 + 2) = 3.333
Answer
ess = 3.333
4.Block diagrams
Answer: T(0) = 0.7273
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).
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 = 77 → ωₙ = 8.775 rad/s
2ζωₙ = 6 → ζ = 6/(2·8.775) = 0.3419
Result
ζωₙ = 3 → ts ≈ 4/3 = 1.333 s
Answer
ts = 1.333 s
6.Step response
Answer: vC,max = 12.51 V
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 + 11s² + 24s − 10 = 0
Routh array
s³: 1, 24
s²: 11, −10
s¹: 24.91
s⁰: −10
Sign changes in the first column
1 → 11 → 24.91 → −10
N = 1
Answer
N = 1
Positive feedback makes the constant coefficient negative: one real pole always moves into the right half-plane.
8.Stability: Routh–Hurwitz & root locus
Answer: K = 4.125
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 + 2|·|s + 6|
Evaluate at the point
K = 0.5·1.5·5.5 = 4.125
Answer
K = 4.125
Check: s(s + 2)(s + 6) + K is zero at s = −0.5.
9.Steady-state error & PID control
Answer: ess = 4.255 %
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 = 9·2.5 = 22.5
Step error
ess = 100/(1 + 22.5) = 4.255 %
Answer
ess = 4.255 %
10.Steady-state error & PID control
Answer: ts = 1 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 + 8) + 25·2 = s² + 8s + 50
ωₙ and ζ
ωₙ = √50 = 7.071 rad/s
ζ = 8/(2·7.071) = 0.5657
Settling time
ζωₙ = 8/2 = 4
ts ≈ 4/4 = 1 s
Answer
ts = 1 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) = −101.9 °
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.