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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) = 40/(s(s + 2)(s + 4)) by partial fractions, and evaluate f(0.5).
F(s) = 40/(s(s + 2)(s + 4))t = 0.5 s
f(0.5) =
2.Laplace transformsHard
A system obeys y″ + 5y′ + 4y = 16·u(t), with y(0) = 4 and y′(0) = 0. At t = 0 a step input u(t) = 10 is applied. Use the Laplace transform to find y(1).
A unity negative feedback loop has G(s) = 5(s + 1)/((s + 3)(s + 5)). A step of size 1 is applied at the reference. Find the steady-state error ess.
G(s) = 5(s + 1)/((s + 3)(s + 5))R = 1
ess =
4.Block diagramsHard
A proportional controller K drives the plant G(s) = 9/(s + 3) in a unity negative feedback loop. A disturbance step of size 2 enters at the plant input while R = 0. Find the smallest gain K that keeps the steady-state output at or below 0.6.
G(s) = 9/(s + 3)D = 2css ≤ 0.6
K =
5.Step responseMedium
A system has the transfer function G(s) = 93/(s² + 3s + 93) and receives a unit step input. Find the percent overshoot %OS of the step response.
G(s) = 93/(s² + 3s + 93)
%OS = %
6.Step responseHard
A unity negative feedback loop has G(s) = 105/(s(s + 8)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).
G(s) = 105/(s(s + 8))
ts = s
7.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram runs with K = 42 and L(s) = K/(s(s + 3)(s + 4)). By what factor m can the gain be raised before the closed loop becomes unstable?
L(s) = K/(s(s + 3)(s + 4))K = 42
m =
8.Stability: Routh–Hurwitz & root locusHard
For the loop L(s) = K/(s(s + 3)(s + 10)), find the gain K that puts a closed-loop pole at s = −1.
L(s) = K/(s(s + 3)(s + 10))s = −1
K =
9.Steady-state error & PID controlMedium
A unity-feedback loop has G(s) = K/((s + 2)(s + 5)). Find the gain K that gives a steady-state error of 25 % for a step input.
G(s) = K/((s + 2)(s + 5))ess = 25 %
K =
10.Steady-state error & PID controlHard
A proportional controller K = 25 drives the plant G(s) = 4/(s(s + 8)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 4/(s(s + 8))K = 25
ts = s
11.Bode plots & marginsMedium
Find the exact magnitude |G(jω)| of G(s) = 42/((s + 3)(s + 7)) at ω = 1 rad/s, in dB.
G(s) = 42/((s + 3)(s + 7))ω = 1 rad/s
|G(jω)| = dB
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 640/(s(s + 8)). Find the phase margin PM in degrees.
L(s) = 640/(s(s + 8))
PM = °
3Solutions
1.Laplace transforms
Answer: f(0.5) = 1.998
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).
Partial fractions
F(s) = A/s + B/(s + 2) + C/(s + 4)
A = 40/(2·4) = 5
B = 40/((−2)·2) = −10
C = 40/((−4)·(−2)) = 5
Inverse transform
f(t) = 5 − 10·e−2t + 5·e−4t
Evaluate
f(0.5) = 5 − 10·e−2·0.5 + 5·e−4·0.5 = 1.998
Answer
f(0.5) = 1.998
2.Laplace transforms
Answer: y(1) = 22.56
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) = 4s + 20 + 160/s
Y(s) = (4s² + 20s + 160)/(s(s + 1)(s + 4))
Partial fractions
A = 160/(1·4) = 40
B = 144/((−1)·3) = −48
C = 144/((−4)·(−3)) = 12
Result
y(t) = 40 − 48·e−t + 12·e−4t
y(1) = 22.56
Answer
y(1) = 22.56
Check: y(0) = A + B + C = 4.
3.Block diagrams
Answer: ess = 0.75
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) = 5·1/(3·5) = 0.3333
Steady-state error
ess = A/(1 + Kₚ) = 1/(1 + 0.3333) = 0.75
Answer
ess = 0.75
4.Block diagrams
Answer: K = 3
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).
Disturbance response
css = D·G(0)/(1 + K·G(0)) = 2·9/(3 + 9K)
Solve for K
3 + 9K = 2·9/0.6 = 30
K = (30 − 3)/9 = 3
Answer
K = 3
5.Step response
Answer: %OS = 60.98 %
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 = 93 → ωₙ = 9.644 rad/s
2ζωₙ = 3 → ζ = 3/(2·9.644) = 0.1555
Result
%OS = 100·e−ζπ/√(1−ζ²) = 100·e−0.4947 = 60.98 %
Answer
%OS = 60.98 %
6.Step response
Answer: ts = 1 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 = 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 + 7s² + 12s + K = 0
s³: 1, 12
s²: 7, K
s¹: (7·12 − K)/7
s⁰: K
Largest stable gain
Kmax = 7·12 = 84
Factor
m = 84/42 = 2
Answer
m = 2
This factor is the gain margin as a ratio: 6.021 dB.
8.Stability: Routh–Hurwitz & root locus
Answer: K = 18
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 + 3|·|s + 10|
Evaluate at the point
K = 1·2·9 = 18
Answer
K = 18
Check: s(s + 3)(s + 10) + K is zero at s = −1.
9.Steady-state error & PID control
Answer: K = 30
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 error constant
1/(1 + Kp) = 0.25 → Kp = 3
Gain
K/10 = 3 → K = 30
Answer
K = 30
Check the closed-loop stability as well: for this second-order loop any K > 0 is stable.
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) + 25·4 = s² + 8s + 100
ωₙ and ζ
ωₙ = √100 = 10 rad/s
ζ = 8/(2·10) = 0.4
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)| = 5.475 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.
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.