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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) = (4s + 18)/(s² + 9s + 20). Its inverse transform has the form f(t) = A·e−4t + B·e−5t. Find the coefficient A by partial fractions.
F(s) = (4s + 18)/(s² + 9s + 20)
A =
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) = 4 is applied. Use the Laplace transform to find y(1.2).
The plant G(s) = 9/(s + 4) has an inner (minor) negative feedback loop with gain H₁ = 0.5. A controller gain K = 18 and an outer unity negative feedback loop close the system. Find the closed-loop pole p.
G(s) = 9/(s + 4)H₁ = 0.5K = 18
p =
4.Block diagramsHard
A proportional controller K drives the plant G(s) = 3/(s + 8) in a unity negative feedback loop. A disturbance step of size 0.5 enters at the plant input while R = 0. Find the smallest gain K that keeps the steady-state output at or below 0.001875.
G(s) = 3/(s + 8)D = 0.5css ≤ 0.001875
K =
5.Step responseMedium
A system has the transfer function G(s) = 51/(s² + 4s + 51) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 51/(s² + 4s + 51)
ts = s
6.Step responseHard
A mass m = 2 kg hangs on a spring k = 500 N/m with a viscous damper c = 9 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 10 N is suddenly applied. Find the largest displacement xmax in mm.
m = 2 kgk = 500 N/mc = 9 N·s/mF = 10 N
xmax = mm
7.Stability: Routh–Hurwitz & root locusMedium
A unity-feedback loop has L(s) = K/((s + 4)(s + 5)(s + 6)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.
L(s) = K/((s + 4)(s + 5)(s + 6))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
A unity-feedback loop has L(s) = K/(s(s + 1)(s + 2)). Find the point σa where the asymptotes of the root locus cross the real axis.
L(s) = K/(s(s + 1)(s + 2))
σa =
9.Steady-state error & PID controlMedium
A unity-feedback loop has G(s) = K/((s + 5)(s + 10)). Find the gain K that gives a steady-state error of 25 % for a step input.
G(s) = K/((s + 5)(s + 10))ess = 25 %
K =
10.Steady-state error & PID controlHard
A proportional controller K = 25 drives the plant G(s) = 4/(s(s + 6)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 4/(s(s + 6))K = 25
ts = s
11.Bode plots & marginsMedium
The RC low-pass filter shown has R = 10 kΩ, C = 22 nF. A sinusoidal input of 1.4 kHz is applied. Find the gain |H| = |vout/vin| in dB.
R = 10 kΩC = 22 nFf = 1.4 kHz
|H| = dB
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 16/(s(s + 1)(s + 5)). Find the gain margin GM in dB.
L(s) = 16/(s(s + 1)(s + 5))
GM = dB
3Solutions
1.Laplace transforms
Answer: A = 2
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 + 20 = (s + 4)(s + 5)
F(s) = A/(s + 4) + B/(s + 5)
Cover-up rule
Multiply by (s + 4) and set s = −4: the B-term vanishes.
A = (4s + 18)/(s + 5) |s = −4 = 2/1 = 2
Answer
A = 2
(B = 2, so f(t) = 2·e−4t + 2·e−5t)
2.Laplace transforms
Answer: y(1.2) = 11.21
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 + 64/s
Y(s) = (4s² + 20s + 64)/(s(s + 1)(s + 4))
Partial fractions
A = 64/(1·4) = 16
B = 48/((−1)·3) = −16
C = 48/((−4)·(−3)) = 4
Result
y(t) = 16 − 16·e−t + 4·e−4t
y(1.2) = 11.21
Answer
y(1.2) = 11.21
Check: y(0) = A + B + C = 4.
3.Block diagrams
Answer: p = −170.5
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).
Disturbance response
css = D·G(0)/(1 + K·G(0)) = 0.5·3/(8 + 3K)
Solve for K
8 + 3K = 0.5·3/0.001875 = 800
K = (800 − 8)/3 = 264
Answer
K = 264
5.Step response
Answer: ts = 2 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).
Compare with the standard form
The denominator must match s² + 2ζωₙs + ωₙ².
ωₙ² = 51 → ωₙ = 7.141 rad/s
2ζωₙ = 4 → ζ = 4/(2·7.141) = 0.2801
Result
ζωₙ = 2 → ts ≈ 4/2 = 2 s
Answer
ts = 2 s
6.Step response
Answer: xmax = 32.73 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) = √(500/2) = 15.81 rad/s
ζ = c/(2√(km)) = 9/(2·√(500·2)) = 0.1423
Result
xst = F/k = 10/500 m = 20 mm
%OS = 63.66 %
xmax = 20·(1 + 0.6366) = 32.73 mm
Answer
xmax = 32.73 mm
7.Stability: Routh–Hurwitz & root locus
Answer: Kmax = 990
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 + 15s² + 74s + 120 + K = 0
Routh array
s³: 1, 74
s²: 15, 120 + K
s¹: (15·74 − 120 − K)/15
s⁰: 120 + K
Condition
The s¹ entry must stay positive.
K < 15·74 − 120 = 990
Answer
Kmax = 990
At the limit the loop oscillates at ω = √74 = 8.602 rad/s.
8.Stability: Routh–Hurwitz & root locus
Answer: σa = −1
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.
Poles and zeros
poles: 0, −1, −2
n − m = 3 − 0 = 3
Centroid
σa = (−3)/3 = −1
Answer
σa = −1
The 3 asymptotes leave the centroid at angles of 60°, 180°, 300°.
9.Steady-state error & PID control
Answer: K = 150
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/50 = 3 → K = 150
Answer
K = 150
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 = 1.333 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 + 6) + 25·4 = s² + 6s + 100
ωₙ and ζ
ωₙ = √100 = 10 rad/s
ζ = 6/(2·10) = 0.3
Settling time
ζωₙ = 6/2 = 3
ts ≈ 4/3 = 1.333 s
Answer
ts = 1.333 s
The settling time does not depend on K here: raising K only makes the response oscillate faster.
11.Bode plots & margins
Answer: |H| = −6.762 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/(10000·2.2 × 10⁻⁸) = 4545 rad/s
fc = 723.4 Hz → f/fc = 1400/723.4 = 1.935
Gain
|H| = 1/√(1 + 1.935²) = 0.4591
20·log₁₀(0.4591) = −6.762 dB
Answer
|H| = −6.762 dB
12.Bode plots & margins
Answer: GM = 5.46 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.