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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) = (3s + 17)/(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) = (3s + 17)/(s² + 9s + 18)
A =
2.Laplace transformsHard
Solve y″ + 7y′ + 6y = 0 with y(0) = 8 and y′(0) = 14 using the Laplace transform, and find y(0.3).
y″ + 7y′ + 6y = 0y(0) = 8y′(0) = 14t = 0.3 s
y(0.3) =
3.Block diagramsMedium
The plant G(s) = 2/(s + 7) has an inner (minor) negative feedback loop with gain H₁ = 3. A controller gain K = 17 and an outer unity negative feedback loop close the system. Find the closed-loop pole p.
G(s) = 2/(s + 7)H₁ = 3K = 17
p =
4.Block diagramsHard
The plant G(s) = 60/(s(s + 4)) is in a negative feedback loop with sensor gain H = 2. The closed-loop poles are complex. Find their real part σ.
G(s) = 60/(s(s + 4))H = 2
σ =
5.Step responseMedium
A system has the transfer function G(s) = 12/(s² + 5s + 24) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 12/(s² + 5s + 24)
ts = s
6.Step responseHard
A series RLC circuit with R = 67 Ω, L = 100 mH and C = 22 µF is connected to a 5 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the percent overshoot %OS of the capacitor voltage.
R = 67 ΩL = 100 mHC = 22 µFV = 5 V
%OS = %
7.Stability: Routh–Hurwitz & root locusMedium
A unity-feedback loop has L(s) = K(s + 10)/(s(s + 1)(s + 3)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.
L(s) = K(s + 10)/(s(s + 1)(s + 3))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
For L(s) = K/((s + 7)(s² + 2s + 5)), find the angle of departure θd of the root locus from the complex pole in the upper half-plane, measured from the positive real axis.
L(s) = K/((s + 7)(s² + 2s + 5))
θd = °
9.Steady-state error & PID controlMedium
A unity-feedback loop with G(s) = K/(s(s + 4)) must follow the ramp r(t) = 1t with a steady-state error of 0.01. Find the gain K.
G(s) = K/(s(s + 4))r(t) = 1tess = 0.01
K =
10.Steady-state error & PID controlHard
A proportional controller K drives the plant G(s) = 2/(s(s + 6)) in a unity-feedback loop. Find the gain K that gives the closed loop a damping ratio ζ = 0.6.
G(s) = 2/(s(s + 6))ζ = 0.6
K =
11.Bode plots & marginsMedium
The RC high-pass filter shown has R = 22 kΩ, C = 22 nF. A sinusoidal input of 160 Hz is applied. Find the gain |H| = |vout/vin| in dB.
R = 22 kΩC = 22 nFf = 160 Hz
|H| = dB
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 9/(s(s + 1)). Find the gain crossover frequency ωc, where |L(jω)| = 1.
L(s) = 9/(s(s + 1))
ωc = rad/s
3Solutions
1.Laplace transforms
Answer: A = 2.667
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 = (3s + 17)/(s + 6) |s = −3 = 8/3 = 2.667
Answer
A = 2.667
(B = 0.3333, so f(t) = 2.667·e−3t + 0.3333·e−6t)
2.Laplace transforms
Answer: y(0.3) = 8.459
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
s²Y − 8s − 14 + 7(sY − 8) + 6Y = 0
Y(s) = (8s + 70)/((s + 1)(s + 6))
Partial fractions
A = 62/5 = 12.4
B = 22/(−5) = −4.4
Result
y(t) = 12.4·e−t − 4.4·e−6t
y(0.3) = 8.459
Answer
y(0.3) = 8.459
Check: y(0) = A + B = 8.
3.Block diagrams
Answer: p = −47
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).
Characteristic equation
s(s + 4) + 60·2 = 0 → s² + 4s + 120 = 0
Roots
s = −2 ± j·√(120 − 4) = −2 ± j10.77
σ = −2
Answer
σ = −2
5.Step response
Answer: ts = 1.6 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 = 24 → ωₙ = 4.899 rad/s
2ζωₙ = 5 → ζ = 5/(2·4.899) = 0.5103
Result
ζωₙ = 2.5 → ts ≈ 4/2.5 = 1.6 s
Answer
ts = 1.6 s
6.Step response
Answer: %OS = 16.55 %
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.
With the zero to the right of s = −4 the loop would stay stable for every K > 0.
8.Stability: Routh–Hurwitz & root locus
Answer: θd = 71.57 °
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.
Pole locations
s² + 2s + 5 = 0 → s = −1 ± j2
s = −7
Angles from the other poles
∠(−1 + j2 + 7) = atan2(2, 6) = 18.43°
∠(−1 + j2 − (−1 − j2)) = ∠(j4) = 90°
Angle of departure
θd = 180° − 18.43° − 90° = 71.57°
Answer
θd = 71.57°
9.Steady-state error & PID control
Answer: K = 400
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 velocity constant
Kv = A/ess = 1/0.01 = 100
Gain
K = 4·Kv = 400
Answer
K = 400
A large gain makes this loop very lightly damped: ζ = 0.1. That conflict is why compensators are used.
10.Steady-state error & PID control
Answer: K = 12.5
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).
Characteristic polynomial
s(s + 6) + 2K = s² + 6s + 2K
Natural frequency
2·0.6·ωₙ = 6 → ωₙ = 5 rad/s
Gain
2K = ωₙ² = 25 → K = 12.5
Answer
K = 12.5
11.Bode plots & margins
Answer: |H| = −7.18 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/(22000·2.2 × 10⁻⁸) = 2066 rad/s
fc = 328.8 Hz → f/fc = 160/328.8 = 0.4866
Gain
|H| = 0.4866/√(1 + 0.4866²) = 0.4375
20·log₁₀(0.4375) = −7.18 dB
Answer
|H| = −7.18 dB
12.Bode plots & margins
Answer: ωc = 2.918 rad/s
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.