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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) = (5s + 39)/(s² + 2s + 17) by completing the square, and evaluate f(1).
F(s) = (5s + 39)/(s² + 2s + 17)t = 1 s
f(1) =
2.Laplace transformsHard
A system obeys y″ + 9y′ + 20y = 10·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 1 for t ≥ 0. Use the Laplace transform to find the output y(0.1).
The plant G(s) = K/((s + 3)(s + 9)) is in a unity negative feedback loop. Find the gain K for which the closed loop is critically damped (two equal real poles).
G(s) = K/((s + 3)(s + 9))unity negative feedback
K =
4.Block diagramsHard
Reduce the block diagram with G₁ = 2/(s + 5), G₂ = 8/(s + 7), an inner negative feedback H₁ = 1 around G₂ and an outer negative feedback H₂ = 0.25. Find the DC gain T(0) of C(s)/R(s).
G₁ = 2/(s + 5)G₂ = 8/(s + 7)H₁ = 1H₂ = 0.25
T(0) =
5.Step responseMedium
A system has the transfer function G(s) = 177/(s² + 3s + 59) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.
G(s) = 177/(s² + 3s + 59)
σ =
6.Step responseHard
A unity negative feedback loop has G(s) = 205/(s(s + 8)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).
G(s) = 205/(s(s + 8))
ts = s
7.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram uses positive feedback, with L(s) = 10/(s(s + 1)(s + 3)). How many closed-loop poles N lie in the right half-plane?
L(s) = 10/(s(s + 1)(s + 3))
N =
8.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K(s + 6)/(s(s + 3)) leaves the real axis between s = 0 and s = −3, circles round and joins the real axis again to the left of the zero. Find the break-in point si.
L(s) = K(s + 6)/(s(s + 3))
si =
9.Steady-state error & PID controlMedium
A unity-feedback loop has G(s) = K/((s + 1)(s + 2)). Find the gain K that gives a steady-state error of 20 % for a step input.
G(s) = K/((s + 1)(s + 2))ess = 20 %
K =
10.Steady-state error & PID controlHard
A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 5/(s + 1) in a unity-feedback loop. The closed loop must have ωₙ = 4 rad/s and ζ = 0.8. Find the integral gain Ki.
G(s) = 5/(s + 1)ωₙ = 4 rad/sζ = 0.8
Ki =
11.Bode plots & marginsMedium
In the series RLC band-pass filter shown the output is taken across R = 5 Ω, with L = 1 mH and C = 1 µF. Find the centre (resonant) frequency f0 in Hz.
R = 5 ΩL = 1 mHC = 1 µF
f0 = Hz
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 1560/(s(s + 20)). Find the gain crossover frequency ωc, where |L(jω)| = 1.
L(s) = 1560/(s(s + 20))
ωc = rad/s
3Solutions
1.Laplace transforms
Answer: f(1) = −3.569
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).
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
Y(s) = 10/(s(s + 4)(s + 5))
Partial fractions
A = 10/(4·5) = 0.5
B = 10/((−4)·1) = −2.5
C = 10/((−5)·(−1)) = 2
Output
y(t) = 0.5 − 2.5·e−4t + 2·e−5t
y(0.1) = 0.03726
Answer
y(0.1) = 0.03726
3.Block diagrams
Answer: K = 9
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).
A quadratic s² + βs + γ has a double root when β² = 4γ.
12² = 4(27 + K) → K = 144/4 − 27 = 9
Answer
K = 9
Both closed-loop poles are then at s = −6.
4.Block diagrams
Answer: T(0) = 0.2025
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 = 59 → ωₙ = 7.681 rad/s
2ζωₙ = 3 → ζ = 3/(2·7.681) = 0.1953
Result
σ = −ζωₙ = −3/2 = −1.5
Answer
σ = −1.5
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: N = 1
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 + 4s² + 3s − 10 = 0
Routh array
s³: 1, 3
s²: 4, −10
s¹: 5.5
s⁰: −10
Sign changes in the first column
1 → 4 → 5.5 → −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: si = −10.24
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 + 3s)/(s + 6)
Stationary points
Differentiate with the quotient rule and set the numerator to zero.
(2s + 3)(s + 6) − (s² + 3s) = s² + 12s + 18 = 0
s = −6 ± √18 = −1.757, −10.24
Break-in point
si = −10.24
Answer
si = −10.24
Between the two break points the locus is a circle centred on the zero, with radius √18 = 4.243.
9.Steady-state error & PID control
Answer: K = 8
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) = 0.2 → Kp = 4
Gain
K/2 = 4 → K = 8
Answer
K = 8
Check the closed-loop stability as well: for this second-order loop any K > 0 is stable.
10.Steady-state error & PID control
Answer: Ki = 3.2
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
1 + C(s)G(s) = 0, multiplied by s(s + 1).
s² + (1 + 5Kp)s + 5Ki
Target polynomial
s² + 2·0.8·4s + 4² = s² + 6.4s + 16
Match the coefficients
5Ki = 16 → Ki = 3.2
Answer
Ki = 3.2
The integral term makes the loop type 1, so a step is followed without steady-state error.
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: ωc = 37.05 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.