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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 + 20)/(s² + 7s + 10). Its inverse transform has the form f(t) = A·e−2t + B·e−5t. Find the coefficient A by partial fractions.
F(s) = (3s + 20)/(s² + 7s + 10)
A =
2.Laplace transformsHard
Solve y″ + 8y′ + 15y = 0 with y(0) = 7 and y′(0) = 14 using the Laplace transform, and find y(0.1).
y″ + 8y′ + 15y = 0y(0) = 7y′(0) = 14t = 0.1 s
y(0.1) =
3.Block diagramsMedium
A plant G(s) = 32/(s + 8) is controlled in a negative feedback loop whose sensor has the gain H = 2. Find the time constant τ of the closed loop.
G(s) = 32/(s + 8)H = 2
τ = s
4.Block diagramsHard
The plant G(s) = 72/(s(s + 4)) is in a negative feedback loop with sensor gain H = 1. The closed-loop poles are complex. Find their imaginary part ωd (the positive value), in rad/s.
G(s) = 72/(s(s + 4))H = 1
ωd = rad/s
5.Step responseMedium
A system has the transfer function G(s) = 10/(s² + 5s + 20) and receives a unit step input. Find the damping ratio ζ.
G(s) = 10/(s² + 5s + 20)
ζ =
6.Step responseHard
A mass m = 2 kg hangs on a spring k = 5000 N/m with a viscous damper c = 80 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 100 N is suddenly applied. Find the percent overshoot %OS of the displacement.
m = 2 kgk = 5000 N/mc = 80 N·s/mF = 100 N
%OS = %
7.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram has negative feedback and L(s) = 168/(s(s + 3)(s + 4)). How many closed-loop poles N lie in the right half-plane?
L(s) = 168/(s(s + 3)(s + 4))
N =
8.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K/((s + 6)(s + 8)) is shown. Find the gain K at the break-away point sb, where the two closed-loop poles meet on the real axis.
L(s) = K/((s + 6)(s + 8))
K =
9.Steady-state error & PID controlMedium
The input r(t) = 2 + 1t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 10/(s(s + 5)). Find the steady-state error ess.
G(s) = 10/(s(s + 5))r(t) = 2 + 1t
ess =
10.Steady-state error & PID controlHard
A PI controller C(s) = 5 + 5/s drives the plant G(s) = 4/((s + 1)(s + 10)) in a unity-feedback loop. Find the steady-state error ess for the ramp input r(t) = 2t.
C(s) = 5 + 5/sG(s) = 4/((s + 1)(s + 10))r(t) = 2t
ess =
11.Bode plots & marginsMedium
The RL high-pass filter shown has R = 47 Ω, L = 47 mH. A sinusoidal input of 40 Hz is applied. Find the gain |H| = |vout/vin| in dB.
R = 47 ΩL = 47 mHf = 40 Hz
|H| = dB
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 30/(s(s + 2)). Find the gain crossover frequency ωc, where |L(jω)| = 1.
L(s) = 30/(s(s + 2))
ωc = rad/s
3Solutions
1.Laplace transforms
Answer: A = 4.667
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 + 7s + 10 = (s + 2)(s + 5)
F(s) = A/(s + 2) + B/(s + 5)
Cover-up rule
Multiply by (s + 2) and set s = −2: the B-term vanishes.
A = (3s + 20)/(s + 5) |s = −2 = 14/3 = 4.667
Answer
A = 4.667
(B = −1.667, so f(t) = 4.667·e−2t − 1.667·e−5t)
2.Laplace transforms
Answer: y(0.1) = 7.536
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 − 7s − 14 + 8(sY − 7) + 15Y = 0
Y(s) = (7s + 70)/((s + 3)(s + 5))
Partial fractions
A = 49/2 = 24.5
B = 35/(−2) = −17.5
Result
y(t) = 24.5·e−3t − 17.5·e−5t
y(0.1) = 7.536
Answer
y(0.1) = 7.536
Check: y(0) = A + B = 7.
3.Block diagrams
Answer: τ = 0.01389 s
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) + 72·1 = 0 → s² + 4s + 72 = 0
Roots
s = −2 ± j·√(72 − 4) = −2 ± j8.246
ωd = 8.246 rad/s
Answer
ωd = 8.246 rad/s
5.Step response
Answer: ζ = 0.559
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 = 20 → ωₙ = 4.472 rad/s
2ζωₙ = 5 → ζ = 5/(2·4.472) = 0.559
Result
ζ = 0.559
Answer
ζ = 0.559
6.Step response
Answer: %OS = 25.38 %
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) = √(5000/2) = 50 rad/s
ζ = c/(2√(km)) = 80/(2·√(5000·2)) = 0.4
Result
%OS = 100·e−ζπ/√(1−ζ²) = 25.38 %
Answer
%OS = 25.38 %
7.Stability: Routh–Hurwitz & root locus
Answer: N = 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.
Characteristic polynomial
s³ + 7s² + 12s + 168 = 0
Routh array
s³: 1, 12
s²: 7, 168
s¹: −12
s⁰: 168
Sign changes in the first column
1 → 7 → −12 → 168
N = 2
Answer
N = 2
The gain is above the limit K = 84, so a complex pair has crossed into the right half-plane.
8.Stability: Routh–Hurwitz & root locus
Answer: K = 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.
Break-away point
dK/ds = 0 for K = −(s + 6)(s + 8) gives 2s + 14 = 0.
sb = −7
Magnitude condition
K = |−7 + 6|·|−7 + 8| = 1·1 = 1
Answer
K = 1
For this gain the closed loop is critically damped (ζ = 1).
9.Steady-state error & PID control
Answer: ess = 0.5
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.
Velocity constant
Kv = lim s·G(s) = 2
Step part
Kp = ∞ → estep = 0
Ramp error
ess = 1/2 = 0.5
Answer
ess = 0.5
10.Steady-state error & PID control
Answer: ess = 1
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).
Velocity constant
s·C(s) = 5s + 5 → 5 as s → 0.
Kv = 5·4/10 = 2
Ramp error
ess = 2/2 = 1
Answer
ess = 1
The PI controller removes the step error completely; a ramp is followed with a constant lag.
11.Bode plots & margins
Answer: |H| = −12.26 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 = R/L = 47/0.047 = 1000 rad/s
fc = 159.2 Hz → f/fc = 40/159.2 = 0.2513
Gain
|H| = 0.2513/√(1 + 0.2513²) = 0.2437
20·log₁₀(0.2437) = −12.26 dB
Answer
|H| = −12.26 dB
12.Bode plots & margins
Answer: ωc = 5.298 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.