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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 + 40)/(s² + 6s + 8). Its inverse transform has the form f(t) = A·e−2t + B·e−4t. Find the coefficient A by partial fractions.
F(s) = (4s + 40)/(s² + 6s + 8)
A =
2.Laplace transformsHard
A system is described by the differential equation y″ + 2y′ + 2y = u′ + 8u, with input u(t) and output y(t). Find its transfer function G(s) = Y(s)/U(s) (zero initial conditions) and evaluate the impulse response g(t) at t = 1.5 s. Find g(1.5).
y″ + 2y′ + 2y = u′ + 8ut = 1.5 s
g(1.5) =
3.Block diagramsMedium
A unity negative feedback loop has G(s) = 7(s + 5)/((s + 3)(s + 7)). A step of size 1 is applied at the reference. Find the steady-state error ess.
G(s) = 7(s + 5)/((s + 3)(s + 7))R = 1
ess =
4.Block diagramsHard
A proportional controller K drives the plant G(s) = 2/(s + 9) in a unity negative feedback loop. A disturbance step of size 1 enters at the plant input while R = 0. Find the smallest gain K that keeps the steady-state output at or below 0.04444.
G(s) = 2/(s + 9)D = 1css ≤ 0.04444
K =
5.Step responseMedium
A system has the transfer function G(s) = 80/(s² + 7s + 80) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 80/(s² + 7s + 80)
ts = s
6.Step responseHard
A second-order system without zeros must have a step response with 50 % overshoot. Find the damping ratio ζ.
%OS = 50 %
ζ =
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
The root locus of L(s) = K/(s(s + 4)) is shown. Find the gain K for which the closed-loop poles have the damping ratio ζ = 0.5.
L(s) = K/(s(s + 4))ζ = 0.5
K =
9.Steady-state error & PID controlMedium
A ramp r(t) = 1t is applied to a unity-feedback loop with G(s) = 50/(s(s + 1)). Find the steady-state error ess.
G(s) = 50/(s(s + 1))r(t) = 1t
ess =
10.Steady-state error & PID controlHard
A proportional controller K = 10 drives the plant G(s) = 2/(s(s + 4)) in a unity-feedback loop. Find the percent overshoot %OS of the step response.
G(s) = 2/(s(s + 4))K = 10
%OS = %
11.Bode plots & marginsMedium
The RL low-pass filter shown has R = 1000 Ω, L = 4.7 mH. A sinusoidal input of 340 kHz is applied. Find the gain |H| = |vout/vin| in dB.
R = 1000 ΩL = 4.7 mHf = 340 kHz
|H| = dB
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = K/(s(s + 3)(s + 10)). Find the gain K that gives a gain margin of 12 dB.
L(s) = K/(s(s + 3)(s + 10))GM = 12 dB
K =
3Solutions
1.Laplace transforms
Answer: A = 16
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 + 6s + 8 = (s + 2)(s + 4)
F(s) = A/(s + 2) + B/(s + 4)
Cover-up rule
Multiply by (s + 2) and set s = −2: the B-term vanishes.
A = (4s + 40)/(s + 4) |s = −2 = 32/2 = 16
Answer
A = 16
(B = −12, so f(t) = 16·e−2t − 12·e−4t)
2.Laplace transforms
Answer: g(1.5) = 1.574
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).
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) = 7·5/(3·7) = 1.667
Steady-state error
ess = A/(1 + Kₚ) = 1/(1 + 1.667) = 0.375
Answer
ess = 0.375
4.Block diagrams
Answer: K = 18
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)) = 1·2/(9 + 2K)
Solve for K
9 + 2K = 1·2/0.04444 = 45
K = (45 − 9)/2 = 18
Answer
K = 18
5.Step response
Answer: ts = 1.143 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 = 80 → ωₙ = 8.944 rad/s
2ζωₙ = 7 → ζ = 7/(2·8.944) = 0.3913
Result
ζωₙ = 3.5 → ts ≈ 4/3.5 = 1.143 s
Answer
ts = 1.143 s
6.Step response
Answer: ζ = 0.2155
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: K = 16
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.
Compare with the standard form
s² + 4s + K = s² + 2ζωₙs + ωₙ²
2·0.5·ωₙ = 4 → ωₙ = 4 rad/s
Gain
K = ωₙ² = 4² = 16
Answer
K = 16
The closed-loop poles sit where the ζ line crosses the vertical branch at s = −2.
9.Steady-state error & PID control
Answer: ess = 0.02
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) = 50
Ramp error
ess = 1/50 = 0.02
Answer
ess = 0.02
10.Steady-state error & PID control
Answer: %OS = 20.79 %
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 + 4) + 10·2 = s² + 4s + 20
ωₙ and ζ
ωₙ = √20 = 4.472 rad/s
ζ = 4/(2·4.472) = 0.4472
Overshoot
%OS = 100·e−0.4472π/√(1 − 0.4472²) = 20.79 %
Answer
%OS = 20.79 %
11.Bode plots & margins
Answer: |H| = −20.08 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 = 1000/0.0047 = 212766 rad/s
fc = 33863 Hz → f/fc = 340000/33863 = 10.04
Gain
|H| = 1/√(1 + 10.04²) = 0.09911
20·log₁₀(0.09911) = −20.08 dB
Answer
|H| = −20.08 dB
12.Bode plots & margins
Answer: K = 97.96
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.