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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) = 9/(s² + 8s + 25) by completing the square, and evaluate f(0.2).
F(s) = 9/(s² + 8s + 25)t = 0.2 s
f(0.2) =
2.Laplace transformsHard
A system is described by the differential equation y″ + 3y′ + 2y = u′ + 15u, 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 = 0.8 s. Find g(0.8).
y″ + 3y′ + 2y = u′ + 15ut = 0.8 s
g(0.8) =
3.Block diagramsMedium
A unity negative feedback loop has G(s) = 38/(s(s + 7)). The reference is a ramp r(t) = 10·t. Find the steady-state error ess.
G(s) = 38/(s(s + 7))r(t) = 10·t
ess =
4.Block diagramsHard
The plant G(s) = K/(s(s + 4)(s + 5)) is in a unity negative feedback loop. At the largest stable gain the loop oscillates. Find the frequency of oscillation ω in rad/s.
G(s) = K/(s(s + 4)(s + 5))unity negative feedback
ω = rad/s
5.Step responseMedium
A system has the transfer function G(s) = 36/(s² + 4s + 36) and receives a unit step input. Find the peak time tp of the step response.
G(s) = 36/(s² + 4s + 36)
tp = s
6.Step responseHard
The step response of a second-order system without zeros shows 8 % overshoot, and the first peak occurs at t = 0.5 s. Find the natural frequency ωn.
%OS = 8 %tp = 0.5 s
ωn = rad/s
7.Stability: Routh–Hurwitz & root locusMedium
The plant G(s) = (s + 8)/((s − 3)(s + 2)) has an unstable pole at s = 3. A proportional gain K closes a unity-feedback loop around it. Find the smallest gain Kmin that makes the closed loop stable.
G(s) = (s + 8)/((s − 3)(s + 2))
Kmin =
8.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K/(s(s + 8)) is shown. Find the gain K for which the closed-loop poles have the damping ratio ζ = 0.8.
L(s) = K/(s(s + 8))ζ = 0.8
K =
9.Steady-state error & PID controlMedium
A unity-feedback loop has G(s) = K/((s + 1)(s + 10)). Find the gain K that gives a steady-state error of 4 % for a step input.
G(s) = K/((s + 1)(s + 10))ess = 4 %
K =
10.Steady-state error & PID controlHard
A proportional controller K = 25 drives the plant G(s) = 4/(s(s + 4)) in a unity-feedback loop. Find the percent overshoot %OS of the step response.
G(s) = 4/(s(s + 4))K = 25
%OS = %
11.Bode plots & marginsMedium
The RC high-pass filter shown has R = 1 kΩ, C = 100 nF. A sinusoidal input of 800 Hz is applied. Find the phase shift φ of vout relative to vin, in degrees.
R = 1 kΩC = 100 nFf = 800 Hz
φ = °
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 7.5/(s(s + 5)). Find the gain crossover frequency ωc, where |L(jω)| = 1.
L(s) = 7.5/(s(s + 5))
ωc = rad/s
3Solutions
1.Laplace transforms
Answer: f(0.2) = 0.7611
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).
Complete the square
s² + 8s + 25 = (s + 4)² + 9 = (s + 4)² + 3²
Match the numerator
F(s) = (9/3)·3/((s + 4)² + 3²)
Inverse transform
f(t) = e−4t·(3·sin 3t)
f(0.2) = e−4·0.2·(3·sin(3·0.2)) = 0.7611
Answer
f(0.2) = 0.7611
2.Laplace transforms
Answer: g(0.8) = 3.666
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).
Transfer function
G(s) = Y(s)/U(s) = (s + 15)/(s² + 3s + 2)
Partial fractions
s² + 3s + 2 = (s + 1)(s + 2)
A = 14, B = −13
Impulse response
g(t) = 14·e−t − 13·e−2t
g(0.8) = 3.666
Answer
g(0.8) = 3.666
3.Block diagrams
Answer: ess = 1.842
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).
Velocity error constant
Kᵥ = lims→0 s·G(s) = 38/7 = 5.429
Steady-state error
ess = A/Kᵥ = 10/5.429 = 1.842
Answer
ess = 1.842
4.Block diagrams
Answer: ω = 4.472 rad/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).
Characteristic equation
s³ + 9s² + 20s + K = 0
Routh array
s³ | 1 20
s² | 9 K
s¹ | (9·20 − K)/9
s⁰ | K
Stability limit
All entries of the first column must be positive.
Kmax = 9·20 = 180
Frequency of oscillation
9·(jω)² + 180 = 0 → ω² = 180/9 = 20
ω = 4.472 rad/s
Answer
ω = 4.472 rad/s
5.Step response
Answer: tp = 0.5554 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).
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.
For s² + c₁s + c₀ all roots lie in the left half-plane exactly when c₁ > 0 and c₀ > 0.
-1 + K > 0 → K > 1
8K − 6 > 0 → K > 0.75
Kmin = 1
Answer
Kmin = 1
8.Stability: Routh–Hurwitz & root locus
Answer: K = 25
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 + 8s + K = s² + 2ζωₙs + ωₙ²
2·0.8·ωₙ = 8 → ωₙ = 5 rad/s
Gain
K = ωₙ² = 5² = 25
Answer
K = 25
The closed-loop poles sit where the ζ line crosses the vertical branch at s = −4.
9.Steady-state error & PID control
Answer: K = 240
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) = 0.04 → Kp = 24
Gain
K/10 = 24 → K = 240
Answer
K = 240
Check the closed-loop stability as well: for this second-order loop any K > 0 is stable.
10.Steady-state error & PID control
Answer: %OS = 52.66 %
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) + 25·4 = s² + 4s + 100
ωₙ and ζ
ωₙ = √100 = 10 rad/s
ζ = 4/(2·10) = 0.2
Overshoot
%OS = 100·e−0.2π/√(1 − 0.2²) = 52.66 %
Answer
%OS = 52.66 %
11.Bode plots & margins
Answer: φ = 63.31 °
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/(1000·1 × 10⁻⁷) = 10000 rad/s
fc = 1592 Hz → f/fc = 800/1592 = 0.5027
Phase
φ = 90° − atan(0.5027) = 63.31°
Answer
φ = 63.31°
12.Bode plots & margins
Answer: ωc = 1.441 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.