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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) = 30/(s(s + 1)(s + 4)) by partial fractions, and evaluate f(1).
F(s) = 30/(s(s + 1)(s + 4))t = 1 s
f(1) =
2.Laplace transformsHard
A system obeys y″ + 7y′ + 10y = 15·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 10 for t ≥ 0. Use the Laplace transform to find the output y(0.4).
The plant G(s) = K/((s + 1)(s + 5)) 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 + 1)(s + 5))unity negative feedback
K =
4.Block diagramsHard
A proportional controller K = 11 drives the plant G(s) = 5/(s + 6) in a unity negative feedback loop. A disturbance D enters at the plant input (it is added to the controller output). A reference step of size 5 and a disturbance step of size 1 are applied together. Find the steady-state output css.
K = 11G(s) = 5/(s + 6)D = 1R = 5
css =
5.Step responseMedium
A system has the transfer function G(s) = 210/(s² + 6s + 70) and receives a unit step input. Find the percent overshoot %OS of the step response.
G(s) = 210/(s² + 6s + 70)
%OS = %
6.Step responseHard
A unity negative feedback loop has G(s) = K/(s(s + 8)). Find the gain K that gives a step response with 20 % overshoot.
G(s) = K/(s(s + 8))%OS = 20 %
K =
7.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 1)(s + 5)). Find the largest gain Kmax for which the closed loop is stable.
L(s) = K/(s(s + 1)(s + 5))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
A unity-feedback loop has L(s) = K/(s(s + 10)(s + 6)). Find the point σa where the asymptotes of the root locus cross the real axis.
L(s) = K/(s(s + 10)(s + 6))
σa =
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 5 % for a step input.
G(s) = K/((s + 1)(s + 10))ess = 5 %
K =
10.Steady-state error & PID controlHard
A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 1/(s + 2) in a unity-feedback loop. The closed loop must have ωₙ = 8 rad/s and ζ = 1. Find the proportional gain Kp.
G(s) = 1/(s + 2)ωₙ = 8 rad/sζ = 1
Kp =
11.Bode plots & marginsMedium
In the series RLC band-pass filter shown the output is taken across R = 50 Ω, with L = 100 mH and C = 0.47 µF. Find the quality factor Q.
R = 50 ΩL = 100 mHC = 0.47 µF
Q =
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 14/(s(s + 1)(s + 6)). Find the gain margin GM in dB.
L(s) = 14/(s(s + 1)(s + 6))
GM = dB
3Solutions
1.Laplace transforms
Answer: f(1) = 3.867
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).
Partial fractions
F(s) = A/s + B/(s + 1) + C/(s + 4)
A = 30/(1·4) = 7.5
B = 30/((−1)·3) = −10
C = 30/((−4)·(−3)) = 2.5
Inverse transform
f(t) = 7.5 − 10·e−t + 2.5·e−4t
Evaluate
f(1) = 7.5 − 10·e−1 + 2.5·e−4·1 = 3.867
Answer
f(1) = 3.867
2.Laplace transforms
Answer: y(0.4) = 5.12
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) = 150/(s(s + 2)(s + 5))
Partial fractions
A = 150/(2·5) = 15
B = 150/((−2)·3) = −25
C = 150/((−5)·(−3)) = 10
Output
y(t) = 15 − 25·e−2t + 10·e−5t
y(0.4) = 5.12
Answer
y(0.4) = 5.12
3.Block diagrams
Answer: K = 4
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
1 + K/((s + 1)(s + 5)) = 0 → s² + 6s + 5 + K = 0
Repeated root
A quadratic s² + βs + γ has a double root when β² = 4γ.
6² = 4(5 + K) → K = 36/4 − 5 = 4
Answer
K = 4
Both closed-loop poles are then at s = −3.
4.Block diagrams
Answer: css = 4.59
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).
Transfer functions
C/D = G/(1 + K·G) = 5/(s + 6 + 55)
C/R = K·G/(1 + K·G) = 55/(s + 6 + 55)
Steady state
css = 5·55/61 + 1·5/61 = 4.59
Answer
css = 4.59
A larger K reduces the effect of the disturbance, but never to zero with a proportional controller.
5.Step response
Answer: %OS = 29.92 %
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 = 70 → ωₙ = 8.367 rad/s
2ζωₙ = 6 → ζ = 6/(2·8.367) = 0.3586
Result
%OS = 100·e−ζπ/√(1−ζ²) = 100·e−1.207 = 29.92 %
Answer
%OS = 29.92 %
6.Step response
Answer: K = 76.96
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).
Closed-loop transfer function
T(s) = K/(s² + 8s + K)
Damping ratio from the overshoot
ζ = −ln(0.2)/√(π² + ln²(0.2)) = 0.4559
Gain
ωₙ = 8/(2ζ) = 8.773 rad/s
K = ωₙ² = 76.96
Answer
K = 76.96
7.Stability: Routh–Hurwitz & root locus
Answer: Kmax = 30
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 equation
s³ + 6s² + 5s + K = 0
Routh array
s³: 1, 5
s²: 6, K
s¹: (6·5 − K)/6
s⁰: K
Conditions
Every entry of the first column must be positive.
K > 0
6·5 − K > 0 → K < 30
Answer
Kmax = 30
At K = 30 the closed loop oscillates at ω = √5 = 2.236 rad/s.
8.Stability: Routh–Hurwitz & root locus
Answer: σa = −5.333
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.
Poles and zeros
poles: 0, −10, −6
n − m = 3 − 0 = 3
Centroid
σa = (−16)/3 = −5.333
Answer
σa = −5.333
The 3 asymptotes leave the centroid at angles of 60°, 180°, 300°.
9.Steady-state error & PID control
Answer: K = 190
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) = 0.05 → Kp = 19
Gain
K/10 = 19 → K = 190
Answer
K = 190
Check the closed-loop stability as well: for this second-order loop any K > 0 is stable.
10.Steady-state error & PID control
Answer: Kp = 14
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 + 2).
s² + (2 + 1Kp)s + 1Ki
Target polynomial
s² + 2·1·8s + 8² = s² + 16s + 64
Match the coefficients
2 + 1Kp = 16 → Kp = 14
Answer
Kp = 14
The integral term makes the loop type 1, so a step is followed without steady-state error.
11.Bode plots & margins
Answer: Q = 9.225
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/√(0.1·4.7 × 10⁻⁷) = 4613 rad/s
f₀ = ω₀/(2π) = 734.1 Hz
Quality factor
BW = R/L = 50/0.1 = 500 rad/s = 79.58 Hz
Q = ω₀/BW = 4613/500 = 9.225
Answer
Q = 9.225
12.Bode plots & margins
Answer: GM = 9.542 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.