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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) = 10/(s(s + 1)(s + 2)) by partial fractions, and evaluate f(1.2).
F(s) = 10/(s(s + 1)(s + 2))t = 1.2 s
f(1.2) =
2.Laplace transformsHard
Solve y″ + 3y′ + 2y = 0 with y(0) = 8 and y′(0) = 13 using the Laplace transform, and find y(0.6).
y″ + 3y′ + 2y = 0y(0) = 8y′(0) = 13t = 0.6 s
y(0.6) =
3.Block diagramsMedium
The plant G(s) = 8/(s + 8) has an inner (minor) negative feedback loop with gain H₁ = 3. A controller gain K = 19 and an outer unity negative feedback loop close the system. Find the closed-loop pole p.
G(s) = 8/(s + 8)H₁ = 3K = 19
p =
4.Block diagramsHard
A proportional controller K drives the plant G(s) = 7/(s + 7) in a unity negative feedback loop. A disturbance step of size 2 enters at the plant input while R = 0. Find the smallest gain K that keeps the steady-state output at or below 0.02.
G(s) = 7/(s + 7)D = 2css ≤ 0.02
K =
5.Step responseMedium
A system has the transfer function G(s) = 45/(s² + 6s + 15) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.
G(s) = 45/(s² + 6s + 15)
σ =
6.Step responseHard
A unity negative feedback loop has G(s) = 255/(s(s + 5)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).
G(s) = 255/(s(s + 5))
ts = s
7.Stability: Routh–Hurwitz & root locusMedium
A unity-feedback loop has L(s) = K/((s + 1)(s + 2)(s + 6)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.
L(s) = K/((s + 1)(s + 2)(s + 6))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
A unity-feedback loop has L(s) = K/((s + 3)(s + 6)(s + 10)). Find the point σa where the asymptotes of the root locus cross the real axis.
L(s) = K/((s + 3)(s + 6)(s + 10))
σa =
9.Steady-state error & PID controlMedium
A unity-feedback loop with G(s) = K/(s(s + 2)) must follow the ramp r(t) = 1t with a steady-state error of 0.01. Find the gain K.
G(s) = K/(s(s + 2))r(t) = 1tess = 0.01
K =
10.Steady-state error & PID controlHard
A proportional controller K = 25 drives the plant G(s) = 1/(s(s + 2)) in a unity-feedback loop. Find the peak time tp of the step response.
G(s) = 1/(s(s + 2))K = 25
tp = s
11.Bode plots & marginsMedium
In the series RLC band-pass filter shown the output is taken across R = 10 Ω, with L = 5 mH and C = 0.1 µF. Find the quality factor Q.
R = 10 ΩL = 5 mHC = 0.1 µF
Q =
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 32/(s(s + 8)). Find the gain crossover frequency ωc, where |L(jω)| = 1.
L(s) = 32/(s(s + 8))
ωc = rad/s
3Solutions
1.Laplace transforms
Answer: f(1.2) = 2.442
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 + 2)
A = 10/(1·2) = 5
B = 10/((−1)·1) = −10
C = 10/((−2)·(−1)) = 5
Inverse transform
f(t) = 5 − 10·e−t + 5·e−2t
Evaluate
f(1.2) = 5 − 10·e−1.2 + 5·e−2·1.2 = 2.442
Answer
f(1.2) = 2.442
2.Laplace transforms
Answer: y(0.6) = 9.59
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 − 8s − 13 + 3(sY − 8) + 2Y = 0
Y(s) = (8s + 37)/((s + 1)(s + 2))
Partial fractions
A = 29/1 = 29
B = 21/(−1) = −21
Result
y(t) = 29·e−t − 21·e−2t
y(0.6) = 9.59
Answer
y(0.6) = 9.59
Check: y(0) = A + B = 8.
3.Block diagrams
Answer: p = −184
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).
Disturbance response
css = D·G(0)/(1 + K·G(0)) = 2·7/(7 + 7K)
Solve for K
7 + 7K = 2·7/0.02 = 700
K = (700 − 7)/7 = 99
Answer
K = 99
5.Step response
Answer: σ = −3
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 = 15 → ωₙ = 3.873 rad/s
2ζωₙ = 6 → ζ = 6/(2·3.873) = 0.7746
Result
σ = −ζωₙ = −6/2 = −3
Answer
σ = −3
6.Step response
Answer: ts = 1.6 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: Kmax = 168
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 + 9s² + 20s + 12 + K = 0
Routh array
s³: 1, 20
s²: 9, 12 + K
s¹: (9·20 − 12 − K)/9
s⁰: 12 + K
Condition
The s¹ entry must stay positive.
K < 9·20 − 12 = 168
Answer
Kmax = 168
At the limit the loop oscillates at ω = √20 = 4.472 rad/s.
8.Stability: Routh–Hurwitz & root locus
Answer: σa = −6.333
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: −3, −6, −10
n − m = 3 − 0 = 3
Centroid
σa = (−19)/3 = −6.333
Answer
σa = −6.333
The 3 asymptotes leave the centroid at angles of 60°, 180°, 300°.
9.Steady-state error & PID control
Answer: K = 200
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 velocity constant
Kv = A/ess = 1/0.01 = 100
Gain
K = 2·Kv = 200
Answer
K = 200
A large gain makes this loop very lightly damped: ζ = 0.07071. That conflict is why compensators are used.
10.Steady-state error & PID control
Answer: tp = 0.6413 s
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 + 2) + 25·1 = s² + 2s + 25
ωₙ and ζ
ωₙ = √25 = 5 rad/s
ζ = 2/(2·5) = 0.2
Peak time
ωd = 5·√(1 − 0.2²) = 4.899 rad/s
tp = π/4.899 = 0.6413 s
Answer
tp = 0.6413 s
11.Bode plots & margins
Answer: Q = 22.36
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/√(0.005·1 × 10⁻⁷) = 44721 rad/s
f₀ = ω₀/(2π) = 7118 Hz
Quality factor
BW = R/L = 10/0.005 = 2000 rad/s = 318.3 Hz
Q = ω₀/BW = 44721/2000 = 22.36
Answer
Q = 22.36
12.Bode plots & margins
Answer: ωc = 3.641 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.