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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) = (6s + 31)/(s² + 13s + 40). Its inverse transform has the form f(t) = A·e−5t + B·e−8t. Find the coefficient A by partial fractions.
F(s) = (6s + 31)/(s² + 13s + 40)
A =
2.Laplace transformsHard
Solve y″ + 10y′ + 24y = 0 with y(0) = −8 and y′(0) = 2 using the Laplace transform, and find y(0.2).
y″ + 10y′ + 24y = 0y(0) = −8y′(0) = 2t = 0.2 s
y(0.2) =
3.Block diagramsMedium
The plant G(s) = K/((s + 4)(s + 11)) 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 + 4)(s + 11))unity negative feedback
K =
4.Block diagramsHard
A proportional controller K drives the plant G(s) = 2/(s + 4) in a unity negative feedback loop. A disturbance step of size 10 enters at the plant input while R = 0. Find the smallest gain K that keeps the steady-state output at or below 0.1.
G(s) = 2/(s + 4)D = 10css ≤ 0.1
K =
5.Step responseMedium
A system has the transfer function G(s) = 37.5/(s² + 12s + 75) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 37.5/(s² + 12s + 75)
ts = s
6.Step responseHard
A unity negative feedback loop has G(s) = K/(s(s + 16)). Find the gain K that gives a step response with 30 % overshoot.
G(s) = K/(s(s + 16))%OS = 30 %
K =
7.Stability: Routh–Hurwitz & root locusMedium
A unity-feedback loop has L(s) = K/((s + 2)(s + 4)(s + 5)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.
L(s) = K/((s + 2)(s + 4)(s + 5))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K/((s + 4)(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 + 4)(s + 8))
K =
9.Steady-state error & PID controlMedium
A unity-feedback loop has G(s) = K/((s + 2)(s + 5)). Find the gain K that gives a steady-state error of 20 % for a step input.
G(s) = K/((s + 2)(s + 5))ess = 20 %
K =
10.Steady-state error & PID controlHard
A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 2/(s + 1) in a unity-feedback loop. The closed loop must have ωₙ = 2 rad/s and ζ = 0.5. Find the proportional gain Kp.
G(s) = 2/(s + 1)ωₙ = 2 rad/sζ = 0.5
Kp =
11.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 28(s + 12)/(s(s + 4)) at ω = 15 rad/s, in degrees.
G(s) = 28(s + 12)/(s(s + 4))ω = 15 rad/s
∠G(jω) = °
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 334/(s(s + 7)(s + 8)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.
L(s) = 334/(s(s + 7)(s + 8))
ω180 = rad/s
3Solutions
1.Laplace transforms
Answer: A = 0.3333
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 + 13s + 40 = (s + 5)(s + 8)
F(s) = A/(s + 5) + B/(s + 8)
Cover-up rule
Multiply by (s + 5) and set s = −5: the B-term vanishes.
A = (6s + 31)/(s + 8) |s = −5 = 1/3 = 0.3333
Answer
A = 0.3333
(B = 5.667, so f(t) = 0.3333·e−5t + 5.667·e−8t)
2.Laplace transforms
Answer: y(0.2) = −5.817
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 − (−8)s − 2 + 10(sY − (−8)) + 24Y = 0
Y(s) = (−8s − 78)/((s + 4)(s + 6))
Partial fractions
A = −46/2 = −23
B = −30/(−2) = 15
Result
y(t) = −23·e−4t + 15·e−6t
y(0.2) = −5.817
Answer
y(0.2) = −5.817
Check: y(0) = A + B = −8.
3.Block diagrams
Answer: K = 12.25
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).
A quadratic s² + βs + γ has a double root when β² = 4γ.
15² = 4(44 + K) → K = 225/4 − 44 = 12.25
Answer
K = 12.25
Both closed-loop poles are then at s = −7.5.
4.Block diagrams
Answer: K = 98
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)) = 10·2/(4 + 2K)
Solve for K
4 + 2K = 10·2/0.1 = 200
K = (200 − 4)/2 = 98
Answer
K = 98
5.Step response
Answer: ts = 0.6667 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 = 75 → ωₙ = 8.66 rad/s
2ζωₙ = 12 → ζ = 12/(2·8.66) = 0.6928
Result
ζωₙ = 6 → ts ≈ 4/6 = 0.6667 s
Answer
ts = 0.6667 s
6.Step response
Answer: K = 499.8
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 + 16s + K)
Damping ratio from the overshoot
ζ = −ln(0.3)/√(π² + ln²(0.3)) = 0.3579
Gain
ωₙ = 16/(2ζ) = 22.36 rad/s
K = ωₙ² = 499.8
Answer
K = 499.8
7.Stability: Routh–Hurwitz & root locus
Answer: Kmax = 378
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 + 11s² + 38s + 40 + K = 0
Routh array
s³: 1, 38
s²: 11, 40 + K
s¹: (11·38 − 40 − K)/11
s⁰: 40 + K
Condition
The s¹ entry must stay positive.
K < 11·38 − 40 = 378
Answer
Kmax = 378
At the limit the loop oscillates at ω = √38 = 6.164 rad/s.
8.Stability: Routh–Hurwitz & root locus
Answer: K = 4
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 + 4)(s + 8) gives 2s + 12 = 0.
sb = −6
Magnitude condition
K = |−6 + 4|·|−6 + 8| = 2·2 = 4
Answer
K = 4
For this gain the closed loop is critically damped (ζ = 1).
9.Steady-state error & PID control
Answer: K = 40
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) = 0.2 → Kp = 4
Gain
K/10 = 4 → K = 40
Answer
K = 40
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 = 0.5
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 + 1).
s² + (1 + 2Kp)s + 2Ki
Target polynomial
s² + 2·0.5·2s + 2² = s² + 2s + 4
Match the coefficients
1 + 2Kp = 2 → Kp = 0.5
Answer
Kp = 0.5
The integral term makes the loop type 1, so a step is followed without steady-state error.
11.Bode plots & margins
Answer: ∠G(jω) = −113.7 °
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.
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.