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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) = (5s + 5)/(s² + 8s + 52) by completing the square, and evaluate f(0.5).
F(s) = (5s + 5)/(s² + 8s + 52)t = 0.5 s
f(0.5) =
2.Laplace transformsHard
A system obeys y″ + 5y′ + 4y = 20·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 2 for t ≥ 0. Use the Laplace transform to find the output y(1.2).
A unity negative feedback loop has G(s) = 8(s + 5)/((s + 4)(s + 9)). A step of size 1 is applied at the reference. Find the steady-state error ess.
G(s) = 8(s + 5)/((s + 4)(s + 9))R = 1
ess =
4.Block diagramsHard
The plant G(s) = K/((s + 4)(s + 5)(s + 6)) is in a unity negative feedback loop. Use the Routh–Hurwitz criterion to find the largest gain Kmax for which the closed loop is stable.
A system has the transfer function G(s) = 22/(s² + 7s + 22) and receives a unit step input. Find the damping ratio ζ.
G(s) = 22/(s² + 7s + 22)
ζ =
6.Step responseHard
A mass m = 1 kg hangs on a spring k = 200 N/m with a viscous damper c = 3 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 10 N is suddenly applied. Find the time tp at which the displacement is largest.
m = 1 kgk = 200 N/mc = 3 N·s/mF = 10 N
tp = s
7.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 1)(s + 8)). The gain is raised until the closed loop is on the edge of stability. Find the frequency ω of the sustained oscillation.
L(s) = K/(s(s + 1)(s + 8))
ω = rad/s
8.Stability: Routh–Hurwitz & root locusHard
For L(s) = K/((s + 7)(s² + 8s + 17)), find the angle of departure θd of the root locus from the complex pole in the upper half-plane, measured from the positive real axis.
L(s) = K/((s + 7)(s² + 8s + 17))
θd = °
9.Steady-state error & PID controlMedium
A unity-feedback loop has G(s) = K/((s + 1)(s + 2)). Find the gain K that gives a steady-state error of 2 % for a step input.
G(s) = K/((s + 1)(s + 2))ess = 2 %
K =
10.Steady-state error & PID controlHard
A proportional controller K = 20 drives the plant G(s) = 2/(s(s + 5)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 2/(s(s + 5))K = 20
ts = s
11.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 2(s + 2)/(s(s + 10)) at ω = 20 rad/s, in degrees.
G(s) = 2(s + 2)/(s(s + 10))ω = 20 rad/s
∠G(jω) = °
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 42/(s(s + 4)(s + 6)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.
L(s) = 42/(s(s + 4)(s + 6))
ω180 = rad/s
3Solutions
1.Laplace transforms
Answer: f(0.5) = −0.7177
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).
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) = 40/(s(s + 1)(s + 4))
Partial fractions
A = 40/(1·4) = 10
B = 40/((−1)·3) = −13.33
C = 40/((−4)·(−3)) = 3.333
Output
y(t) = 10 − 13.33·e−t + 3.333·e−4t
y(1.2) = 6.012
Answer
y(1.2) = 6.012
3.Block diagrams
Answer: ess = 0.4737
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) = 8·5/(4·9) = 1.111
Steady-state error
ess = A/(1 + Kₚ) = 1/(1 + 1.111) = 0.4737
Answer
ess = 0.4737
4.Block diagrams
Answer: Kmax = 990
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 + 15s² + 74s + 120 + K = 0
Routh array
s³ | 1 74
s² | 15 120 + K
s¹ | (15·74 − 120 − K)/15
s⁰ | 120 + K
Stability limit
All entries of the first column must be positive.
Kmax = 15·74 − 120 = 990
Answer
Kmax = 990
5.Step response
Answer: ζ = 0.7462
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 = 22 → ωₙ = 4.69 rad/s
2ζωₙ = 7 → ζ = 7/(2·4.69) = 0.7462
Result
ζ = 0.7462
Answer
ζ = 0.7462
6.Step response
Answer: tp = 0.2234 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).
Natural frequency and damping ratio
ωₙ = √(k/m) = √(200/1) = 14.14 rad/s
ζ = c/(2√(km)) = 3/(2·√(200·1)) = 0.1061
Result
ωd = 14.06 rad/s
tp = π/ωd = 0.2234 s
Answer
tp = 0.2234 s
7.Stability: Routh–Hurwitz & root locus
Answer: ω = 2.828 rad/s
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.
Routh array
s³ + 9s² + 8s + K = 0
s³: 1, 8
s²: 9, K
s¹: (9·8 − K)/9
s⁰: K
Critical gain
K = 9·8 = 72
Auxiliary equation
At the critical gain the s¹ row is zero; the s² row gives the poles on the imaginary axis.
9s² + 72 = 0 → s = ±j√8
ω = 2.828 rad/s
Answer
ω = 2.828 rad/s
8.Stability: Routh–Hurwitz & root locus
Answer: θd = 71.57 °
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.
Pole locations
s² + 8s + 17 = 0 → s = −4 ± j1
s = −7
Angles from the other poles
∠(−4 + j1 + 7) = atan2(1, 3) = 18.43°
∠(−4 + j1 − (−4 − j1)) = ∠(j2) = 90°
Angle of departure
θd = 180° − 18.43° − 90° = 71.57°
Answer
θd = 71.57°
9.Steady-state error & PID control
Answer: K = 98
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) = 0.02 → Kp = 49
Gain
K/2 = 49 → K = 98
Answer
K = 98
Check the closed-loop stability as well: for this second-order loop any K > 0 is stable.
10.Steady-state error & PID control
Answer: ts = 1.6 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 + 5) + 20·2 = s² + 5s + 40
ωₙ and ζ
ωₙ = √40 = 6.325 rad/s
ζ = 5/(2·6.325) = 0.3953
Settling time
ζωₙ = 5/2 = 2.5
ts ≈ 4/2.5 = 1.6 s
Answer
ts = 1.6 s
The settling time does not depend on K here: raising K only makes the response oscillate faster.
11.Bode plots & margins
Answer: ∠G(jω) = −69.15 °
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.