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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) = (2s + 17)/(s² + 6s + 9) and evaluate f(0.1).
F(s) = (2s + 17)/(s² + 6s + 9)t = 0.1 s
f(0.1) =
2.Laplace transformsHard
A system is described by the differential equation y″ + 6y′ + 5y = 2u′ + 14u, 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 = 0.25 s. Find g(0.25).
y″ + 6y′ + 5y = 2u′ + 14ut = 0.25 s
g(0.25) =
3.Block diagramsMedium
A unity negative feedback loop has G(s) = 6(s + 8)/((s + 5)(s + 9)). A step of size 5 is applied at the reference. Find the steady-state error ess.
G(s) = 6(s + 8)/((s + 5)(s + 9))R = 5
ess =
4.Block diagramsHard
The plant G(s) = K/(s(s + 1)(s + 7)) 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 + 1)(s + 7))unity negative feedback
ω = rad/s
5.Step responseMedium
A system has the transfer function G(s) = 6/(s² + 3s + 6) and receives a unit step input. Find the peak time tp of the step response.
G(s) = 6/(s² + 3s + 6)
tp = s
6.Step responseHard
A mass m = 5 kg hangs on a spring k = 1000 N/m with a viscous damper c = 42 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 25 N is suddenly applied. Find the time tp at which the displacement is largest.
m = 5 kgk = 1000 N/mc = 42 N·s/mF = 25 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 + 3)). 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 + 3))
ω = rad/s
8.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K(s + 10)/(s(s + 2)) leaves the real axis between s = 0 and s = −2, circles round and joins the real axis again to the left of the zero. Find the break-in point si.
L(s) = K(s + 10)/(s(s + 2))
si =
9.Steady-state error & PID controlMedium
A process G(s) = 4/(2s + 1) is controlled by a proportional gain K = 9 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.
G(s) = 4/(2s + 1)K = 9
ess = %
10.Steady-state error & PID controlHard
A PD controller C(s) = 20 + Kd·s drives the plant G(s) = 4/(s(s + 3)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.6.
G(s) = 4/(s(s + 3))Kp = 20ζ = 0.6
Kd = s
11.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 26(s + 10)/(s(s + 4)) at ω = 10 rad/s, in degrees.
G(s) = 26(s + 10)/(s(s + 4))ω = 10 rad/s
∠G(jω) = °
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 35/(s(s + 10)). Find the phase margin PM in degrees.
L(s) = 35/(s(s + 10))
PM = °
3Solutions
1.Laplace transforms
Answer: f(0.1) = 2.297
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 + 6s + 9 = (s + 3)²
Partial fractions
2s + 17 = 2(s + 3) + 11
F(s) = 2/(s + 3) + 11/(s + 3)²
Inverse transform
f(t) = 2·e−3t + 11·t·e−3t
f(0.1) = (2 + 11·0.1)·e−3·0.1 = 2.297
Answer
f(0.1) = 2.297
2.Laplace transforms
Answer: g(0.25) = 2.05
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) = (2s + 14)/(s² + 6s + 5)
Partial fractions
s² + 6s + 5 = (s + 1)(s + 5)
A = 3, B = −1
Impulse response
g(t) = 3·e−t − e−5t
g(0.25) = 2.05
Answer
g(0.25) = 2.05
3.Block diagrams
Answer: ess = 2.419
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) = 6·8/(5·9) = 1.067
Steady-state error
ess = A/(1 + Kₚ) = 5/(1 + 1.067) = 2.419
Answer
ess = 2.419
4.Block diagrams
Answer: ω = 2.646 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 + 8s² + 7s + K = 0
Routh array
s³ | 1 7
s² | 8 K
s¹ | (8·7 − K)/8
s⁰ | K
Stability limit
All entries of the first column must be positive.
Kmax = 8·7 = 56
Frequency of oscillation
8·(jω)² + 56 = 0 → ω² = 56/8 = 7
ω = 2.646 rad/s
Answer
ω = 2.646 rad/s
5.Step response
Answer: tp = 1.622 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).
Natural frequency and damping ratio
ωₙ = √(k/m) = √(1000/5) = 14.14 rad/s
ζ = c/(2√(km)) = 42/(2·√(1000·5)) = 0.297
Result
ωd = 13.5 rad/s
tp = π/ωd = 0.2326 s
Answer
tp = 0.2326 s
7.Stability: Routh–Hurwitz & root locus
Answer: ω = 1.732 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 + 4s² + 3s + K = 0
s³: 1, 3
s²: 4, K
s¹: (4·3 − K)/4
s⁰: K
Critical gain
K = 4·3 = 12
Auxiliary equation
At the critical gain the s¹ row is zero; the s² row gives the poles on the imaginary axis.
4s² + 12 = 0 → s = ±j√3
ω = 1.732 rad/s
Answer
ω = 1.732 rad/s
8.Stability: Routh–Hurwitz & root locus
Answer: si = −18.94
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.
Gain along the real axis
K(s) = −(s² + 2s)/(s + 10)
Stationary points
Differentiate with the quotient rule and set the numerator to zero.
(2s + 2)(s + 10) − (s² + 2s) = s² + 20s + 20 = 0
s = −10 ± √80 = −1.056, −18.94
Break-in point
si = −18.94
Answer
si = −18.94
Between the two break points the locus is a circle centred on the zero, with radius √80 = 8.944.
9.Steady-state error & PID control
Answer: ess = 2.703 %
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.
Position constant
Kp = 9·4 = 36
Step error
ess = 100/(1 + 36) = 2.703 %
Answer
ess = 2.703 %
10.Steady-state error & PID control
Answer: Kd = 1.933 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 + 3) + 4(20 + Kds) = s² + (3 + 4Kd)s + 80
Natural frequency
ωₙ = √80 = 8.944 rad/s
Derivative gain
3 + 4Kd = 2·0.6·8.944 = 10.73
Kd = 1.933 s
Answer
Kd = 1.933 s
The derivative term adds damping without changing ωₙ.
11.Bode plots & margins
Answer: ∠G(jω) = −113.2 °
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.