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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) = (4s + 28)/(s² + 5s + 6). Its inverse transform has the form f(t) = A·e−2t + B·e−3t. Find the coefficient A by partial fractions.
F(s) = (4s + 28)/(s² + 5s + 6)
A =
2.Laplace transformsHard
A system is described by the differential equation y″ + 5y′ + 6y = 4u′ + 12u, 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 = 0.4 s. Find g(0.4).
y″ + 5y′ + 6y = 4u′ + 12ut = 0.4 s
g(0.4) =
3.Block diagramsMedium
The plant G(s) = 7/(s + 6) has an inner (minor) negative feedback loop with gain H₁ = 2. A controller gain K = 3 and an outer unity negative feedback loop close the system. Find the closed-loop DC gain T(0).
G(s) = 7/(s + 6)H₁ = 2K = 3
T(0) =
4.Block diagramsHard
A proportional controller K = 9 drives the plant G(s) = 9/(s + 9) 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 0.5 are applied together. Find the steady-state output css.
K = 9G(s) = 9/(s + 9)D = 0.5R = 5
css =
5.Step responseMedium
A system has the transfer function G(s) = 26.5/(s² + 7s + 53) and receives a unit step input. Find the damping ratio ζ.
G(s) = 26.5/(s² + 7s + 53)
ζ =
6.Step responseHard
A unity negative feedback loop has G(s) = 220/(s(s + 5)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).
G(s) = 220/(s(s + 5))
ts = s
7.Stability: Routh–Hurwitz & root locusMedium
The plant G(s) = (s + 2)/((s − 2)(s + 2)) has an unstable pole at s = 2. A proportional gain K closes a unity-feedback loop around it. Find the smallest gain Kmin that makes the closed loop stable.
G(s) = (s + 2)/((s − 2)(s + 2))
Kmin =
8.Stability: Routh–Hurwitz & root locusHard
For the loop L(s) = K/(s(s + 3)(s + 8)), find the gain K that puts a closed-loop pole at s = −2.5.
L(s) = K/(s(s + 3)(s + 8))s = −2.5
K =
9.Steady-state error & PID controlMedium
A process G(s) = 5/(2s + 1) is controlled by a proportional gain K = 4 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.
G(s) = 5/(2s + 1)K = 4
ess = %
10.Steady-state error & PID controlHard
A proportional controller K drives the plant G(s) = 2/(s(s + 8)) in a unity-feedback loop. Find the gain K that gives the closed loop a damping ratio ζ = 0.6.
G(s) = 2/(s(s + 8))ζ = 0.6
K =
11.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 19(s + 16)/(s(s + 6)) at ω = 30 rad/s, in degrees.
G(s) = 19(s + 16)/(s(s + 6))ω = 30 rad/s
∠G(jω) = °
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = K/(s(s + 7)(s + 10)). Find the gain K that gives a gain margin of 12 dB.
L(s) = K/(s(s + 7)(s + 10))GM = 12 dB
K =
3Solutions
1.Laplace transforms
Answer: A = 20
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 + 5s + 6 = (s + 2)(s + 3)
F(s) = A/(s + 2) + B/(s + 3)
Cover-up rule
Multiply by (s + 2) and set s = −2: the B-term vanishes.
A = (4s + 28)/(s + 3) |s = −2 = 20/1 = 20
Answer
A = 20
(B = −16, so f(t) = 20·e−2t − 16·e−3t)
2.Laplace transforms
Answer: g(0.4) = 1.797
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) = (4s + 12)/(s² + 5s + 6)
Partial fractions
s² + 5s + 6 = (s + 2)(s + 3)
A = 4, B = 0
Impulse response
g(t) = 4·e−2t
g(0.4) = 1.797
Answer
g(0.4) = 1.797
3.Block diagrams
Answer: T(0) = 0.5122
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).
Transfer functions
C/D = G/(1 + K·G) = 9/(s + 9 + 81)
C/R = K·G/(1 + K·G) = 81/(s + 9 + 81)
Steady state
css = 5·81/90 + 0.5·9/90 = 4.55
Answer
css = 4.55
A larger K reduces the effect of the disturbance, but never to zero with a proportional controller.
5.Step response
Answer: ζ = 0.4808
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 = 53 → ωₙ = 7.28 rad/s
2ζωₙ = 7 → ζ = 7/(2·7.28) = 0.4808
Result
ζ = 0.4808
Answer
ζ = 0.4808
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: Kmin = 2
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.
For s² + c₁s + c₀ all roots lie in the left half-plane exactly when c₁ > 0 and c₀ > 0.
0 + K > 0 → K > 0
2K − 4 > 0 → K > 2
Kmin = 2
Answer
Kmin = 2
8.Stability: Routh–Hurwitz & root locus
Answer: K = 6.875
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.
Magnitude condition
K = |s|·|s + 3|·|s + 8|
Evaluate at the point
K = 2.5·0.5·5.5 = 6.875
Answer
K = 6.875
Check: s(s + 3)(s + 8) + K is zero at s = −2.5.
9.Steady-state error & PID control
Answer: ess = 4.762 %
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 = 4·5 = 20
Step error
ess = 100/(1 + 20) = 4.762 %
Answer
ess = 4.762 %
10.Steady-state error & PID control
Answer: K = 22.22
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 + 8) + 2K = s² + 8s + 2K
Natural frequency
2·0.6·ωₙ = 8 → ωₙ = 6.667 rad/s
Gain
2K = ωₙ² = 44.44 → K = 22.22
Answer
K = 22.22
11.Bode plots & margins
Answer: ∠G(jω) = −106.8 °
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.