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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) = (2s + 21)/(s² + 7s + 10). Its inverse transform has the form f(t) = A·e−2t + B·e−5t. Find the coefficient A by partial fractions.
F(s) = (2s + 21)/(s² + 7s + 10)
A =
2.Laplace transformsHard
A system is described by the differential equation y″ + 9y′ + 20y = 2u′ + 18u, 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 = 0.2 s. Find g(0.2).
y″ + 9y′ + 20y = 2u′ + 18ut = 0.2 s
g(0.2) =
3.Block diagramsMedium
A plant G(s) = 15/(s + 1) is controlled in a negative feedback loop whose sensor has the gain H = 0.2. Find the time constant τ of the closed loop.
G(s) = 15/(s + 1)H = 0.2
τ = s
4.Block diagramsHard
The plant G(s) = K/((s + 4)(s + 5)(s + 8)) 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) = 47/(s² + 11s + 47) and receives a unit step input. Find the percent overshoot %OS of the step response.
G(s) = 47/(s² + 11s + 47)
%OS = %
6.Step responseHard
A unity negative feedback loop has G(s) = 30/(s(s + 4)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).
G(s) = 30/(s(s + 4))
ts = s
7.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 2)(s + 4)). 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 + 2)(s + 4))
ω = rad/s
8.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K/(s(s + 6)) is shown. Find the gain K for which the closed-loop poles have the damping ratio ζ = 0.707.
L(s) = K/(s(s + 6))ζ = 0.707
K =
9.Steady-state error & PID controlMedium
A process G(s) = 0.5/(2s + 1) is controlled by a proportional gain K = 19 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.
G(s) = 0.5/(2s + 1)K = 19
ess = %
10.Steady-state error & PID controlHard
A PI controller C(s) = 1 + 1/s drives the plant G(s) = 10/((s + 1)(s + 10)) in a unity-feedback loop. Find the steady-state error ess for the ramp input r(t) = 2t.
The RL high-pass filter shown has R = 10 Ω, L = 47 mH. A sinusoidal input of 6.8 Hz is applied. Find the phase shift φ of vout relative to vin, in degrees.
R = 10 ΩL = 47 mHf = 6.8 Hz
φ = °
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 220/(s(s + 20)). Find the phase margin PM in degrees.
L(s) = 220/(s(s + 20))
PM = °
3Solutions
1.Laplace transforms
Answer: A = 5.667
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 + 7s + 10 = (s + 2)(s + 5)
F(s) = A/(s + 2) + B/(s + 5)
Cover-up rule
Multiply by (s + 2) and set s = −2: the B-term vanishes.
A = (2s + 21)/(s + 5) |s = −2 = 17/3 = 5.667
Answer
A = 5.667
(B = −3.667, so f(t) = 5.667·e−2t − 3.667·e−5t)
2.Laplace transforms
Answer: g(0.2) = 1.55
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 + 18)/(s² + 9s + 20)
Partial fractions
s² + 9s + 20 = (s + 4)(s + 5)
A = 10, B = −8
Impulse response
g(t) = 10·e−4t − 8·e−5t
g(0.2) = 1.55
Answer
g(0.2) = 1.55
3.Block diagrams
Answer: τ = 0.25 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).
Closed-loop transfer function
T(s) = G/(1 + G·H) = 15/(s + 1 + 3) = 15/(s + 4)
Result
τ = 1/4 = 0.25 s
Answer
τ = 0.25 s
4.Block diagrams
Answer: Kmax = 1404
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 + 17s² + 92s + 160 + K = 0
Routh array
s³ | 1 92
s² | 17 160 + K
s¹ | (17·92 − 160 − K)/17
s⁰ | 160 + K
Stability limit
All entries of the first column must be positive.
Kmax = 17·92 − 160 = 1404
Answer
Kmax = 1404
5.Step response
Answer: %OS = 1.467 %
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 = 47 → ωₙ = 6.856 rad/s
2ζωₙ = 11 → ζ = 11/(2·6.856) = 0.8023
Result
%OS = 100·e−ζπ/√(1−ζ²) = 100·e−4.222 = 1.467 %
Answer
%OS = 1.467 %
6.Step response
Answer: ts = 2 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: 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 + 6s² + 8s + K = 0
s³: 1, 8
s²: 6, K
s¹: (6·8 − K)/6
s⁰: K
Critical gain
K = 6·8 = 48
Auxiliary equation
At the critical gain the s¹ row is zero; the s² row gives the poles on the imaginary axis.
6s² + 48 = 0 → s = ±j√8
ω = 2.828 rad/s
Answer
ω = 2.828 rad/s
8.Stability: Routh–Hurwitz & root locus
Answer: K = 18.01
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.
Compare with the standard form
s² + 6s + K = s² + 2ζωₙs + ωₙ²
2·0.707·ωₙ = 6 → ωₙ = 4.243 rad/s
Gain
K = ωₙ² = 4.243² = 18.01
Answer
K = 18.01
The closed-loop poles sit where the ζ line crosses the vertical branch at s = −3.
9.Steady-state error & PID control
Answer: ess = 9.524 %
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 = 19·0.5 = 9.5
Step error
ess = 100/(1 + 9.5) = 9.524 %
Answer
ess = 9.524 %
10.Steady-state error & PID control
Answer: ess = 2
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).
Velocity constant
s·C(s) = 1s + 1 → 1 as s → 0.
Kv = 1·10/10 = 1
Ramp error
ess = 2/1 = 2
Answer
ess = 2
The PI controller removes the step error completely; a ramp is followed with a constant lag.
11.Bode plots & margins
Answer: φ = 78.65 °
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.
Corner frequency
ωc = R/L = 10/0.047 = 212.8 rad/s
fc = 33.86 Hz → f/fc = 6.8/33.86 = 0.2008
Phase
φ = 90° − atan(0.2008) = 78.65°
Answer
φ = 78.65°
12.Bode plots & margins
Answer: PM = 63.74 °
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.