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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) = (3s + 25)/(s² + 8s + 20) by completing the square, and evaluate f(0.25).
F(s) = (3s + 25)/(s² + 8s + 20)t = 0.25 s
f(0.25) =
2.Laplace transformsHard
A system obeys y″ + 5y′ + 4y = 10·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) = 3(s + 2)/((s + 6)(s + 7)). A step of size 10 is applied at the reference. Find the steady-state error ess.
G(s) = 3(s + 2)/((s + 6)(s + 7))R = 10
ess =
4.Block diagramsHard
The plant G(s) = K/((s + 1)(s + 3)(s + 5)) 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) = 59/(s² + 11s + 59) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 59/(s² + 11s + 59)
ts = s
6.Step responseHard
A series RLC circuit with R = 80 Ω, L = 20 mH and C = 2 µF is connected to a 5 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the percent overshoot %OS of the capacitor voltage.
R = 80 ΩL = 20 mHC = 2 µFV = 5 V
%OS = %
7.Stability: Routh–Hurwitz & root locusMedium
A unity-feedback loop has L(s) = K/((s + 2)(s + 5)(s + 6)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.
L(s) = K/((s + 2)(s + 5)(s + 6))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K/((s + 1)(s + 3)) 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 + 1)(s + 3))
K =
9.Steady-state error & PID controlMedium
The parabolic input r(t) = 2·t²/2 is applied to a unity-feedback loop with G(s) = 5(s + 1)/(s²(s + 5)). Find the steady-state error ess.
G(s) = 5(s + 1)/(s²(s + 5))r(t) = 2·t²/2
ess =
10.Steady-state error & PID controlHard
A proportional controller K = 1 drives the plant G(s) = 10/(s(s + 5)) in a unity-feedback loop. Find the percent overshoot %OS of the step response.
G(s) = 10/(s(s + 5))K = 1
%OS = %
11.Bode plots & marginsMedium
The RC low-pass filter shown has R = 22 kΩ, C = 47 nF. A sinusoidal input of 1.5 kHz is applied. Find the gain |H| = |vout/vin| in dB.
R = 22 kΩC = 47 nFf = 1.5 kHz
|H| = dB
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 0.7/(s(s + 1)). Find the phase margin PM in degrees.
L(s) = 0.7/(s(s + 1))
PM = °
3Solutions
1.Laplace transforms
Answer: f(0.25) = 2.115
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) = 20/(s(s + 1)(s + 4))
Partial fractions
A = 20/(1·4) = 5
B = 20/((−1)·3) = −6.667
C = 20/((−4)·(−3)) = 1.667
Output
y(t) = 5 − 6.667·e−t + 1.667·e−4t
y(1.2) = 3.006
Answer
y(1.2) = 3.006
3.Block diagrams
Answer: ess = 8.75
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) = 3·2/(6·7) = 0.1429
Steady-state error
ess = A/(1 + Kₚ) = 10/(1 + 0.1429) = 8.75
Answer
ess = 8.75
4.Block diagrams
Answer: Kmax = 192
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 + 9s² + 23s + 15 + K = 0
Routh array
s³ | 1 23
s² | 9 15 + K
s¹ | (9·23 − 15 − K)/9
s⁰ | 15 + K
Stability limit
All entries of the first column must be positive.
Kmax = 9·23 − 15 = 192
Answer
Kmax = 192
5.Step response
Answer: ts = 0.7273 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 = 59 → ωₙ = 7.681 rad/s
2ζωₙ = 11 → ζ = 11/(2·7.681) = 0.716
Result
ζωₙ = 5.5 → ts ≈ 4/5.5 = 0.7273 s
Answer
ts = 0.7273 s
6.Step response
Answer: %OS = 25.38 %
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 = 1/√(LC) = 1/√(0.02·2 × 10⁻⁶) = 5000 rad/s
ζ = (R/2)·√(C/L) = (80/2)·√(2 × 10⁻⁶/0.02) = 0.4
Result
%OS = 100·e−ζπ/√(1−ζ²) = 25.38 %
Answer
%OS = 25.38 %
7.Stability: Routh–Hurwitz & root locus
Answer: Kmax = 616
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 + 13s² + 52s + 60 + K = 0
Routh array
s³: 1, 52
s²: 13, 60 + K
s¹: (13·52 − 60 − K)/13
s⁰: 60 + K
Condition
The s¹ entry must stay positive.
K < 13·52 − 60 = 616
Answer
Kmax = 616
At the limit the loop oscillates at ω = √52 = 7.211 rad/s.
8.Stability: Routh–Hurwitz & root locus
Answer: K = 1
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 + 1)(s + 3) gives 2s + 4 = 0.
sb = −2
Magnitude condition
K = |−2 + 1|·|−2 + 3| = 1·1 = 1
Answer
K = 1
For this gain the closed loop is critically damped (ζ = 1).
9.Steady-state error & PID control
Answer: ess = 2
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.
Acceleration constant
Ka = 5·1/5 = 1
Parabola error
ess = 2/1 = 2
Answer
ess = 2
10.Steady-state error & PID control
Answer: %OS = 1.732 %
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) + 1·10 = s² + 5s + 10
ωₙ and ζ
ωₙ = √10 = 3.162 rad/s
ζ = 5/(2·3.162) = 0.7906
Overshoot
%OS = 100·e−0.7906π/√(1 − 0.7906²) = 1.732 %
Answer
%OS = 1.732 %
11.Bode plots & margins
Answer: |H| = −19.82 dB
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 = 1/(RC) = 1/(22000·4.7 × 10⁻⁸) = 967.1 rad/s
fc = 153.9 Hz → f/fc = 1500/153.9 = 9.745
Gain
|H| = 1/√(1 + 9.745²) = 0.1021
20·log₁₀(0.1021) = −19.82 dB
Answer
|H| = −19.82 dB
12.Bode plots & margins
Answer: PM = 59.03 °
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.