In the print window, choose “Save as PDF” to keep a copy on your device.
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) = (6s + 39)/(s² + 10s + 25) and evaluate f(0.3).
F(s) = (6s + 39)/(s² + 10s + 25)t = 0.3 s
f(0.3) =
2.Laplace transformsHard
A system obeys y″ + 7y′ + 12y = 24·u(t), with y(0) = 4 and y′(0) = 0. At t = 0 a step input u(t) = 4 is applied. Use the Laplace transform to find y(0.3).
The plant G(s) = 7/(s + 2) has an inner (minor) negative feedback loop with gain H₁ = 3. A controller gain K = 8 and an outer unity negative feedback loop close the system. Find the closed-loop pole p.
G(s) = 7/(s + 2)H₁ = 3K = 8
p =
4.Block diagramsHard
The plant G(s) = 52/(s(s + 5)) is in a negative feedback loop with sensor gain H = 0.5. The closed-loop poles are complex. Find their imaginary part ωd (the positive value), in rad/s.
G(s) = 52/(s(s + 5))H = 0.5
ωd = rad/s
5.Step responseMedium
A system has the transfer function G(s) = 36/(s² + 6s + 18) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 36/(s² + 6s + 18)
ts = s
6.Step responseHard
A series RLC circuit with R = 14 Ω, L = 500 mH and C = 100 µF is connected to a 12 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 = 14 ΩL = 500 mHC = 100 µFV = 12 V
%OS = %
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)). Find the largest gain Kmax for which the closed loop is stable.
L(s) = K/(s(s + 1)(s + 8))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
For L(s) = K/((s + 8)(s² + 6s + 25)), 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 + 8)(s² + 6s + 25))
θd = °
9.Steady-state error & PID controlMedium
A ramp r(t) = 0.5t is applied to a unity-feedback loop with G(s) = 10/(s(s + 1)). Find the steady-state error ess.
G(s) = 10/(s(s + 1))r(t) = 0.5t
ess =
10.Steady-state error & PID controlHard
A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 10/(s + 1) in a unity-feedback loop. The closed loop must have ωₙ = 5 rad/s and ζ = 0.8. Find the proportional gain Kp.
G(s) = 10/(s + 1)ωₙ = 5 rad/sζ = 0.8
Kp =
11.Bode plots & marginsMedium
Find the exact magnitude |G(jω)| of G(s) = 30/((s + 1)(s + 15)) at ω = 1 rad/s, in dB.
G(s) = 30/((s + 1)(s + 15))ω = 1 rad/s
|G(jω)| = dB
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 200/(s(s + 10)). Find the phase margin PM in degrees.
L(s) = 200/(s(s + 10))
PM = °
3Solutions
1.Laplace transforms
Answer: f(0.3) = 1.941
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 + 10s + 25 = (s + 5)²
Partial fractions
6s + 39 = 6(s + 5) + 9
F(s) = 6/(s + 5) + 9/(s + 5)²
Inverse transform
f(t) = 6·e−5t + 9·t·e−5t
f(0.3) = (6 + 9·0.3)·e−5·0.3 = 1.941
Answer
f(0.3) = 1.941
2.Laplace transforms
Answer: y(0.3) = 5.109
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
(s² + 7s + 12)·Y(s) = 4s + 28 + 96/s
Y(s) = (4s² + 28s + 96)/(s(s + 3)(s + 4))
Partial fractions
A = 96/(3·4) = 8
B = 48/((−3)·1) = −16
C = 48/((−4)·(−1)) = 12
Result
y(t) = 8 − 16·e−3t + 12·e−4t
y(0.3) = 5.109
Answer
y(0.3) = 5.109
Check: y(0) = A + B + C = 4.
3.Block diagrams
Answer: p = −79
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).
Characteristic equation
s(s + 5) + 52·0.5 = 0 → s² + 5s + 26 = 0
Roots
s = −2.5 ± j·√(26 − 6.25) = −2.5 ± j4.444
ωd = 4.444 rad/s
Answer
ωd = 4.444 rad/s
5.Step response
Answer: ts = 1.333 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 = 18 → ωₙ = 4.243 rad/s
2ζωₙ = 6 → ζ = 6/(2·4.243) = 0.7071
Result
ζωₙ = 3 → ts ≈ 4/3 = 1.333 s
Answer
ts = 1.333 s
6.Step response
Answer: %OS = 73.16 %
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).
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 equation
s³ + 9s² + 8s + K = 0
Routh array
s³: 1, 8
s²: 9, K
s¹: (9·8 − K)/9
s⁰: K
Conditions
Every entry of the first column must be positive.
K > 0
9·8 − K > 0 → K < 72
Answer
Kmax = 72
At K = 72 the closed loop oscillates at ω = √8 = 2.828 rad/s.
8.Stability: Routh–Hurwitz & root locus
Answer: θd = 51.34 °
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 + 6s + 25 = 0 → s = −3 ± j4
s = −8
Angles from the other poles
∠(−3 + j4 + 8) = atan2(4, 5) = 38.66°
∠(−3 + j4 − (−3 − j4)) = ∠(j8) = 90°
Angle of departure
θd = 180° − 38.66° − 90° = 51.34°
Answer
θd = 51.34°
9.Steady-state error & PID control
Answer: ess = 0.05
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.
Velocity constant
Kv = lim s·G(s) = 10
Ramp error
ess = 0.5/10 = 0.05
Answer
ess = 0.05
10.Steady-state error & PID control
Answer: Kp = 0.7
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
1 + C(s)G(s) = 0, multiplied by s(s + 1).
s² + (1 + 10Kp)s + 10Ki
Target polynomial
s² + 2·0.8·5s + 5² = s² + 8s + 25
Match the coefficients
1 + 10Kp = 8 → Kp = 0.7
Answer
Kp = 0.7
The integral term makes the loop type 1, so a step is followed without steady-state error.
11.Bode plots & margins
Answer: |G(jω)| = 2.991 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.
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.