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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) = (3s + 2)/(s² + 5s + 4). Its inverse transform has the form f(t) = A·e−t + B·e−4t. Find the coefficient A by partial fractions.
F(s) = (3s + 2)/(s² + 5s + 4)
A =
2.Laplace transformsHard
A system is described by the differential equation y″ + 9y′ + 20y = 2u′ + 13u, 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″ + 9y′ + 20y = 2u′ + 13ut = 0.25 s
g(0.25) =
3.Block diagramsMedium
The plant G(s) = 2/(s + 7) has an inner (minor) negative feedback loop with gain H₁ = 1. A controller gain K = 19 and an outer unity negative feedback loop close the system. Find the closed-loop DC gain T(0).
G(s) = 2/(s + 7)H₁ = 1K = 19
T(0) =
4.Block diagramsHard
The plant G(s) = K/(s(s + 4)(s + 10)) 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) = 21/(s² + 3s + 7) and receives a unit step input. Find the natural frequency ωn in rad/s.
G(s) = 21/(s² + 3s + 7)
ωn = rad/s
6.Step responseHard
A unity negative feedback loop has G(s) = K/(s(s + 8)). Find the gain K that gives a step response with 10 % overshoot.
G(s) = K/(s(s + 8))%OS = 10 %
K =
7.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram has negative feedback and L(s) = 780/(s(s + 3)(s + 10)). How many closed-loop poles N lie in the right half-plane?
L(s) = 780/(s(s + 3)(s + 10))
N =
8.Stability: Routh–Hurwitz & root locusHard
For L(s) = K/((s + 2)(s² + 8s + 20)), 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 + 2)(s² + 8s + 20))
θd = °
9.Steady-state error & PID controlMedium
The parabolic input r(t) = 0.5·t²/2 is applied to a unity-feedback loop with G(s) = 4(s + 3)/(s²(s + 2)). Find the steady-state error ess.
G(s) = 4(s + 3)/(s²(s + 2))r(t) = 0.5·t²/2
ess =
10.Steady-state error & PID controlHard
A proportional controller K = 25 drives the plant G(s) = 1/(s(s + 5)) in a unity-feedback loop. Find the peak time tp of the step response.
G(s) = 1/(s(s + 5))K = 25
tp = s
11.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 26(s + 12)/(s(s + 4)) at ω = 1 rad/s, in degrees.
G(s) = 26(s + 12)/(s(s + 4))ω = 1 rad/s
∠G(jω) = °
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 60/(s(s + 2)(s + 10)). Find the gain margin GM in dB.
L(s) = 60/(s(s + 2)(s + 10))
GM = dB
3Solutions
1.Laplace transforms
Answer: A = −0.3333
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 + 4 = (s + 1)(s + 4)
F(s) = A/(s + 1) + B/(s + 4)
Cover-up rule
Multiply by (s + 1) and set s = −1: the B-term vanishes.
A = (3s + 2)/(s + 4) |s = −1 = −1/3 = −0.3333
Answer
A = −0.3333
(B = 3.333, so f(t) = −0.3333·e−t + 3.333·e−4t)
2.Laplace transforms
Answer: g(0.25) = 0.9799
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 + 13)/(s² + 9s + 20)
Partial fractions
s² + 9s + 20 = (s + 4)(s + 5)
A = 5, B = −3
Impulse response
g(t) = 5·e−4t − 3·e−5t
g(0.25) = 0.9799
Answer
g(0.25) = 0.9799
3.Block diagrams
Answer: T(0) = 0.8085
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
s³ + 14s² + 40s + K = 0
Routh array
s³ | 1 40
s² | 14 K
s¹ | (14·40 − K)/14
s⁰ | K
Stability limit
All entries of the first column must be positive.
Kmax = 14·40 = 560
Answer
Kmax = 560
5.Step response
Answer: ωn = 2.646 rad/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 = 7 → ωₙ = 2.646 rad/s
2ζωₙ = 3 → ζ = 3/(2·2.646) = 0.5669
Result
ωₙ = 2.646 rad/s
Answer
ωn = 2.646 rad/s
6.Step response
Answer: K = 45.78
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).
Closed-loop transfer function
T(s) = K/(s² + 8s + K)
Damping ratio from the overshoot
ζ = −ln(0.1)/√(π² + ln²(0.1)) = 0.5912
Gain
ωₙ = 8/(2ζ) = 6.766 rad/s
K = ωₙ² = 45.78
Answer
K = 45.78
7.Stability: Routh–Hurwitz & root locus
Answer: N = 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.
Characteristic polynomial
s³ + 13s² + 30s + 780 = 0
Routh array
s³: 1, 30
s²: 13, 780
s¹: −30
s⁰: 780
Sign changes in the first column
1 → 13 → −30 → 780
N = 2
Answer
N = 2
The gain is above the limit K = 390, so a complex pair has crossed into the right half-plane.
8.Stability: Routh–Hurwitz & root locus
Answer: θd = −45 °
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 + 8s + 20 = 0 → s = −4 ± j2
s = −2
Angles from the other poles
∠(−4 + j2 + 2) = atan2(2, -2) = 135°
∠(−4 + j2 − (−4 − j2)) = ∠(j4) = 90°
Angle of departure
θd = 180° − 135° − 90° = −45°
Answer
θd = −45°
9.Steady-state error & PID control
Answer: ess = 0.08333
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 = 4·3/2 = 6
Parabola error
ess = 0.5/6 = 0.08333
Answer
ess = 0.08333
10.Steady-state error & PID control
Answer: tp = 0.7255 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 + 5) + 25·1 = s² + 5s + 25
ωₙ and ζ
ωₙ = √25 = 5 rad/s
ζ = 5/(2·5) = 0.5
Peak time
ωd = 5·√(1 − 0.5²) = 4.33 rad/s
tp = π/4.33 = 0.7255 s
Answer
tp = 0.7255 s
11.Bode plots & margins
Answer: ∠G(jω) = −99.27 °
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.