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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) = 40/(s(s + 3)(s + 4)) by partial fractions, and evaluate f(0.5).
F(s) = 40/(s(s + 3)(s + 4))t = 0.5 s
f(0.5) =
2.Laplace transformsHard
A system obeys y″ + 6y′ + 8y = 16·u(t), with y(0) = 1 and y′(0) = 0. At t = 0 a step input u(t) = 1 is applied. Use the Laplace transform to find y(0.1).
A unity negative feedback loop has G(s) = 6/(s(s + 4)). The reference is a ramp r(t) = 10·t. Find the steady-state error ess.
G(s) = 6/(s(s + 4))r(t) = 10·t
ess =
4.Block diagramsHard
The plant G(s) = 16/(s(s + 3)) 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) = 16/(s(s + 3))H = 0.5
ωd = rad/s
5.Step responseMedium
A system has the transfer function G(s) = 17.5/(s² + 9s + 35) and receives a unit step input. Find the percent overshoot %OS of the step response.
G(s) = 17.5/(s² + 9s + 35)
%OS = %
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 plant G(s) = (s + 4)/((s − 4)(s + 3)) has an unstable pole at s = 4. 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 + 4)/((s − 4)(s + 3))
Kmin =
8.Stability: Routh–Hurwitz & root locusHard
For L(s) = K/((s + 5)(s² + 2s + 17)), 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 + 5)(s² + 2s + 17))
θd = °
9.Steady-state error & PID controlMedium
A process G(s) = 2.5/(5s + 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) = 2.5/(5s + 1)K = 19
ess = %
10.Steady-state error & PID controlHard
A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 5/(s + 2) in a unity-feedback loop. The closed loop must have ωₙ = 10 rad/s and ζ = 1. Find the integral gain Ki.
G(s) = 5/(s + 2)ωₙ = 10 rad/sζ = 1
Ki =
11.Bode plots & marginsMedium
Find the exact magnitude |G(jω)| of G(s) = 798/((s + 3)(s + 19)) at ω = 3 rad/s, in dB.
G(s) = 798/((s + 3)(s + 19))ω = 3 rad/s
|G(jω)| = dB
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 20/(s(s + 10)). Find the gain crossover frequency ωc, where |L(jω)| = 1.
L(s) = 20/(s(s + 10))
ωc = rad/s
3Solutions
1.Laplace transforms
Answer: f(0.5) = 1.712
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
(s² + 6s + 8)·Y(s) = 1s + 6 + 16/s
Y(s) = (s² + 6s + 16)/(s(s + 2)(s + 4))
Partial fractions
A = 16/(2·4) = 2
B = 8/((−2)·2) = −2
C = 8/((−4)·(−2)) = 1
Result
y(t) = 2 − 2·e−2t + e−4t
y(0.1) = 1.033
Answer
y(0.1) = 1.033
Check: y(0) = A + B + C = 1.
3.Block diagrams
Answer: ess = 6.667
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).
Velocity error constant
Kᵥ = lims→0 s·G(s) = 6/4 = 1.5
Steady-state error
ess = A/Kᵥ = 10/1.5 = 6.667
Answer
ess = 6.667
4.Block diagrams
Answer: ωd = 2.398 rad/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).
Characteristic equation
s(s + 3) + 16·0.5 = 0 → s² + 3s + 8 = 0
Roots
s = −1.5 ± j·√(8 − 2.25) = −1.5 ± j2.398
ωd = 2.398 rad/s
Answer
ωd = 2.398 rad/s
5.Step response
Answer: %OS = 2.52 %
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 = 35 → ωₙ = 5.916 rad/s
2ζωₙ = 9 → ζ = 9/(2·5.916) = 0.7606
Result
%OS = 100·e−ζπ/√(1−ζ²) = 100·e−3.681 = 2.52 %
Answer
%OS = 2.52 %
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: Kmin = 3
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.
-1 + K > 0 → K > 1
4K − 12 > 0 → K > 3
Kmin = 3
Answer
Kmin = 3
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 + 2s + 17 = 0 → s = −1 ± j4
s = −5
Angles from the other poles
∠(−1 + j4 + 5) = atan2(4, 4) = 45°
∠(−1 + j4 − (−1 − j4)) = ∠(j8) = 90°
Angle of departure
θd = 180° − 45° − 90° = 45°
Answer
θd = 45°
9.Steady-state error & PID control
Answer: ess = 2.062 %
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 = 19·2.5 = 47.5
Step error
ess = 100/(1 + 47.5) = 2.062 %
Answer
ess = 2.062 %
10.Steady-state error & PID control
Answer: Ki = 20
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 + 2).
s² + (2 + 5Kp)s + 5Ki
Target polynomial
s² + 2·1·10s + 10² = s² + 20s + 100
Match the coefficients
5Ki = 100 → Ki = 20
Answer
Ki = 20
The integral term makes the loop type 1, so a step is followed without steady-state error.
11.Bode plots & margins
Answer: |G(jω)| = 19.81 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.