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
F(s) = (2s + 7)/(s² + 8s + 12). Its inverse transform has the form f(t) = A·e−2t + B·e−6t. Find the coefficient A by partial fractions.
F(s) = (2s + 7)/(s² + 8s + 12)
A =
2.Laplace transformsHard
Solve y″ + 5y′ + 6y = 0 with y(0) = 2 and y′(0) = 8 using the Laplace transform, and find y(0.2).
y″ + 5y′ + 6y = 0y(0) = 2y′(0) = 8t = 0.2 s
y(0.2) =
3.Block diagramsMedium
The plant G(s) = K/(s(s + 2)) is in a unity negative feedback loop. Find the gain K for which the closed loop is critically damped (two equal real poles).
G(s) = K/(s(s + 2))unity negative feedback
K =
4.Block diagramsHard
A proportional controller K drives the plant G(s) = 2/(s + 2) in a unity negative feedback loop. A disturbance step of size 10 enters at the plant input while R = 0. Find the smallest gain K that keeps the steady-state output at or below 0.1.
G(s) = 2/(s + 2)D = 10css ≤ 0.1
K =
5.Step responseMedium
A system has the transfer function G(s) = 198/(s² + 13s + 66) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.
G(s) = 198/(s² + 13s + 66)
σ =
6.Step responseHard
The step response of a second-order system without zeros shows 15 % overshoot, and the first peak occurs at t = 0.2 s. Find the natural frequency ωn.
%OS = 15 %tp = 0.2 s
ωn = rad/s
7.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram uses positive feedback, with L(s) = 20/(s(s + 4)(s + 5)). How many closed-loop poles N lie in the right half-plane?
L(s) = 20/(s(s + 4)(s + 5))
N =
8.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K(s + 10)/(s(s + 7)) leaves the real axis between s = 0 and s = −7, circles round and joins the real axis again to the left of the zero. Find the break-in point si.
L(s) = K(s + 10)/(s(s + 7))
si =
9.Steady-state error & PID controlMedium
A unity-feedback loop has G(s) = K/((s + 2)(s + 5)). Find the gain K that gives a steady-state error of 10 % for a step input.
G(s) = K/((s + 2)(s + 5))ess = 10 %
K =
10.Steady-state error & PID controlHard
A PI controller C(s) = 2 + 0.5/s drives the plant G(s) = 20/((s + 1)(s + 2)) in a unity-feedback loop. Find the steady-state error ess for the ramp input r(t) = 5t.
The RL low-pass filter shown has R = 470 Ω, L = 100 mH. A sinusoidal input of 3 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.
R = 470 ΩL = 100 mHf = 3 kHz
φ = °
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 50/(s(s + 2)(s + 7)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.
L(s) = 50/(s(s + 2)(s + 7))
ω180 = rad/s
3Solutions
1.Laplace transforms
Answer: A = 0.75
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 + 8s + 12 = (s + 2)(s + 6)
F(s) = A/(s + 2) + B/(s + 6)
Cover-up rule
Multiply by (s + 2) and set s = −2: the B-term vanishes.
A = (2s + 7)/(s + 6) |s = −2 = 3/4 = 0.75
Answer
A = 0.75
(B = 1.25, so f(t) = 0.75·e−2t + 1.25·e−6t)
2.Laplace transforms
Answer: y(0.2) = 2.799
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
s²Y − 2s − 8 + 5(sY − 2) + 6Y = 0
Y(s) = (2s + 18)/((s + 2)(s + 3))
Partial fractions
A = 14/1 = 14
B = 12/(−1) = −12
Result
y(t) = 14·e−2t − 12·e−3t
y(0.2) = 2.799
Answer
y(0.2) = 2.799
Check: y(0) = A + B = 2.
3.Block diagrams
Answer: K = 1
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
1 + K/(s(s + 2)) = 0 → s² + 2s + K = 0
Repeated root
A quadratic s² + βs + γ has a double root when β² = 4γ.
2² = 4K → K = 4/4 = 1
Answer
K = 1
Both closed-loop poles are then at s = −1.
4.Block diagrams
Answer: K = 99
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).
Disturbance response
css = D·G(0)/(1 + K·G(0)) = 10·2/(2 + 2K)
Solve for K
2 + 2K = 10·2/0.1 = 200
K = (200 − 2)/2 = 99
Answer
K = 99
5.Step response
Answer: σ = −6.5
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 = 66 → ωₙ = 8.124 rad/s
2ζωₙ = 13 → ζ = 13/(2·8.124) = 0.8001
Result
σ = −ζωₙ = −13/2 = −6.5
Answer
σ = −6.5
6.Step response
Answer: ωn = 18.35 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).
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 + 9s² + 20s − 20 = 0
Routh array
s³: 1, 20
s²: 9, −20
s¹: 22.22
s⁰: −20
Sign changes in the first column
1 → 9 → 22.22 → −20
N = 1
Answer
N = 1
Positive feedback makes the constant coefficient negative: one real pole always moves into the right half-plane.
8.Stability: Routh–Hurwitz & root locus
Answer: si = −15.48
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.
Gain along the real axis
K(s) = −(s² + 7s)/(s + 10)
Stationary points
Differentiate with the quotient rule and set the numerator to zero.
(2s + 7)(s + 10) − (s² + 7s) = s² + 20s + 70 = 0
s = −10 ± √30 = −4.523, −15.48
Break-in point
si = −15.48
Answer
si = −15.48
Between the two break points the locus is a circle centred on the zero, with radius √30 = 5.477.
9.Steady-state error & PID control
Answer: K = 90
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.
Required error constant
1/(1 + Kp) = 0.1 → Kp = 9
Gain
K/10 = 9 → K = 90
Answer
K = 90
Check the closed-loop stability as well: for this second-order loop any K > 0 is stable.
10.Steady-state error & PID control
Answer: ess = 1
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) = 2s + 0.5 → 0.5 as s → 0.
Kv = 0.5·20/2 = 5
Ramp error
ess = 5/5 = 1
Answer
ess = 1
The PI controller removes the step error completely; a ramp is followed with a constant lag.
11.Bode plots & margins
Answer: φ = −76 °
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 = 470/0.1 = 4700 rad/s
fc = 748 Hz → f/fc = 3000/748 = 4.011
Phase
φ = − atan(4.011) = −76°
Answer
φ = −76°
12.Bode plots & margins
Answer: ω180 = 3.742 rad/s
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.