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) = 20/(s(s + 1)(s + 6)) by partial fractions, and evaluate f(0.6).
F(s) = 20/(s(s + 1)(s + 6))t = 0.6 s
f(0.6) =
2.Laplace transformsHard
A system obeys y″ + 6y′ + 5y = 2.5·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 5 for t ≥ 0. Use the Laplace transform to find the output y(0.5).
A unity negative feedback loop has G(s) = 28/(s(s + 4)). The reference is a ramp r(t) = 1·t. Find the steady-state error ess.
G(s) = 28/(s(s + 4))r(t) = 1·t
ess =
4.Block diagramsHard
The plant G(s) = K/(s(s + 1)(s + 3)) 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.
G(s) = K/(s(s + 1)(s + 3))unity negative feedback
Kmax =
5.Step responseMedium
A system has the transfer function G(s) = 24/(s² + 6s + 24) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.
G(s) = 24/(s² + 6s + 24)
σ =
6.Step responseHard
A unity negative feedback loop has G(s) = K/(s(s + 5)). Find the gain K that gives a step response with 20 % overshoot.
G(s) = K/(s(s + 5))%OS = 20 %
K =
7.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 1)(s + 4)). Find the largest gain Kmax for which the closed loop is stable.
L(s) = K/(s(s + 1)(s + 4))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K/(s(s + 5)) is shown. Find the gain K for which the closed-loop poles have the damping ratio ζ = 0.8.
L(s) = K/(s(s + 5))ζ = 0.8
K =
9.Steady-state error & PID controlMedium
The input r(t) = 5 + 1t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 10(s + 2)/(s(s + 1)(s + 2)). Find the steady-state error ess.
G(s) = 10(s + 2)/(s(s + 1)(s + 2))r(t) = 5 + 1t
ess =
10.Steady-state error & PID controlHard
A PI controller C(s) = 1 + 4/s drives the plant G(s) = 10/((s + 1)(s + 4)) in a unity-feedback loop. Find the steady-state error ess for the ramp input r(t) = 2t.
C(s) = 1 + 4/sG(s) = 10/((s + 1)(s + 4))r(t) = 2t
ess =
11.Bode plots & marginsMedium
The RC low-pass filter shown has R = 22 kΩ, C = 2.2 nF. A sinusoidal input of 660 Hz is applied. Find the phase shift φ of vout relative to vin, in degrees.
R = 22 kΩC = 2.2 nFf = 660 Hz
φ = °
12.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = K/(s(s + 4)(s + 6)). Find the gain K that gives a gain margin of 4 dB.
L(s) = K/(s(s + 4)(s + 6))GM = 4 dB
K =
3Solutions
1.Laplace transforms
Answer: f(0.6) = 1.156
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).
Partial fractions
F(s) = A/s + B/(s + 1) + C/(s + 6)
A = 20/(1·6) = 3.333
B = 20/((−1)·5) = −4
C = 20/((−6)·(−5)) = 0.6667
Inverse transform
f(t) = 3.333 − 4·e−t + 0.6667·e−6t
Evaluate
f(0.6) = 3.333 − 4·e−0.6 + 0.6667·e−6·0.6 = 1.156
Answer
f(0.6) = 1.156
2.Laplace transforms
Answer: y(0.5) = 0.6559
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) = 12.5/(s(s + 1)(s + 5))
Partial fractions
A = 12.5/(1·5) = 2.5
B = 12.5/((−1)·4) = −3.125
C = 12.5/((−5)·(−4)) = 0.625
Output
y(t) = 2.5 − 3.125·e−t + 0.625·e−5t
y(0.5) = 0.6559
Answer
y(0.5) = 0.6559
3.Block diagrams
Answer: ess = 0.1429
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) = 28/4 = 7
Steady-state error
ess = A/Kᵥ = 1/7 = 0.1429
Answer
ess = 0.1429
4.Block diagrams
Answer: Kmax = 12
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 + 4s² + 3s + K = 0
Routh array
s³ | 1 3
s² | 4 K
s¹ | (4·3 − K)/4
s⁰ | K
Stability limit
All entries of the first column must be positive.
Kmax = 4·3 = 12
Answer
Kmax = 12
5.Step response
Answer: σ = −3
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 = 24 → ωₙ = 4.899 rad/s
2ζωₙ = 6 → ζ = 6/(2·4.899) = 0.6124
Result
σ = −ζωₙ = −6/2 = −3
Answer
σ = −3
6.Step response
Answer: K = 30.06
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 + 5s + K)
Damping ratio from the overshoot
ζ = −ln(0.2)/√(π² + ln²(0.2)) = 0.4559
Gain
ωₙ = 5/(2ζ) = 5.483 rad/s
K = ωₙ² = 30.06
Answer
K = 30.06
7.Stability: Routh–Hurwitz & root locus
Answer: Kmax = 20
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 + 5s² + 4s + K = 0
Routh array
s³: 1, 4
s²: 5, K
s¹: (5·4 − K)/5
s⁰: K
Conditions
Every entry of the first column must be positive.
K > 0
5·4 − K > 0 → K < 20
Answer
Kmax = 20
At K = 20 the closed loop oscillates at ω = √4 = 2 rad/s.
8.Stability: Routh–Hurwitz & root locus
Answer: K = 9.766
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.
Compare with the standard form
s² + 5s + K = s² + 2ζωₙs + ωₙ²
2·0.8·ωₙ = 5 → ωₙ = 3.125 rad/s
Gain
K = ωₙ² = 3.125² = 9.766
Answer
K = 9.766
The closed-loop poles sit where the ζ line crosses the vertical branch at s = −2.5.
9.Steady-state error & PID control
Answer: ess = 0.1
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
Step part
Kp = ∞ → estep = 0
Ramp error
ess = 1/10 = 0.1
Answer
ess = 0.1
10.Steady-state error & PID control
Answer: ess = 0.2
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) = 1s + 4 → 4 as s → 0.
Kv = 4·10/4 = 10
Ramp error
ess = 2/10 = 0.2
Answer
ess = 0.2
The PI controller removes the step error completely; a ramp is followed with a constant lag.
11.Bode plots & margins
Answer: φ = −11.35 °
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/(22000·2.2 × 10⁻⁹) = 20661 rad/s
fc = 3288 Hz → f/fc = 660/3288 = 0.2007
Phase
φ = − atan(0.2007) = −11.35°
Answer
φ = −11.35°
12.Bode plots & margins
Answer: K = 151.4
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.