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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.Step responseMedium
A system has the transfer function G(s) = 70/(s² + 8s + 70) and receives a unit step input. Find the peak time tp of the step response.
G(s) = 70/(s² + 8s + 70)
tp = s
2.Step responseHard
The step response of a second-order system without zeros shows 20 % overshoot, and the first peak occurs at t = 0.25 s. Find the natural frequency ωn.
%OS = 20 %tp = 0.25 s
ωn = rad/s
3.Bode plots & marginsMedium
The RL low-pass filter shown has R = 10 Ω, L = 4.7 mH. A sinusoidal input of 1 kHz is applied. Find the gain |H| = |vout/vin| in dB.
R = 10 ΩL = 4.7 mHf = 1 kHz
|H| = dB
4.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = K/(s(s + 6)(s + 10)). Find the gain K that gives a gain margin of 8 dB.
L(s) = K/(s(s + 6)(s + 10))GM = 8 dB
K =
5.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 6)(s + 8)). The gain is raised until the closed loop is on the edge of stability. Find the frequency ω of the sustained oscillation.
L(s) = K/(s(s + 6)(s + 8))
ω = rad/s
6.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K/((s + 1)(s + 7)) is shown. Find the gain K at the break-away point sb, where the two closed-loop poles meet on the real axis.
L(s) = K/((s + 1)(s + 7))
K =
7.Steady-state error & PID controlMedium
A process G(s) = 5/(2s + 1) is controlled by a proportional gain K = 9 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.
G(s) = 5/(2s + 1)K = 9
ess = %
8.Steady-state error & PID controlHard
A proportional controller K = 25 drives the plant G(s) = 10/(s(s + 6)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 10/(s(s + 6))K = 25
ts = s
9.Block diagramsMedium
A unity negative feedback loop has G(s) = 8(s + 8)/((s + 3)(s + 4)). A step of size 5 is applied at the reference. Find the steady-state error ess.
G(s) = 8(s + 8)/((s + 3)(s + 4))R = 5
ess =
10.Block diagramsHard
A proportional controller K drives the plant G(s) = 10/(s + 10) in a unity negative feedback loop. A disturbance step of size 1 enters at the plant input while R = 0. Find the smallest gain K that keeps the steady-state output at or below 0.05.
G(s) = 10/(s + 10)D = 1css ≤ 0.05
K =
3Solutions
1.Step response
Answer: tp = 0.4275 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).
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).
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 = 10/0.0047 = 2128 rad/s
fc = 338.6 Hz → f/fc = 1000/338.6 = 2.953
Gain
|H| = 1/√(1 + 2.953²) = 0.3207
20·log₁₀(0.3207) = −9.877 dB
Answer
|H| = −9.877 dB
4.Bode plots & margins
Answer: K = 382.2
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.
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.
Routh array
s³ + 14s² + 48s + K = 0
s³: 1, 48
s²: 14, K
s¹: (14·48 − K)/14
s⁰: K
Critical gain
K = 14·48 = 672
Auxiliary equation
At the critical gain the s¹ row is zero; the s² row gives the poles on the imaginary axis.
14s² + 672 = 0 → s = ±j√48
ω = 6.928 rad/s
Answer
ω = 6.928 rad/s
6.Stability: Routh–Hurwitz & root locus
Answer: K = 9
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.
Break-away point
dK/ds = 0 for K = −(s + 1)(s + 7) gives 2s + 8 = 0.
sb = −4
Magnitude condition
K = |−4 + 1|·|−4 + 7| = 3·3 = 9
Answer
K = 9
For this gain the closed loop is critically damped (ζ = 1).
7.Steady-state error & PID control
Answer: ess = 2.174 %
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 = 9·5 = 45
Step error
ess = 100/(1 + 45) = 2.174 %
Answer
ess = 2.174 %
8.Steady-state error & PID control
Answer: ts = 1.333 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 + 6) + 25·10 = s² + 6s + 250
ωₙ and ζ
ωₙ = √250 = 15.81 rad/s
ζ = 6/(2·15.81) = 0.1897
Settling time
ζωₙ = 6/2 = 3
ts ≈ 4/3 = 1.333 s
Answer
ts = 1.333 s
The settling time does not depend on K here: raising K only makes the response oscillate faster.
9.Block diagrams
Answer: ess = 0.7895
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).
Position error constant
Kₚ = G(0) = 8·8/(3·4) = 5.333
Steady-state error
ess = A/(1 + Kₚ) = 5/(1 + 5.333) = 0.7895
Answer
ess = 0.7895
10.Block diagrams
Answer: K = 19
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).