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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) = 35/(s² + 8s + 70) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.
G(s) = 35/(s² + 8s + 70)
σ =
2.Step responseHard
A second-order system without zeros must have a step response with 10 % overshoot. Find the damping ratio ζ.
%OS = 10 %
ζ =
3.Bode plots & marginsMedium
The RL low-pass filter shown has R = 470 Ω, L = 4.7 mH. A sinusoidal input of 8 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.
R = 470 ΩL = 4.7 mHf = 8 kHz
φ = °
4.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 76/(s(s + 3)(s + 5)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.
L(s) = 76/(s(s + 3)(s + 5))
ω180 = rad/s
5.Stability: Routh–Hurwitz & root locusMedium
A unity-feedback loop has L(s) = K(s + 10)/(s(s + 2)(s + 3)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.
L(s) = K(s + 10)/(s(s + 2)(s + 3))
Kmax =
6.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K(s + 7)/(s(s + 3)) leaves the real axis between s = 0 and s = −3, circles round and joins the real axis again to the left of the zero. Find the break-away point sb.
L(s) = K(s + 7)/(s(s + 3))
sb =
7.Steady-state error & PID controlMedium
A unity-feedback loop has G(s) = K/((s + 4)(s + 10)). Find the gain K that gives a steady-state error of 4 % for a step input.
G(s) = K/((s + 4)(s + 10))ess = 4 %
K =
8.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 = 4 rad/s and ζ = 1. Find the proportional gain Kp.
G(s) = 5/(s + 2)ωₙ = 4 rad/sζ = 1
Kp =
9.Block diagramsMedium
A plant G(s) = 6/(s + 1) is controlled in a negative feedback loop whose sensor has the gain H = 0.2. A step of size 12 is applied at R. Find the steady-state output css.
G(s) = 6/(s + 1)H = 0.2R = 12
css =
10.Block diagramsHard
A proportional controller K = 28 drives the plant G(s) = 1/(s + 7) in a unity negative feedback loop. A disturbance D enters at the plant input (it is added to the controller output). With R = 0, a disturbance step of size 1 is applied. Find the steady-state output css.
K = 28G(s) = 1/(s + 7)D = 1R = 0
css =
3Solutions
1.Step response
Answer: σ = −4
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 = 70 → ωₙ = 8.367 rad/s
2ζωₙ = 8 → ζ = 8/(2·8.367) = 0.4781
Result
σ = −ζωₙ = −8/2 = −4
Answer
σ = −4
2.Step response
Answer: ζ = 0.5912
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 = 470/0.0047 = 100000 rad/s
fc = 15915 Hz → f/fc = 8000/15915 = 0.5027
Phase
φ = − atan(0.5027) = −26.69°
Answer
φ = −26.69°
4.Bode plots & margins
Answer: ω180 = 3.873 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.
Phase crossover
(ω/3)·(ω/5) = 1 → ω180 = √(3·5) = 3.873 rad/s
Answer
ω180 = 3.873 rad/s
5.Stability: Routh–Hurwitz & root locus
Answer: Kmax = 6
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.
With the zero to the right of s = −5 the loop would stay stable for every K > 0.
6.Stability: Routh–Hurwitz & root locus
Answer: sb = −1.708
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 + 3s)/(s + 7)
Stationary points
Differentiate with the quotient rule and set the numerator to zero.
(2s + 3)(s + 7) − (s² + 3s) = s² + 14s + 21 = 0
s = −7 ± √28 = −1.708, −12.29
Break-away point
sb = −1.708
Answer
sb = −1.708
Between the two break points the locus is a circle centred on the zero, with radius √28 = 5.292.
7.Steady-state error & PID control
Answer: K = 960
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) = 0.04 → Kp = 24
Gain
K/40 = 24 → K = 960
Answer
K = 960
Check the closed-loop stability as well: for this second-order loop any K > 0 is stable.
8.Steady-state error & PID control
Answer: Kp = 1.2
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).
Characteristic polynomial
1 + C(s)G(s) = 0, multiplied by s(s + 2).
s² + (2 + 5Kp)s + 5Ki
Target polynomial
s² + 2·1·4s + 4² = s² + 8s + 16
Match the coefficients
2 + 5Kp = 8 → Kp = 1.2
Answer
Kp = 1.2
The integral term makes the loop type 1, so a step is followed without steady-state error.
9.Block diagrams
Answer: css = 32.73
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).
Transfer functions
C/D = G/(1 + K·G) = 1/(s + 7 + 28)
Steady state
css = 1·1/35 = 0.02857
Answer
css = 0.02857
A larger K reduces the effect of the disturbance, but never to zero with a proportional controller.