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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) = 33/(s² + 2s + 33) and receives a unit step input. Find the percent overshoot %OS of the step response.
G(s) = 33/(s² + 2s + 33)
%OS = %
2.Step responseHard
A unity negative feedback loop has G(s) = K/(s(s + 16)). Find the gain K that gives a step response with 5 % overshoot.
G(s) = K/(s(s + 16))%OS = 5 %
K =
3.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 45(s + 15)/(s(s + 5)) at ω = 1 rad/s, in degrees.
G(s) = 45(s + 15)/(s(s + 5))ω = 1 rad/s
∠G(jω) = °
4.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 74/(s(s + 5)(s + 7)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.
L(s) = 74/(s(s + 5)(s + 7))
ω180 = rad/s
5.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 2)(s + 5)). 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 + 2)(s + 5))
ω = rad/s
6.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K(s + 12)/(s(s + 5)) leaves the real axis between s = 0 and s = −5, circles round and joins the real axis again to the left of the zero. Find the break-in point si.
L(s) = K(s + 12)/(s(s + 5))
si =
7.Steady-state error & PID controlMedium
The parabolic input r(t) = 10·t²/2 is applied to a unity-feedback loop with G(s) = 2(s + 3)/(s²(s + 4)). Find the steady-state error ess.
G(s) = 2(s + 3)/(s²(s + 4))r(t) = 10·t²/2
ess =
8.Steady-state error & PID controlHard
A PD controller C(s) = 20 + Kd·s drives the plant G(s) = 5/(s(s + 3)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.9.
G(s) = 5/(s(s + 3))Kp = 20ζ = 0.9
Kd = s
9.Block diagramsMedium
The plant G(s) = K/((s + 5)(s + 11)) 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 + 5)(s + 11))unity negative feedback
K =
10.Block diagramsHard
A proportional controller K drives the plant G(s) = 10/(s + 1) 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 1.
G(s) = 10/(s + 1)D = 1css ≤ 1
K =
3Solutions
1.Step response
Answer: %OS = 57.39 %
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 = 33 → ωₙ = 5.745 rad/s
2ζωₙ = 2 → ζ = 2/(2·5.745) = 0.1741
Result
%OS = 100·e−ζπ/√(1−ζ²) = 100·e−0.5554 = 57.39 %
Answer
%OS = 57.39 %
2.Step response
Answer: K = 134.4
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).
Closed-loop transfer function
T(s) = K/(s² + 16s + K)
Damping ratio from the overshoot
ζ = −ln(0.05)/√(π² + ln²(0.05)) = 0.6901
Gain
ωₙ = 16/(2ζ) = 11.59 rad/s
K = ωₙ² = 134.4
Answer
K = 134.4
3.Bode plots & margins
Answer: ∠G(jω) = −97.5 °
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.
Phase crossover
(ω/5)·(ω/7) = 1 → ω180 = √(5·7) = 5.916 rad/s
Answer
ω180 = 5.916 rad/s
5.Stability: Routh–Hurwitz & root locus
Answer: ω = 3.162 rad/s
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 + 7s² + 10s + K = 0
s³: 1, 10
s²: 7, K
s¹: (7·10 − K)/7
s⁰: K
Critical gain
K = 7·10 = 70
Auxiliary equation
At the critical gain the s¹ row is zero; the s² row gives the poles on the imaginary axis.
7s² + 70 = 0 → s = ±j√10
ω = 3.162 rad/s
Answer
ω = 3.162 rad/s
6.Stability: Routh–Hurwitz & root locus
Answer: si = −21.17
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 + 5s)/(s + 12)
Stationary points
Differentiate with the quotient rule and set the numerator to zero.
(2s + 5)(s + 12) − (s² + 5s) = s² + 24s + 60 = 0
s = −12 ± √84 = −2.835, −21.17
Break-in point
si = −21.17
Answer
si = −21.17
Between the two break points the locus is a circle centred on the zero, with radius √84 = 9.165.
7.Steady-state error & PID control
Answer: ess = 6.667
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.
Acceleration constant
Ka = 2·3/4 = 1.5
Parabola error
ess = 10/1.5 = 6.667
Answer
ess = 6.667
8.Steady-state error & PID control
Answer: Kd = 3 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 + 3) + 5(20 + Kds) = s² + (3 + 5Kd)s + 100
Natural frequency
ωₙ = √100 = 10 rad/s
Derivative gain
3 + 5Kd = 2·0.9·10 = 18
Kd = 3 s
Answer
Kd = 3 s
The derivative term adds damping without changing ωₙ.
9.Block diagrams
Answer: K = 9
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).
A quadratic s² + βs + γ has a double root when β² = 4γ.
16² = 4(55 + K) → K = 256/4 − 55 = 9
Answer
K = 9
Both closed-loop poles are then at s = −8.
10.Block diagrams
Answer: K = 0.9
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).