Control Systems II: Stability, Root Locus, Bode and Nyquist, Lag–Lead Compensation, and P, PI and PID Controllers
1. Stability and the Routh–Hurwitz criterion
A linear system is BIBO stable exactly when every pole of its transfer function lies in the open left half of the s-plane; poles on the imaginary axis (simple) give marginal stability and a sustained oscillation. The Routh–Hurwitz criterion counts the closed-loop poles in the right half-plane without finding them. For aₙsⁿ + … + a₀ = 0, a necessary condition is that every coefficient be present and of the same sign; then build the Routh array, two rows from the coefficients and each later row from the two above it, and the number of sign changes in the first column equals the number of right-half-plane roots.
| Row | First column | Second column |
|---|---|---|
| s³ | 1 | 3 |
| s² | 2 | 10 |
| s¹ | (2·3 − 1·10)/2 = −2 | 0 |
| s⁰ | 10 | — |
The first column 1, 2, −2, 10 changes sign twice, so the polynomial has two roots in the right half-plane. With a gain parameter the same array gives the range for stability. For the loop K/[(s + 1)(s + 2)(s + 3)] the characteristic equation is s³ + 6s² + 11s + (6 + K) = 0: the s¹ entry is (66 − 6 − K)/6 = (60 − K)/6, which stays positive for K < 60, and the s⁰ entry 6 + K needs K > −6. At K = 60 the s¹ row is zero and the row above gives the auxiliary equation 6s² + 66 = 0, so the loop oscillates at ω = √11 ≈ 3.32 rad/s.
- A zero in the first column with non-zero entries after it: replace the zero by a small ε, complete the array and take the limit ε → 0⁺ to count the sign changes.
- A whole row of zeros: the row above gives an auxiliary polynomial whose roots, symmetric about the origin, are roots of the original; differentiate it to continue. Roots on the imaginary axis mean marginal stability.
2. Root loci
The root locus is the set of closed-loop poles, the roots of 1 + KG(s)H(s) = 0, as K goes from 0 to ∞. Its rules follow from the angle and magnitude conditions: the branches start at the open-loop poles (K = 0) and end at the open-loop zeros or at infinity (K → ∞); the number of branches equals the number of poles; a point on the real axis is on the locus if the count of real poles and zeros to its right is odd; the P − Z branches going to infinity follow asymptotes at angles (2q + 1)·180°/(P − Z) meeting at the centroid σ = (Σ poles − Σ zeros)/(P − Z); breakaway and break-in points satisfy dK/ds = 0; and the locus crosses the imaginary axis at the gain and frequency given by Routh’s auxiliary equation.
Worked: G(s) = K/[s(s + 2)(s + 4)], three poles and no zeros. The real-axis locus is [−2, 0] and (−∞, −4]. There are three asymptotes at 60°, 180° and 300°, meeting at σ = (0 − 2 − 4)/3 = −2. The breakaway point solves dK/ds = 0 for K = −s(s + 2)(s + 4) = −(s³ + 6s² + 8s): 3s² + 12s + 8 = 0, s = −0.845 (the root −3.155 lies off the locus). The imaginary-axis crossing comes from s³ + 6s² + 8s + K: the s¹ entry (48 − K)/6 is zero at K = 48, and 6s² + 48 = 0 gives ω = √8 = 2.83 rad/s. Beyond K = 48 two branches are in the right half-plane.
- Adding a pole to the open loop pushes the locus to the right, towards instability; adding a zero pulls it to the left, towards stability. This is the whole idea of the lead compensator.
- A closed-loop pole with damping ratio ζ lies on the line at angle cos⁻¹ζ from the negative real axis, so a required overshoot is a wedge on the locus plot.
3. Frequency response, Bode plots and stability margins
The sinusoidal steady state of a stable G(s) is G(jω): the magnitude and the phase at each frequency. A Bode plot draws 20 log₁₀|G(jω)| in dB and ∠G(jω) against log ω, and a product of factors becomes a sum of straight-line asymptotes. A constant K is a horizontal line at 20 log K; an integrator 1/s falls at −20 dB/decade with phase −90°, passing 0 dB at ω = 1; a simple pole 1/(1 + s/ω_c) is flat below the corner ω_c and falls at −20 dB/decade above it, its phase going from 0° to −90° through −45° at the corner; a zero does the opposite, at +20 dB/decade and up to +90°. A DC gain of 10 is 20 dB; a gain of 100 is 40 dB.
The gain crossover ω_gc is where |G| = 1 (0 dB) and the phase crossover ω_pc where ∠G = −180°. The phase margin is PM = 180° + ∠G(jω_gc), and the gain margin is GM = −20 log|G(jω_pc)| in dB: how much more phase lag, or gain, the loop can take before it reaches the point −1. A stable loop has both positive. Worked: for G = 10/[s(1 + 0.1s)], |G| = 1 gives ω²(1 + 0.01ω²) = 100, so ω_gc = 7.86 rad/s, ∠G = −90° − tan⁻¹(0.786) = −128.2°, and PM = 51.8°. For G = K/[s(s + 1)(s + 2)] with K = 2, ω_pc = √2 and |G| = K/6 = 1/3 there, so GM = 20 log 3 = 9.54 dB.
4. The Nyquist criterion
The Nyquist criterion decides closed-loop stability from the open-loop frequency response. Plot G(jω)H(jω) for ω from −∞ to ∞, closed at infinity so that it covers the whole right half-plane, and let N be the number of clockwise encirclements of the critical point −1 + j0 and P the number of open-loop poles in the right half-plane. The number of closed-loop poles in the right half-plane is Z = N + P, and the closed loop is stable exactly when Z = 0. For an open-loop stable system (P = 0) the plot must not encircle −1; for P > 0 it must encircle −1 anticlockwise P times.
The margins are read off the same plot: the phase margin at the point where the locus crosses the unit circle, and the gain margin at the point where it crosses the negative real axis, at −1/GM. Worked: for K/[s(s + 1)(s + 2)] the plot crosses the negative real axis at ω = √2 with magnitude K/6, so it reaches −1 at K = 6, agreeing with Routh: the loop is stable for 0 < K < 6. A pole at the origin makes the plot start at infinity, so the contour is indented round it, which adds a large arc at infinity whose direction is part of the count.
5. Lag, lead and lead–lag compensators, and P, PI and PID control
A lead compensator C(s) = (1 + aTs)/(1 + Ts), a > 1, puts a zero at −1/(aT) closer to the origin than its pole at −1/T. It supplies phase lead, greatest at ω_m = 1/(T√a) where φ_m = sin⁻¹[(a − 1)/(a + 1)], raises the phase margin, widens the bandwidth and speeds the response, but raises the high-frequency gain by a and so amplifies noise. To obtain a lead of φ_m: a = (1 + sin φ_m)/(1 − sin φ_m), so 30° needs a = 3, and a = 10 gives 54.9°. One stage rarely gives more than about 60°.
A lag compensator C(s) = (s + z)/(s + p) with z > p, both near the origin, gives a DC gain of z/p and almost no change at higher frequency, so it raises K_v (or K_p) by z/p, improving steady-state accuracy, without shifting the locus near the dominant poles. Its price is phase lag, which is why the pair is placed well below the gain crossover, and a lower bandwidth and slower response. A lead–lag compensator combines both, for a loop that needs accuracy and speed together.
| Feature | Lead | Lag |
|---|---|---|
| Main purpose | Transient response and phase margin | Steady-state accuracy |
| Pole and zero | Zero nearer the origin than the pole | Pole nearer the origin than the zero |
| Phase contribution | Positive (lead) | Negative (lag) |
| Bandwidth | Increased | Reduced |
| Similar controller | PD | PI |
The PID controller acts on the error e: u = K_p e + K_i ∫e dt + K_d de/dt, C(s) = K_p + K_i/s + K_d s. P gives an action proportional to the error, and leaves a steady-state error for a type-0 plant. I removes the steady-state error by integrating it, at the cost of a pole at the origin and less stability. D responds to the rate of change, adds damping and anticipates, and amplifies high-frequency noise, so a real derivative is filtered. PI is a lag-type action and PD a lead-type action. A PI controller on a first-order plant makes the loop type 1, so it follows a step with zero error; the integral term has no effect on the transient beyond changing the characteristic equation.
| Term | Effect on the loop |
|---|---|
| K_p | Faster response, smaller error, less damping |
| K_i | Zero steady-state error to a step; can destabilise |
| K_d | More damping, less overshoot; amplifies noise |
Key takeaways
- Routh: sign changes in the first column count right-half-plane roots; a row of zeros gives the auxiliary equation and the marginal frequency.
- Root locus: starts at poles, ends at zeros or infinity; asymptote centroid (Σp − Σz)/(P − Z); imaginary-axis crossing from Routh.
- Bode: −20 dB/decade per pole and −45° at its corner; PM = 180° + ∠G at gain crossover; GM = −20 log|G| at phase crossover.
- Nyquist: Z = N + P; stable iff Z = 0, i.e. P anticlockwise encirclements of −1.
- Lead: φ_m = sin⁻¹[(a − 1)/(a + 1)] at 1/(T√a), better transient; lag raises K_v by z/p; PID: I removes error, D adds damping.
Practice questions (18)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A linear time-invariant system is BIBO stable if and only if all the poles of its transfer function lie:
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Answer: A — in the open left half of the s-plane
Each pole gives a term ept, which decays only if Re p < 0. Poles on the imaginary axis give a marginal, not a bounded-output, response to a suitable bounded input. Complex poles in the left half-plane are stable too, so the negative real axis alone is too narrow.The characteristic equation of a feedback system is s³ + 2s² + 3s + 10 = 0. The number of roots in the right half of the s-plane is ____.
Numerical answer — type the value.
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Answer: 2
The Routh first column is 1, 2, (2·3 − 10)/2 = −2, 10, which changes sign twice (2 to −2, and −2 to 10), so there are two right-half-plane roots. Counting the negative entry alone would give one.A unity-feedback loop has G(s) = K/[(s + 1)(s + 2)(s + 3)]. The largest value of K for which the closed loop is stable is ____.
Numerical answer — type the value.
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Answer: 60
The characteristic equation is s³ + 6s² + 11s + 6 + K = 0. The s¹ entry is (6·11 − (6 + K))/6 = (60 − K)/6, which is positive for K < 60, and the s⁰ entry 6 + K is positive for K > −6. So K = 60 is marginal, where 6s² + 66 = 0 gives ω = √11 ≈ 3.32 rad/s.For the open-loop transfer function G(s)H(s) = K/[s(s + 2)(s + 4)], the centroid of the asymptotes of the root locus lies at s = ____.
Numerical answer — type the value.
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Answer: -2
σ = (Σ poles − Σ zeros)/(P − Z) = (0 − 2 − 4 − 0)/3 = −2, with three asymptotes at 60°, 180° and 300°. Dividing by 2 (the number of finite poles other than the origin) instead of P − Z = 3 gives −3.For the loop K/[s(s + 2)(s + 4)] under unity negative feedback, the root-locus branches cross the imaginary axis when K equals ____.
Numerical answer — type the value.
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Answer: 48
The characteristic equation is s³ + 6s² + 8s + K = 0. The s¹ Routh entry is (6·8 − K)/6, zero at K = 48, and the auxiliary equation 6s² + 48 = 0 gives s = ±j√8 = ±j2.83. For K > 48 two roots are in the right half-plane.A point on the real axis lies on the root locus of a negative-feedback system if the number of real open-loop poles and zeros to its right is:
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Answer: A — odd
The angle condition requires ∠G = ±180° (odd multiples), and each real pole or zero to the right of the test point contributes 180°, so an odd number puts the total at 180°. The even count is the rule for the complementary (negative-K) locus.A pure integrator 1/s has a Bode magnitude plot with a slope of:
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Answer: A — −20 dB/decade, with a constant phase of −90°
|1/(jω)| = 1/ω falls by a factor of 10, that is 20 dB, per decade, and ∠(1/jω) = −90° at every frequency. The +20 dB/decade, +90° pair is a differentiator.The low-frequency magnitude of G(s) = 100/(s + 10) on a Bode plot, in decibels, is ____ dB.
Numerical answer — type the value.
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Answer: 20
At low frequency G(0) = 100/10 = 10, and 20 log₁₀ 10 = 20 dB. The corner is at 10 rad/s, and above it the magnitude falls at −20 dB/decade. Using 10 log (power) gives 10 dB, the wrong dB convention for an amplitude ratio.A unity-feedback loop has G(s) = 10/[s(1 + 0.1s)]. Its phase margin, correct to one decimal place, is ____ degrees.
Numerical answer — type the value.
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Answer: 51.8
|G(jω)| = 10/[ω√(1 + 0.01ω²)] = 1 gives 0.01ω⁴ + ω² − 100 = 0, so ω² = 61.8 and ω_gc = 7.86 rad/s. ∠G = −90° − tan⁻¹(0.1 × 7.86) = −90° − 38.2° = −128.2°, and PM = 180° − 128.2° = 51.8°.A unity-feedback loop has G(s) = 2/[s(s + 1)(s + 2)]. Its gain margin, correct to two decimal places, is ____ dB.
Numerical answer — type the value.
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Answer: 9.54
The phase is −180° at ω_pc = √2 rad/s, where |G| = 2/[√2 × √3 × √6] = 2/6 = 1/3. So GM = −20 log(1/3) = 20 log 3 = 9.54 dB; equivalently, Routh gives marginal stability at K = 6, a factor of 3 above K = 2.According to the Nyquist criterion, with P open-loop poles in the right half-plane and N clockwise encirclements of the point −1 by the open-loop frequency response, the number of closed-loop poles in the right half-plane is:
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Answer: A — Z = N + P
The closed-loop poles are the zeros of 1 + GH, and the argument principle gives N = Z − P for clockwise encirclements of the origin by 1 + GH, that is of −1 by GH: Z = N + P. A loop with P = 1 that encircles −1 once anticlockwise (N = −1) has Z = 0 and is stable.A phase-lead compensator is to provide a maximum phase lead of 30°. The ratio a of its zero-to-pole time constants, in C(s) = (1 + aTs)/(1 + Ts), must be ____.
Numerical answer — type the value.
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Answer: 3
a = (1 + sin φ_m)/(1 − sin φ_m) = (1 + 0.5)/(1 − 0.5) = 3; check: sin⁻¹(2/4) = 30°. Using a = 1 + 2 sin φ_m = 2 would not satisfy the formula.A lead compensator has a = 10. Its maximum phase lead, correct to one decimal place, is ____ degrees.
Numerical answer — type the value.
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Answer: 54.9
φ_m = sin⁻¹[(a − 1)/(a + 1)] = sin⁻¹(9/11) = sin⁻¹(0.8182) = 54.9°. The lead is a function of a alone; the frequency at which it occurs is 1/(T√10).A lag compensator is used mainly to:
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Answer: A — improve the steady-state accuracy while leaving the transient response nearly unchanged
A pole–zero pair close to the origin, pole nearer, raises the low-frequency gain by z/p and so K_v, while the loop near the crossover is hardly touched. Faster response and added damping belong to the lead compensator.Which of the following statements about compensators and PID control are true?
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Answer: A — A lead compensator increases the phase margin and the bandwidth; B — A lag compensator increases the low-frequency loop gain; C — Integral action removes the steady-state error to a step for a type-0 plant
(A) Lead adds phase near crossover and shifts the crossover up. (B) The pole–zero pair z > p gives DC gain z/p. (C) The integrator makes the loop type 1. (D) False: differentiation amplifies noise, which is why a practical derivative term is filtered.In a PID controller, the integral term is mainly responsible for:
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Answer: A — eliminating the steady-state error to a constant reference
The integrator keeps growing while any error remains, so the loop can only settle where the error is zero; the price is an extra pole at the origin and less phase margin. Anticipation is the derivative term, and no term filters noise.Which of the following statements about stability and root loci are true?
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Answer: A — A polynomial with a missing power of s, between its highest and lowest, cannot have all its roots in the left half-plane; B — A row of zeros in the Routh array signals roots placed symmetrically about the origin; C — Adding an open-loop zero in the left half-plane tends to pull the root locus towards the left
(A) A necessary condition: all coefficients present with the same sign. (B) The auxiliary polynomial is even in s, so its roots come in ± pairs. (C) A zero attracts the branches. (D) False: a stable loop has both margins positive; a negative phase margin means the loop is unstable.For the loop G(s)H(s) = K/[s(s + 1)(s + 2)], the Nyquist plot crosses the negative real axis at −K/6. The closed loop is stable for:
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Answer: A — 0 < K < 6
The loop has no right-half-plane poles (P = 0), so the plot must not encircle −1, which needs the crossing −K/6 to lie to the right of −1, that is K < 6. At K = 6 it passes through −1 and the loop is marginal, agreeing with the Routh limit.