Lead and lag compensators

Why it matters

Raising the gain alone usually trades accuracy against stability: more gain cuts steady-state error but erodes phase margin. Lead and lag compensators are simple pole–zero networks (an RC circuit, an op-amp stage or a few lines of firmware) that reshape the loop's frequency response so that you can have both — a lead network for speed and damping, a lag network for accuracy. They are the frequency-domain cousins of PD and PI control.

Key ideas

Lead compensator. Gc(s) = Kc·(1 + T·s)/(1 + α·T·s) with 0 < α < 1.

• Zero at s = −1/T, pole at s = −1/(αT): the zero is closer to the origin than the pole.
• It adds positive phase (lead) between the two corners, with a maximum φm at the geometric mean of the corners, ωm.
• It raises high-frequency gain by 1/α (20·log₁₀(1/α) dB), so the gain crossover moves up: wider bandwidth, faster response, but more noise sensitivity.
• Use: when the uncompensated loop has too little phase margin (too much overshoot) or too slow a response. In root-locus terms it pulls the locus to the left. It behaves like a filtered PD controller.
• Limit: one stage gives at most about 60–65° of lead in practice (α not much below 0.1); for more, use two stages.

Lag compensator. Gc(s) = Kc·(1 + T·s)/(1 + β·T·s) with β > 1.

• Pole at s = −1/(βT), zero at s = −1/T: the pole is closer to the origin than the zero.
• It adds phase lag between the corners and attenuates high frequencies by a factor 1/β (−20·log₁₀β dB), while keeping the low-frequency gain.
• Use: when the transient response is acceptable but the steady-state error is too large. Raise Kc to meet the error constant (Kv increases by β relative to a loop with the same crossover), then place the lag corners about a decade below the new gain crossover so the extra phase lag there is only about 5°. The crossover moves to a lower frequency, so the response becomes slower, but the phase margin is restored.
• It behaves like an approximate PI controller; the pole–zero pair near the origin creates a slow "tail" in the step response.

Lead–lag. A cascade of both: the lead section supplies phase near crossover, the lag section supplies low-frequency gain. Like PID, it improves transient and steady-state behaviour together.

Design steps for lead (Bode method).

1. Choose the gain K to meet the steady-state error requirement.
2. Find the uncompensated phase margin.
3. Required lead φm = required PM − present PM + 5° to 12° safety (the crossover will shift right, where the plant phase is lower).
4. α = (1 − sin φm)/(1 + sin φm).
5. Place ωm at the frequency where the uncompensated magnitude is −10·log₁₀(1/α) dB; this becomes the new crossover.
6. T = 1/(ωm·√α). Check the final PM; iterate if needed.

Formulas

Gc(s) = Kc·(1 + T·s)/(1 + α·T·s), 0 < α < 1 — lead. Gc(s) = Kc·(1 + T·s)/(1 + β·T·s), β > 1 — lag.

ωm = 1/(T·√α) — frequency of maximum phase lead (rad/s); geometric mean of the corners 1/T and 1/(αT).

sin φm = (1 − α)/(1 + α), equivalently α = (1 − sin φm)/(1 + sin φm)

|(1 + jωmT)/(1 + jωmαT)| = 1/√α — magnitude of the lead factor at ωm, i.e. 10·log₁₀(1/α) dB.

High-frequency attenuation of the lag factor: 1/β, i.e. −20·log₁₀β dB.

Pole–zero form: a lead (s + z)/(s + p) with p > z has α = z/p; a lag (s + z)/(s + p) with z > p has β = z/p.

Worked examples

Example 1 (standard — properties of a lead network). For Gc(s) = (s + 2)/(s + 10), find α, the maximum phase lead, the frequency where it occurs, and the magnitude of Gc there.

1. Time-constant form: Gc = (2/10)·(1 + 0.5s)/(1 + 0.1s), so T = 0.5 s and αT = 0.1 s → α = 0.2.
2. sin φm = (1 − 0.2)/(1 + 0.2) = 0.667 → φm = 41.8°.
3. ωm = 1/(0.5 × √0.2) = √(2 × 10) = 4.47 rad/s.
4. The lead factor gives 1/√α = 2.236 (+6.99 dB) at ωm; with the 0.2 DC factor, |Gc(jωm)| = 0.2 × 2.236 = 0.447 (−6.99 dB).
5. Because the network's DC gain is 0.2, an amplifier of gain 5 is normally added so that the error constant is not reduced.

Example 2 (GATE level — effect on phase margin). A unity-feedback loop has G(s) = 10/(s(s + 1)). A lead compensator Gc(s) = (1 + 0.4s)/(1 + 0.1s) is added in cascade. Find the phase margin and gain crossover before and after.

1. Uncompensated: |G| = 10/(ω√(1 + ω²)) = 1 → ω⁴ + ω² − 100 = 0 → ω² = 9.51 → ω_gc = 3.08 rad/s.
2. PM = 180° − 90° − tan⁻¹(3.08) = 90° − 72.0° = 18.0° — more than 50 % overshoot.
3. Lead data: α = 0.25, φm = sin⁻¹(0.75/1.25) = 36.9° at ωm = 1/(0.4 × 0.5) = 5 rad/s.
4. Compensated crossover: solve 10·√(1 + 0.16ω²) / (ω·√(1 + ω²)·√(1 + 0.01ω²)) = 1 numerically → ω_gc = 4.18 rad/s.
5. PM = 90° − tan⁻¹(4.18) + tan⁻¹(1.672) − tan⁻¹(0.418) = 90° − 76.55° + 59.13° − 22.69° = 49.9°.
6. The lead raised the PM by about 32° (less than φm = 36.9°, because the crossover is not exactly at ωm) and raised the crossover frequency — a faster, much better-damped loop. Kv = 10 s⁻¹ is unchanged, since Gc(0) = 1.

Common mistakes

• Swapping the pole and zero order: for lead the zero is nearer the origin; for lag the pole is.
• Forgetting the lead network's DC attenuation (α in a passive network) and so losing steady-state accuracy.
• Placing ωm at the old gain crossover — the lead's gain moves the crossover to the right, so place it at the new one.
• Placing lag corners close to crossover, which eats the phase margin.
• Expecting a lag network to speed up the response; it slows it.
• Using sin φm = (1 − α)/(1 + α) with α > 1 (that is a lag network).

For GATE ME

Expect: identifying lead or lag from a transfer function or pole–zero plot; maximum phase lead and ωm for a given network; α for a required φm; magnitude at ωm; qualitative effects of lead and lag on bandwidth, PM, steady-state error and noise; and the equivalence lead ≈ PD, lag ≈ PI. Practise the sin φm formula both ways.

Quick check

1. A network has zero at −1 and pole at −3. Is it lead or lag, and what is its maximum phase?
2. At what frequency does that maximum occur?
3. Which compensator increases bandwidth?
4. What α gives φm = 30°?
5. Which compensator mainly improves steady-state error?

Answers: 1. Lead (zero nearer origin); α = 1/3, φm = sin⁻¹(0.5) = 30°. 2. √(1 × 3) = 1.73 rad/s. 3. Lead. 4. α = (1 − 0.5)/(1 + 0.5) = 1/3. 5. Lag.