Closed-loop response to servo and regulator problems

Why it matters

A loop is judged by two jobs: following a new set point (servo) and holding the set point against disturbances (regulator). Most chemical-plant loops are regulators, so knowing how much a load upset still shows up in the controlled variable, whether an offset remains and how fast the loop recovers is what separates a loop that "works" from one that keeps operators busy.

Key ideas

Servo and regulator problems. From the block diagram (previous topic), with G_OL = G_c·G_v·G_p·G_m:

• Servo: L = 0, set point changes. C/R = G_c·G_v·G_p/(1 + G_OL) (unity set-point gain).
• Regulator: R = 0, load changes. C/L = G_d/(1 + G_OL).
• Both share the characteristic equation 1 + G_OL = 0, so their speed and damping are the same; their numerators, and so their offsets and initial responses, differ.

First-order process under P control. Take G_p = K_p/(τs + 1), G_v = G_m = 1, G_c = K_c and write K = K_c·K_p.

• Servo: C/R = [K/(1 + K)]/(τ_CL·s + 1), with τ_CL = τ/(1 + K). Faster than open loop, but the final value is K/(1 + K) instead of 1: offset = 1/(1 + K).
• Regulator (with G_d = K_d/(τs + 1)): C/L = [K_d/(1 + K)]/(τ_CL·s + 1). The load effect is reduced by the factor 1 + K but never removed.
• Raising K_c shrinks offset and speeds the loop; for a first-order process it never causes instability.

First-order process under PI control.

• Characteristic equation: τ_I·τ·s² + τ_I(1 + K)·s + K = 0, a second-order closed loop with τ_CL = √(τ_I·τ/K) and ζ = (1 + K)/2·√(τ_I/(K·τ)).
• Servo: final value 1 (no offset). Regulator: C/L has a zero at s = 0, so the final change in C is zero. Integral action removes offset for both problems.
• Smaller τ_I lowers ζ (more oscillation); larger K_c raises ζ for this process.

Higher-order processes under P control. For two lags in series, increasing K_c lowers the closed-loop ζ: the loop becomes underdamped, with overshoot, while offset still remains. For three or more lags, or with dead time, a high enough K_c makes the loop unstable (Routh, Bode topics). This gain–offset–stability trade-off is why integral action and careful tuning are needed.

Integrating process (pumped-out tank) under P control. G_p = K_p/(As). Servo offset is zero (the process itself integrates), but a step load still leaves an offset under P control.

Derivative action adds damping; on a first-order process under PD control the closed-loop time constant becomes (τ + K·τ_D)/(1 + K), and the offset is the same as under P control.

Formulas

C/R = G_c·G_v·G_p / (1 + G_c·G_v·G_p·G_m) (servo); C/L = G_d / (1 + G_c·G_v·G_p·G_m) (regulator)

τ_CL = τ/(1 + K), K = K_c·K_p

• P control of K_p/(τs + 1); τ and τ_CL in s or min.

Offset (servo, unit step) = 1/(1 + K); Final load effect = K_d/(1 + K) (unit load step)

• P control, self-regulating process, unity valve and measurement gains.

τ_I·τ·s² + τ_I·(1 + K)·s + K = 0

• PI control of a first-order process.

ζ = (1 + K)/2 · √(τ_I/(K·τ)), τ_CL = √(τ_I·τ/K)

• Closed-loop damping and natural period under PI control.

τ₁τ₂s² + (τ₁ + τ₂)s + 1 + K = 0

• P control of K_p/((τ₁s + 1)(τ₂s + 1)); ζ = (τ₁ + τ₂)/(2√(τ₁τ₂(1 + K))).

Worked examples

Example 1 (standard): P control, servo and regulator. G_p = G_d = 2/(5s + 1) (min), G_v = G_m = 1, K_c = 4. Find (a) the closed-loop time constant, (b) the servo offset and output at 1 min after a unit set-point step, (c) the final change in C for a unit load step.

1. K = K_c·K_p = 8; τ_CL = 5/(1 + 8) = 0.556 min.
2. Servo final value K/(1 + K) = 8/9 = 0.889, so offset = 0.111.
3. C(1 min) = 0.889·(1 − e^(−1/0.556)) = 0.889·(1 − e^(−1.8)) = 0.889·0.8347 = 0.742.
4. Regulator: C(∞) = K_d/(1 + K) = 2/9 = 0.222 (without control it would be 2).

τ_CL ≈ 0.556 min; servo offset 0.111, C(1 min) ≈ 0.742; load effect 0.222 (9 times smaller than open loop)

Example 2 (GATE level): gain for a target damping. G_p = 2/((s + 1)(0.5s + 1)) (min) is under P control with G_v = G_m = 1. Find K_c for a closed-loop ζ = 0.7, and the resulting offset and overshoot for a unit set-point step.

1. Characteristic equation: (s + 1)(0.5s + 1) + 2K_c = 0.5s² + 1.5s + (1 + 2K_c) = 0.
2. Normalise: τ² = 0.5/(1 + 2K_c), 2ζτ = 1.5/(1 + 2K_c), so ζ = 1.5/(2·√(0.5·(1 + 2K_c))).
3. For ζ = 0.7: √(0.5·(1 + 2K_c)) = 1.5/1.4 = 1.0714, so 0.5·(1 + 2K_c) = 1.148, 1 + 2K_c = 2.296.
4. K_c = 0.648.
5. Offset = 1/(1 + 2K_c) = 1/2.296 = 0.436.
6. Overshoot = exp(−π·0.7/√(1 − 0.49)) = exp(−3.079) = 0.046 (4.6 %) of the final value.

K_c ≈ 0.648; offset ≈ 0.436; overshoot ≈ 4.6 %. A 44 % offset is unacceptable, which shows why integral action is added rather than raising K_c further.

Common mistakes

• Assuming the regulator response has the same final value as the servo response; numerators differ.
• Saying P control removes the load effect; it only divides it by 1 + K.
• Forgetting that K in the offset formula is the product of all steady-state gains around the loop.
• Believing a higher K_c always makes a loop more stable; for two or more lags it reduces damping.
• Claiming integral action removes offset only for set-point changes; it removes it for loads too.

For GATE CH

This is a regular source of numericals: offset for set-point or load steps under P control, closed-loop time constant, the controlled variable at a given time, and the gain giving a specified damping factor. PI questions ask for closed-loop ζ or period. Always write the characteristic equation first, normalise it, then read off τ and ζ.

Quick check

1. Under P control with K_cK_p = 3, what is the servo offset for a set-point step of 2 units?
2. Does integral action remove offset in the regulator problem?
3. For a first-order process with τ = 6 min and K_cK_p = 5, what is τ_CL?
4. For two lags under P control, what happens to ζ as K_c increases?

Answers: 1. 2·1/(1 + 3) = 0.5 unit. 2. Yes. 3. 1 min. 4. It decreases (more oscillatory).