Open-loop and closed-loop control systems

Why it matters

Every mechatronic product, from a washing machine to a CNC feed drive or a drone, is either running a fixed command (open loop) or measuring what it actually did and correcting itself (closed loop). Choosing between the two decides cost, accuracy, how the system copes with load changes and wear, and whether it can become unstable.

Key ideas

• Control system: an interconnection of components (controller, actuator, plant, sensor) arranged so that an output variable follows a desired reference.
• Open-loop system: the control action does not depend on the output. The controller sends a pre-set command and has no way of knowing whether the result was achieved. Examples: a toaster or washing-machine timer, a traffic signal on a fixed cycle, a stepper motor driven by counted pulses without an encoder.
  • Accuracy depends entirely on calibration. Any disturbance (load torque, supply change) or parameter drift (wear, temperature) appears directly in the output.
  • Simple, cheap, and cannot become unstable through feedback (a stable plant stays stable).
• Closed-loop (feedback) system: a sensor measures the output, the measured value is compared with the reference, and the difference (the actuating error) drives the controller. Examples: room thermostat, cruise control, servo motor with encoder, voltage regulator.
• Signals in a feedback loop: reference r, output c, feedback signal b = H·c, actuating error e = r − b. With unity feedback (H = 1) the actuating error equals the true error r − c.
• What negative feedback buys (for a large loop gain G·H):
  • Reduced sensitivity to plant parameter changes: the closed-loop gain tends to 1/H, set by the sensor, not by the plant.
  • Disturbance rejection: a disturbance entering at the output is divided by (1 + G·H).
  • Higher bandwidth / faster response: for a first-order plant, feedback moves the pole further left.
  • Reduced steady-state error (studied with error constants).
• What feedback costs: lower overall gain (more amplification needed), a sensor (cost, noise, calibration), and the risk of instability — high gain with phase lag can make a stable plant oscillate. So "closed loop is always more stable" is false; feedback can stabilise an unstable plant or destabilise a stable one.
• Positive feedback (b added to r) increases gain and is used in oscillators and latches, rarely in regulators.
• Servomechanism vs regulator: a servo tracks a changing reference (position control); a regulator holds a constant reference against disturbances (speed or temperature control).
• This topic connects to transfer functions and block diagrams (next), steady-state error (error constants) and stability (Routh, root locus, Nyquist).

Formulas

• T(s) = C(s)/R(s) = G(s) / (1 + G(s)·H(s)) — closed-loop transfer function for negative feedback.
  • G(s) forward-path transfer function, H(s) feedback-path transfer function, C(s), R(s) Laplace transforms of output and reference. Units of G·H are dimensionless (loop gain); G alone carries output-unit/input-unit.
• T(s) = G(s) / (1 − G(s)·H(s)) — positive feedback.
• E(s) = R(s) − H(s)·C(s) = R(s) / (1 + G(s)·H(s)) — actuating error.
• S_G^T = (∂T/T)/(∂G/G) = 1 / (1 + G·H) — sensitivity of closed-loop gain to forward-path changes (dimensionless). Open loop: S = 1.
• S_H^T = −G·H / (1 + G·H) — sensitivity to feedback-path changes (≈ −1 for large loop gain: the sensor must be accurate).
• C_d(s) = D(s) / (1 + G(s)·H(s)) — effect of a disturbance D(s) entering at the output.
• Static (DC) values: put s = 0 when the system is stable and the input is a step.

Worked examples

Example 1 (standard). A plant G(s) = 10/(s + 2) is placed in a unity negative feedback loop. Find the closed-loop transfer function, its time constant, and compare with the open-loop plant.

1. Formula: T(s) = G/(1 + G·H), H = 1.
2. T(s) = [10/(s + 2)] / [1 + 10/(s + 2)] = 10/(s + 12).
3. Open-loop pole at s = −2, time constant τ_ol = 1/2 = 0.5 s. Closed-loop pole at s = −12, τ_cl = 1/12 = 0.0833 s.
4. DC gain: open loop G(0) = 10/2 = 5; closed loop T(0) = 10/12 = 0.833.
5. Result: T(s) = 10/(s + 12), the response is 6 times faster (τ = 0.0833 s) but the DC gain drops from 5 to 0.833, so a step of 1 settles at 0.833 and the steady-state error is 1 − 0.833 = 0.167.

Example 2 (GATE level). An amplifier has forward gain K = 1000 and feedback factor H = 0.099 (negative feedback). (a) Find the closed-loop gain. (b) If K drops by 20 % due to ageing, find the new closed-loop gain and its percentage change. (c) Compare with an open-loop amplifier of the same nominal gain.

1. T = K/(1 + K·H). Loop gain K·H = 1000 × 0.099 = 99.
2. (a) T = 1000/(1 + 99) = 1000/100 = 10.0.
3. (b) New K = 800: K·H = 800 × 0.099 = 79.2; T' = 800/80.2 = 9.975.
4. Percentage change = (9.975 − 10)/10 × 100 = −0.25 %.
5. Check with sensitivity: S = 1/(1 + K·H) = 1/100 = 0.01; predicted change ≈ 0.01 × (−20 %) = −0.2 % (the small-change estimate; the exact figure is −0.25 % because a 20 % change is not small).
6. (c) An open-loop amplifier of gain 10 losing 20 % of its gain changes by the full −20 %.
7. Result: closed-loop gain 10.0, falling to 9.975 (−0.25 %) — feedback reduced the effect of the gain change by a factor of about 80.

Common mistakes

• Calling a system "closed loop" because it has a sensor that only displays the value; the loop is closed only if the measurement changes the control action.
• Using + in the denominator for positive feedback. Negative feedback gives 1 + GH, positive feedback gives 1 − GH.
• Computing the error as r − c when H ≠ 1; the actuating signal is r − H·c.
• Claiming feedback always improves stability. It can destabilise a system if the loop gain is too high.
• Forgetting that feedback reduces overall gain, so closed-loop DC gain is less than the open-loop plant gain.
• Applying s = 0 (final values) to an unstable loop.

For GATE ME

Expect conceptual MCQs on open- versus closed-loop examples and on the effects of feedback (sensitivity, bandwidth, disturbance rejection, instability risk), and short numericals on closed-loop gain, sensitivity 1/(1 + GH) and percentage change of gain. Practise reducing a single loop quickly and estimating the effect of a parameter change with and without feedback.

Quick check

1. Is a traffic signal running on a fixed timer open- or closed-loop?
2. Write the closed-loop transfer function for a positive feedback loop.
3. With G = 50, H = 0.2 (negative feedback), what is the closed-loop gain?
4. What is the sensitivity of T to G in Q3?
5. Can feedback make a stable plant unstable?

Answers: 1. Open loop; 2. G/(1 − GH); 3. 50/11 = 4.55; 4. 1/11 = 0.0909; 5. Yes, with high gain and phase lag.