Open-loop and closed-loop systems and effects of feedback

Why it matters

Every instrument loop in a plant — a temperature controller on a furnace, a level loop on a tank, a speed loop on a motor — is a closed-loop system built around a sensor. Knowing exactly what feedback buys you (accuracy, insensitivity to drift, disturbance rejection, speed) and what it costs (gain, complexity, the risk of instability) is the starting point for the whole subject.

Key ideas

Open-loop system

• The control action does not depend on the output. The input is applied through a controller and plant, and nobody checks the result.
• Examples: a toaster on a timer, a washing machine on a fixed programme, a stepper motor driven by a counted pulse train.
• Accuracy depends entirely on calibration. Any change in plant parameters (ageing, temperature drift) or any disturbance appears directly in the output.
• Advantages: simple, cheap, no sensor needed, cannot become unstable through feedback (an open-loop system is unstable only if the plant itself is).

Closed-loop (feedback) system

• A sensor measures the output c(t), the feedback element H(s) turns it into a feedback signal b(t), and a comparator forms the actuating (error) signal e(t) = r(t) − b(t). The controller acts on e(t).
• Examples: a room thermostat, an automatic voltage regulator, a pressure loop with a transmitter and control valve.
• With negative feedback the loop works to drive e(t) toward zero, so the output follows the reference even when the plant changes.

Positive vs negative feedback

• Negative feedback (e = r − b) is used in control: it reduces error and sensitivity. Positive feedback (e = r + b) increases gain and is used in oscillators and latches, not in regulators.

Effects of negative feedback (with loop gain G(s)H(s) large over the frequencies of interest)

1. Overall gain is reduced by the factor 1 + GH.
2. Sensitivity to forward-path parameter changes is reduced by the factor 1 + GH. The closed-loop gain becomes ≈ 1/H, so accuracy now depends on the sensor and feedback element, not the plant. This is why a good transmitter matters more than a perfect valve.
3. Sensitivity to the feedback element is not reduced: S_H ≈ −1 at high loop gain. An error in the sensor goes straight into the output.
4. Disturbances entering the forward path or at the output are attenuated by 1 + GH. Sensor noise, however, is not attenuated — it is treated like a reference signal.
5. Speed and bandwidth increase. For a first-order plant, the closed-loop pole moves further left, so the time constant shrinks by 1 + K.
6. Stability can be improved or destroyed. Feedback can stabilise an unstable plant, but a loop with high gain and enough phase lag (several poles, dead time) can oscillate. Stability is studied in the Routh, root-locus and Nyquist topics.

Formulas

• C(s)/R(s) = T(s) = G(s) / (1 + G(s)H(s)) — closed-loop transfer function, negative feedback.
• T(s) = G(s) / (1 − G(s)H(s)) — positive feedback.
• E(s) = R(s) − B(s) = R(s) / (1 + G(s)H(s)) — actuating signal, with B(s) = H(s)C(s).
• L(s) = G(s)H(s) — open-loop (loop) transfer function; its roots of 1 + L(s) = 0 are the closed-loop poles (the characteristic equation).
• S_G = (∂T/T) / (∂G/G) = 1 / (1 + GH) — sensitivity of T to the forward path.
• S_H = (∂T/T) / (∂H/H) = −GH / (1 + GH) — sensitivity of T to the feedback element.
• C(s) = [G/(1 + GH)]·R(s) + [1/(1 + GH)]·D(s) — output with a disturbance D(s) added at the output.

Symbols: R(s) reference input, C(s) controlled output, G(s) forward-path transfer function, H(s) feedback-path transfer function. Each transfer function has the units of (its output)/(its input), e.g. °C/V for a heater plant or V/°C for a sensor; GH is dimensionless. All relations assume linear, time-invariant (LTI) blocks and zero initial conditions. The sensitivity formulas are small-change (differential) results.

Worked examples

Example 1 — first-order plant with unity feedback Given: G(s) = 10/(s + 2), H(s) = 1.

1. Open-loop: DC gain G(0) = 10/2 = 5; time constant τ = 1/2 = 0.5 s.
2. Closed-loop: T(s) = G/(1 + G) = [10/(s + 2)] / [(s + 12)/(s + 2)] = 10/(s + 12).
3. Closed-loop pole at s = −12, which is in the left half-plane, so the loop is stable.
4. Closed-loop DC gain T(0) = 10/12 = 0.833; time constant τ = 1/12 = 0.0833 s.
5. Steady-state error to a unit step: e_ss = 1/(1 + G(0)) = 1/(1 + 5) = 0.167.

Result: T(s) = 10/(s + 12); feedback cut the gain from 5 to 0.833 but made the response 6 times faster (0.5 s → 0.0833 s). The 6 is exactly 1 + G(0).

Example 2 — sensitivity (GATE level) Given: amplifier with forward gain K = 100 and feedback factor H = 0.1. K drifts up by 10 %. Find the change in closed-loop gain.

1. Nominal: T = K/(1 + KH) = 100/(1 + 10) = 9.091.
2. Sensitivity: S_K = 1/(1 + KH) = 1/11 = 0.0909.
3. Predicted (differential) change: ΔT/T ≈ S_K·(ΔK/K) = 0.0909 × 10 % = 0.909 %.
4. Exact check: K = 110 gives T = 110/(1 + 11) = 9.167; ΔT/T = (9.167 − 9.091)/9.091 = 0.833 %.

Result: a 10 % change in K becomes only about 0.83 % (exact) or 0.91 % (first-order estimate) in the closed-loop gain, while a 10 % change in H would change T by nearly 10 % (S_H = −10/11 = −0.909).

Common mistakes

• Using 1 − GH in the denominator for a negative-feedback loop (or 1 + GH for positive feedback). Check the sign at the summing junction.
• Calling G(s) alone the "open-loop transfer function" when H ≠ 1. The loop transfer function that decides stability is G(s)H(s).
• Thinking feedback always improves stability. More loop gain with more lag eventually causes oscillation.
• Assuming feedback cancels sensor errors. The output tracks R/H, so a wrong H gives a wrong output.
• Mixing up error e = r − b (actuating signal) with true error r − c. They are equal only for unity feedback.
• Treating the differential sensitivity result as exact for large changes.

For GATE IN

Expect conceptual MCQs on the effects of feedback (gain, sensitivity, bandwidth, disturbance and noise), numerical questions on closed-loop gain and its percentage change when a parameter drifts, and questions that combine a first-order plant with feedback to ask for the new time constant or steady-state error. Practise deriving S_G and S_H, and reading the sign of the summing junction from a block diagram.

Quick check

1. A plant G = 50 is placed in a loop with H = 0.02. What is the closed-loop gain?
2. For the loop in Q1, what is the sensitivity of T to G?
3. Does negative feedback reduce the effect of sensor noise on the output?
4. A first-order plant K/(τs + 1) with K = 4 is put in unity feedback. By what factor does the time constant shrink?

Answers: 1. 50/(1 + 1) = 25; 2. 1/(1 + 1) = 0.5; 3. No — sensor noise enters like a reference and is not attenuated; 4. By 1 + K = 5.