On-off and time-delay systems

Why it matters

The cheapest controller in a plant is a switch: a thermostat on an oven, a pressure switch on a compressor, a level switch on a sump pump. And the hardest plant to control is one with dead time — a conveyor, a long pipeline, a stack analyser at the end of a sample line. Instrumentation engineers meet both every day, so you need to predict how an on-off loop cycles and how much gain a loop with transport lag can tolerate.

Key ideas

On-off (two-position) control

• The controller output takes only two values: u = U_on when the error is positive and u = U_off when it is negative (heater on/off, valve open/shut).
• An ideal on-off controller would chatter at very high frequency around the set point, so practical controllers have a differential gap (hysteresis, dead band): the output switches on at SP − h/2 and off at SP + h/2 (for a gap h centred on the set point). Some thermostats place the gap differently — always read how the band is defined.
• Result: the controlled variable cycles continuously between the limits — a limit cycle, not a steady state. A wider gap means fewer switching cycles (less wear on contactors and valves) but a larger swing.
• With a first-order process, the variable rises exponentially toward the "on" final value and falls toward the "off" final value; the times on each half-cycle follow from the exponential.
• With dead time θ, the variable keeps moving for θ after each switching, so the actual swing exceeds the gap by about (rate of change) × θ at each end.
• Used where some cycling is acceptable and the process has a large capacity: domestic heating, refrigeration, sump levels, alarm-type actions.

Describing function view: an on-off element is nonlinear, but for a sinusoidal input of amplitude A its fundamental output can be approximated by a gain N(A). For an ideal relay with output ±M, N = 4M/(πA). A limit cycle is predicted where the Nyquist plot of the linear part G(jω) meets −1/N(A).

Time delay (dead time, transport lag)

• A pure delay θ shifts a signal in time: y(t) = x(t − θ), Y(s) = e^(−θs)·X(s). Typical sources: material on a conveyor (θ = L/v), fluid in a pipe (θ = pipe volume / flow), sampling lines, and digital computation.
• In frequency: |e^(−jωθ)| = 1, ∠e^(−jωθ) = −ωθ rad. No gain change, unlimited phase lag growing with frequency.
• Consequences: reduced phase margin and lower usable gain; Routh cannot be applied directly (the characteristic equation is not a polynomial); the root locus has infinitely many branches.
• Approximations: first-order Padé e^(−θs) ≈ (1 − θs/2)/(1 + θs/2) (a non-minimum-phase zero), useful for Routh or root-locus sketches; or e^(−θs) ≈ 1/(1 + θs) for small θ.
• Many processes are modelled as first-order plus dead time (FOPDT): G(s) = K·e^(−θs)/(τs + 1); the ratio θ/τ measures how difficult the loop is. Tuning rules such as Ziegler–Nichols and Cohen–Coon are built on this model. For large θ/τ, a Smith predictor (model-based dead-time compensator) lets the controller act on a delay-free prediction.

Formulas

• θ = L/v — conveyor dead time; θ = V/Q — plug-flow pipe dead time.
• L[x(t − θ)] = e^(−θs)·X(s).
• ∠e^(−jωθ) = −ωθ rad (−57.3·ωθ degrees).
• G(s) = K·e^(−θs)/(τs + 1) — FOPDT model.
• −ωpc·θ − tan⁻¹(ωpc·τ) = −π — phase crossover of an FOPDT loop.
• Ku = √(1 + (ωpc·τ)²)/K, Pu = 2π/ωpc — ultimate gain and period under P control.
• t = τ·ln[(y_final − y_start)/(y_final − y_end)] — time for a first-order variable to move from y_start to y_end.
• N(A) = 4M/(πA) — describing function of an ideal relay.

Symbols: θ dead time (s); L length (m); v speed (m/s); V volume (m³); Q flow (m³/s); τ time constant (s); K process gain; ω rad/s; M relay output level; A input amplitude (same unit as the error signal).

Worked examples

Example 1 — cycle period of an on-off temperature loop Given: oven with first-order behaviour, τ = 10 min. With the heater on, temperature would settle at 100 °C; with it off, at the ambient 20 °C. Set point 60 °C, differential gap 4 °C centred on the set point (switch on at 58 °C, off at 62 °C). Neglect dead time.

1. Heating (from 58 toward 100, stop at 62): t_on = 10·ln[(100 − 58)/(100 − 62)] = 10·ln(42/38) = 1.00 min.
2. Cooling (from 62 toward 20, stop at 58): t_off = 10·ln[(62 − 20)/(58 − 20)] = 10·ln(42/38) = 1.00 min.
3. Period = t_on + t_off = 2.00 min; duty cycle 50 % because the set point is midway between 20 and 100 °C.

Result: the oven cycles between 58 °C and 62 °C with a period of about 2.0 min. Doubling the gap to 8 °C roughly doubles the period and halves the number of switching operations.

Example 2 — ultimate gain of a dead-time process (GATE level) Given: FOPDT process G(s) = e^(−2s)/(10s + 1) (K = 1, θ = 2 s, τ = 10 s) under proportional control Kc.

1. Phase crossover: −2ω − tan⁻¹(10ω) = −π. Solving numerically: ωpc = 0.844 rad/s (check: 2 × 0.844 = 1.689 rad = 96.8° and tan⁻¹(8.44) = 83.2°; sum 180°).
2. |G(jωpc)| = 1/√(1 + 8.44²) = 1/8.50.
3. Ultimate gain Ku = 8.50; ultimate period Pu = 2π/0.844 = 7.44 s.
4. Ziegler–Nichols PI starting values: Kc = 0.45 × 8.50 = 3.83, Ti = 7.44/1.2 = 6.2 s.

Result: Ku = 8.50 and Pu = 7.44 s. Without the 2 s dead time, this first-order loop could not be destabilised by any gain — the delay alone limits it.

Common mistakes

• Assuming the on-off swing equals the gap even when the process has dead time — overshoot past the switching points adds to it.
• Mixing gap conventions (centred on the set point versus starting at the set point).
• Treating dead time as a first-order lag: a lag attenuates and is limited to −90°; a delay has unlimited phase lag and no attenuation.
• Applying Routh directly to an equation containing e^(−θs).
• Forgetting to convert ωθ from radians to degrees when combining with tan⁻¹ terms.
• Computing conveyor delay with mismatched units (m and m/min).

For GATE IN

Expect questions on dead time of conveyors and pipelines, the phase contribution of e^(−sT) at a given frequency, the gain or phase margin of a loop with delay, on-off control with a differential gap (switching points, cycle times for a first-order process), and the describing function of a relay used to predict a limit cycle. Practise solving ωθ + tan⁻¹(ωτ) = π by iteration.

Quick check

1. A belt 150 m long moves at 3 m/s. What is the dead time?
2. What phase lag does a 0.1 s delay add at 10 rad/s?
3. A thermostat has a 2 °C gap centred on 22 °C. At what temperature does heating switch off?
4. For an ideal relay with output ±5 V and input amplitude 2 V, what is the describing function gain?

Answers: 1. 50 s; 2. 1 rad = 57.3°; 3. 23 °C; 4. 4 × 5/(π × 2) = 3.18.