Calculation
y(t)=0 for t<Td; y(t)=K·Δu·[1−e^(−(t−Td)/T)] for t≥Td
Dead time shifts the complete first-order response: initially the output does not react, then it approaches its final value exponentially.
Zacher/Reuter 2024, Abschnitte 3.2 und 8.2.1: PT1-Strecke mit Totzeit
First-order plus dead-time step response
Dead time shifts the complete first-order response: initially the output does not react, then it approaches its final value exponentially.
MINTSI
02
Result
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Calculated step response
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Understand the inputs
- Output change y(t) — Change in process output relative to its value before the step. It is zero before dead time has elapsed; afterwards use it for simulation, measured-response checks or estimating a response instant.
- Static process gain K — Final output change divided by input step. Determine it from two steady operating points: K=(y∞−y0)/(u1−u0). Scale consistently if input and output use different physical quantities.
- Input step Δu — New minus old process input at the start instant, such as a change in valve opening or heater power. Do not enter the new absolute value.
- Time constant T — Time scale of the exponential rise after the response begins. After T, 63.2% of the final change is reached. Obtain it from a step test, identification or process model.
- Dead time Td — Time between the input step and the first detectable output response, for example due to transport distance, sensor location or processing. Determine it from step-test timestamps.
- Time since the step t — Observation time measured from the input change. For t<Td the result must be zero; at t=Td the exponential rise begins.
Example
K=2, Δu=3, T=5 s, Td=2 s and t=7 s give an effective rise time of 5 s and y≈3.7927. At t=1 s, y=0.
Assumptions and limits
Linear time-invariant first-order-plus-dead-time model, initially at rest and driven by an ideal step. Additional lags, input ramping, saturation, disturbances and measurement noise are excluded.
Understand First-order plus dead-time step response
This calculator combines two effects that commonly occur together in real processes: a dead time during which no output response is visible, followed by a first-order lag. It gives the output at any instant and plots the complete step response.
What does this quantity describe?
A first-order-plus-dead-time model represents a process whose response starts late and then approaches its final value exponentially without overshoot. Its transfer function is G(s)=K·e^(−sTd)/(1+sT), where Td is pure dead time and T is the first-order time constant.
Formula and variables
y(t)=0 for t<Td; y(t)=K·Δu·[1−e^(−(t−Td)/T)] for t≥Td
t<Td: y(t)=0t≥Td: y(t)=K·Δu·[1−exp(−(t−Td)/T)]y∞=K·Δu
| Symbol / input | Meaning |
|---|---|
| Output change y(t) | Change in process output relative to its value before the step. It is zero before dead time has elapsed; afterwards use it for simulation, measured-response checks or estimating a response instant. |
| Static process gain K | Final output change divided by input step. Determine it from two steady operating points: K=(y∞−y0)/(u1−u0). Scale consistently if input and output use different physical quantities. |
| Input step Δu | New minus old process input at the start instant, such as a change in valve opening or heater power. Do not enter the new absolute value. |
| Time constant T | Time scale of the exponential rise after the response begins. After T, 63.2% of the final change is reached. Obtain it from a step test, identification or process model. |
| Dead time Td | Time between the input step and the first detectable output response, for example due to transport distance, sensor location or processing. Determine it from step-test timestamps. |
| Time since the step t | Observation time measured from the input change. For t<Td the result must be zero; at t=Td the exponential rise begins. |
Choose the inputs correctly
K is the static process gain from two steady states. Δu is only the input change. T is the time from the start of the response to 63.2% of the final change. Td is the time from the input step to the first detectable output response. t is the timestamp to evaluate. T and Td normally come from the same recorded step test.
How to use the calculator
Reference the output to its value before the test, change the input as a safe step, and record input and output on one time base. Determine K from final output change divided by input step, read Td at response onset, and measure T from that onset to 63.2%. Enter a timestamp and compare the plotted model with the measurements.
Worked example
With K=2, Δu=3, T=5 s and Td=2 s, output change stays zero for 2 s. At t=7 s the exponential part has run for exactly one time constant, so y=6·(1−e⁻¹)=3.7927.
Understand the result and units
y(t) is the change from the pre-step output, not necessarily the absolute sensor value. Add the old operating point for an absolute value. A large Td/T ratio means the controller receives no process feedback for a long time, making aggressive tuning especially risky.
All time quantities must have the same dimension; the interface converts seconds, minutes and hours internally to seconds. Physically K has output units per input unit, but is entered dimensionlessly for consistently scaled signals.
Typical applications
Use the model for thermal, flow, concentration and transport processes, controller-tuning models and checking a recorded step response.
Assumptions, limits and common mistakes
The model assumes an ideal input step, constant parameters and one dominant time constant. Input ramping, several comparable lags, overshoot, nonlinearity, saturation or changing transport velocity are excluded.
Common mistake: Do not subtract Td manually and then also enter a shifted time. Do not measure T from the original input step: the 63.2% interval starts only after dead time has elapsed.
Frequently asked questions
What is “First-order process with dead time” used for?
Use the model for thermal, flow, concentration and transport processes, controller-tuning models and checking a recorded step response.
Where do the input values come from?
K is the static process gain from two steady states. Δu is only the input change. T is the time from the start of the response to 63.2% of the final change. Td is the time from the input step to the first detectable output response. t is the timestamp to evaluate. T and Td normally come from the same recorded step test.
What does the result not cover?
The model assumes an ideal input step, constant parameters and one dominant time constant. Input ramping, several comparable lags, overshoot, nonlinearity, saturation or changing transport velocity are excluded.