Calculation
ωc=√[(Kp·Ks)²−1]/T; φm=π−atan(ωcT)−ωcTd
Dead time adds increasing phase lag with frequency; phase margin shows how far the simplified loop remains from its stability boundary at crossover.
Zacher/Reuter 2024, Abschnitte 6.3.2 und 8.2.1: Durchtrittsfrequenz und Phasenreserve
First-order process with dead time: phase margin
Dead time adds increasing phase lag with frequency; phase margin shows how far the simplified loop remains from its stability boundary at crossover.
MINTSI
02
Result
Select a target and calculate.
Understand the inputs
- Phase margin φm — Angular distance from −180° phase at gain crossover. Positive values are necessary in this simplified model; use it to compare tuning and robustness targets, not as the sole approval criterion.
- Controller gain Kp — Proportional controller gain in the chosen signal scaling. Take it from controller settings or design; changing it shifts crossover frequency.
- Static process gain Ks — Final output change per input step from two steady operating points. Together with Kp it must give a positive loop-gain magnitude here.
- Time constant T — First-order time constant from response onset to 63.2% of final change. Obtain it from a step test or identified model.
- Dead time Td — Time between input change and visible response onset. Transport, sensor and processing delays can contribute; determine it from the same step test as T.
Example
Kp=2, Ks=1, T=5 s and Td=1 s give ωc≈0.3464 rad/s. The first-order lag and dead time together contribute about −79.85° phase, leaving approximately 100.15° phase margin.
Assumptions and limits
Negative unity feedback; ideal proportional controller; open-loop model G₀(s)=Kp·Ks·e^(−sTd)/(1+sT); positive gains and Kp·Ks>1 so a finite positive 0 dB crossover exists. Additional poles, zeros, filters, sampling and actuator dynamics are excluded.
Understand First-order process with dead time: phase margin
This calculator evaluates a simple loop in the frequency domain: a proportional controller drives a first-order process with dead time. From four directly interpretable model values it determines crossover internally and returns the remaining angular distance to the stability boundary.
What does this quantity describe?
A first-order lag has proportional static behaviour and one time constant. Dead time Td shifts a signal without attenuating magnitude but adds frequency-dependent phase lag −ωTd. Phase margin φm is defined at the frequency where open-loop magnitude equals one or 0 dB. It is the distance from phase there to −180°.
Formula and variables
ωc=√[(Kp·Ks)²−1]/T; φm=π−atan(ωcT)−ωcTd
Open loop: G₀(s)=Kp·Ks·e^(−sTd)/(1+sT)Crossover: ωc=√[(Kp·Ks)²−1]/TPhase margin: φm=π−atan(ωcT)−ωcTd
| Symbol / input | Meaning |
|---|---|
| Phase margin φm | Angular distance from −180° phase at gain crossover. Positive values are necessary in this simplified model; use it to compare tuning and robustness targets, not as the sole approval criterion. |
| Controller gain Kp | Proportional controller gain in the chosen signal scaling. Take it from controller settings or design; changing it shifts crossover frequency. |
| Static process gain Ks | Final output change per input step from two steady operating points. Together with Kp it must give a positive loop-gain magnitude here. |
| Time constant T | First-order time constant from response onset to 63.2% of final change. Obtain it from a step test or identified model. |
| Dead time Td | Time between input change and visible response onset. Transport, sensor and processing delays can contribute; determine it from the same step test as T. |
Choose the inputs correctly
Kp is the active proportional controller gain. Ks is static process gain from final output change divided by input step. T is first-order time constant and Td is dead time, normally identified from the same step test. Kp and Ks must use the same signal normalisation so their product is dimensionless loop gain.
How to use the calculator
First identify the first-order-plus-dead-time model from a sufficiently small step test. Enter Kp from planned or active tuning. The calculator applies only if Kp·Ks>1; otherwise this simplified model has no positive 0 dB crossover. Compare several Kp values, then simulate or measure the complete real model.
Worked example
For Kp=2, Ks=1 and T=5 s, crossover is ωc=√3/5≈0.3464 rad/s. Without dead time margin would be 120°. Td=1 s adds about 19.85° lag, leaving φm≈100.15°.
Understand the result and units
Positive phase margin is a necessary stability condition for this model; negative margin warns of closed-loop instability. More margin generally means greater robustness but often slower control. One value does not replace gain-margin assessment, time simulation, actuator saturation and model uncertainty.
T and Td convert internally to seconds. Crossover frequency uses rad/s, where rad means radians. φm is displayed in degrees. Kp·Ks must be dimensionless.
Typical applications
Useful for preliminary comparison of proportional gains, studying transport-delay effects, teaching Bode diagrams and checking a simplified process model.
Assumptions, limits and common mistakes
The result applies only to open loop Kp·Ks·exp(−sTd)/(1+sT) with negative unity feedback. Additional lags, zeros, filters, sampling, sensor and actuator dynamics change crossover and phase. Do not commission a real plant based on this result alone.
Common mistake: Do not treat dead time as another time constant; it changes phase differently. Do not mix degrees and radians. For Kp·Ks≤1, do not reinterpret ω=0 as an ordinary crossover. Ensure signs actually produce negative feedback.
Frequently asked questions
What is “Phase margin of a first-order process with dead time” used for?
Useful for preliminary comparison of proportional gains, studying transport-delay effects, teaching Bode diagrams and checking a simplified process model.
Where do the input values come from?
Kp is the active proportional controller gain. Ks is static process gain from final output change divided by input step. T is first-order time constant and Td is dead time, normally identified from the same step test. Kp and Ks must use the same signal normalisation so their product is dimensionless loop gain.
What does the result not cover?
The result applies only to open loop Kp·Ks·exp(−sTd)/(1+sT) with negative unity feedback. Additional lags, zeros, filters, sampling, sensor and actuator dynamics change crossover and phase. Do not commission a real plant based on this result alone.