Calculation
Âout = K·Âin / √[1+(2πfT)²]
A first-order system passes slow sinusoidal signals close to its static gain and increasingly attenuates faster variations.
Zacher/Reuter 2024, Abschnitte 3.2 und 5.1.4: Frequenzgang eines P-T1-Glieds
First-order frequency response: output amplitude under sinusoidal excitation
A first-order system passes slow sinusoidal signals close to its static gain and increasingly attenuates faster variations.
MINTSI
02
Result
Select a target and calculate.
Understand the inputs
- Steady output amplitude Âout — Half the peak-to-peak range of the steady sinusoidal output. Compare with permitted measurement range or required useful-signal amplitude; startup transient is not included.
- Static gain K — Ratio of a very slow or steady output change to input change. Obtain it from two steady states or the first-order model.
- Input amplitude Âin — Distance from mean to positive peak of the sinusoidal input. Divide a peak-to-peak value by two; do not use RMS amplitude.
- Time constant T — First-order time constant from a step test, datasheet or model identification. It sets corner frequency fc=1/(2πT).
- Excitation frequency f — Number of complete sinusoidal cycles per second. Obtain it from a signal generator, measurement or expected process excitation; f=0 is the static limiting case.
Example
K=2, Âin=3, T=0.1 s and f=1 Hz give 2πfT≈0.628 and Âout≈5.081. Without dynamic attenuation, 6 units would be expected.
Assumptions and limits
Linear time-invariant first-order system in sinusoidal steady state, constant parameters and no DC component in the amplitude calculation. Dead time, saturation, additional poles, noise and transient response are excluded.
Understand First-order frequency response: output amplitude under sinusoidal excitation
This calculator moves the first-order model from time to frequency domain. It shows the amplitude remaining after transients when the input varies sinusoidally, for example a temperature wave, speed fluctuation or smoothed sensor signal.
What does this quantity describe?
A first-order system has proportional static behaviour and one lag. Its frequency response is G(jω)=K/(1+jωT). Magnitude |G(jω)| gives the factor applied to input amplitude at angular frequency ω=2πf. Here j denotes the imaginary unit used in complex frequency calculations.
Formula and variables
Âout = K·Âin / √[1+(2πfT)²]
Frequency response: G(jω)=K/(1+jωT)Magnitude: |G(jω)|=K/√[1+(ωT)²]Angular frequency: ω=2πfOutput amplitude: Âout=|G(jω)|·Âin
| Symbol / input | Meaning |
|---|---|
| Steady output amplitude Âout | Half the peak-to-peak range of the steady sinusoidal output. Compare with permitted measurement range or required useful-signal amplitude; startup transient is not included. |
| Static gain K | Ratio of a very slow or steady output change to input change. Obtain it from two steady states or the first-order model. |
| Input amplitude Âin | Distance from mean to positive peak of the sinusoidal input. Divide a peak-to-peak value by two; do not use RMS amplitude. |
| Time constant T | First-order time constant from a step test, datasheet or model identification. It sets corner frequency fc=1/(2πT). |
| Excitation frequency f | Number of complete sinusoidal cycles per second. Obtain it from a signal generator, measurement or expected process excitation; f=0 is the static limiting case. |
Choose the inputs correctly
K is static gain from two steady operating points. Âin is peak amplitude relative to mean, half a peak-to-peak value. T comes from a step test or model identification. f is complete cycles per second and comes from the excitation source or measurement.
How to use the calculator
First verify the system is approximately linear and reaches a stationary sinusoid after startup. Evaluate input relative to its mean and determine peak amplitude. Use K and T from the same operating point. Compare output amplitude with measurement range, disturbance limit or required useful signal.
Worked example
For K=2, Âin=3, T=0.1 s and f=1 Hz, ωT≈0.628. Magnitude is 2/√(1+0.628²)≈1.694, giving Âout≈5.081 instead of static value 6.
Understand the result and units
As f→0, Âout approaches K·Âin. At corner frequency fc=1/(2πT), amplitude falls to 70.7% of that static value, corresponding to −3 dB. Above the corner frequency, faster fluctuations are increasingly attenuated.
f is entered in hertz; 1 Hz means one cycle per second. T converts internally to seconds, making 2πfT dimensionless. Input and output amplitudes may carry physical units; this general calculator treats them on a consistent normalised scale.
Typical applications
Useful for estimating sensor and process smoothing, filter action, periodic disturbance transmission, test design and learning Bode diagrams.
Assumptions, limits and common mistakes
The result covers only the steady sinusoidal component of a linear time-invariant first-order system. Mean value, startup transient, phase shift, dead time, additional poles, nonlinearity and saturation are outside the output. Non-sinusoidal signals require separate frequency components.
Common mistake: Do not enter peak-to-peak height or RMS value as peak amplitude. Do not confuse hertz with angular frequency in rad/s; the calculator applies ω=2πf. Dead time does not change ideal magnitude but does change phase and possible loop stability.
Frequently asked questions
What is “First-order output amplitude under sinusoidal excitation” used for?
Useful for estimating sensor and process smoothing, filter action, periodic disturbance transmission, test design and learning Bode diagrams.
Where do the input values come from?
K is static gain from two steady operating points. Âin is peak amplitude relative to mean, half a peak-to-peak value. T comes from a step test or model identification. f is complete cycles per second and comes from the excitation source or measurement.
What does the result not cover?
The result covers only the steady sinusoidal component of a linear time-invariant first-order system. Mean value, startup transient, phase shift, dead time, additional poles, nonlinearity and saturation are outside the output. Non-sinusoidal signals require separate frequency components.