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A linear second-order system is characterized by static gain, undamped natural frequency and damping ratio.
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Formula
K depends on input/output units; ωn in rad/s; s in s−1; ζ dimensionless
| Symbol | Meaning | Unit |
|---|---|---|
| Transfer function | dimension-dependent | |
| Static gain | dimension-dependent | |
| Undamped natural angular frequency | rad/s | |
| Damping ratio | dimensionless |
G(s)=\frac{K\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}This describes a linear time-invariant system with a pole pair and no dead time. For ζ<1, the step response is underdamped. Extra poles, zeros, saturation and initial conditions are omitted.
This describes a linear time-invariant system with a pole pair and no dead time. For ζ<1, the step response is underdamped. Extra poles, zeros, saturation and initial conditions are omitted.
A linear second-order system is characterized by static gain, undamped natural frequency and damping ratio.
Base equation for the target quantity
Preserve signs, boundary conditions and units when rearranging.
Work through the numerical example for “Normalized second-order transfer function” and check its unit and magnitude.
For K=1, ωn=10 rad/s and ζ=0.5, the denominator is s2+10s+100.
Substituting a related quantity that shares the same symbol.
Use variable definitions and check: K depends on input/output units; ωn in rad/s; s in s−1; ζ dimensionless.
Formula cluster
A linear second-order system is characterized by static gain, undamped natural frequency and damping ratio.
For K=1, ωn=10 rad/s and ζ=0.5, the denominator is s2+10s+100.
Model oscillatory sensor, actuator or mechanical plant dynamics.
Here is how to work through a typical Normalized second-order transfer function (G(s)=Kωn²/(s²+2ζωns+ωn²)) task step by step:
Task
Work through the numerical example for “Normalized second-order transfer function” and check its unit and magnitude.
Solution path
For K=1, ωn=10 rad/s and ζ=0.5, the denominator is s2+10s+100.
Poles solve s2+2ζωn s+ωn2=0. For ζ<1 they are complex conjugates with real part −ζωn; for ζ≥1 they are real. Their locations shape decay and overshoot. This canonical model assumes a linear time-invariant second-order system without dead time. Application: Model oscillatory sensor, actuator or mechanical plant dynamics. The numerical example For K=1, ωn=10 rad/s and ζ=0.5, the denominator is s2+10s+100 anchors the idea in a concrete measurement case. The corresponding units are K depends on input/output units; ωn in rad/s; s in s−1; ζ dimensionless. For K=1, ωn=10 rad/s and ζ=0.5, the denominator is s2+10s+100. This checks that the reference quantities are compatible.
An ideal system with 0<ζ<1 is underdamped and typically overshoots after a step. Overshoot decreases as ζ rises and is absent for ζ≥1. Zeros and extra dynamics can alter the response. This canonical model assumes a linear time-invariant second-order system without dead time. The corresponding units are K depends on input/output units; ωn in rad/s; s in s−1; ζ dimensionless. For K=1, ωn=10 rad/s and ζ=0.5, the denominator is s2+10s+100. This checks that the reference quantities are compatible. Scope: This describes a linear time-invariant system with a pole pair and no dead time. For ζ<1, the step response is underdamped. Extra poles, zeros, saturation and initial conditions are omitted. A practical application is Model oscillatory sensor, actuator or mechanical plant dynamics.
For this normalized form, G(0)=K. K is the static gain, with units determined by output over input. K=1 is not automatic for every sensor or actuator. This canonical model assumes a linear time-invariant second-order system without dead time. Scope: This describes a linear time-invariant system with a pole pair and no dead time. For ζ<1, the step response is underdamped. Extra poles, zeros, saturation and initial conditions are omitted. A practical application is Model oscillatory sensor, actuator or mechanical plant dynamics.
Write the quantity and unit from the variable list next to every value: K depends on input/output units; ωn in rad/s; s in s−1; ζ dimensionless. Substitute only after converting to a consistent unit system, and state sign or reference direction explicitly. The example For K=1, ωn=10 rad/s and ζ=0.5, the denominator is s2+10s+100. is a calculation check: result, unit and order of magnitude must fit the described situation. Round after intermediate calculations so small contributions are not lost prematurely. This canonical model assumes a linear time-invariant second-order system without dead time.
The background states this starting condition: This describes a linear time-invariant system with a pole pair and no dead time. For ζ<1, the step. That limits when the result is a valid model prediction. Compare the problem’s boundary conditions with this assumption before substituting values; real components, material properties or loads can introduce additional effects. The cited reference supports the relation used here but does not remove the need to check whether the specific case fits the model. This canonical model assumes a linear time-invariant second-order system without dead time.
Quanta exam set
Curated exam set for "Normalized second-order transfer function":
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Which formula describes Normalized second-order transfer function?
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How do you rearrange Normalized second-order transfer function for Base equation for the target quantity?
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Which common mistake happens with Normalized second-order transfer function?
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Create a curated FSRS exam set for G(s)=Kωn²/(s²+2ζωns+ωn²): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.
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