Frequency-domain and state-space descriptions

Describe memory with first-order equations

For m d²x/dt²+c dx/dt+kx=F, x is displacement, m mass, c viscous damping coefficient and k spring stiffness. Choose states x₁=x and x₂=dx/dt. Then dx₁/dt=x₂ and dx₂/dt=(F−cx₂−kx₁)/m. An output equation specifies which combination is measured. Position measurement is y=x₁; it does not imply the velocity state is directly observed.

The compact form dx/dt=Ax+Bu and y=Cx+Du collects these relations into matrices. Here bold-state-vector meaning must be distinguished from scalar displacement; coefficients carry the appropriate units. A state estimate uses a model and available observations to infer unmeasured information. Its accuracy depends on the sensor information and model, rather than on a label such as observer.

Change the question from a trajectory to a sinusoidal response

For a linear time-invariant system, a transfer function represents input-to-output behaviour with zero initial conditions. Replacing the differentiation symbol by a complex frequency s gives force-to-displacement transfer G(s)=1/(ms²+cs+k). It complements the state model rather than replacing initial-state information. Hidden internal behaviour requires care when using a reduced transfer representation.

For a stable system after transients decay, a sinusoidal force at angular frequency ω produces a sinusoid at the same frequency with changed amplitude and phase. G(jω) describes that ratio, where j²=−1. Magnitude is the amplitude ratio and phase is the shift relative to the input. Angular frequency in rad/s equals 2π times frequency in hertz.

Read gain and phase together

A Bode display uses a logarithmic frequency axis, commonly magnitude in 20log₁₀ of an amplitude ratio and phase in degrees. The accompanying figure instead uses an explicitly linear frequency axis and physical compliance units. A peak means this force frequency produces a relatively large displacement under the stated linear assumptions.

Loop stability margins concern the open-loop transfer around the complete feedback loop. A large plant-response magnitude alone does not give a phase margin. Delays can leave a sinusoid’s amplitude unchanged while shifting phase. Nonlinear limits, contact changes and large departures from a linearization point require additional analysis.

One model seen through a sinusoidal test

Supply m=2 kg, c=4 N·s/m and k=8 N/m. At ω=2 rad/s, k−mω²=0 and cω=8 N/m. Thus G(j 2)=1/(j 8)=−j/8 m/N: magnitude 0.125 m/N and phase −90°. A 0.4 N force amplitude gives a 0.050 m displacement amplitude after startup effects decay.

The states still specify any additional transient due to initial displacement or velocity. The same final sinusoidal response can follow different initial histories. These assumed coefficients describe an ideal linear example; the displacement amplitude does not certify a real spring’s working range.

Try the reasoning

1. What is 2 rad/s in hertz?

   Show the answer and reasoning

   2/(2π)=1/π, approximately 0.318 Hz. Treating 2 rad/s as 2 Hz changes the forcing frequency.

2. Does knowing displacement at one instant establish velocity?

   Show the answer and reasoning

   No. The mass could be moving in either direction or resting. Another observation or a justified state estimate is needed.

3. A pure delay adds−ωT radians of phase. At ω=2 rad/s and T=.1 s, what is it?

   Show the answer and reasoning

   −.2rad, about −11.46°. An unchanged amplitude therefore need not imply an unchanged feedback effect.

What to carry forward

Use states for internal evolution and frequency response for sustained linear input-output behaviour.