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Mechatronics Engineering DISCUSSION
My position loop is modelled as the standard second-order system ω_n² / (s² + 2ζω_n s + ω_n²) with ω_n = 40 rad/s and ζ = 0.5. The specification I have to meet says the closed-loop bandwidth must be at least 8 Hz.
Is the bandwidth just ω_n converted to hertz (6.4 Hz), or is it something else? And how does it relate to the rise time I see in a step response?
Bandwidth is normally defined as the frequency where the closed-loop gain has fallen 3 dB below its low-frequency value, that is to 0.707 of it. For the standard second-order form, setting the magnitude equal to 0.707 gives
ω_BW = ω_n × sqrt(1 - 2ζ² + sqrt(4ζ⁴ - 4ζ² + 2)).
With ζ = 0.5 the factor is sqrt(0.5 + sqrt(1.25)) = sqrt(1.618) = 1.27, so ω_BW = 1.27 × 40 = 50.9 rad/s, or 8.1 Hz. Your loop just meets the 8 Hz requirement even though ω_n is only 6.4 Hz.
The factor depends strongly on damping, which is why ω_n alone is not the answer: about 1.45 at ζ = 0.3, 1.27 at ζ = 0.5, exactly 1.00 at ζ = 0.707 and 0.64 at ζ = 1. A lightly damped loop therefore shows a wide bandwidth on paper, but it gets it from a resonant peak. For ζ below 0.707 the peak gain is 1 / (2ζ × sqrt(1 - ζ²)) at ω_n × sqrt(1 - 2ζ²). In your case that is 1.15 (about +1.2 dB) at 28 rad/s, and the step response overshoots by about 16 percent. A bandwidth figure should always be read together with the peaking or the overshoot.
For the link to the time domain, a first-order system has a 10 to 90 percent rise time of 0.35 / f_BW, and that remains a fair estimate for a reasonably damped second-order loop: 0.35 / 8.1 Hz is about 43 ms. A commonly used second-order estimate for damping near 0.5 is t_r ≈ 1.8 / ω_n, which gives 45 ms here, so the two agree.
Settling is governed by the real part of the poles instead: the 2 percent settling time is roughly 4 / (ζ × ω_n) = 4 / 20 = 0.2 s. Doubling bandwidth by raising ω_n halves both figures.
Check which definition your specification uses before declaring a pass. Some motion-control documents quote bandwidth as the frequency where the closed-loop phase reaches -90°, not where the gain is -3 dB. For this transfer function the phase is -90° exactly at ω_n, so by that definition your loop has 6.4 Hz and fails the 8 Hz requirement.
To measure it on the real axis, command a small sine wave, sweep the frequency and log command and encoder position; the ratio of amplitudes gives the gain and the time shift gives the phase. Keep the amplitude small enough that the drive never reaches its current or voltage limit, otherwise the measured bandwidth depends on amplitude.
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