The mechatronics project: checkpoints before results
Introduit| English | Sens |
|---|---|
| Anti-windup | A method that stops the integral term growing while the actuator is saturated. |
| control/kənˈtrəʊl/ | Making a system's output follow a desired value, usually by measuring it and correcting the input. |
| overshoot/ˌəʊvəˈʃuːt/ | How far a step response goes past its final value, usually given as a percentage of the step. |
Show it meets the spec
- The course ends with a valve positioner that must move 90° quickly, stop exactly and hold against the water. The project shows, with traceable numbers, that it meets its spec, and where it does not.
- Write a spec with measurable metrics, simulate the full nonlinear model, compare candidates against every metric, and check the simulation first.
Put the project method in order.
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Checkpoints come before any reported result.
Checkpoints for the simulation
- Three results are known without simulating: the saturated speed of 3.58 rad/s; P control stopping $3.0/134 = 1.29°$ short; and 4.6% overshoot for a small unloaded step at $\zeta = 0.7$.
- The simulation reproduces all three, so its other numbers are worth reporting.
P-only shortfall under load: 3.0/134 rad in degrees (two decimals)?
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0.0224 rad.
Saturated speed 12 x 0.398 - 3.0 x 0.398 (two decimals, rad/s)?
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9 V of useful drive.
The spec table
- Spec: 90° within $\pm0.5°$ by 0.6 s, overshoot below 5%, under a 5 N·m load.
- P only stops 1.29° short: fails. PI without anti-windup overshoots to 159°: fails.
- PI with clamping is inside the band at $\mathbf{0.52}$ s with no overshoot: passes, also when sampled at 1 kHz.
Pass, fail or limit?
Classify each result in the project report.
Which candidate passes the spec?
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It reaches the band at 0.52 s without overshoot.
Limits, stated
- Without load, a small 5° step overshoots by 18%: the integral zero adds overshoot the load normally hides.
- Only 0.08 s of margin, and it depends on the supply: at 11 V the valve reaches the band at 0.58 s. Untested: encoder faults, sticking, and copper's +0.4%/K resistance.
Time margin of a 0.52 s result against a 0.6 s spec (two decimals, s)?
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0.6 minus 0.52.
A passing design needs no stated limits.
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The report must say where it stops being valid.
The move time by hand
- Slewing at the loaded rate limit, 90° takes $1.571/3.58 = \mathbf{0.44}$ s. The simulation's 0.52 s adds 0.08 s for the final, linear approach into the band.
- At 11 V the motor runs at $(11 - 3.04)/0.05 = 159$ rad/s, 3.18 at the valve: $0.49 + 0.08 = 0.57$ s against the simulated 0.58. A hand estimate this close is a second check on the simulation.
A 90° move at a rate limit of 3.58 rad/s: slewing time (two decimals, s)?
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$1.571/3.58 = 0.439$ s.
Worked example: a hot motor
- After a long duty the windings are 50 K warmer. Copper's +0.4%/K raises $R$ by 20%, to 1.44 Ω.
- Under the same load the motor turns at $(12 - 2.54 \times 1.44)/0.05 = 167$ rad/s, 3.34 at the valve: the slew takes 0.47 s and the move about $\mathbf{0.55}$ s. It still passes, with 0.05 s of margin. Name the case in the report: it is a worst case the spec did not list.
Winding resistance 1.2 Ω at +0.4%/K, 50 K warmer: new resistance (two decimals, Ω)?
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$1.2 \times (1 + 0.004 \times 50) = 1.44$ Ω.
With 1.44 Ω, 12 V and 2.54 A: valve speed through 50:1 (two decimals, rad/s)?
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$(12 - 2.54 \times 1.44)/0.05 = 166.9$ rad/s; $/50 = 3.34$.
Three project slips
- A simulation never checked against a known result. Its numbers cannot be trusted.
- "Meets spec" without the margin. State how close it is.
- Only the passing candidate reported. Show every one.
Recap
- Checkpoints first: 3.58 rad/s, 1.29° short, 4.6% overshoot.
- Only PI with anti-windup passes: 0.52 s, no overshoot.
- Margins and limits are part of the result.