buildWave10LutForMaxBaud()

Wave10Lut fl::buildWave10LutForMaxBaud	(	const ChipsetTimingConfig &	timing,
		u32	max_baud_rate )

Build a UART waveform LUT with a backend-specific baud ceiling.

Definition at line 221 of file uart_wave_encoder.cpp.hpp.

                                                                   {
    Wave10Lut result = {};
 
    // Evaluate both frame geometries (FastLED#3572 follow-up):
    //   P=5 -- wave10: 8 data bits, baud = 5/period (the classic shape)
    //   P=4 -- wave8-frame: 6 data bits, baud = 4/period (lower baud ->
    //         reaches faster chipsets under the 5 Mbps cap; different
    //         quantization grid -- e.g. SK6812 is exact at period/4)
    // Pick the feasible geometry with the smaller total quantization
    // error. P=5 wins ties and near-ties (50 ns hysteresis) so the
    // long-proven wave10 shapes stay stable for existing chipsets.
    const UartWaveFit fit5 = fitUartWave(timing, 5, max_baud_rate);
    const UartWaveFit fit4 = fitUartWave(timing, 4, max_baud_rate);
 
    u8 P = 0;
    UartWaveFit fit = {};
    constexpr u32 kHysteresisNs = 50;
    // A symbol further than this from nominal is one a receiver may not
    // classify at all. WS281x parts quote about +/-150 ns, and AutoResearch
    // decodes at +/-170; `fitUartWave` on its own admits half a pulse width,
    // which at a 2500 ns period is 250 ns -- wider than either, which is how a
    // geometry emitting an unclassifiable symbol got selected. See #4379.
    constexpr u32 kMaxSymbolErrorNs = 150;
    if (fit5.ok && fit4.ok) {
        const bool fit5_classifiable = fit5.max_err_ns <= kMaxSymbolErrorNs;
        const bool fit4_classifiable = fit4.max_err_ns <= kMaxSymbolErrorNs;
        if (fit5_classifiable != fit4_classifiable) {
            // Exactly one geometry keeps every symbol inside a receiver's
            // tolerance. Take it, whatever the totals say -- a smaller sum
            // bought by putting one symbol out of range is not a better fit.
            // WS2811 at 400 kHz is the case: P=5 nails T0H and misses T1H by
            // 200 ns, P=4 spends 125 ns on T0H to bring T1H within 50 ns.
            P = fit5_classifiable ? 5 : 4;
            fit = fit5_classifiable ? fit5 : fit4;
        } else if (fit4.total_err_ns + kHysteresisNs < fit5.total_err_ns) {
            P = 4;
            fit = fit4;
        } else {
            P = 5;
            fit = fit5;
        }
    } else if (fit5.ok) {
        P = 5;
        fit = fit5;
    } else if (fit4.ok) {
        P = 4;
        fit = fit4;
    } else {
        return result; // infeasible -- pulses_per_bit stays 0
    }
 
    result.pulses_per_bit = P;
    result.lut[0] = buildUartByte(fit.pulses_0, fit.pulses_0, P); // "00"
    result.lut[1] = buildUartByte(fit.pulses_0, fit.pulses_1, P); // "01"
    result.lut[2] = buildUartByte(fit.pulses_1, fit.pulses_0, P); // "10"
    result.lut[3] = buildUartByte(fit.pulses_1, fit.pulses_1, P); // "11"
 
    return result;
}

References FL_NO_EXCEPT, and P().

Referenced by buildWave10Lut(), and uartWireTiming().

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