用c ++进行准确采样 [英] accurate sampling in c++

问题描述

我想对每秒从gpio中获得的值进行4000次采样，目前我正在执行类似的操作:

I want to sample values I get from a gpio 4000 times per second, currently I do something like that:

std::vector<int> sample_a_chunk(unsigned int rate, unsigned int block_size_in_seconds) {
    std::vector<std::int> data;
    constexpr unsigned int times = rate * block_size_in_seconds;
    constexpr unsigned int delay = 1000000 / rate; // microseconds
    for (int j=0; j<times; j++) {
      data.emplace_back(/* read the value from the gpio */);
      std::this_thread::sleep_for(std::chrono::microseconds(delay));
    }
    return data;
}

根据参考sleep_for，仍然可以保证至少指定时间的等待时间.

yet according to the reference sleep_for is guaranteed to wait for at least the specified amount of time.

我如何让我的系统等待 exact 的时间，或者至少达到最佳精度?如何确定系统的时间分辨率?

How can I have my system wait for the exact amount of time, or at least achieve the best possible accuracy? How can I be sure of the time resolution of my system?

推荐答案

我认为您可能可以实现的最佳效果是使用绝对计时以避免漂移.

I think the best you can probably achieve is to use absolute timing so as to avoid drift.

类似这样的东西:

std::vector<int> sample_a_chunk(unsigned int rate,
    unsigned int block_size_in_seconds)
{
    using clock = std::chrono::steady_clock;

    std::vector<int> data;

    const auto times = rate * block_size_in_seconds;
    const auto delay = std::chrono::microseconds{1000000 / rate};

    auto next_sample = clock::now() + delay;

    for(int j = 0; j < times; j++)
    {
        data.emplace_back(/* read the value from the gpio */);

        std::this_thread::sleep_until(next_sample);

        next_sample += delay; // don't refer back to clock, stay absolute
    }
    return data;
}

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