Apple的讲话背后的数学说明这里的例子

Question

我有一个关于苹果正在使用中它的数学问题在这里说话例子.

一点背景:我知道AVAudioRecorder和AVAudioPlayer返回的平均功率和峰值功率以dB为单位.我也理解为什么RMS功率以dB为单位,需要将其转换为放大器pow(10, (0.5 * avgPower)).

我的问题是:

Apple使用这个公式来创建它的"Meter Table"

MeterTable::MeterTable(float inMinDecibels, size_t inTableSize, float inRoot)
    : mMinDecibels(inMinDecibels),
    mDecibelResolution(mMinDecibels / (inTableSize - 1)), 
    mScaleFactor(1. / mDecibelResolution)
{
    if (inMinDecibels >= 0.)
    {
        printf("MeterTable inMinDecibels must be negative");
        return;
    }

    mTable = (float*)malloc(inTableSize*sizeof(float));

    double minAmp = DbToAmp(inMinDecibels);
    double ampRange = 1. - minAmp;
    double invAmpRange = 1. / ampRange;

    double rroot = 1. / inRoot;
    for (size_t i = 0; i < inTableSize; ++i) {
        double decibels = i * mDecibelResolution;
        double amp = DbToAmp(decibels);
        double adjAmp = (amp - minAmp) * invAmpRange;
        mTable[i] = pow(adjAmp, rroot);
    }
}

什么是所有计算 - 或者更确切地说,这些步骤的作用是什么？我认为mDecibelResolution并且mScaleFactor用于绘制超过400个值的80dB范围(除非我弄错了).然而,有什么的意义inRoot,ampRange,invAmpRange和adjAmp？另外,为什么仪表表中的第i个条目" mTable[i] = pow(adjAmp, rroot);"？

任何帮助深表感谢!:)

在此先感谢并欢呼!

Answer 1

自从我问过这个问题已经过去了一个月,感谢Geebs,感谢您的回复!:)

所以,这与我一直在研究的项目有关,并且基于此的功能在提出问题后约2天实施.很明显,我已经发布了一个结束回复(对不起).我也在1月7日发表评论,但是回过头来看,似乎我对var名称感到困惑.> _ <.以为我会给这个问题提供一个完整的逐行答案(带图片).:)

所以,这里是:

//mDecibelResolution is the "weight" factor of each of the values in the meterTable.
//Here, the table is of size 400, and we're looking at values 0 to 399.
//Thus, the "weight" factor of each value is minValue / 399.


MeterTable::MeterTable(float inMinDecibels, size_t inTableSize, float inRoot)
    : mMinDecibels(inMinDecibels),
    mDecibelResolution(mMinDecibels / (inTableSize - 1)), 
    mScaleFactor(1. / mDecibelResolution)
{
    if (inMinDecibels >= 0.)
    {
        printf("MeterTable inMinDecibels must be negative");
        return;
    }

    //Allocate a table to store the 400 values
    mTable = (float*)malloc(inTableSize*sizeof(float));

    //Remember, "dB" is a logarithmic scale.
    //If we have a range of -160dB to 0dB, -80dB is NOT 50% power!!!
    //We need to convert it to a linear scale. Thus, we do pow(10, (0.05 * dbValue)), as stated in my question.

    double minAmp = DbToAmp(inMinDecibels);

    //For the next couple of steps, you need to know linear interpolation.
    //Again, remember that all calculations are on a LINEAR scale.
    //Attached is an image of the basic linear interpolation formula, and some simple equation solving.

    //As per the image, and the following line, (y1 - y0) is the ampRange - 
    //where y1 = maxAmp and y0 = minAmp.
    //In this case, maxAmp = 1amp, as our maxDB is 0dB - FYI: 0dB = 1amp.
    //Thus, ampRange = (maxAmp - minAmp) = 1. - minAmp
    double ampRange = 1. - minAmp;

    //As you can see, invAmpRange is the extreme right hand side fraction on our image's "Step 3"
    double invAmpRange = 1. / ampRange;

    //Now, if we were looking for different values of x0, x1, y0 or y1, simply substitute it in that equation and you're good to go. :)
    //The only reason we were able to get rid of x0 was because our minInterpolatedValue was 0.

    //I'll come to this later.
    double rroot = 1. / inRoot;

    for (size_t i = 0; i < inTableSize; ++i) {
        //Thus, for each entry in the table, multiply that entry with it's "weight" factor.
        double decibels = i * mDecibelResolution;

        //Convert the "weighted" value to amplitude using pow(10, (0.05 * decibelValue));
        double amp = DbToAmp(decibels);

        //This is linear interpolation - based on our image, this is the same as "Step 3" of the image.
        double adjAmp = (amp - minAmp) * invAmpRange;

        //This is where inRoot and rroot come into picture.
        //Linear interpolation gives you a "straight line" between 2 end-points.
       //rroot =  0.5
       //If I raise a variable, say myValue by 0.5, it is essentially taking the square root of myValue.
       //So, instead of getting a "straight line" response, by storing the square root of the value,
       //we get a curved response that is similar to the one drawn in the image (note: not to scale).
        mTable[i] = pow(adjAmp, rroot);
    }
}

响应曲线图像:如您所见,"线性曲线"不完全是曲线.> _ <

希望这能以某种方式帮助社区.:)