iOS FFT绘制光谱 [英] iOS FFT Draw spectrum

问题描述

我已经阅读了这些问题：

I've read these question:

使用Apple FFT和加速框架

如何使用Accelerate框架进行FFT时设置缓冲区？

iOS FFT Accerelate.framework在播放期间绘制频谱

它们都描述了如何使用加速框架设置fft。在他们的帮助下，我能够设置fft并获得一个基本的频谱分析仪。现在，我正在显示我从fft获得的所有值。但是，我只想显示10-15或一个可变数量的条形图来确定某些频率。就像iTunes或WinAmp Level Meter一样。
1.我是否需要从一系列频率中平均幅度值？或者他们只是向您展示特定频率条的幅度？
2.另外，我是否需要将我的幅度值转换为db？
3.如何将数据映射到特定范围。我是否会根据声音bitdepth的最大db范围进行映射？获取bin的最大值将导致跳转最大映射值。

They all describe how to setup fft with the accelerate framework. With their help I was able to setup fft and get a basic spectrum analyzer. Right now, I'm displaying all values I got from the fft. However, I only want to show 10-15, or a variable number, of bars respreseting certain frequencies. Just like the iTunes or WinAmp Level Meter. 1. Do I need to average magnitude values from a range of frequencies? Or do they just show you a magnitude for the specific frequency bar? 2. Also, do I need to convert my magnitude values to db? 3. How do I map my data to a certain range. Do I map against the max db range for my sounds bitdepth? Getting the max Value for a bin will lead to jumping max mapping values.

我的RenderCallback：

My RenderCallback:

static OSStatus PlaybackCallback(void *inRefCon,
                                 AudioUnitRenderActionFlags *ioActionFlags,
                                 const AudioTimeStamp *inTimeStamp,
                                 UInt32 inBusNumber,
                                 UInt32 inNumberFrames,
                                 AudioBufferList *ioData)
{
    UInt32 maxSamples = kAudioBufferNumFrames;

    UInt32 log2n = log2f(maxSamples); //bins
    UInt32 n = 1 << log2n;

    UInt32 stride = 1;
    UInt32 nOver2 = n/2;

    COMPLEX_SPLIT   A;
    float          *originalReal, *obtainedReal, *frequencyArray, *window, *in_real;

    in_real = (float *) malloc(maxSamples * sizeof(float));

    A.realp = (float *) malloc(nOver2 * sizeof(float));
    A.imagp = (float *) malloc(nOver2 * sizeof(float));
    memset(A.imagp, 0, nOver2 * sizeof(float));

    obtainedReal = (float *) malloc(n * sizeof(float));
    originalReal = (float *) malloc(n * sizeof(float));
    frequencyArray = (float *) malloc(n * sizeof(float));

    //-- window

    UInt32 windowSize = maxSamples;
    window = (float *) malloc(windowSize * sizeof(float));

    memset(window, 0, windowSize * sizeof(float));
    //    vDSP_hann_window(window, windowSize, vDSP_HANN_DENORM);

    vDSP_blkman_window(window, windowSize, 0);

    vDSP_vmul(ioBuffer, 1, window, 1, in_real, 1, maxSamples);

    //-- window

    vDSP_ctoz((COMPLEX*)in_real, 2, &A, 1, maxSamples/2);

    vDSP_fft_zrip(fftSetup, &A, stride, log2n, FFT_FORWARD);
    vDSP_fft_zrip(fftSetup, &A, stride, log2n, FFT_INVERSE);

    float scale = (float) 1.0 / (2 * n);

    vDSP_vsmul(A.realp, 1, &scale, A.realp, 1, nOver2);
    vDSP_vsmul(A.imagp, 1, &scale, A.imagp, 1, nOver2);

    vDSP_ztoc(&A, 1, (COMPLEX *) obtainedReal, 2, nOver2);
    vDSP_zvmags(&A, 1, obtainedReal, 1, nOver2);

    Float32 one = 1;
    vDSP_vdbcon(obtainedReal, 1, &one, obtainedReal, 1, nOver2, 0);

    for (int i = 0; i < nOver2; i++) {
        frequencyArray[i] = obtainedReal[i];
    }


    // Extract the maximum value
    double fftMax = 0.0;
    vDSP_maxmgvD((double *)obtainedReal, 1, &fftMax, nOver2);

    float max = sqrt(fftMax);
}

播放一些音乐，我得到的值从-96db到0db。
绘制一个点：

Playing some music, I get values from -96db to 0db. Plotting a point at:

CGPointMake(i, kMaxSpectrumHeight * (1 - frequencyArray[i]/-96.));

给出一条相当圆的曲线：

is giving my a rather rounded curve:

plot1

如果我不要转换为db我可以通过将我的数组值乘以10000得到很好的峰值。

If I don't convert to db I can plot by multiplying my array value by 10000 and get nice peaks.

plot2

我做错了什么？如何显示可变数量的柱？

Am I doing something totally wrong? And how do I get to showing a variable number of bars?

推荐答案

1. 我是否需要从一系列频率中平均幅度值？或者它们只显示特定频率条的幅度？

是的，你肯定需要平均你定义的波段的平均值。只显示一个FFT bin是疯狂的。

Yes, you definitely need to average values across the bands you've defined. Showing just one FFT bin is madness.

1. 此外，我是否需要转换我的幅度值到db？

是：dB是对数刻度。并非巧合的是，人类听觉也可以（大致）以对数尺度工作。因此，如果你在绘制它们之前采用值的log2（），那么这些值看起来会更自然。

Yes: dB is a log scale. Not coincidentally, human hearing also works (roughly) on a log scale. The values will therefore look more natural to humans if you take log2() of the values before plotting them.

1. 如何将数据映射到特定范围。我是否会根据声音bitdepth的最大db范围进行映射？获取bin的最大值将是
   导致跳转最大映射值。

我发现最简单的事情（概念上至少）是将你的值从任何格式转换为 0..1 ，即'normalized and scaled'浮点值。然后，从那里你可以转换为你需要绘制的东西。例如

I find the easiest thing to do (conceptually at least) is to convert your values from whatever format into a 0..1, i.e. 'normalised and scaled' float value. Then from there you can convert if necessary to something you need to plot. For example

SInt16 rawValue = fft[0]; // let's say this comes back as 12990

float scaledValue = rawValue/32767.; // This is MAX_INT for 16-bit;
        // dividing we get .396435438 which is much easier for most people
        // to see conceptually as 39% of our max possible value

float displayValue = log2(scaledValue);

my_fft[0] = displayValue;

这篇关于iOS FFT绘制光谱的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！