手动重新实现Biquad IIR过滤器的vDSP_deq22

时间:2015-12-16 12:55:54

标签: filtering signal-processing accelerate-framework

我正在移植一个过滤器库,该过滤器库目前使用Apple特定的(加速)vDSP功能vDSP_deq22到Android(其中Accelerate不可用)。滤波器组是一组带通滤波器,每个滤波器滤波器返回其各自频带的RMS幅度。目前代码(ObjectiveC ++,改编自NVDSP)如下所示:

- (float) filterContiguousData: (float *)data numFrames:(UInt32)numFrames channel:(UInt32)channel {

    // Init float to store RMS volume
    float rmsVolume = 0.0f;

    // Provide buffer for processing
    float tInputBuffer[numFrames + 2];
    float tOutputBuffer[numFrames + 2];

    // Copy the two frames we stored into the start of the inputBuffer, filling the rest with the current buffer data 
    memcpy(tInputBuffer, gInputKeepBuffer[channel], 2 * sizeof(float));
    memcpy(tOutputBuffer, gOutputKeepBuffer[channel], 2 * sizeof(float));
    memcpy(&(tInputBuffer[2]), data, numFrames * sizeof(float));

    // Do the processing
    vDSP_deq22(tInputBuffer, 1, coefficients, tOutputBuffer, 1, numFrames);
    vDSP_rmsqv(tOutputBuffer, 1, &rmsVolume, numFrames);

    // Copy the last two data points of each array to be put at the start of the next buffer.
    memcpy(gInputKeepBuffer[channel], &(tInputBuffer[numFrames]), 2 * sizeof(float));
    memcpy(gOutputKeepBuffer[channel], &(tOutputBuffer[numFrames]), 2 * sizeof(float));

    return rmsVolume;
}

here所示,deq22通过递归函数对给定的输入向量实现双二阶滤波器。这是文档中函数的描述: vDSP_deq22

  • A =:单精度实输入向量
  • IA =:大步为A.
  • B =:5个单精度输入(滤波器系数),步幅为1.
  • C =:单精度实数输出向量。
  • IC =:跨越C.
  • N =:要生成的新输出元素的数量。

这是我到目前为止(它在Swift中,就像我已在Android上运行的其余代码库一样):

// N is fixed on init to be the same size as buffer.count, below
// 'input' and 'output' are initialised with (N+2) length and filled with 0s

func getFilteredRMSMagnitudeFromBuffer(var buffer: [Float]) -> Float {
    let inputStride = 1 // hardcoded for now
    let outputStride = 1

    input[0] = input[N]
    input[1] = input[N+1]
    output[0] = output[N]
    output[1] = output[N+1]

    // copy the current buffer into input
    input[2 ... N+1] = buffer[0 ..< N]

    // Not sure if this is neccessary, just here to duplicate NVDSP behaviour:
    output[2 ... N+1] = [Float](count: N, repeatedValue: 0)[0 ..< N]

    // Again duplicating NVDSP behaviour, can probably just start at 0:
    var sumOfSquares = (input[0] * input[0]) + (input[1] * input[1])

    for n in (2 ... N+1) {
        let sumG = (0...2).reduce(Float(0)) { total, p in
            return total + input[(n - p) * inputStride] * coefficients[p]
        }

        let sumH = (3...4).reduce(Float(0)) { total, p in
            return total + output[(n - p + 2) * outputStride] * coefficients[p]
        }

        let filteredFrame = sumG - sumH
        output[n] = filteredFrame
        sumOfSquares = filteredFrame * filteredFrame
    }

    let meanSquare = sumOfSquares / Float(N + 2) // we added 2 values by hand, before the loop
    let rootMeanSquare = sqrt(meanSquare)
    return rootMeanSquare
}

滤波器为deq22提供了不同的幅度输出,并且似乎在其中具有循环波动圆形“噪声”(具有恒定的输入音调,该频率的幅度上下波动)。

我已经检查过以确保每个实现之间的系数数组是相同的。每个滤波器实际上似乎“工作”,因为它拾取正确的频率(并且只有那个频率),它只是这个泵浦,并且RMS幅度输出比vDSP更安静,通常是数量级:

   Naive    |    vDSP
3.24305e-06   0.000108608
1.57104e-06   5.53645e-05
1.96445e-06   4.33506e-05
2.05422e-06   2.09781e-05
1.44778e-06   1.8729e-05
4.28997e-07   2.72648e-05

任何人都可以看到我的逻辑问题吗?

编辑:这是结果的gif视频,具有恒定的440Hz音调。各种绿色条是各个滤带。第3个频段(此处显示)是调谐到440Hz的频段。

Pumping

NVDSP版本只是按预期显示与输入音量成比例的恒定(非波动)幅度读数。

1 个答案:

答案 0 :(得分:4)

好的,行sumOfSquares = filteredFrame * filteredFrame应该是+=,而不是作业。所以只计算了最后一帧,解释了很多;)

如果你想在Swift中进行一些双二阶过滤,请随意使用它。麻省理工学院许可证就像之前的NVDSP一样。