处理线性上采样音频阵列

Question

我有一个实时信号，一个样品一个样品，我需要对它进行4倍的重采样。我有musicdsp.org的这门课：

#ifndef TD_INTERPOLATOR_H_INCLUDED
#define TD_INTERPOLATOR_H_INCLUDED

/************************************************************************
*    Linear interpolator class                                            *
************************************************************************/

class interpolator_linear
{
public:
    interpolator_linear() {
        reset_hist();
    }

    // reset history
    void reset_hist() {
        d1 = 0.f;
    }


    // 4x interpolator
    // out: pointer to float[4]
    inline void process4x(float const in, float *out) {
        float y = in-d1;
        out[0] = d1 + 0.25f*y;    // interpolate
        out[1] = d1 + 0.5f*y;
        out[2] = d1 + 0.75f*y;
        out[3] = in;
        d1 = in; // store delay
    }


    }

private:
    float d1; // previous input
};

#endif // TD_INTERPOLATOR_H_INCLUDED

我想上面的说法是正确的。现在的问题是如何单独返回数组元素？

void TD_OSclip::subProcessClip4( int bufferOffset, int sampleFrames )
{

    float* in  = bufferOffset + pinInput.getBuffer();
    float* outputt = bufferOffset + pinOutput.getBuffer();



    for( int s = sampleFrames; s > 0; --s )
    {


        float input = *in;


//upsample 4x Linear --How should I call it here?

interpolator_linear::process4(input, what should be here??);

////I need all the seperate out arrays elements for the next stage

//do process
float clip = 0.5f;
float neg_clip = -0.5f;

float out0 = std::max( neg_clip, std::min( clip, out[0] ) );
float out1 = std::max( neg_clip, std::min( clip, out[1] ) );
float out2 = std::max( neg_clip, std::min( clip, out[2] ) );
float out3 = std::max( neg_clip, std::min( clip, out[3] ) );


//lowpass filter ommitted for briefness

float out0f = out0;
float out1f = out0;
float out2f = out0;
float out3f = out0;



//downsample

float output1 = ( out0f + out1f + out2f + out3f ) / 4.f;


                *outputt = output1;

            ++in;
                ++outputt;


    }

    }

我很清楚线性插值是不好的，但这是我见过的最简单的。我对编码还很陌生，所以在这个阶段，“易于实现”超过了性能。

问候安德鲁

Answer 1

您需要向process4提供一个缓冲区，它可以容纳4个浮点值。其次，您需要实例化interpolater_linear才能使用它。

interpolator_linear interp; // you'll want to make this a member of TD_OSclip.

float out[4];
for( int s = sampleFrames; s > 0; --s )
{
    float input = *in;

    interp.process4(input, out);
    ...
}