Goertzel算法，以获得相位θ [英] Goertzel algorithm to get the phase?

问题描述

我使用Goertzel算法来获得一定频率的幅度。 我想现在从它那里得到的阶段，我不知道怎么办。

能否有人解释，并告诉我怎么得到一定-F从这个code的阶段？

此外，我用它来16kHz的，与采样率44.1。有什么样的最小长度，我可以运行它？

 双AlgorithmGoertzel（int16_t *样品，诠释的采样率，双频率，INT LEN）
{
    双realW = 2.0 * COS（2.0 * M_PI *频率/采样率）;
    双imagW = 2.0 * SIN（2.0 * M_PI *频率/采样率）;
    双D1 = 0;
    双D2 = 0;
    双Y;
    的for（int i = 0; I＆LT; LEN;我++）{
        Y =（双）（签订短期）样本[I] + realW * D1  -  D2;
        D2 = D1;
        D1 = Y;
    }
    双RR = 0.5 * realW * D1-D2;
    双RI = 0.5 * imagW * D1-D2;

    返程（SQRT（PO​​W（RR，2）+ POW（RI，2）））/ LEN;
}
 

解决方案

我不认为算法由一个常数乘以序列的，而是由复杂的信号 EXP（N * I *二皮*频率/采样率）; 0℃= N＆LT; =长度，并获得了平均幅度（或动力的信号）

由于复杂的输出为R * EXP（ⅰTHETA）中，R给出在给定频率的功率和θ给出的相位。 （THETA == ATAN2（IMAG，真正的））

I am using Goertzel algorithm to get the amplitude of a certain frequency. I am trying now to get the phase from it, and I don't know how.

Can some one explain, and show me how to get the phase of a certain-f from this code?

Also, I am using it to 16khz, with sample rate 44.1. What's the smallest length of samples that I can run it on?

double AlgorithmGoertzel( int16_t *sample,int sampleRate, double Freq, int len )
{
    double realW = 2.0 * cos(2.0 * M_PI * Freq / sampleRate);
    double imagW = 2.0 * sin(2.0 * M_PI * Freq / sampleRate);
    double d1 = 0;
    double d2 = 0;
    double y;
    for (int i = 0; i < len; i++) {
        y=(double)(signed short)sample[i] +realW * d1 - d2;
        d2 = d1;
        d1 = y;
    }
    double rR = 0.5 * realW *d1-d2;
    double rI = 0.5 * imagW *d1-d2;

    return (sqrt(pow(rR, 2)+pow(rI,2)))/len;
}

解决方案

I don't think the algorithm consists of multiplying the sequence by a constant, but by the complex signal exp(n*i*2pi*freq/samplerate); 0<=n<=length, and getting the average magnitude (or power of the signal).

As the complex output is R*exp(i theta), R gives the power at the given frequency and theta gives the phase. (theta == atan2 ( imag, real))

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