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我现在正在工作的项目正在调整到 PMR 频率并将音频保存到 WAV 文件，但是当我从 SDR 设备接收 IQ 样本并解调到 AM 时 - 我只听到噪音。

编辑，这是写入 WAV 文件的 RAW IQ： https ://voca.ro/iSRELps4JBg

对于解调，我使用了这种方法：

https://github.com/simonyiszk/minidemod/blob/master/minidemod-wfm-atan.c

然而声音似乎是噪音......

  while(get_samples(buffer, buffer_size) == 1)
    {


        // Demodulate to AM
        uint8_t uiAMBuff[buffer_size];

        double i1=0, i2, q1=0, q2;
        short s; // 2 bytes
        double sum=0;
        double dbAmp=0;
        for (int i = 0; i < buffer_size; i+=2)
        {
            // I / Q

            i2=((unsigned char)buffer[i]-127);
            q2=((unsigned char)buffer[i+1]-127);


            double phi1=get_phase(i1,q1);
            double phi2=get_phase(i2,q2);

            double dphi=phi2-phi1;
            if(dphi<-PI) dphi+=2*PI; //dphi below -180°? Let's interpret it in the other direction!
            if(dphi>PI) dphi-=2*PI;  //dphi above 180°? Let's interpret it in the other direction!

            // Now let's use signed 16 bit output rather than 8 bit unsigned.
            s=((SHRT_MAX-1)/PI)*dphi; //Okay so let's get the actual sample. dphi should be between -PI and +PI.

            // Store AM in the buffer that gets written to a WAV file
            uiAMBuff[i]=s;
            uiAMBuff[i+1]=s>>8;

            i1=i2;
            q1=q2;
            // This is to calculate DB
            double y = s/32768.0;
            sum += y * y; 

        }

请问如何正确解调IQ到AM？

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2 回答 2

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您的代码用于 FM 解调器（频率作为相位的导数，或离散情况下的相位增量）。对于 AM 解调，您只需要获取 IQ 分量的幅度 (abs()) 和低通滤波器。

于 2020-04-08T23:11:35.980 回答
1

我想自己发布一个答案。

这是我如何将 IQ (RTLSDR) 解调为 AM，从 SDR 中采样。

uint8_t uiAMBuff[buffer_size];

float sum=0;

for (int i = 0; i < buffer_size; i+=2)
{

    // Get 0 - 255
    int _i=buffer[i];
    int _q=buffer[i+1];

    // This will range : 0 - 32768 (short) 2 bytes
    short amplitude = sqrt((_i*_i)+(_q*_q));
    if(amplitude > 32768) amplitude = 32768;

    // Store in separate buffer (serialize) 2 bytes
    uiAMBuff[i]=amplitude>>0;
    uiAMBuff[i+1]=amplitude>>8;

    // Calculate AM Amplitude which is always 0..1 thats why division by 32768.0
    sum += amplitude / 32768.0;

}
于 2020-04-10T23:23:33.537 回答