DSP 库 - RFFT - 奇怪的结果 [英] DSP libraries - RFFT - strange results

问题描述

最近我一直在尝试在我的 STM32F4-Discovery 评估板上进行 FFT 计算，然后将其发送到 PC.我已经调查了我的问题 - 我认为制造商提供的 FFT 函数有问题.

Recently I've been trying to do FFT calculations on my STM32F4-Discovery evaluation board then send it to PC. I have looked into my problem - I think that I'm doing something wrong with FFT functions provided by manufacturer.

我正在使用 CMSIS-DSP 库.目前我一直在用代码生成样本(如果工作正常，我将通过麦克风进行采样).

I'm using CMSIS-DSP libraries. For now I've have been generating samples with code (if that works correct I'll do sampling by microphone).

我正在使用 arm_rfft_fast_f32 因为我的数据将来会是浮点数，但是我在输出数组中得到的结果是疯狂的(我认为) - 我频率低于 0.

I'm using arm_rfft_fast_f32 as my data are going to be floats in the future, but results I get in my output array are insane (I think) - I'm getting frequencies below 0.

number_of_samples = 512; (l_probek in code)
dt = 1/freq/number_of_samples

这是我的代码

float32_t buffer_input[l_probek];
uint16_t i;
uint8_t mode;
float32_t dt;
float32_t freq;
bool DoFlag = false;
bool UBFlag = false;
uint32_t rozmiar = 4*l_probek;

union
{
    float32_t f[l_probek];
    uint8_t b[4*l_probek];
}data_out;


union
{
    float32_t f[l_probek];
    uint8_t b[4*l_probek];
}data_mag;

union
{
    float32_t f;
    uint8_t b[4];
}czest_rozdz;


/* Pointers ------------------------------------------------------------------*/
arm_rfft_fast_instance_f32 S;
arm_cfft_radix4_instance_f32 S_CFFT;
uint16_t output;
/* ---------------------------------------------------------------------------*/
int main(void)
{
    freq = 5000;
    dt = 0.000000390625;


    _GPIO();
    _LED();
    _NVIC();    
    _EXTI(0);

    arm_rfft_fast_init_f32(&S, l_probek);
    GPIO_SetBits(GPIOD, LED_Green);

    mode = 2;


    //----------------- Infinite loop
  while (1)
    {
        if(true)//(UBFlag == true)

                    for(i=0; i<l_probek; ++i)
                    {
                        buffer_input[i] = (float32_t) 15*sin(2*PI*freq*i*dt);
                    }

            //Obliczanie FFT
            arm_rfft_fast_f32(&S, buffer_input, data_out.f, 0);
            //Obliczanie modulow
            arm_cmplx_mag_f32(data_out.f, data_mag.f, l_probek);

            USART_putdata(USART1, data_out.b, data_mag.b, rozmiar);
            //USART_putdata(USART1, czest_rozdz.b, data_mag.b, rozmiar);
            GPIO_ToggleBits(GPIOD, LED_Orange);
            //mode++;
            //UBFlag = false;

        }

    }
}

推荐答案

我正在使用 arm_rfft_fast_f32 因为我的数据将来会是浮点数，但是我在输出数组中得到的结果是疯狂的(我认为) - 我频率低于 0.

I'm using arm_rfft_fast_f32 as my data are going to be floats in the future, but results I get in my output array are insane (I think) - I'm getting frequencies below 0.

arm_rfft_fast_f32 函数不返回频率，而是返回使用 快速傅立叶变换 (FFT).因此，这些系数为负是完全合理的.更具体地说，幅度为 15 的单周期 sin 测试音输入的预期系数为:

The arm_rfft_fast_f32 function does not return frequencies, but rather complex-valued coefficients computed using the Fast Fourier Transform (FFT). It is thus perfectly reasonable for those coefficients to be negative. More specifically, the expected coefficients for your single-cycle sin test tone input with an amplitude of 15 would be:

0.0,     0.0; // special case packing real-valued X[0] and X[N/2]
0.0, -3840.0; // X[1]
0.0,     0.0; // X[2]
0.0,     0.0; // X[3]
...
0.0,     0.0; // X[255]

请注意，如文档中所示 前两个输出对应于纯实数系数 X[0] 和 X[N/2](您在后续调用中应特别注意这种特殊情况arm_cmplx_mag_f32；见下面最后一点).

Note that as indicated in the documentation the first two outputs correspond to the purely real coefficients X[0] and X[N/2] (you should be particularly careful about this special case in your subsequent call to arm_cmplx_mag_f32; see last point below).

每个频率分量的频率由 k*fs/N 给出，其中 N 是样本数(在您的情况下 l_probek) 和 fs = 1/dt 是采样率(在你的情况下 freq*l_probek):

The frequency of each of those frequency components are given by k*fs/N, where N is the number of samples (in your case l_probek) and fs = 1/dt is the sampling rate (in your case freq*l_probek):

X[0] -> 0*freq*l_probek/l_probek =              0
X[1] -> 1*freq*l_probek/l_probek =   freq =  5000
X[2] -> 2*freq*l_probek/l_probek = 2*freq = 10000
X[3] -> 3*freq*l_probek/l_probek = 2*freq = 15000
...

最后，由于前两个值的特殊打包，在计算N/2+1个星等时需要小心:

Finally, due to the special packing of the first two values, you need to be careful when computing the N/2+1 magnitudes:

// General case for the magnitudes
arm_cmplx_mag_f32(data_out.f+2, data_mag.f+1, l_probek/2 - 1);
// Handle special cases
data_mag.f[0]          = data_out.f[0];
data_mag.f[l_probek/2] = data_out.f[1];

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