c++ - 带通 FIR 滤波器

Question

我需要制作一个简单的带通音频滤波器。现在我使用了这个简单的 C++ 类:http://www.cardinalpeak.com/blog/a-c-class-to-implement-low-pass-high-pass-and-band-pass-filters

效果很好，可以切断所需的 strip 。但是，当我尝试以较小的步幅更改上限或下限时，在某些限制值上我听到了错误的结果 - 频率衰减或偏移(与当前限制不对应)的声音。

计算脉冲响应的函数:

void Filter::designBPF()
{
    int n;
    float mm;

    for(n = 0; n < m_num_taps; n++){
        mm = n - (m_num_taps - 1.0) / 2.0;
        if( mm == 0.0 ) m_taps[n] = (m_phi - m_lambda) / M_PI;
        else m_taps[n] = (   sin( mm * m_phi ) -
                             sin( mm * m_lambda )   ) / (mm * M_PI);
    }

    return;
}

在哪里

m_lambda = M_PI * Fl / (Fs/2);
m_phi = M_PI * Fu / (Fs/2);

Fs - 采样率 (44.100)Fl - 下限傅-上限

以及简单的过滤功能:

float Filter::do_sample(float data_sample)
{
    int i;
    float result;

    if( m_error_flag != 0 ) return(0);

    for(i = m_num_taps - 1; i >= 1; i--){
        m_sr[i] = m_sr[i-1];
    }   
    m_sr[0] = data_sample;

    result = 0;
    for(i = 0; i < m_num_taps; i++) result += m_sr[i] * m_taps[i];

    return result;
}

我是否需要使用任何窗口函数(Blackman 等)？如果是，我该怎么做？我试图将我的脉冲响应乘以布莱克曼窗口:

m_taps[n] *= 0.42 - 0.5 * cos(2.0 * M_PI * n / double(N - 1)) +
                0.08 * cos(4.0 * M_PI * n / double(N - 1));

但是结果是错误的。我需要标准化拍子吗？

Answer 1

我找到了一个不错的 FIR 滤波器免费实现: http://www.iowahills.com/A7ExampleCodePage.html

...This Windowed FIR Filter C Code has two parts, the first is the calculation of the impulse response for a rectangular window (low pass, high pass, band pass, or notch). Then a window (Kaiser, Hanning, etc) is applied to the impulse response. There are several windows to choose from...