Matlab firpm 对于大型 AFR 数据阵列失败

发布于 2024-12-10 09:02:10 字数 978 浏览 2 评论 0 原文

这是一个快速&尝试创建高精度均衡器的肮脏代码：

bandPoints = 355;
for n = 1:bandPoints
         x = (n / (bandPoints + 2));
         f = (x*x)*(22000-20)+20; % 20...22000
         freqs(n) = f;
         niqfreqs(n) = f/22050.0;
         amps(n) = 0;        
end

amps(bandPoints+1) = 0; % firpm needs even numbers
niqfreqs(bandPoints+1) = 1; % firpm needs even numbers

% set some point to have a high amplitude

amps(200) = 1;
fircfs = firpm(101,niqfreqs,amps);

[h,w] = freqz(fircfs,1,512);
plot(w/pi,abs(h));
legend('firpm Design')

但它给了我

Warning:
  *** FAILURE TO CONVERGE ***
  Probable cause is machine rounding error.

所有 FIR 系数均为 0。

如果我将 n 参数从 101 降低到 91，firpm 工作时没有错误，但频率响应远未达到预期。考虑到我想计算硬件DSP FIR模块的FIR系数，该模块最多支持12288个抽头，我怎样才能让Matlab计算所需的系数？ firpm 是否能够执行此操作，或者我是否需要在 Matlab 以及稍后的应用程序 C++ 代码中使用另一种方法（逆 FFT）？

原文

Here is a quick & dirty code for trying to create a high precision equalizer:

bandPoints = 355;
for n = 1:bandPoints
         x = (n / (bandPoints + 2));
         f = (x*x)*(22000-20)+20; % 20...22000
         freqs(n) = f;
         niqfreqs(n) = f/22050.0;
         amps(n) = 0;        
end

amps(bandPoints+1) = 0; % firpm needs even numbers
niqfreqs(bandPoints+1) = 1; % firpm needs even numbers

% set some point to have a high amplitude

amps(200) = 1;
fircfs = firpm(101,niqfreqs,amps);

[h,w] = freqz(fircfs,1,512);
plot(w/pi,abs(h));
legend('firpm Design')

but it gives me

Warning:
  *** FAILURE TO CONVERGE ***
  Probable cause is machine rounding error.

and all FIR coefficients are 0.

If I lower the n parameter from 101 to 91, firpm works without errors, but the frequency response is far from desired. And taking into account, that I want to calculate FIR coefficients for a hardware DSP FIR module, which supports up to 12288 taps, how can I make Matlab calculate the needed coefficients? Is firpm capable of doing this or do I need to use another approach (inverse FFT) in both Matlab and later in my application C++ code?