Question

我有一组简单低通滤波器的幅度和相位数据。我可以使用Octave中的invfreqz成功地将滤波器拟合到复杂的频率响应中。但是，如果我希望以更高的采样率（例如4x）使用相同的数据来安装相同的滤波器呢？但是，在新的4x奈奎斯特频率之前，没有可用的幅度和频率数据。磁场和相位数据不能在较高频率下收集，因此必须通过其他方法或近似值进行加法。

将填充量和相位数据“入侵”到新的奈奎斯特频率的最简单方法是什么，以便invfreqz能够以新的采样率最有可能获得所收集的数据？

Answer 1

这是我前一段时间写的一个函数，它可用于改变低通/带通滤波器的采样率（我从不打扰将它扩展到高通）。一般来说，你应该选择滤波器的响应几乎没有作为“切断”频率的点（我犹豫地说'截止'，因为它不是正常意义上的截止）;在此频率之上，响应设置为零。

function [b, a, fNew, HNew, fOld, HOld, HCut] = ...
    shiftFilterRate(b, a, fs1, fs2, order, fCut)
% SHIFTFILTERRATE   Shift sampling rate of filter
%
%   [bNew, aNew] = shiftFilterRate(b, a, fs1, fs2, order, fCut) designs a
%   FIR filter of the given order to operate at sampling frequency fs2, in
%   Hz, whose frequency magnitude characteristics match those of the filter
%   whose coefficients are b, a, that operates at sampling rate fs1, in Hz.
%   The function will only try to match the filter's magnitude in the
%   region [0 fCut], in Hz.
%
%   [bNew, aNew, fNew, HNew, fOld, HOld, HCut] = shiftFilterRate(...)
%   returns additional parameters:
%       fNew:   The frequencies at which the designed filter's transfer
%               function was evaluated (with resolution of 1 Hz)
%       HNew:   The transfer function of the designed filter
%       fOld:   The frequencies at which the existing filter's transfer
%               function was evaluated (with a resolution of 1 Hz)
%       HOld:   The transfer function of the existing filter.
%       HCut:   The desired frequency response of the filter used as input
%               to the function fir2.
%
%   FIXME: Make this work for high pass filters.
%

if nargin < 5, fCut = inf; end
if nargin < 4, order = 50; end

%% Zero padding in frequency domain
res = 1; % Hz resolution
N1 = fs1 / res; % points at resolution before padding

% Original freq response
[H, f] = freqz(b, a, N1);
nanInd = find(isnan(H));
% Stabilise. NOTE: Will break if nanInd contains last elem or there are
% multiple NaNs in a row
H(nanInd) = H(nanInd + 1); 
f = f / pi;

% Normalise cutoff freq
fCutNorm = fCut / (fs1 / 2);

% Cut frequencies above fCut, we don't really need them and it makes the
% FIR filter nasty
HCut = H;
HCut(f > fCutNorm) = 0;

% Create new freq response
NNew = ceil(fs2 / fs1 * length(HCut));
fNew = linspace(0, 1, NNew)';
HNew = [HCut; zeros(NNew - N1, 1)];

% Design filter
b = fir2(order, fNew, abs(HNew));
a = 1;

HOld = H;
fOld = f;

if nargout > 3
    HNew = freqz(b, a, length(fNew));
end

请注意，在您的情况下，您可能需要将其设置为-20 dB re 1，因为这似乎是滤镜提供的最大衰减。在这一点上：它看起来不是一个非常好的过滤器......你有什么理由拥有来匹配这个响应吗？你可能更好的只是设计，例如，Butterworth具有相同的截止值（你肯定会在阻带中获得更多的衰减和更多的线性相位）。

Octave或Matlab中invfreqz的策略

1 个答案: