我正在尝试用另一种语言重现一段冗长的MATLAB代码，该代码没有内置等效的无相位滤波器filtfilt()。我试图在简单的过滤（或卷积）操作方面重新构建函数，以便我可以轻松地重现它。
我知道这个过滤操作相当于前向过滤，然后是反向过滤，但我发现数据边缘的差异很小。具体做法是：
data = [1 1 1 2 2 3 5 7 1 1 1 1 1];
ker = [2 1 1];
a = filtfilt(ker,1,data)
b = fliplr( filter( ker, 1, fliplr( filter(ker, 1, data) ) ) )
% a =
%
% 16 18 21 29 39 57 66 68 42 28 16 16 16
%
% b =
%
% 11 16 21 29 39 57 66 68 42 28 16 12 8
我尝试在一个或两个过滤操作之前，在一端或两端用零填充数据。我想我可能会遗漏一些明显的东西，但无法发现它。
答案 0 :(得分：15)
这是我对Matlab的filtfilt函数的个人实现。我用FIR和IIR滤波器系数测试了它，这个实现在MATLAB实现时给出了相同的结果。我使用Eigen库来计算滤波器的初始条件，就像在MATLAB中完成一样，但是如果你愿意实现矩阵乘法和逆运算，你就可以摆脱它，代码只依赖于STL {{ 1}}和<vector>标头。
<algorithm>要测试它，您可以使用以下代码：
#include <vector>
#include <exception>
#include <algorithm>
#include <Eigen/Dense>
typedef std::vector<int> vectori;
typedef std::vector<double> vectord;
void add_index_range(vectori &indices, int beg, int end, int inc = 1)
{
for (int i = beg; i <= end; i += inc)
{
indices.push_back(i);
}
}
void add_index_const(vectori &indices, int value, size_t numel)
{
while (numel--)
{
indices.push_back(value);
}
}
void append_vector(vectord &vec, const vectord &tail)
{
vec.insert(vec.end(), tail.begin(), tail.end());
}
vectord subvector_reverse(const vectord &vec, int idx_end, int idx_start)
{
vectord result(&vec[idx_start], &vec[idx_end+1]);
std::reverse(result.begin(), result.end());
return result;
}
inline int max_val(const vectori& vec)
{
return std::max_element(vec.begin(), vec.end())[0];
}
void filter(vectord B, vectord A, const vectord &X, vectord &Y, vectord &Zi)
{
if (A.empty())
{
throw std::domain_error("The feedback filter coefficients are empty.");
}
if (std::all_of(A.begin(), A.end(), [](double coef){ return coef == 0; }))
{
throw std::domain_error("At least one of the feedback filter coefficients has to be non-zero.");
}
if (A[0] == 0)
{
throw std::domain_error("First feedback coefficient has to be non-zero.");
}
// Normalize feedback coefficients if a[0] != 1;
auto a0 = A[0];
if (a0 != 1.0)
{
std::transform(A.begin(), A.end(), A.begin(), [a0](double v) { return v / a0; });
std::transform(B.begin(), B.end(), B.begin(), [a0](double v) { return v / a0; });
}
size_t input_size = X.size();
size_t filter_order = std::max(A.size(), B.size());
B.resize(filter_order, 0);
A.resize(filter_order, 0);
Zi.resize(filter_order, 0);
Y.resize(input_size);
const double *x = &X[0];
const double *b = &B[0];
const double *a = &A[0];
double *z = &Zi[0];
double *y = &Y[0];
for (size_t i = 0; i < input_size; ++i)
{
size_t order = filter_order - 1;
while (order)
{
if (i >= order)
{
z[order - 1] = b[order] * x[i - order] - a[order] * y[i - order] + z[order];
}
--order;
}
y[i] = b[0] * x[i] + z[0];
}
Zi.resize(filter_order - 1);
}
void filtfilt(vectord B, vectord A, const vectord &X, vectord &Y)
{
using namespace Eigen;
int len = X.size(); // length of input
int na = A.size();
int nb = B.size();
int nfilt = (nb > na) ? nb : na;
int nfact = 3 * (nfilt - 1); // length of edge transients
if (len <= nfact)
{
throw std::domain_error("Input data too short! Data must have length more than 3 times filter order.");
}
// set up filter's initial conditions to remove DC offset problems at the
// beginning and end of the sequence
B.resize(nfilt, 0);
A.resize(nfilt, 0);
vectori rows, cols;
//rows = [1:nfilt-1 2:nfilt-1 1:nfilt-2];
add_index_range(rows, 0, nfilt - 2);
if (nfilt > 2)
{
add_index_range(rows, 1, nfilt - 2);
add_index_range(rows, 0, nfilt - 3);
}
//cols = [ones(1,nfilt-1) 2:nfilt-1 2:nfilt-1];
add_index_const(cols, 0, nfilt - 1);
if (nfilt > 2)
{
add_index_range(cols, 1, nfilt - 2);
add_index_range(cols, 1, nfilt - 2);
}
// data = [1+a(2) a(3:nfilt) ones(1,nfilt-2) -ones(1,nfilt-2)];
auto klen = rows.size();
vectord data;
data.resize(klen);
data[0] = 1 + A[1]; int j = 1;
if (nfilt > 2)
{
for (int i = 2; i < nfilt; i++)
data[j++] = A[i];
for (int i = 0; i < nfilt - 2; i++)
data[j++] = 1.0;
for (int i = 0; i < nfilt - 2; i++)
data[j++] = -1.0;
}
vectord leftpad = subvector_reverse(X, nfact, 1);
double _2x0 = 2 * X[0];
std::transform(leftpad.begin(), leftpad.end(), leftpad.begin(), [_2x0](double val) {return _2x0 - val; });
vectord rightpad = subvector_reverse(X, len - 2, len - nfact - 1);
double _2xl = 2 * X[len-1];
std::transform(rightpad.begin(), rightpad.end(), rightpad.begin(), [_2xl](double val) {return _2xl - val; });
double y0;
vectord signal1, signal2, zi;
signal1.reserve(leftpad.size() + X.size() + rightpad.size());
append_vector(signal1, leftpad);
append_vector(signal1, X);
append_vector(signal1, rightpad);
// Calculate initial conditions
MatrixXd sp = MatrixXd::Zero(max_val(rows) + 1, max_val(cols) + 1);
for (size_t k = 0; k < klen; ++k)
{
sp(rows[k], cols[k]) = data[k];
}
auto bb = VectorXd::Map(B.data(), B.size());
auto aa = VectorXd::Map(A.data(), A.size());
MatrixXd zzi = (sp.inverse() * (bb.segment(1, nfilt - 1) - (bb(0) * aa.segment(1, nfilt - 1))));
zi.resize(zzi.size());
// Do the forward and backward filtering
y0 = signal1[0];
std::transform(zzi.data(), zzi.data() + zzi.size(), zi.begin(), [y0](double val){ return val*y0; });
filter(B, A, signal1, signal2, zi);
std::reverse(signal2.begin(), signal2.end());
y0 = signal2[0];
std::transform(zzi.data(), zzi.data() + zzi.size(), zi.begin(), [y0](double val){ return val*y0; });
filter(B, A, signal2, signal1, zi);
Y = subvector_reverse(signal1, signal1.size() - nfact - 1, nfact);
}
答案 1 :(得分：6)
filtfilt算法匹配过滤器上的初始条件，以最小化开始和结束瞬变（来自doc filtfilt）。如果您输入edit filtfilt，则可以看到代码 - 有一个函数getCoeffsAndInitialConditions(b,a,Npts)会显示详细信息。