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我正在尝试通过以下链接制作原型音频识别系统：http ://www.ifp.illinois.edu/~minhdo/teaching/speaker_recognition/ 。这很简单，所以几乎没有什么可担心的。但我的问题是梅尔频率功能。这是网站上提供的代码：

function m = melfb(p, n, fs)
% MELFB         Determine matrix for a mel-spaced filterbank
%
% Inputs:       p   number of filters in filterbank
%               n   length of fft
%               fs  sample rate in Hz
%
% Outputs:      x   a (sparse) matrix containing the filterbank amplitudes
%                   size(x) = [p, 1+floor(n/2)]
%
% Usage:        For example, to compute the mel-scale spectrum of a
%               colum-vector signal s, with length n and sample rate fs:
%
%               f = fft(s);
%               m = melfb(p, n, fs);
%               n2 = 1 + floor(n/2);
%               z = m * abs(f(1:n2)).^2;
%
%               z would contain p samples of the desired mel-scale spectrum
%
%               To plot filterbanks e.g.:
%
%               plot(linspace(0, (12500/2), 129), melfb(20, 256, 12500)'),
%               title('Mel-spaced filterbank'), xlabel('Frequency (Hz)');

f0 = 700 / fs;
fn2 = floor(n/2);

lr = log(1 + 0.5/f0) / (p+1);

% convert to fft bin numbers with 0 for DC term
bl = n * (f0 * (exp([0 1 p p+1] * lr) - 1));

b1 = floor(bl(1)) + 1;
b2 = ceil(bl(2));
b3 = floor(bl(3));
b4 = min(fn2, ceil(bl(4))) - 1;

pf = log(1 + (b1:b4)/n/f0) / lr;
fp = floor(pf);
pm = pf - fp;

r = [fp(b2:b4) 1+fp(1:b3)];
c = [b2:b4 1:b3] + 1;
v = 2 * [1-pm(b2:b4) pm(1:b3)];

m = sparse(r, c, v, p, 1+fn2);

但它给了我一个错误：

错误使用 * 内矩阵尺寸必须一致。

MFFC 中的错误（第 17 行）z = m * abs(f(1:n2)).^2;

当我在第 17 行之前包含这两行时：

size(m)
size(abs(f(1:n2)).^2)

它给了我：

ans =

    20    65


ans =

     1    65

那么我应该转置第二个矩阵吗？或者我应该将其解释为逐行乘法并修改代码？

编辑：这是主要功能（我只是运行 MFCC()）：

function result = MFFC()
[y Fs] = audioread('s1.wav');
% sound(y,Fs)

Frames = Frame_Blocking(y,128);
Windowed = Windowing(Frames);
spectrum = FFT_After_Windowing(Windowed);
%imagesc(mag2db(abs(spectrum)))

p = 20;
S = size(spectrum);
n = S(2);

f = spectrum;
m = melfb(p, n, Fs);
n2 = 1 + floor(n/2);
size(m)
size(abs(f(1:n2)).^2)
z = m * abs(f(1:n2)).^2;

result = z;

以下是辅助功能：

function f = Frame_Blocking(y,N)
% Parameters: M = 100, N = 256
% Default : M = 100; N = 256;
M = fix(N/3);

Frames = [];
first = 1; last = N;
len = length(y);
while last <= len
    Frames = [Frames; y(first:last)'];
    first = first + M;
    last  = last + M;
end;
if last < len
    first = first + M;
    Frames = [Frames; y(first : len)];
end
f = Frames;

function f = Windowing(Frames)
S = size(Frames);
N = S(2);
M = S(1);
Windowed = zeros(M,N);
nn = 1:N;
wn = 0.54 - 0.46*cos(2*pi/(N-1)*(nn-1));
for ii = 1:M
    Windowed(ii,:) = Frames(ii,:).*wn;
end;
f = Windowed;

function f = FFT_After_Windowing(Windowed)
spectrum = fft(Windowed);
f = spectrum;
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2 回答 2

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转s置或转置结果f（这只是一个约定问题）。

您正在使用的函数没有任何问题melfb，仅与您尝试运行的示例中的信号尺寸有关（在注释的第 14-17 行中）。

%  f = fft(s);
%  m = melfb(p, n, fs);
%  n2 = 1 + floor(n/2);
%  z = m * abs(f(1:n2)).^2;

该示例假设您使用的是"colum-vector signal s"。从您的傅立叶变换的大小f（通过fft尊重输入信号尺寸完成），您的输入信号s是一个行向量信号。

给出错误的部分是实际的滤波操作，需要将p x n2矩阵与n2 x 1列向量相乘（即，每个滤波器的响应逐点乘以输入信号的傅立叶）。由于您的输入s是1 x n，因此您的f意志1 x n和向量乘法的最终矩阵z将给出错误。

于 2016-01-04T14:49:57.307 回答
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感谢gevang的回答，我能够找出我的错误。这是我修改代码的方式：

function result = MFFC()
[y Fs] = audioread('s2.wav');
% sound(y,Fs)

Frames = Frame_Blocking(y,128);
Windowed = Windowing(Frames);
%spectrum = FFT_After_Windowing(Windowed');
%imagesc(mag2db(abs(spectrum)))

p = 20;
%S = size(spectrum);
%n = S(2);

%f = spectrum;

S1 = size(Windowed);

n  = S1(2);
n2 = 1 + floor(n/2);
%z  = zeros(S1(1),n2);
z = zeros(20,S1(1));
for ii=1: S1(1)

    s = (FFT_After_Windowing(Windowed(ii,:)'));

    f = fft(s);
    m = melfb(p,n,Fs);
    % n2 = 1 + floor(n/2);
    z(:,ii) = m * abs(f(1:n2)).^2;
end;


%f = FFT_After_Windowing(Windowed');
%S = size(f);
%n = S(2);
%size(f)
%m = melfb(p, n, Fs);
%n2 = 1 + floor(n/2);
%size(m)
%size(abs(f(1:n2)).^2)
%z = m * abs(f(1:n2)).^2;

result = z;

如您所见，我天真地假设该函数处理逐行矩阵，但实际上它处理的是列向量（也可能是逐列矩阵）。所以我遍历输入矩阵的每一列，然后组合结果。但我不认为这是高效的矢量化代码。此外，我仍然无法弄清楚如何对输入矩阵进行按列操作（窗口化 - 在窗口化步骤之后），而不是使用循环。

于 2016-01-04T16:55:30.400 回答