Question

我在信号处理方面遇到了一些问题。这个想法是分析一个声音信号，包含有人在老式模拟电话线上拨打电话号码的声音。函数 dtmf_84125P 将信号的文件名作为输入，读取它并返回包含所拨号码的字符串，例如

number = dtmf_84125P('dtmf_84125P.wav')

然后例如，number ='0504063452'。

我试图做的事情如下所示：

function [output_seq] = dtmf_84125P(input_file)
%dtmf_84125P Decodes DTMF-signal to numbers.
%   input_file: string with filename of DTMF-signal
%   output_seq: string with decoded numbers from the DTMF-signal
%
%   Mikael Eriksson
%   84125P
%   mikael.eriksson@aalto.fi

[x,fs] = wavread(input_file);
N = 205;    % N-point Fourier transform
L = 5; %Frame length for energy calculations
MINPEAK = 20;
f = linspace(0,fs/2,N);
f2 = linspace(0,fs/2,ceil(N/L));
soundlength = 0.07;     
breaklength = 0.04; 
%soundsc(x)
output_seq = [];
harm1 = [697,770,852,941];
harm2 = [1209,1336,1477];
%           1209 Hz     1336 Hz     1477 Hz
%   697 Hz  1           2           3       
%   770 Hz  4           5           6 
%   852 Hz  7           8           9
%   941 Hz              0    


%Test
% t = linspace(0,1,8000);
% fr1 = 500;
% fr2 = 900;
% x = 5*sin(2*pi*fr1*t)+2*sin(2*pi*fr2*t);
% soundsc(x) 

for i = 1:round((soundlength+breaklength)*fs):length(x)-round((soundlength+breaklength)*fs)

    %Choose subset of signal i time-domain, moving the window forward in
    %every loop. Perform Fourier-transform on this subsignal...
    num = '';
    s = x(i:i+round((soundlength+breaklength)*fs)-1);
    S = fft(s,N);
    S_abs = abs(S);        
    E = zeros(ceil(N/L),1);
    ind  = 0;

    %...and calculate its energy.
    for k = 1 : L : N-L
        ind = ind+1;
        E(ind) = sum(S_abs(k:k+L-1).^2)/L;
    end;

   %Plot the Fourier-transform and the energy of the signal in the current
   %window.
   plot(f,S_abs,f2,E,'ro')
   soundsc(s)
   %pause(0.2)

    % See if there are energy peaks at certain frequencies. First find all
    % peaks, then find the two different frequerencies, and finally the
    % number of this frequency-pair.
    [pks,locs] = findpeaks(E(2:end),'MINPEAKHEIGHT', min([MINPEAK, 0.99*max(E(2:end))]));
    freq = f2(locs);

    if length(pks) >= 2
        [amplitude_sorted, index] = sort(pks,'descend');
        freq = freq(index);

        if freq(1) < freq(2) 
            [tmp,ind] = min(abs(harm1-freq(1)));
            freq1 = harm1(ind);
            [tmp,ind] = min(abs(harm2-freq(2)));
            freq2 = harm2(ind);
        else
            [tmp,ind] = min(abs(harm1-freq(2)));
            freq1 = harm1(ind);
            [tmp,ind] = min(abs(harm2-freq(1)));
            freq2 = harm2(ind);
        end

        switch freq1
            case 697
                switch freq2
                    case 1209
                        num = '1';                
                    case 1336
                        num = '2';
                    case 1477
                        num = '3';
                end
            case 770
                switch freq2
                    case 1209
                        num = '4';                
                    case 1336
                        num = '5';
                    case 1477
                        num = '6';
                end
            case 852
                switch freq2
                    case 1209
                        num = '7';                
                    case 1336
                        num = '8';
                    case 1477
                        num = '9';
                end
        end
    elseif length(pks) == 1
        num = '0';
    end

    % If a number was dialed in this window of the signal and this was
    % detected, add it to the output sequence.
    if num ~= ''
        output_seq = [output_seq,num];
    end          

end

end

因此，我们的想法是分析小部分的信号，知道在0.07 + 0.04秒的窗口中永远不会有多个拨打的号码。所以我遍历原始信号x，并且每次信号的分析部分具有这个长度并命名为s。我将其傅里叶变换为S，并分析光谱不同点的能量。如果存在显着能量的峰值，我会尝试根据代码开头给出的表格将这些峰值分类为数字。

但是，此代码生成的输出只是一个空字符串。我对每个窗口绘制的光谱持怀疑态度，因为它通常包含4个峰而不是2个。此外，低峰似乎处于略低的频率，而高峰则处于太高的频率。表。我实际上不确定这种信号能量方式是否是最好的方法，所以我愿意接受建议，但是如果已经实现了代码，那么如果可以使用它当然会很方便。它至少可以节省一些时间。

我现在不再提出这个问题，但请不要犹豫，询问您是否需要其他信息。谢谢你的帮助！

-Mikael

Matlab中声音信号的频率分析与分类

0 个答案: