How can i do detection in DTMF ? i want to enter a tone and get the numbers of it

i want to enter a tone and get the numbers that consist it ....

[Star Strider]

What do you want to do?

How does your Question here differ from: How can i detect frequency of dtmf tone (.wav )file?

[Eng Abeer]

how can i use this code to do this ?

fs=8000;                                        % Audio Format
ts=1/fs;
t=0:ts:0.3;
[R,fs]=wavread('dtmf.wav');
TL=[12 15 21 25 29 34 38 43 47 52 56 61]*10^3;  %The beginnings time of every tone
TH=[15 19 24 28 33 37 42 46 51 55 59 64]*10^3;  %The ends time of every tone
d=1;                                            % Index
for i=1:4
    for j=1:3;
   B=R(TL(d):TH(d))  ;                          % Clip the DTMF tone to generate single tone 
   sound(B)
   Bf=fftshift(fft(B))*ts;                      % To Find Fourier Transform            
   delta=fs/length(Bf);
   f=-fs/2:delta:fs/2-delta;
   figure(i); set(gcf,'Color',[0.7,0.78,1])     % Figure & it's background color 
   subplot(3,1,j)
   stem(f,abs(Bf))                              %  To plot magnitude spectrum 

     axis([-2000 2000 0 0.05])                    %  And find FL&FH in Frequency domain to know the number
     title('Frequency domain representation')
     N=int2str(d);
     xlabel(['FIG(9.3.',N,')'])                   % To add label to plots
     d=d+1;
      end 
  end
  % NOTE(9.3.2) : AS SHOWN IN FIGURES 1,2,3,4
  %      1) FIG(9,3,1)  FL=697 and FH=1209 the number is  '1'
  %      2) FIG(9,3,2)  FL=697 and FH=1336 the number is  '2'
  %      3) FIG(9,3,3)  FL=697 and FH=1477 the number is  '3'
  %      4) FIG(9,3,4)  FL=770 and FH=1209 the number is  '4'
  %      5) FIG(9,3,5)  FL=770 and FH=1336 the number is  '5'
  %      6) FIG(9,3,6)  FL=770 and FH=1477 the number is  '6'
  %      7) FIG(9,3,7)  FL=852 and FH=1209 the number is  '7'
  %      8) FIG(9,3,8)  FL=852 and FH=1336 the number is  '8'
  %      9) FIG(9,3,9)  FL=852 and FH=1477 the number is  '9'
  %      10)FIG(9,3,10) FL=941 and FH=1336 the number is  '0'
  %      11)FIG(9,3,11) FL=941 and FH=1209 the tone   is  '*'
  %      12)FIG(9,3,12) FL=941 and FH=1477 the tone   is  '#'

please help

[Star Strider]

Attach your ‘dtmf.wav’ file. You will most likely have to save ‘R’ and ‘fs’ as a .mat file because it is no longer possible to attach .wav files.

First, save your sound and sampling frequency variables to a .mat file:

save('dtmf.mat', 'R', 'fs');

Second, use the ‘paperclip’ icon to upload the file. Be sure to complete both the ‘Choose file’ and ‘Attach file’ steps.

[Eng Abeer]

Star Strider

[Star Strider]

I cannot read that file. (I get an error that it is not a binary .mat file. Using the '-ascii' option also fails.)

Please do this:

[R,fs]=wavread('dtmf.wav');
save('dtmf.mat', 'R', 'fs');

and upload it again.

[Eng Abeer]

i do like this exactly ...

[Eng Abeer]

if it isnt open .. replace it with any (.wav) tone . include numbers on it ...

[Deema Hani]

>> n=[0:30]

>>xtilde =ss_per(x,n)

>>stem(n<xtilde)

Answer 1

[Star Strider]

Success!

Sort of.

I had to kludge it a bit because there are problems getting good time separation as well as good frequency separation, but — other than producing duplicate results — this works and gives reasonable answers. You will likely want to experiment with it. It is quite likely not robust, and may only work with the file you sent. (I left the spectrogram plot in but commented it out. Un-comment it if you want to see what it does.) The ‘TL’ and ‘TH’ vectors seem to make no sense in the context of the length of the signal and the sampling frequency, since your entire signal is only about 3.63 seconds in length.

The code:

dtmf_tones = load('Eng Abeer dtmf.mat');
xr  = dtmf_tones.R;
fs = dtmf_tones.fs;
x = xr - mean(xr);                                  % Remove d-c Component

[S,F,T] = spectrogram(x, 1536, 64, 512, fs, 'yaxis');

% Fm = repmat(F, 1, length(T));
% Tm = repmat(T, length(F), 1);
% figure(2)
% surf(Fm, Tm, abs(S))
% xlabel('Frequency')
% ylabel('Time')
% axis([0  2000    ylim    zlim])
% grid on

Sa = abs(S);
[r, c] = find(Sa >= 20);
Fr = F(r);
Tc = T(c)';
FT = [Tc  Fr];
[C, ia, ic] = unique(FT(:,1));                                              % Find Unique Times
for k1 = 1:size(C,1)                                                        % Create Cell Array By Time
    FrqTime{k1} = FT(FT(:,1) == C(k1),:);                                   % Time & Frequency Cell
    if size(FrqTime{k1},1) > 2
        FrqTime{k1} = [FrqTime{k1}(1,:); mean(FrqTime{k1}(2:end,:))];       % If MoreThan 2 Rows, Second Row Is Mean Of Last 2 Rows
    end
end
FrqTime{1:10}
original_f = [697  770  852  941  1209  1336  1477];                        % DTMF Frequencies
dtmf_dcd = [1 5; 1 6; 1 7; 2 5; 2 6; 2 7; 3 5; 3 6; 3 7; 4 5; 4 6; 4 7];    % Combination Codes w.r.t. ‘original_f’
nbr_map = ['1' '2' '3' '4' '5' '6' '7' '8' '9' '*' '0' '#'];                % Number Key Map
for k1 = 1:size(C,1)
    freq_dist = abs(bsxfun(@minus, FrqTime{k1}(:,2), original_f));          % Distance Of ‘FrqTime’ Frequencies From ‘original_f’ Frequencies
    [~,freq_pos(:,k1)] = min(freq_dist,[],2);                               % Frequency Positions Of ‘FrqTime’ In ‘original_f’
    num_pad(k1) = nbr_map(ismember(dtmf_dcd, freq_pos(:,k1)', 'rows'));     % Map To Number Key Pad
end

The only other way to approach this would be to create bandpass filters for each DTMF frequency, then program the logic to decode each one of those frequency pairs. The spectrogram approach has the virtue of being easier.

[Eng Abeer]

thanks a lot ....

[Star Strider]

My pleasure!

Answer 2

[M.Manu Prakash]

dtmf_tones = load('Eng Abeer dtmf.mat');

xr = dtmf_tones.R;

fs = dtmf_tones.fs;

x = xr - mean(xr); % Remove d-c Component

[S,F,T] = spectrogram(x, 1536, 64, 512, fs, 'yaxis');

% Fm = repmat(F, 1, length(T));

% Tm = repmat(T, length(F), 1);

% figure(2)

% surf(Fm, Tm, abs(S))

% xlabel('Frequency')

% ylabel('Time')

% axis([0 2000 ylim zlim])

% grid on

Sa = abs(S);

[r, c] = find(Sa >= 20);

Fr = F(r);

Tc = T(c)';

FT = [Tc Fr];

[C, ia, ic] = unique(FT(:,1)); % Find Unique Times

for k1 = 1:size(C,1) % Create Cell Array By Time

FrqTime{k1} = FT(FT(:,1) == C(k1),:); % Time & Frequency Cell

if size(FrqTime{k1},1) > 2

FrqTime{k1} = [FrqTime{k1}(1,:); mean(FrqTime{k1}(2:end,:))]; % If MoreThan 2 Rows, Second Row Is Mean Of Last 2 Rows

end

end

FrqTime{1:10}

original_f = [697 770 852 941 1209 1336 1477]; % DTMF Frequencies

dtmf_dcd = [1 5; 1 6; 1 7; 2 5; 2 6; 2 7; 3 5; 3 6; 3 7; 4 5; 4 6; 4 7]; % Combination Codes w.r.t. ‘original_f’

nbr_map = ['1' '2' '3' '4' '5' '6' '7' '8' '9' '*' '0' '#']; % Number Key Map

for k1 = 1:size(C,1)

freq_dist = abs(bsxfun(@minus, FrqTime{k1}(:,2), original_f)); % Distance Of ‘FrqTime’ Frequencies From ‘original_f’ Frequencies

[~,freq_pos(:,k1)] = min(freq_dist,[],2); % Frequency Positions Of ‘FrqTime’ In ‘original_f’

num_pad(k1) = nbr_map(ismember(dtmf_dcd, freq_pos(:,k1)', 'rows')); % Map To Number Key Pad

end