there is a way to recognize vocals from voice?
i'm wondering if i can recognize not all speach, but only vocal letters...
do i have to use FFT?
using oscilloscope from apple a letter "e" seems equal to "a"
do anyone knows how to recognize letters?
thanks
do i have to use FFT?
using oscilloscope from apple a letter "e" seems equal to "a"
do anyone knows how to recognize letters?
thanks
That would be pretty hard. I know in animation, there are apps that do automatic lip-syncing. But that is basically changing a mouth shape to the level of the db. Trying to point out actual letters from wave form would be like describing a mans hair color by judging his foot print.
Speech recognition is a difficult problem, an active area of research.
OS X has some speech recognition built in (see Speech system preferences.) I've only played with it a little; it seemed hit or miss. It might get better with training. But perhaps Apple gives you a system API you can use instead of writing something yourself.
One way to turn it into a more tractable problem is to limit the legal input. For example, when you call a company's voicemail prompt system it's often designed to recognize spoken digits, and a few words like "yes", "no", "main menu".
OS X has some speech recognition built in (see Speech system preferences.) I've only played with it a little; it seemed hit or miss. It might get better with training. But perhaps Apple gives you a system API you can use instead of writing something yourself.
One way to turn it into a more tractable problem is to limit the legal input. For example, when you call a company's voicemail prompt system it's often designed to recognize spoken digits, and a few words like "yes", "no", "main menu".
Measure twice, cut once, curse three or four times.
(Sep 3, 2011 05:00 PM)MattDiamond Wrote: For example, when you call a company's voicemail prompt system it's often designed to recognize spoken digits, and a few words like "yes", "no", "main menu".Comparing wave forms can work but has issues. I'm sure we have all heard, "I did not understand your request. Please say bla or bla or bla again." Or something like that.
seems pretty difficult, thanks for advices
It looks like this team from Rice University did exactly what you are trying to do:
http://cnx.org/content/m11734/latest/?co...l10223/1.5
They do use FFT and it turns out that vowel formants are pretty distinctive...
And they include MATLAB code (looks pretty easy to port to C++) actually used to detect vowels:
( http://cnx.org/content/m11734/latest/formants.txt )
http://cnx.org/content/m11734/latest/?co...l10223/1.5
They do use FFT and it turns out that vowel formants are pretty distinctive...
And they include MATLAB code (looks pretty easy to port to C++) actually used to detect vowels:
( http://cnx.org/content/m11734/latest/formants.txt )
Code:
function answer = formants(x)
%input is a file at 8000 Hz at 8 bits
%windows input into sample sections, finding formants for each section
%after formants are found, it associates each set with a vowel or consonant
%returns a string which indicates the order of consonant and vowel sounds in the word
%example: formants('mexico.wav') = CeCiCoC
%note: Consonant will be returned at beginning and end regardless of actually content of input
%constants: window size is size of window; n is order of AR model
winsize=256;
n=10;
%reads input file into vector, establishes how many windows will be utilized
x=wavread(x);
[b1,b2]=size(x);
j = (b1 - rem(b1,winsize))/winsize;
%vowel database
%values shown are values of frquency of first and second formant
a=[.650; 1.150];
e=[.430; 1.650];
i=[.250; 1.950];
o=[.420; 1.075];
u=[.325; 1.350]; for k=1:j
%windowing/normalization process
c = x(1+winsize*(k-1):winsize*k,1);
c = (c-mean(c))/max(abs((c-mean(c))));
c = c.*hamming(winsize);
%AR model
fn=ar(c,n);
[num,den]=tfdata(fn,'v');
%freq. response of AR model
[h,w]=freqz(num,den);
f=w.*8000/(2000*pi);
%finds all formant frequency and magnitude values
hnorm=abs(h);
z=zeros(1,2);
for(d=1:510)
if max(hnorm(d:d+2,1)) == hnorm(d+1,1)
z=[z; [f(d+1,1) hnorm(d+1,1)]];
end
end
%generates graphs of frequency response of each window for troubleshooting purposes
%figure
%semilogy(f,hnorm)
%xlabel('Frequency(kHz)')
%ylabel('Response')
%title('Vocal Model')
%k
%makes v vector which contains 0 for definite consonant, 1 for possible vowel
[blip,blop]=size(z);
g = z(2:blip,1:2);
if g(1,1) >= .225 & g(1,2) >= 5
v(k,1)=1;
else v(k,1)=0;
end
%running log of all formant frequency and magnitude values
t(1:blip-1,k)=g(1:blip-1,1);
mag(1:blip-1,k)=g(1:blip-1,2);
end
%initial smoother, eliminates 1 long strings of possible vowels as being anomolous
for k=2:j-1
if (v(k,1) ~= v(k-1,1)) & (v(k,1) ~= v(k+1,1))
v(k,1)=0;
end
end
%v(:,2) will label row numbers of output for troubleshooting reference
v(:,2)=(1:j)';
v(:,3) = 0; %systematic elimination method
for k=1:j;
if v(k,1) ~= 0
flaga = 1;
flage = 1;
flagi = 1;
flago = 1;
flagu = 1; %weeds out negative matches by first formant
if abs(a(1,1)-t(1,k)) > .15
flaga = 0;
end
if abs(e(1,1)-t(1,k)) > .125
flage = 0;
end
if abs(i(1,1)-t(1,k)) > .05
flagi = 0;
end
if abs(o(1,1)-t(1,k)) > .100
flago = 0;
end
if abs(u(1,1)-t(1,k)) > .075
flagu = 0;
end %weeds out negative matches by second formant
if flaga == 1 & abs(a(2,1)-t(2,k)) > .250
flaga = 0;
end
if flage == 1 & abs(e(2,1)-t(2,k)) > .200
flage = 0;
end
if flagi == 1 & t(2,k) < 1.6
flagi = 0;
end
if flago == 1 & abs(o(2,1)-t(2,k)) > .150
flago = 0;
end
if flagu == 1 & abs(u(2,1)-t(2,k)) > .200
flagu = 0;
end %assesses what vowel(s) it thinks it has
if flaga == 1
v(k,3) = v(k,3) + 1;
end
if flage == 1
v(k,3) = v(k,3) + 10;
end
if flagi == 1
v(k,3) = v(k,3) + 100;
end
if flago == 1
v(k,3) = v(k,3) + 1000;
end
if flagu == 1
v(k,3) = v(k,3) + 10000;
end
end end %final smoother, eliminates 1 long strings of specific vowels
for k=2:j-1
if v(k,3) ~= v(k-1,3) & v(k,3) ~= v(k+1,3)
v(k,3) = 0;
end
end %converts the numeric values into letters
answer = 'C';
for k=2:j
if v(k,3) == 0 & v(k-1,3) ~= 0
answer = [answer 'C'];
elseif v(k,3) == 1 & v(k-1,3) ~= 1
answer = [answer 'a'];
elseif v(k,3) == 10 & v(k-1,3) ~= 10
answer = [answer 'e'];
elseif v(k,3) == 100 & v(k-1,3) ~= 100
answer = [answer 'i'];
elseif v(k,3) == 1000 & v(k-1,3) ~= 1000
answer = [answer 'o'];
elseif v(k,3) == 10000 & v(k-1,3) ~= 10000
answer = [answer 'u'];
end
end
thanks i'll try to read something!
Possibly Related Threads...
Thread: | Author | Replies: | Views: | Last Post | |
mic blow detection vs very high voice | sefiroths | 4 | 6,840 |
Dec 14, 2010 01:09 AM Last Post: sefiroths |