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