How do I get the Time information after STFT? I'm tring to code a STFT program in Matlab. I got results but I'm confused a bit.
I have a audio signal of 900000 pts sampled at 1000Hz (15 min signal). 
I've taken fft with following parameters =>


hop size of H = 1400 pts            
the window length M = H * 6


So, 5/6 of the window overlaps with the next one. 
The program runs for 637 iterations, as (637*1400 + M = 900200)

Now, I get a 637 pt vector with frequency in them. I should get a frequency vs time plot. But it only has 637 frequency bins, for a signal of 900000 time sample. 


How, can I find the change of frequency with Time (900000 sample) or
15 min?
Moreover, how does changing the window length and overlap affect the outcome of STFT? The code is below.




%hop size
L = 1400;

%window length &amp; Formation
D = 6;                        %User Defined Parameter in Cooper's paper 
M = L*D;                             
w = hanning(M, 'periodic');    %window formation

k = 1; % k =index
j = 1; % j =result vector index  
while k + M &lt;= xlen

  %data chunk
  %x = x(k:1400);

  %window length &amp; Window
  xw = x(k+1:k+M) .*w;                  %windowed signal

  %zero padding
  b = 5;                           %Zero Padding Factor 'b'
  xw = [xw; zeros(M*5,1)];
  [nft,c] = size(xw);

  %plot(abs(fft(xw)))
  X = abs(fft(xw));
  XF = X(1:length(X)/2);
  [q,ind] = max(XF);
  v(j) = ((fs/2)/(nft/2)) * (ind-1);  %convert Bin to frequency 
  k = k+L;
  j = j+1;
end

figure(1);
plot(v)
axis([0 1400 0 100])

figure(2);   %zoom version
plot(v)
How, can I find the change of frequency with Time (900000 sample) or 15 min?


OK, so you need 900000 frequencies.  Well, one way to do that is to take each element of your 637 length vector and expand it to (up to) $H$ samples.


  Moreover, how does changing the window length and overlap affect the outcome of STFT? 


The longer the window length, the better the frequency resolution is. The shorter the length, the worse the frequency resolution will be.

Before we can answer your question, we will need to know what form your signal takes.

If it's a single tone in noise, then you may be able to use one of these or one of these or this approach.