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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.

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

%hop size L = 1400; %window length & 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 <= xlen %data chunk %x = x(k:1400); %window length & 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; %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) 

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.

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

%hop size L = 1400; %window length & 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 <= xlen %data chunk %x = x(k:1400); %window length & 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) 

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(1400)*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.

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

%hop size L = 1400; %window length & 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 <= xlen %data chunk %x = x(k:1400); %window length & 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; %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) 

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.

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

%hop size L = 1400; %window length & 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 <= xlen %data chunk %x = x(k:1400); %window length & 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; %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) 

• hop size of H = 1400 pts

• the window length M = H(1400)*6

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

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

%hop size L = 1400;

%window length & 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 <= xlen %data chunk %x = x(k:1400); %window length & 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; %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) 

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

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

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

%hop size L = 1400; %window length & 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 <= xlen %data chunk %x = x(k:1400); %window length & 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; %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)