how to find the number of points in audio?

point=200;
 %Calculate the number of frames  
n=floor(length(x)/point);  %n takes the closest largest integer  
enframe=zeros(point,n);%Initialize, one frame per column

how to find number of points in audio?

Answer 1

[Walter Roberson]

Audio files do not inherently have "points".

The code you are using appears to be part of dividing a single channel of audio into a number of fixed-sized windows, and you are asking how to determine the window size. The appropriate size for a window depends upon the sampling frequency and upon what kind of operations you are doing with the windows.

Also, for audio, it is common to use overlapping windows in order to better match phases.

I recommend that you look at https://www.mathworks.com/help/signal/ref/buffer.html buffer() to do the work of creating windows from your signal, once you have figured out how big the window should be.

[john]

%[x,fs]=wavread('D:\speech_signal\pena.wav'); %Read audio file

the whole program is like that

[x,fs]= audioread('pena.wav');

%Normalized processing

c=max(abs(x));

x=x/c;

N=length(x);

t=(0:N-1)/fs;

subplot(211);

plot(t,x,'k');

xlabel('time/s');

ylabel('Amplitude');

title('Signal waveform');

axis([0 max(t) -1 1]);

grid;

point=200;

%Calculate the number of frames

n=floor(length(x)/point); %n takes the closest largest integer

enframe=zeros(point,n);%Initialize, one frame per column

% framing, 400 points per frame

for i=1:n;

for j=1:point;

enframe(j,i)=x(j+(i-1)*point);

end

end

%Calculate the short-term energy of each frame

for i=1:n;

energy_short(i)=sum(enframe(:,i).^2);%Take the ith column of all rows of the matrix

end

%Short-term energy normalization

% a=max(energy_short);

% energy_short=energy_short/a;

% finds the starting point of the voice. If the short-term energy is greater than 0.2, the frame at this time is used as the starting frame.

for i=1:n;

if energy_short(i)>0.2 break;

end

end

frame_begin=i;

begin=(i-1)*point+1%Starting point

%Find the end of speech

for i=n:-1:1;

if energy_short(i)>0.2 break;

end

end

last=i*point

x1=x(begin:last);

N=length(x1);

t=(0:N-1)/fs;

subplot(212);

plot(t,x1,'k');

xlabel('Time/s');

ylabel('Amplitude');

title('Waveform after endpoint detection');

axis([0 max(t) -1 1]);

grid;

[Walter Roberson]

The 200 (400 in the comments) and 0.2 cutoff are arbitrary, especially since you commented out the normalization. To get something that was not arbitrary you would need to have a calibrated system that you could calculate SPL (sound pressure level) from and you would combine that with information on studies on audibility in humans.