## divide audio signal into frames

on 21 Dec 2011

### Naz (view profile)

hi all,

i've 2second audio signal, i want to divide it into 20 frames and each frame is 100ms length.

```for i = 1:100:2000
%do process
end
```

did i write the right code? or is there any other way to divide it? really need ur help...

thank u

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### Naz (view profile)

on 21 Dec 2011

If you know for sure that your signal is 2 seconds long, then the sampling frequency can be found from dt*N=2, where dt=1/fs is sampling period and N is number of samples, which you don't need.

IF you already have an array of sampled signal, it has particular length (number of samples). In addition, you know that the 'duration' of that array is 2 seconds, so, you must break your array onto 2000ms/100ms=20 pieces:

```x=[your array of N samples];
n=round(length(x)/20); %find how many samples will each frame contain
P=zeros(n,20); %preallocate the matrix for 20 colums of Nsamples/20 in each
```
```for k=0:19
P(:,k+1)=x(1+n*k:n*(k+1));
end
```

Not sure if this works, but the idea is there. There could be an easier way to break your signals into frames: just know that the array of samples must be broken onto 20 pieces.

Naz

### Naz (view profile)

on 21 Dec 2011

If your signal is stereo, let me know. Some code modifications are needed

Rusmaya Luthfina

### Rusmaya Luthfina (view profile)

on 22 Dec 2011

thx for your answer, i get the idea of your answer, actually i'm extracting audio features in time domain such as average energy, zcr, and silence ratio. i've implemented ur idea but with simple code i understand.. anyway,, thank u so much for ur answer..

__maya__

### Walter Roberson (view profile)

on 21 Dec 2011

That code is probably not correct, as it does not take in to account the sampling frequency and instead implicitly assumes that the data was sampled at 1 ms per sample which would be a sampling rate of 1000 Hz.

When Fs designates your sampling frequency, 100 ms would be Fs/10 . That will probably be an integer, but better would not be to assume that.

But being lax for a moment,

```windowsize = Fs/10;
trailingsamples = mod(length(YourSignal), windowsize);
sampleframes = reshape( YourSignal(1:end-trailingsamples), windowsize, []);
```

Now the columns of sampleframes will be the individual frames, such as sampleframes(:,3) for the third frame.

prasanna patil

### prasanna patil (view profile)

on 12 Mar 2013

sir, i am getting trouble in last line of ur code.

```sampleframes = reshape( YourSignal(1:end-trailingsamples), windowsize, []);
```

the error is ->> Error using reshape & Size arguments must be real integers.

and then i tried

```>>sampleframes = reshape(x(1:end-trailingsamples), windowsize, [:,3]);
```

then i got this error.

```Error: Unexpected MATLAB operator.
```

can u plz help me sir?

Walter Roberson

### Walter Roberson (view profile)

on 12 Mar 2013

That code is for the case where Fs/10 is an integer. If it is not an integer, then you need to define what it means to divide into 100 ms frames.

prasanna patil

### prasanna patil (view profile)

on 13 Mar 2013

yeah sir, it worked... thank u...

### saibaba (view profile)

on 6 Apr 2013

i am doing a project on "speech enhancement" am also using the same process but am not understanding it clearly what am i using u use round for the length of the signal but i use the floor does it make any difference. i want to post my code can anyone explain what is happening in it......

actually my aim is to reduce the noise using kalman filter and i got the output too... i want how its happening inner view of it... can anyone explain

matlab code:

[x,Fs4,bits4]=wavread('DEKF_white_stat_7db__noisy.wav'); xx=x; N=256; % frame length m=N/2; % of each frame of the moving distance lenth=length(x); % the length of the input signal count=floor(lenth/m)-2; x=x/max(abs(x)); t=(0:length(x)-1)/Fs4; s=1; p=11; a=zeros(1,p); w=hamming(N); y_temp=0; F=zeros(11,11); F(1,2)=1; F(2,3)=1; F(3,4)=1; F(4,5)=1; F(5,6)=1; F(6,7)=1; F(7,8)=1; F(8,9)=1; F(9,10)=1; F(10,11)=1; H=zeros(1,p); S0=zeros(p,1); P0=zeros(p); S=zeros(p); H(11)=1; s=zeros(N,1); G=H'; P=zeros(p); y_temp=cov(x(1:7680)); x_frame=zeros(256,1); x_frame1=zeros(256,1); T=zeros(lenth,1); for r=1:count x_frame=x((r-1)*m+1:(r+1)*m); if r==1 [a,VS]=lpc(x_frame(:),p); else [a,VS]=lpc(T((r-2)*m+1:(r-2)*m+256),p); end if (VS-y_temp>0) VS=VS-y_temp; else VS=0.0005; end

```        F(p,:)=-1*a(p+1:-1:2);
if r==1
S=[zeros(p,1)];  %state vector
P0=[zeros(p,p)];  %error covatiance
else ```
```            P0=P;
end```
```        for j=1:256
if(j==1)
S=F*S0;
Pn=F*P*F'+G*VS*G';
else
S=F*S;
Pn=F*P*F'+G*VS*G';
end
K=Pn*H'*(y_temp+H*P*H').^(-1);
P=(eye(p)-K*H)*Pn;
S=S+K*[x_frame(j)-H*S];
T((r-1)*m+j)=H*S;
end```

% End cycle calculation LPC parameters

end rt=137.78/128; figure(1); subplot(2,1,1); plot(t,x); xlabel('Time'); ylabel('Amplitude'); title('Original'); sound(x,Fs4,bits4); x1=T./rt; wavwrite(x1,Fs4,bits4,'kalman_denosed.wav'); [x1,Fs4,bits4] = wavread('kalman_denosed.wav'); x1=x1/max(abs(x1)); t=(0:length(x1)-1)/Fs4; subplot(2,1,2); plot(t,x1); xlabel('Time'); ylabel('Amplitude'); title('Denoised Kalman'); display('done'); sound(x1,Fs4,bits4);

sr=sum(x.^2) %Speech Power nro=sum((x-x1).^2) %Output Noise Power % nri=sum((speech-x).^2); %Input Noise Power % SNRi=10*log10(sr/nri) %Input SNR SNRo=10*log10(sr/nro) %Output SNR