Sir, when i run my code in matlab it taking toomuch time and always it showing running. I am not getting any output. What's the problem with this. how to rectify this?

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Suchithra K S
Suchithra K S on 5 Feb 2019
Answered: Walter Roberson on 6 Feb 2019
The code is given below:
[audio, fs] = audioread('cryrumble.wav');
%sound(x,fs1);
ts1=1/fs;
N1=length(audio);
Tmax1=(N1-1)*ts1;
t1=(0:ts1:Tmax1);
figure;
plot(t1,audio),xlabel('Time'),title('Original audio');
%Step 1: Pre-Emphasis
a=[1];
b=[1 -0.95];
z=filter(b,a,audio);
subplot(413),plot(t1,z),xlabel('Time'),title('Signal After High Pass Filter - Time Domain');
subplot(414),plot(fs,fftshift(abs(fft(z)))),xlabel('Freq (Hz)'),title('Signal After High Pass Filter - Frequency Spectrum');
nchan = size(audio,2);
for chan = 1 : nchan
%subplot(1, nchan, chan)
spectrogram(audio(:,chan), 256, [], 25, 2000, 'yaxis');
title( sprintf('spectrogram of resampled audio ' ) );
end
% Step 2: Frame Blocking
frameSize=882;
% frameOverlap=128;
% frames=enframe(y,frameSize,frameOverlap);
% NumFrames=size(frames,1);
frame_duration=0.02;
frame_len = frame_duration*fs;
framestep=0.01;
framestep_len=framestep*fs;
% N = length (x);
num_frames =floor(N1/frame_len);
% new_sig =zeros(N,1);
% count=0;
% frame1 =x(1:frame_len);
% frame2 =x(frame_len+1:frame_len*2);
% frame3 =x(frame_len*2+1:frame_len*3);
frames=[];
for j=1:num_frames
frame=z((j-1)*framestep_len + 1: ((j-1)*framestep_len)+frame_len);
% frame=x((j-1)*frame_len +1 :frame_len*j);
% identify the silence by finding frames with max amplitude less than
% 0.025
max_val=max(frame);
if (max_val>0.025)
% count = count+1;
% new_sig((count-1)*frame_len+1:frame_len*count)=frames;
frames=[frames;frame];
end
end
% Step 3: Hamming Windowing
NumFrames=size(frames,1);
hamm=hamming(882);
for i=1:NumFrames
windowed(i,:)=frames(i,:).*hamm;
end
% Step 4: FFT
% Taking only the positive values in the FFT that is the first half of the frame after being computed.
ft = abs( fft(windowed,441, 2) );
plot(ft);
% Step 5: Mel Filterbanks
Lower_Frequency = 100;
Upper_Frequency = fs/2;
% With a total of 22 points we can create 20 filters.
Nofilters=20;
lowhigh=[300 fs/2];
%Here logarithm is of base 'e'
lh_mel=1125*(log(1+lowhigh/700));
mel=linspace(lh_mel(1),lh_mel(2),Nofilters+2);
figure;
plot(mel);
xlabel('frequency in Hertz');ylabel('mels');
title('melscale');
melinhz=700*(exp(mel/1125)-1);
%Converting to frequency resolution
fres=floor(((frameSize)+1)*melinhz/fs2);
%Creating the filters
for m =2:length(mel)-1
for k=1:frameSize/2
if k<fres(m-1)
H(m-1,k) = 0;
elseif (k>=fres(m-1)&&k<=fres(m))
H(m-1,k)= (k-fres(m-1))/(fres(m)-fres(m-1));
elseif (k>=fres(m)&&k<=fres(m+1))
H(m-1,k)= (fres(m+1)-k)/(fres(m+1)-fres(m));
elseif k>fres(m+1)
H(m-1,k) = 0;
end
end
end
%H contains the 20 filterbanks, we now apply it to the processed signal.
for i=1:NumFrames
for j=1:Nofilters
bankans(i,j)=sum((ft(i,:).*H(j,:)).^2);
end
end
figure;
plot(bankans(i,j));
figure;
plot(H);
xlabel('Frequency');ylabel('Magnitude');
title('Mel-Frequency Filter bank');
% Step 6: Nautral Log and DCT
% pkg load signal
%Here logarithm is of base '10'
logged=log10(bankans);
for i=1:NumFrames
mfcc(i,:)=dct2(logged(i,:));
end
%plotting the MFCC
figure
hold on
for i=1:NumFrames
plot(mfcc(i,1:13));
title('mfcc');
end
hold off
% save c5 mfcc
i= mfcc;
save i i
load i.mat
X=i;
k=1;
[IDXi,ci] = kmeans(X,k);
save c41i ci
  3 Comments
Suchithra K S
Suchithra K S on 6 Feb 2019
Sir i tried this now the program showing the error like this,"Error using .*
Matrix dimensions must agree.
Error in mfcheck (line 80)
windowed(i,:)=F(i,:).*hamm;"

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Answers (1)

Walter Roberson
Walter Roberson on 6 Feb 2019
The output of filter applied to aa column vector is aa column vector . You zz is aa column vector . When you index aa column vector you get out aa column vector . Therefore your frame variable is aa column vector . When you vertcat column vector together you get a column vector . Therefore your frames variable is a column vector. You are trying to use .* between the column vector and the hamming window which is not going to give you the result you expect

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