Sir, when I running my codeit showing a error like this"Matrix dimensions must agree. Error in feature(line 98) bankans(i,​j)=sum((ft​(i,:).*h(j​,:)).^2); " .How to rectify the error?

The code is given below,


clc;
clear all;
[x,fs1]=audioread('cryrumble.wav');
 ts1=1/fs1;
    N1=length(x);
    Tmax1=(N1-1)*ts1;
    t1=(0:ts1:Tmax1);
    figure;
  plot(t1,x),xlabel('Time'),title('Original audio');
  fs2 = (20/441)*fs1;
 na=resample(x,2000,44100);
 N2=length(na);
 %framing
 
     frameSize=882;

frame_duration=0.025;
frame_len = frame_duration*fs2;
framestep=0.01;
framestep_len=framestep*fs2;
num_frames =floor(N2/frame_len);

frames=[];
for j=1:num_frames
     frame=na((j-1)*framestep_len + 1: ((j-1)*framestep_len)+frame_len);

max_val=max(frame);
   if (max_val>0.025)

        frames=[frames;frame'];
   end
end
% Step 3: Hamming Windowing

NumFrames=size(frames,1);
hamm=hamming(frame_len)';
windowed=[];
    for i=1:NumFrames
    windowed(i,:)=frames(i,:).*hamm;
    end
    % Step 4: FFT 


    for i=1:NumFrames
    ft(i,:)=abs(fft((windowed(i,:)),frame_len));     
    %plot(ft(i,:))
    end
    
    %Greenwood filterbank
    
filt_num = 45;
kcnst= 500; %nfft
fs=44100;

%elephant sound features between 10hz to 10000hz
fmin= 10;
fmax= 10000;

%A1 and A2 areconstants and x is cochlea position K is cnst equal to 0.88.
greenwoodcnstK= 0.88;
greenwoodcnstA1= fmin/(1-greenwoodcnstK);
greenwoodcnstA2= log10((fmax/greenwoodcnstA1)+greenwoodcnstK);
i=0;
% greenwood frequency
for x = 0 : 0.01 : 1
 i=i+1;
freq_grn(i) = (greenwoodcnstA1)*(10^(greenwoodcnstA2 * x)- (greenwoodcnstK));
end
%perceived frequency 
for f= 10:1:500;
Fp(f)= (1/greenwoodcnstA2)*log10((f/greenwoodcnstA2)+greenwoodcnstK);
end

for i=1:1:filt_num
    f(i)=floor((1024+1)*freq_grn(i)/fs);
end;

% define triangular melbank
for j=2:1:filt_num-1
    for i=1:1:kcnst
        if i<f(j-1)
            h(i,j-1)=0;
        elseif f(j-1)<=i && f(j)>=i
            h(i,j-1)=(i-f(j-1))/(f(j)-f(j-1));
        elseif f(j)<i && f(j+1)>=i
            h(i,j-1)=(f(j+1)-i)/(f(j+1)-f(j));
        else
            h(i,j-1)=0;
        end;
    end;
end;
figure(3);
plot(h);

%processed signal
 for i=1:NumFrames
    for j=1:filt_num
        bankans(i,j)=sum((ft(i,:).*h(j,:)).^2);
    end
    end

Answer 1

[Walter Roberson]

ft is numframes (number of samples * 20/441) by frame_len (fs1/882)

h is kcnst (500) by filt_num-1-1 -> 500 x 43

These are simply unrelated entities that should not be used together.

[Suchithra K S]

Sir, but for my result i have to apply this. In this i tried to find Greenwood frequecny cepstral coefficient(GFCC). The steps for that is like this,

Audiosignal - Framing - Windowing - FFT - Greenwodfilterbank - log - DCT - GFCC.

Sir then how i will correct this error

[Walter Roberson]

In order to do that multiplication successfully, your ft and h would have to have the same number of columns.

By the way, it looks to me as if that code could be phrased something like (ft*h.').^2 if ft and h were compatible sizes.

[Suchithra K S]

Sir , what change i have to do in that code. Can you please tell me that. How to make h anf ft in same size.

[Walter Roberson]

You currently create columns in h up to column ((filt_num-1)-1) . If you arranged so that ((filt_num-1)-1) was the same as frame_len, then the two matrices would have the same number of columns and the multiplication would not fail.

(Whether this is a reasonable thing to do is a different question.)

[Suchithra K S]

sir, where i have make change in the code so that ((filt_num-1)-1) was the same as frame_len, then the two matrices would have the same number of columns and the multiplication would not fail. sir can you tell me the corrected code .

[Walter Roberson]

Your code assigns a constant to frame duration and uses it to calculate frame length. Instead of using a constant you could calculate a duration such that the frame length came out as 43.

Your code assigns 45 to the number of filters and uses that value minus 2 as the number of filters. Instead of assigning a constant you can calculate a filt_num such that after subtracting 2 the results were the frame length.