Error: Unable to perform assignment because the indices on the left side are not compatible with the size of the right side.

close all;
%{Step 0: Reading the File & initializing the Time and Freq.
    [x,fs]=audioread('baby-crying-sleepy.wav');
    x=x(:,1);
    ts=1/fs;
    N=length(x);
    Tmax=(N-1)*ts;
    fsu=fs/(N-1);
    t=(0:ts:Tmax);
    f=(-fs/2:fsu:fs/2);
    figure, subplot(411),plot(t,x),xlabel('Time'),title('Original Speech');
    subplot(412),plot(f,fftshift(abs(fft(x)))),xlabel('Freq (Hz)'),title('Frequency Spectrum');

%%Step 1: Pre-Emphasis

    a=1;
    b=[1 -0.95];
    y=filter(b,a,x);
    subplot(413),plot(t,y),xlabel('Time'),title('Signal After High Pass Filter - Time Domain');
    subplot(414),plot(f,fftshift(abs(fft(y)))),xlabel('Freq (Hz)'),title('Signal After High Pass Filter - Frequency Spectrum');

%%Step 2: Frame Blocking

    frameSize=256;
    frameOverlap=128;
    frames=enframe(y,frameSize,frameOverlap);
    NumFrames=size(frames,1);

%%Step 3: Hamming Windowing

    hamm=hamming(256)';
    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. 

    for i=1:NumFrames
    fft(i,:)=abs(windowed(i,:))*(frameSize/2);     
    end

%%Step 5: Mel Filterbanks
%Lower Frequency = 300Hz
%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);
    melinhz=700*(exp(mel/1125)-1);
    %Converting to frequency resolution
    fres=floor(((frameSize)+1)*melinhz/fs); 
    %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(fft(:,i).* H(j,:));
    end
    end


%%Step 6: Nautral Log and DCT
    pkg load signal
    %Here logarithm is of base '10'
    logged=log(bankans)/log(10);
    for i=1:NumFrames

        lnd(i,:)=dct2(logged(i,:));
    end

%plotting the MFCC

    figure 
    hold on
    for i=1:NumFrames
        plot(lnd(i,:));
    end
    hold off
%

The error is at this line:

bankans(i,j)=sum(fft(:,i).* H(j,:));

Answer 1

[madhan ravi]

bankans=cell(NumFrames,Nofilters);
for i=1:NumFrames
    for j=1:Nofilters
        bankans{i,j}=sum(fft(:,i).* H(j,:));
    end
    end

[Hamizah Mohd Ali]

i tried and it shows this:

Index in position 2 exceeds array bounds (must not exceed 256).

[madhan ravi]

j=1:256 % ?

[Jan]

@Hamizah Mohd Ali: The error message is clear, isn't it? We cannot know, which line is causing the error, if you do not post this detail. But actually you should be able to find out, why the 2nd index is larger than the existing size of the array.

[Hamizah Mohd Ali]

[madhan ravi]

Did the error disappear when you put j=1:256 ?

[Hamizah Mohd Ali]

Not yet, it appears:

Index in position 1 exceeds array bounds (must not exceed 20).

Error in trial (line 75)

bankans{i,j}=sum(fft(:,i).* H(j,:));

[madhan ravi]

size(sum(fft(:,1).* H(1,:));)  % what does it show?
size(fft)  % change fft to someother name becuase there is an inbuilt function named fft()
size(H)

[Hamizah Mohd Ali]

and when I put i=1:20, the error display the same as

Index in position 1 exceeds array bounds (must not exceed 20).

[Hamizah Mohd Ali]

now the error become this