Help on zero padding and understanding of the following example?

To understand the concept and impact of zero padding on computation time, I have created an example below and run the program and note the time taken to complete the program. Using the following program I assume, I will get a lowest time using 3 (with zero padding to array1, see below), but this was not the case. I got the different result than my assumption! Where did I make the mistake? Could you help me?

% Create an example of zero-padding 
close all; clear; clc; workspace

array1 = rand(41,41,41);                   
array2 = ones(256,256,199);

%% 1. direct convolution
tic
A = convn(array1, array2);
toc

%% 2. without zero padding: FFT method
tic
B = fftn(array1);
C = fftn(array2);
D = ifftn(convn(B, C));
toc

%% 3. with zero padding to array1
zero_paded_array1 = zeros(256,256,199);

% insert array1 in the middle of zeros matrix
zero_paded_array1(108:148, 108:148, 79:119)= array1;

tic
E = fftn(zero_paded_array1);
F = fftn(array2);
% Now, E and F are of same size
G = ifftn(convn(E, F));
toc

Elapsed time is 36.725312 seconds.

Elapsed time is 7.637434 seconds.

Elapsed time is 197.981833 seconds. ....>>>> If the zero padding was right then this time should be lower than above two, right?

Answer 1

[KALYAN ACHARJYA]

"If the zero padding was right then this time should be lower than above two, right?"

How?

The only difference between section 2 and 3 are B and E, remain code are exactly same (Within tic toc)

>> whos B
  Name       Size                 Bytes  Class     Attributes
  B         41x41x41            1102736  double    complex   
>> whos E
  Name        Size                     Bytes  Class     Attributes
  E         256x256x199            208666624  double    complex   

E here

Both section (2 and 3) are performing the same operation, except for one variable in segnment 3, which has larger data size (E) as compare to it's counterpart (B) in section 2, therefore expecting higher execution time than section 2

Hello experts and others members, Am I missing something here? Or this is a simple case, as I have mentioned.

[blues]

I may have a wrong belief, but as I understand, the FFT generally requires the zero padding to attain maximum computational efficiency. If the elapsed time in part 3 is greater then time taken in part 2 then what is its advantage? Could someone enlighten me please?

[Allen]

Not sure what the output from these functions is supposed to look like in terms of array size, but the results from method 2 and 3 both produce a resulting array whos size is the sum of the size of each input array minus one. Since the size of E>B, this also means that size of G>D and hence the longer computational time.

>> size(B)+size(C)-1

ans =

   296   296   239

>> size(D)

ans =

   296   296   239

>> size(E)+size(F)-1

ans =

   511   511   397

>> size(G)

ans =

   511   511   397