how to evaluate a convolution integral by fast fourier transform

I need to evaluate a convolution integral by fft. Therefore, I have read somewhere in a paper to first zero-pad two multiplying functions and wrap around one of them. Then, an element-by-element multiplication and inverse transforming back to the spacial domain and then removing the elements corresponding to the added zeros will solve the problem. To give it a try, I started with a simple example. Consider f(x)=1 (for -1<x<1) and 0 (elsewhere). Also, g(x)=1-x.^2 (for -1<x<1) and 0 (elsewhere). I have calculated the convolution integral analytically to compare with my code results. However, my code does not provide acceptable results especially on boundaries. Also, I don't think I know and understand zero-padding and wrapping around order. Can anyone help me with this, please? Here is the simple code I have written:

clc

clear all

% the analytical result

d=[...

   -0.6667

   -0.2678

    0.0868

    0.3970

    0.6630

    0.8846

    1.0619

    1.1948

    1.2835

    1.3278

    1.3278

    1.2835

    1.1948

    1.0619

    0.8846

    0.6630

    0.3970

    0.0868

   -0.2678

   -0.6667];

hold on

plot(linspace(-1,1,20)',d,'o')

N=pow2(10);

dx=2/N;

x=linspace(-1,1,N);

x=x';

f=[1-x.^2;zeros(N,1)];

Ff=fft(f);

g=[ones(N,1);zeros(N,1)];

Fg=fft(g);

FG=Ff.*Fg;

fg=2/N*ifft(FG);

plot(x,fg(N/2+1:3/2*N))

Answer 1

[David Young]

Correct zero-padding to use the FFT for 1-D or 2-D convolutions is implemented in my FEX submission 2-D convolution using the FFT.

[[e-mail]]

Thank you David

But your code doesn't provide the accurate results, either. If you perform your conv_fft2 command in my simple example, you will see the difference.