# How to 'interpret' the Fourier Transform (specifically, of a convolution kernel)

As part of a homework assignment, I had to take the Fourier transform of the kernel I was using to convolve a signal. The kernel was a constant rectangular function, that was 1 within the square $(-1, -1)$, $(1, 1)$ and 0 everywhere else.

I was wondering what this result really means. What have I gotten once I take the Fourier transform of this function?

• This is a two-dimensional kernel, right? Describable as $$h(x,y)=\begin{cases} 1, &-1 \leq x, y \leq +1,\\0, &\text{otherwise.}\end{cases}$$ and not the one-dimensional kernel in @pichenettes answer? Put another way, what formula (or even MATLAB command) did you use to compute the Fourier Transform of the kernel and what did you obtain? A function $H(\omega_1, \omega_2)$ of two arguments or of one argument $H(\omega)$? – Dilip Sarwate Feb 17 '12 at 2:12
• Correct, it is a two dimensional kernel. The formula to compute the FT is integrals in both x and y of $h(x,y)exp(-j2\pi (ux + vy)$ – Steve Feb 17 '12 at 2:31
• So you should have gotten the transform to be something like $H(u,v)=\text{sinc}(2u)\text{sinc}(2v)$, right? So do you see how, given any positive $\epsilon$, no matter how small, $|H(u,v)| < \epsilon$ for all $|u|, |v| > M$ for some $M$ (whose value I will not be able to tell you until you have chosen $\epsilon$)? – Dilip Sarwate Feb 17 '12 at 2:40
• Sorry, I don't follow this. The magnitude of the Fourier transform is smaller than some value, if I make $u$ and $v$ arbitrarily large? – Steve Feb 17 '12 at 2:47
• Yes, as hotpaw2 pointed out to you also, $\text{sinc}(f) = [\sin(\pi x)]/(\pi f)$ decays away as $|f|$ gets large because the numerator is at most $1$ in magnitude while the denominator is increasing without bound. But there is a lot of "ripple" because of the $\sin$ instead of a smooth decaying away to $0$ as $|f|$ increases. – Dilip Sarwate Feb 17 '12 at 2:52

In your case, the filter's impulse response is a rectangular function of width 2 and centered at 0. You can interpret this filter this way: for any $t$, the filter output $y(t)$ will be the average of the input signal $x(t)$ over the interval $[t - 1, t + 1]$. This is an averaging, low-pass filter, and indeed, the Fourier transform of the kernel is a decreasing function, showing that the higher frequencies are attenuated.
• Averaging = summing, up to multiplicative constant :) In your case, the output of your filter at point x will be twice the average of the input signal in $[x - 1, x + 1]$. – pichenettes Feb 17 '12 at 2:12