2
$\begingroup$

The following two images are the ideal low pass filter in the frequency domain. As you can see, the origin (low frequency component), can pass through this filter while the high frequency are blocked. The white value coresponds to 1, the black value coresponds to 0. enter image description here

And corespondingly, its spatial domain representation is (2d sinc function):

enter image description here

However, there is a thing that said: "for the spatial domain, the radius of centre component and number of cycle per unit distance from the origin are inversely proportional to the cutoff frequency of the ideal low pass filter".

So when you increase the cutoff frequency, you skretch in frequency domain, but you compress in spatial domain; how does the "number of cycle per unit distance from the origin" decrease? It should increase.

$\endgroup$
2
  • $\begingroup$ I guess you are right, the larger the cutoff frequency, the more rings you will have in your frequency picture. I guess that's what's meant by cycles. Could you verify how a 'cycle' is defined exactly? In Formuas: $H(a u, a v)$ inverse Fourier transformed becomes $h(x/a, y/a)$ up to some factor for some $a > 0$. $\endgroup$
    – Chris
    Commented Oct 20, 2014 at 7:00
  • $\begingroup$ Is this not just a restatement of the reciprocity between the two domains? If you stretch the frequency domain with $\omega \rightarrow \omega / \alpha$, then the spatial domain becomes $x \rightarrow \alpha x$. So in the spatial domain where prior to scaling you would have, say, $n$ zero-crossings of the sinc function per unit distance, you would now have $\alpha n$ crossings. $\endgroup$
    – AnonSubmitter85
    Commented Oct 21, 2014 at 1:35

2 Answers 2

Reset to default
1
$\begingroup$

If you scale a function by a factor $a$ in the spatial domain (i.e. the coordinates, or input arguments are scaled), the Fourier transform of the function, i.e. the function in the frequency domain, is scaled by a factor $a^{-1}$. So if you make the cutoff-frequency twice as large, which corresponds to scaling the disc (the function in the frequency domain) by a factor 2, the function in the spatial domain is scaled by a factor ½, which confirms what is said about the radius of centre component.

However, the "number of cycle per unit distance" seems to be a frequency, in which case it should be proportional to the cutoff frequency, not inversely proportional to it. so I agree with you that it should increase and not decrease. Probably just a mistake by the one who wrote that piece of text.

By the way, the two-dimensional analogue of the sinc function is called the sombrero function.

$\endgroup$
0
$\begingroup$

The larger the cutoff frequency, you approach the ideal spectrum. If you define the number of cycle "sidelobes" within a particular surface of interest "let us say centered in the origin" then you gain in resolution "The main lobe become narrower" and you suppressed some among of sidelobes ("cycles") when you compress within large cutoff interval. But to rely see this suppression (decrease), the surface of interest have to be large enough. I think this should affect also nearby sidelobes (cycle close to the main lobe).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .