algorithm – What is the idea behind scaling an image using Lanczos?

algorithm – What is the idea behind scaling an image using Lanczos?

The selection of a particular filter for image processing is something of a black art, because the main criterion for judging the result is subjective: in computer graphics, the ultimate question is almost always: does it look good?. There are a lot of good filters out there, and the choice between the best frequently comes down to a judgement call.

That said, I will go ahead with some theory…

Since you are familiar with Fourier analysis for signal processing, you dont really need to know much more to apply it to image processing — all the filters of immediate interest are separable, which basically means you can apply them independently in the x and y directions. This reduces the problem of resampling a (2-D) image to the problem of resampling a (1-D) signal. Instead of a function of time (t), your signal is a function of one of the coordinate axes (say, x); everything else is exactly the same.

Ultimately, the reason you need to use a filter at all is to avoid aliasing: if you are reducing the resolution, you need to filter out high-frequency original data that the new, lower resolution doesnt support, or it will be added to unrelated frequencies instead.

So. While youre filtering out unwanted frequencies from the original, you want to preserve as much of the original signal as you can. Also, you dont want to distort the signal you do preserve. Finally, you want to extinguish the unwanted frequencies as completely as possible. This means — in theory — that a good filter should be a box function in frequency space: with zero response for frequencies above the cutoff, unity response for frequencies below the cutoff, and a step function in between. And, in theory, this response is achievable: as you may know, a straight sinc filter will give you exactly that.

There are two problems with this. First, a straight sinc filter is unbounded, and doesnt drop off very fast; this means that doing a straightforward convolution will be very slow. Rather than direct convolution, it is faster to use an FFT and do the filtering in frequency space…

However, if you actually do use a straight sinc filter, the problem is that it doesnt actually look very good! As the related question says, perceptually there are ringing artifacts, and practically there is no completely satisfactory way to deal with the negative values that result from undershoot.

Finally, then: one way to deal with the problem is to start out with a sinc filter (for its good theoretical properties), and tweak it until you have something that also solves your other problems. Specifically, this will get you something like the Lanczos filter:

Lanczos filter:       L(x)     = sinc(pi x) sinc(pi x/a) box(|x|<a)
frequency response: F[L(x)](f) = box(|f|<1/2) * box(|f|<1/2a) * sinc(2 pi f a)

    [note that * here is convolution, not multiplication]
       [also, I am ignoring normalization completely...]
  • the sinc(pi x) determines the overall shape of the frequency response (for larger a, the frequency response looks more and more like a box function)
  • the box(|x|<a) gives it finite support, so you can use direct convolution
  • the sinc(pi x/a) smooths out the edges of the box and (consequently? equivalently?) greatly improves the rejection of undesirable high frequencies
  • the last two factors (the window) also tone down the ringing; they make a vast improvement in both the perceptual artifact and the practical incidence of undershoot — though without completely eliminating them

Please note that there is no magic about any of this. There are a wide variety of windows available, which work just about as well. Also, for a=1 and 2, the frequency response does not look much like a step function. However, I hope this answers your question why sinc, and gives you some idea about frequency responses and so forth.

algorithm – What is the idea behind scaling an image using Lanczos?

Leave a Reply

Your email address will not be published.