4
$\begingroup$

I am using the InverseRadon function. According to the documentation the function uses a Hann filter and no cutoff ("CutoffFrequency"->1) by default. However, if this is given as an option the result is different from the default.

Is their a way to determine which parameters (Filter and cutoff) are used by default?

I have tested the filters listed in the documentation. The result is closest to the cosine-ramp filter, but not identical.

i = Import["http://i.imgur.com/zxKkFP9.png"];(* Sinogram from InverseRadon Documentation *)
filter = {{Automatic, "Default"}, {#1&, "ramp filter with constant slope"},{#1 Cos[#1 \[Pi]]&, "ramp filter multiplied by cosine function"}, {(1+Cos[#1 \[Pi]])/2&, "Hann filter (default)"}, {0.54` +0.46` Cos[#1 \[Pi]]&, "Hamming filter"}, {Sqrt[1/(1+#^(2n))]&, Row@{"Butterworth filter of order ",n}}, {1&, "no filtering"}};
TableForm@Transpose[{Last@#,ImageAdjust@InverseRadon[i,{300,300},Method->{First@#, "CutoffFrequency"->1}]}&/@(filter/.n->2)]

Different filters for InverseRadon


Edit: I have tested @JM's suggestion to use the scaled cosine-ramp filter.

ImageDifference@@(ImageAdjust@InverseRadon[i, {300, 300}, Method -> #] & /@ {Automatic, (# (1 + Cos[\[Pi] #])/2 &)})

This results a black image image. The two options are almost equal, but not identical. The maximum difference (Max@ImageData) is 0.00249. This is difference image after an ImageAdjust:

enter image description here

$\endgroup$
  • $\begingroup$ Try this: ImageAdjust[InverseRadon[i, {300, 300}, Method -> (# (1 + Cos[π #])/2 &)]]. Use ImageDifference[] to compare with the result of ImageAdjust[InverseRadon[i, {300, 300}]]. $\endgroup$ – J. M. will be back soon Mar 23 '17 at 13:59
  • $\begingroup$ @J.M. I Have tested your suggestion. The difference between the default and the scaled cosine-ramp filter is very minor and for me negligible. This solved my problem. Thank you. $\endgroup$ – jct Mar 24 '17 at 9:35
  • $\begingroup$ @jct In other words, the default setting seems to be a Ram.-Lak. (Ramp) filter multiplied by a Hann window. $\endgroup$ – UDB Nov 28 '18 at 13:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.