Superb explanation, but this video should be titled "Radon transform" to help people find it.

@williamkingston4278

3 жыл бұрын

Interestingly this shows up high up in queue when I search for radon transform

@nopawsclark1longboarder.134

3 жыл бұрын

How would this apply to a biological format?. Excusing the radon classification though.

Thank you is not enough for this amazing elaboration !

Excellent! (y) The illustration is clearly explained, and not just formula.. thumbs up (y)

Perfect VIdeo. The clearest description. Should be called Radon Transformation though

good job!

Are we measuring the intensity of radiation from the projector to the detector to interpret the density of the object?

can you or anyone do tutorial on linear back projection please...

I'm confused by the usage of 2-variable functions f(x,y) (used for 3D planes) to represent 2D objects. Does it imply f(x,y) = 0 for all x & y values?

@giocic94

4 жыл бұрын

the function f(x,y) can be thought as the density of the 2D object. The same as a f(x,y,z) function for a 3D body

I really like the part of your video that mentions that a delta function in the (x,y) plane is transformed to a sine wave in the (t,theta) plane. Does frequency of the sine wave provide information about the position of the delta function? Thanks for providing me with a starting point for how to make sense of this amazing transform!

@AGLubang

Жыл бұрын

No, because the scanning lasts one cycle, thus you only see one sine period. What provides info about the position of delta function would be 1) an angle of detector function where you can find the dot. This angle, along with the "t" where the dot is located, can provide the "actual" angle of the delta with respect to the coordinate system. We still lack the distance of the delta from the origin, so we also need 2) the amplitude of the sine wave. If your range of "t" results in a sine wave that doesn't have both the crest and through of the wave (similar to what is called "clipping" by audio/EE people), then the distance of the delta to origin is already outside the range of "t". Hope this helps. Note that t=0 means that the ray starting from the detector line function (the p_theta(t)) passes though the origin, so the red line in 0:38 IS t=0. Many t values correspond to the dashed red line rays in 1:23. So the "outside the range of "t"" I wrote above means that the location of the delta *can be* "out of range" of the dashed red line rays on some internal values of theta.

@markmcla

Жыл бұрын

@@AGLubang Yes, that's very helpful! Thank you for spending the time to answer my question. My intuition is that since the Radon transform is a linear transform, and since a delta function is transformed into an orthogonal function (like a sine wave), then you can preserve the information about the position of a delta function. The original object that is scanned can be thought of as a superposition of many delta functions. So the inverse transform is possible. -These CT scans look like magic to me. 🙂

@AGLubang

Жыл бұрын

@@markmcla Yes, that's correct. Sorry I made a mistake about the "distance of the delta to origin". See my edit. I think I still cannot fully explain it in words only.

How does the equation of the points on L hold, for the figure that is being shown?

@AGLubang

Жыл бұрын

It is not quite explained, but "x cos(theta) + y sin(theta) = t" can be derived as an equation of a line using trigonometry, with theta being the angle with respect to the y-axis (as shown in the figure), and "t" being the (perpendicular) distance of that line to the origin (0,0). I am sorry this is quite difficult without a visualization. In the figure, one could say that t = 0 there because the line crosses the origin, and change the value of "t" means moving the red line "up" or "down" while still parallel to the red line.

How can the line L(theta, y) be defined implicitly as x cos (theta) + y sin (theta) = t? If theta = 0, we would have a vertical line x = t, but the line L(theta, t) travels along the x-ray beam, not perpendicular to it, doesn't it?

@laurensgoyvaerts8480

4 жыл бұрын

I was thinking exactly the same, I can't figure out how this equation hold based on the geometry of the figure..

@MusicProductionsMSA

3 жыл бұрын

I think that is the parametrization of the detector line, not the red ray, in Hesse normal form

@jaeimp

3 жыл бұрын

@@MusicProductionsMSA Thank you! Now it makes sense: user-images.githubusercontent.com/9312897/105555346-31e8da80-5cd7-11eb-851a-9626447db88b.png

@AGLubang

Жыл бұрын

I think the red line in the figure, L(theta, y), is the x-ray itself, while the p_theta(t) is like a "sheet" where the xray is detected, hence the name "detector function". Note that "t" in "x cos (theta) + y sin (theta) = t" is the (perpendicular) distance from the red line to the origin. In the figure, one would say that t = 0 there, because the red line crosses the origin so the distance is 0.

This should definitely be called Radon transform, or at least include it in the title.

0:49 ok, you lost me already

poor. Inconsistent notation.

poor