8
\$\begingroup\$

An autofocus system is system based on a feedback loop, where the AF sensor error reading if fed back to the lens focus motor. Being a closed loop, then theoretically there should be no focusing errors (at least when lighting is good enough).

So I was wondering, what is the source for this kind of error that requires MFA different for each lens?

The one reason I could think of is that the AF sensor is not perfectly aligned with the image sensor - a thing that will create a constant misfocus. But then, this probably needs to be constant for all lenses.

Whet exactly is being calibrated with MFA? Is is it the AF sensor position, the image sensor, control loop gain, etc.?

Improve this question
\$\endgroup\$
0

2 Answers 2

Reset to default
8
\$\begingroup\$

Your assumption about it being a closed loop may be wrong. I see numerous references to AF systems generally being "open loop". In other words the AF system calculates the amount of movement needed to achieve focus then moves the lens there, but does not recheck. So it may over- or undershoot the target. I could not find an authoritative reference from any manufacturer. For what it's worth, here are some posts regarding this issue:

http://arstechnica.com/civis/viewtopic.php?f=6&t=157055

http://forums.dpreview.com/forums/readflat.asp?forum=1029&message=16985680

Assuming it is "open loop" then the AF system consults some sort of table of adjustments based on the inputs it gets, which could vary lens to lens.

Even in a "closed loop" system, there would have to be tolerances. If you want speed you can't constantly adjust back and forth until "perfect", you want to quickly get within a margin of error, and small adjustments may just not feasible with the build quality and tolerances of many, if not all, lenses and AF motors

Improve this answer
\$\endgroup\$
1
  • \$\begingroup\$ Mike thanks for the links. The arstechnica thread is pretty useful. Assuming that user @ladavacm is knowledgeable as it seems about the internals, then the system is indeed open loop, and it explains the different settings per lens. \$\endgroup\$
    – ysap
    Jul 29, 2011 at 8:18
2
\$\begingroup\$

From what I understand, what is being set is a value that is added or subtracted from the position that the camera is instructing the lens (or focus motor in the body for body driven lenses) to move to. So, if the camera would normally say move to "123" and you have a +3 adjustment, it will actually say move to "126" instead. Note, those are made up numbers just to illustrate. :)

In any event, these are devices with movable parts, so there are variance tolerances in each one in terms of movement range as a result of manufacture. Perfection is hard to achieve and, well, not worth the cost when you can adjust after the fact. So, the camera may have some variance, which will be lens constant, but the lenses also will have variance and that will vary from lens to lens, hence the reason higher end cameras allow for lens specific adjustment.

In the ideal scenario the body variance is the equal reverse of the lens variance and you don't have to adjust.

Improve this answer
\$\endgroup\$
6
  • \$\begingroup\$ John, you would be right if that was an open loop control system (like - take range measurement using a laser rangefinder and set the focus accordingly). But, the beauty of a (negative) feedback control loop is that the controller measures the current state and generates correction commands to the actuator. The only thing that I can think of is that the AF sensor does achieve perfect focus, but the image sensor is at a slightly different position on the optical path, so it misfocuses. But then, why different value for different lenses? \$\endgroup\$
    – ysap
    Jul 28, 2011 at 22:27
  • \$\begingroup\$ @ysap - Lenses have moving parts and moving parts have variance and they won't be the same. I'm also not sure it's a closed loop, but I don't have the knowledge to say for sure. \$\endgroup\$
    – Joanne C
    Jul 28, 2011 at 23:35
  • \$\begingroup\$ Ho yes, sure they have tolerances, but a feedback loop will reduce the error proportionally to the loop's gain (well, at least if my some 20-years control theory memory serves me well). If the loop is open, then this is a whole different game, and the tolerances are indeed the error. \$\endgroup\$
    – ysap
    Jul 29, 2011 at 1:32
  • \$\begingroup\$ I wonder why this answer does not get more reps, as it came to be a pretty correct answer...?! \$\endgroup\$
    – ysap
    Aug 1, 2011 at 0:39
  • \$\begingroup\$ @ysap - I wondered that myself initially, but such is the nature of things. I'm not worried about it. \$\endgroup\$
    – Joanne C
    Aug 1, 2011 at 13:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.