0

I know that the MTF of an entire imaging system can be modeled by multiplying the MTF of each component (possibly because multiplication in spatial frequency domain is convolution in space?)

So, let's say I have an imaging system with an MTF50 of x and I have a lens with MTF50 of y. Is the MTF50 of the entire imaging system necessarily xy? Could I in theory increase the MTF of an existing imaging system by adding a lens with a different MTF in front of it?

Sorry if this question is poorly worded - still learning about all of this fits together!

  • No. Sys_mtf(x) = lens_mtf(x) * sensor_mtf(x); MTF50 is an inversion of that and the math breaks. – Brandon Dube Jan 14 at 19:11
3

The thing you have to remember about MTF is that it is always less than a theoretically perfect "1.000". MTF is like a hitting average in baseball: 283 is actually 0.283.

Any time you multiply two numbers between zero and one by each other, the product is always smaller than either of the two numbers you started with.

So, no. You can not increase the MTF of the entire system by adding another lens with a higher MTF in front of a lens with a lower MTF.

  • But intuitively, wouldn't adding a zoom lens in front of another lens that produces a rectilinear image increase the amount of detail I can resolve? – Carpetfizz Jan 15 at 3:28
  • 1
    Increasing magnification does not necessarily increase detail. Sometimes it just means you've enlarged blur. – Michael C Jan 15 at 4:27
0

I always relate resolution as matching dots. A lens projects dots, the number/size of them depends on the lens quality and aperture setting (and AA filter if one exists). The sensor then records them with dots (pixels) the number/size of them varies with sensor size/MPs. And the recorded dots will then be output with dots, the number/size of them depends on the printer/monitor used.

At whichever point there are the fewest dots, that is the limit of resolution; and it cannot be increased without addressing that limit first.

Your Answer

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

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