Determination of resolution in MTF

Hi ng,

can someone please explain to me why the resolution of a lens is
normally determined at the point of the MTF where the modulation is
only 50%? Is this done due to some perceptual reasons?

Best regards!
Gregory Wool

"Gregory Wool" wrote

can someone please explain to me why the resolution of a lens is
normally determined at the point of the MTF where the modulation is
only 50%?

It's the standard point for determining bandwidth of a signal,
roughly the 3db down point. For a single-pole transfer function
for an electrical signal the -3db frequency is equal to the
time constant of the pole: if 1 - e-ft is the step response of a
system it's 3db down point is f radians/second = f/2pi Hz as the
laplace transform shows:

1 - e^-ft =L= 1/s - 1/(s + f)

When the substitution s = jw is made, where j is the
sqrt (-1) and w is the frequency in radians per second
the transform gives the response in the frequency domain.

http://mathworld.wolfram.com/LaplaceTransform.html

Simple, aren't you glad you asked.

The other answer is "Well it has to _something_ and 50% seems
like a good point everyone can agree on."

[and it's been a long time since University ... so if the
above is wrong then add, subtract of multiply by -1 as needed]
--
Nicholas O. Lindan, Cleveland, Ohio
Darkroom Automation: F-Stop Timers, Enlarging Meters
http://www.nolindan.com/da/index.htm
n o lindan at ix dot netcom dot com

"Nicholas O. Lindan" wrote

1 - e^-ft =L= 1/s - 1/(s + f)

Forgot to add that in the spatial domain the f above
should be read as line pairs / mm (*2pi) and t is
in mm.

If you know the mtf's 50% down frequency then the
edge response of the lens is about
1 - e-(2pi * mtf * distance).

It is possible then to look just at the intensity Vs
distance of a lens's image of a white/black edge and
find the mtf.

If you measure the distance from the start of the image
of the edge to where the intensity has decreased to (1-1/e)
(~63%) then the inverse of this distance is the 50% mtf
(in radians - divide by 2pi to get lp/mm).

Well, that clears it up a lot, doesn't it?

Departmental mid-term on Friday.

--
Nicholas O. Lindan, Cleveland, Ohio
Darkroom Automation: F-Stop Timers, Enlarging Meters
http://www.nolindan.com/da/index.htm
n o lindan at ix dot netcom dot com

It clears it up, oh yeah!
The equations are nothing in can work with, but now I know for sure
that it has_no_perceptual reason. And that´s not too bad either.

Gregory

"Gregory Wool" wrote in message
oups.com...

It clears it up, oh yeah!
The equations are nothing in can work with, but now I know
for sure
that it has_no_perceptual reason. And that´s not too bad
either.

Gregory

I think its arbitrary. Visual acuity is very complicated
and has many variables so I doubt if the 50% response is
related to it in any definite way.
Nicholas is right about the analogous elecrical filters
but lenses and film look like much more complex filters than
monopoles so the slope at any point is not necessarily
predictive of the slopes elsewhere. In general, MTF's must
be combined by convolution. The rule of thumb that total
resolution of, say film and lens, is 1/T = 1/L + 1/F where T
is total, L is lens, F is film resolution holds in a very
approximate manner. Lens MTF's are not smooth due to the
effects of high order aberrations.

--
---
Richard Knoppow
Los Angeles, CA, USA





--
Posted via a free Usenet account from http://www.teranews.com

"Richard Knoppow" wrote

but lenses and film look like much more complex filters than monopoles so
the slope at any point is not necessarily predictive of the slopes
elsewhere.

Oh, absolutely. The 'frequency response'/mtf curve has
pretty close to zilch similarity with a single-pole
low-pass function and for added fun the response curve
also changes across the lens's field. But since it
is so hard to make head or tail of it the mtf
response is treated as if it is a simple function
and that the 50% point has some validity. It's a
'yeah, what the hell, it sounds good to me' point.

There is nothing to say a lens's mtf function doesn't
dip to below the 50% mark and then rise back up again
when the CoC is close to the lp/mm spacing and then
dive to 0.

--
Nicholas O. Lindan, Cleveland, Ohio
Darkroom Automation: F-Stop Timers, Enlarging Meters
http://www.nolindan.com/da/index.htm
n o lindan at ix dot netcom dot com

snip
In general, MTF's must
be combined by convolution. The rule of thumb that total
resolution of, say film and lens, is 1/T = 1/L + 1/F where T
is total, L is lens, F is film resolution holds in a very
approximate manner.
snip

Richard, just to check my understanding of this (school is long time
ago and math has never been my favourite):
If I have a max lens resolution of, say, 160 lp/mm and a film
resolution of 100 lp/mm this reads like 0,00625+0,01=0,01625 of which
the reciprocal is 61,5 lp/mm for the combined system. - Is that
correct?

Best regards!
Gregory Wool

Gregory Wool wrote:
snip
In general, MTF's must
be combined by convolution. The rule of thumb that total
resolution of, say film and lens, is 1/T = 1/L + 1/F where T
is total, L is lens, F is film resolution holds in a very
approximate manner.
snip

Richard, just to check my understanding of this (school is long time
ago and math has never been my favourite):
If I have a max lens resolution of, say, 160 lp/mm and a film
resolution of 100 lp/mm this reads like 0,00625+0,01=0,01625 of which
the reciprocal is 61,5 lp/mm for the combined system. - Is that
correct?

Best regards!
Gregory Wool



The reciprocals rule is only one rule of thumb to compute the combined
resolution. Another rule of thumb is to take the reciprocal of the
square root of the sum of the squares of the reciprocals. In your
example, that would give

1/sqrt(1/160^2 + 1/100^2) which is 84.7998... or about 85 lp/mm

The justification for the second method might be that if you think in
terms of circles of confusion, you should calculate the diameter of the
combination by taking the root mean square of the diameters of the
constituents. Roughly speaking the coc is the reciprocal of the
resolution.

In point of fact, neither method has a convincing theoretical
justification. Indeed, it doesn't even make sense to describe
resolution by a single number, which is why MTFs are used. Both methods
are empirical rules of thumb, and you can use whichever you want.
Hansma's method for optimizing with respect to both diffraction and
defocus, which many people use, is based on a variant of the root mean
square method.

Both methods catch the very important fact that the resolution of a
composite system is necessarily less than that of any constituent of the
system.

Well, as I haven´t studied anything related to optics or photography
I´m just piecing the parts together to get the whole picture. Doing
that I read a lot of stuff about our perception of sharpness and mtfs
as a measure of resolution but wasn´t able to make sense out of the
equations which combine the parts into a system resolution. By now I
understand that one can take the 50% values (explanation by Nicholas)
or the 2% values (because they are related to the threshold of
visibility) and add their reciprocals or the squares of their
reciprocals to get an idea of the combined system resolution. Idea is
important because I´ve also learned that the equations are by far not
accurate. The best way to combine a system would be the convolution of
the curves, but this is a point I haven´t reached yet and so I´ve no
idea how to do that. But: Shouldn´t both equations mentioned by now
give roughly the same result? Or, asked the other way around, is my
calculation faulty because it´s off by more than 20 lp/mm in regard to
Leonard´s?

Gregory Wool

Gregory Wool wrote:
Well, as I haven´t studied anything related to optics or photography
I´m just piecing the parts together to get the whole picture. Doing
that I read a lot of stuff about our perception of sharpness and mtfs
as a measure of resolution but wasn´t able to make sense out of the
equations which combine the parts into a system resolution. By now I
understand that one can take the 50% values (explanation by Nicholas)
or the 2% values (because they are related to the threshold of
visibility) and add their reciprocals or the squares of their
reciprocals to get an idea of the combined system resolution. Idea is
important because I´ve also learned that the equations are by far not
accurate. The best way to combine a system would be the convolution of
the curves,

You actually just multiply the MTFs, not convolve them The way to
think about it is that each component of the system acts like a filter.
The input can be thought of as a 'sum' spatial frequencies (via a
Fourier transform) arising from the different levels of detail in the
scene. The filter acts differently on different frequencies. So for
example, the response at 50 lp/mm might be to reduce intensity to 80
percent in one component and by 60 percent in another. The combined
reduction at 50 lp/mm would be by 80 percent x 60 percent = 48 percent.
Any detail lying mainly at 50 lp/mm would be reduced in intensity by
that amount. And this would happen at every frequency. Convolution is
also involved, but it is a bit difficult to explain why. Multiplication
of MTFs makes quite a lot of sense intuitively.

but this is a point I haven´t reached yet and so I´ve no
idea how to do that. But: Shouldn´t both equations mentioned by now
give roughly the same result? Or, asked the other way around, is my
calculation faulty because it´s off by more than 20 lp/mm in regard to
Leonard´s?


The point is that neither method has any good conceptual or theoretical
basis. The closest you can come is to talk about circles confusion and
what you do with them, but that is considered a limited and generally
faulty approach since it doesn't really describe the physics of what is
happening. The reason they are so different is that neither really has
any justification and are just used as rough rules of thumb. You are
going to miss some important facts about what your system is doing to
input if you use either. So use the one that seems to give you results
that are consistent with your experience. If you use the straight
reciprocal rule, you at least will get a plausible lower bound for how
bad it might be, but in fact the actual performance might be better.


Gregory Wool