Digital BW, The Print

Stephen writes:

> Anthony - can you explain "depth of
> modulation" to me?

I'll using printing for an analogy, since that seems appropriate here.

When you print something (for example), each pixel printed is constructed
from an appropriate number of pixels in the input file.  It is rare that a
one-to-one correspondence exists between input pixels and printed pixels, so
usually some interpolation takes place.  This interpolation reduces the
accuracy of the printed pixels as compared to the input pixels; that is, the
printed pixels won't represent the image quite as accurately as the original
pixels did.

Now, if the number of pixels coming in is the same as the number coming out,
the loss depends on how well they are aligned.  If the number of pixels
coming in is less than the number going out, the true capacity of the
printing device won't be used.  But if the number of pixels coming in is
_greater_ than the number being printed, then the interpolation algorithms
have more raw data to work with, and the resulting pixels going out--even
though they inevitably represent a loss of resolution--are better chosen,
providing a cleaner and apparently sharper image that more accurately
represents the original, to the extent possible.

So the key to printing nice results would seem to be to send in more pixels
than you intend to print; and this is actually how it works.  However, the
principle of diminishing returns rapidly applies.  Up to about 1.5x, adding
pixels on the input side improves the output noticeably, even with its fewer
pixels; but beyond 1.5x, the bang for the buck diminshes, with additional
pixels making less and less difference, until it simply makes no discernable
difference at all to keep increasing input pixels.  Kodak says that 1.5x is
a good ratio, and I tend to agree.  Superstition leads many to believe that
2x is somehow better, but I've not seen any clear evidence of that.
Anything beyond 2x is a waste by just about anyone's standards.

Anyway, this principle of getting better output pixels by having more input
pixels to choose from is depth of modulation.  With more input data, you can
represent the output more precisely--your fewer pixels can be better chosen
(color and intensity) to accurately represent the pixels that would have
been around them at higher resolutions.  Nyquist guarantees that you'll be
able to reproduce a given resolution with only 2x sampling, but visible
image quality often depends on more than just distinguishing between two
lines--more information is required to make the lines look nice, and that's
where depth of modulation and more pixels come in.  However, if you are
going to output x pixels, there isn't much point in having more than 2x
pixels (in linear measure) coming in, and 1.5x is generally fine.

Of course, all of this applies to digital cameras, too.  Having extra pixels
in the CCD beyond those that you output means that your output pixels will
be better chosen, and the image will be cleaner.  The greatest (most
cost-effective) gains are for just a slight increase in pixels, of about
1.5x.  Having ten times more pixels doesn't add enough to justify the cost.
Anyway, the extra pixels provide a depth of modulation that can be useful in
interpolation, for cleaner output.  The difference is visible up to a
certain point, then it becomes difficult or impossible to see beyond.

All of this gets more complicated with digicams that use Bayer patterns, and
digicams with non-square pixels, and printers that use dithering or
stochastic screening, but the basic idea stays the same.

If you want a demonstration, find a digital image and build a mosaic image
based upon it.  Start, say, with an image that contains 100x100 pixels, and
convert that to a mosaic of 10x10 pixels.  On the first pass, use only the
centermost pixel of the original to determine the value of the much larger
pixel in the output mosaic.  On the second pass, use an average of _all_ the
pixels in the original replaced by the mosaic pixel to determine the value
of the output pixel.  You'll see that the second method produces a much more
accurate downsampling than the first method, because all the extra pixels
provide a depth of modulation that a single pixel sample would not provide
in the first method.  This is why image-manipulation programs often give you
a choice of methods for downsampling digital images, and the different
methods yield very visibly different results.

Anthony,

> > Anthony - can you explain "depth of
> > modulation" to me?
>
> I'll using printing for an analogy, since that seems appropriate here.
>
> When you print something (for example), each pixel printed is constructed
> from an appropriate number of pixels in the input file.  It is rare that a
> one-to-one correspondence exists between input pixels and printed
> pixels, so
> usually some interpolation takes place.  This interpolation reduces the
> accuracy of the printed pixels as compared to the input pixels;
> that is, the
> printed pixels won't represent the image quite as accurately as
> the original
> pixels did.

I believe you are misusing the term pixel here.  Printers, at least the
inkjet ones we use here, print dots, they do not print pixels.  There is a
huge difference between a pixel and a dot.

Also, the printer drivers for these inkjet printers do not "interpolate",
that is something entirely different.  They "dither" (amongst other terms).

> Now, if the number of pixels coming in is the same as the number
> coming out,
> the loss depends on how well they are aligned.  If the number of pixels
> coming in is less than the number going out, the true capacity of the
> printing device won't be used.

That's not correct for inkjet printers that dither/halftone.  What matters
is how large an area is used to represent the tones, and that area can be
variable.  Do you really understand the dither/halftone process?

> So the key to printing nice results would seem to be to send in
> more pixels
> than you intend to print; and this is actually how it works.  However, the
> principle of diminishing returns rapidly applies.  Up to about
> 1.5x, adding
> pixels on the input side improves the output noticeably, even
> with its fewer
> pixels; but beyond 1.5x, the bang for the buck diminshes, with additional
> pixels making less and less difference, until it simply makes no
> discernable
> difference at all to keep increasing input pixels.

I don't know what printer you are trying to talk about here, but that's not
the case for inkjet printers that we use here.  They print at 1440 (or
whatever) DPI, and input up to 1/2 that is usable with the Piezo driver, and
probably 1/4th that or so with the Epson driver.

> Kodak says
> that 1.5x is
> a good ratio, and I tend to agree.

Kodak says that about WHAT printer?

> Of course, all of this applies to digital cameras, too.  Having
> extra pixels
> in the CCD beyond those that you output means that your output pixels will
> be better chosen, and the image will be cleaner.

No digital cameras that I am aware of have "extra pixels in the CCD..."
(digital cameras have sensor elements BTW, not pixels...unfortunately,
another commonly misused term).  What cameras do you know do this?

Austin

Anthony - Thanks for the detailed explanation of "depth of modulation"!
Now, are you able to tell me the best way (either theoretically or
empirically) of calculating the number of pixels that a printer would use to
create an image, given a certain image size and printer resolution setting?


Stephen Petegorsky
petegorsky@...
web site - www.spphoto.com

Stephen,

> Now, are you able to tell me the best way (either theoretically or
> empirically) of calculating the number of pixels that a printer
> would use to
> create an image, given a certain image size and printer
> resolution setting?

That is entirely driver (meaning halftone/dither algorithm) and printer
dependant...  It is also image dependant, depending on the halftone/dither
algorithm.  The best you can do is simply experiment for your self.

Regards,

Austin

Stephen writes:

> Now, are you able to tell me the best way
> (either theoretically or empirically) of
> calculating the number of pixels that a
> printer would use to create an image, given
> a certain image size and printer resolution
> setting?

Theoretically it is difficult, because modern printers don't use simple
halftone screen rasters to create their images; some use stochastic
screening and dithering and other tricks to improve resolution (although it
still remains well below the machine-dot pitch).  Empirically, you can
fabricate a test pattern in Photoshop and then print it, and then examine
the pattern under a loupe to see just how well the printer can resolve
details.

Note also that the situation is different for those occasional
dye-sublimation printers out there.  Dye-subs use transparent inks that can
be overlaid one on top of another, and the darkness of each ink can be
controlled by varying how much is deposited.  As a result, they don't
require screening or dithering, and one machine dot = one pixel.  This is
why dye-sub can produce better images at 300 dpi than an inkjet at 2880 dpi.

I don't know if there is such a thing as dye-sub for black and white,
although it would be easy enough to create.

In a message dated 8/1/02 5:31:02 PM, petegorsky@... writes:

>Anthony - Thanks for the detailed explanation of "depth of modulation"!
>Now, are you able to tell me the best way (either theoretically or
>empirically) of calculating the number of pixels that a printer would use
>to
>create an image, given a certain image size and printer resolution setting?

With current dithering algorithms it is not a fixed function... so there is 
no single number. 

C. David Tobie
Design Cooperative
CDTobie@...