To understand the resolution capability of a camera, it is best to visualize how the pixels in the camera interact with the original object to produce an image. Figure 6.5 shows two bars of a three-bar chart. It is composed of ink printed on paper. The bars and the spaces between the bars are the same width. We will work with a camera with a given lens and we will take pho- tos at various distances. In each case it will be necessary to carefully focus the lens. Also, we will assume that the bars are vertical and align perfectly with the columns of pixels in the sensor chip. A very unlikely circumstance, to be sure, but for this exercise it makes it easier to see the relationships. To get started, set the original simple three-bar chart very close to the camera so that the bars take up one-third of the width of the frame. The test target is three inches wide, so the bars portion is one inch wide. Assume that the camera is now fi ve inches from the test target. For simplicity assume that there are 3000 pixels in each row on the sensor chip. In this situation, in the bars section, all three bars and the intervening spaces will be imaged by 1000 pixels. There are fi ve distances of interest in this case: three dark bar widths and two intervening white spaces. So, each distance will be imaged by 200 pixels. That is, there will be 200 pixels for each bar and each space. The camera will have no trouble seeing the bars as separate and distinct.
Now for the second situation, we will move the test target quite a dis- tance away—in fact, 1000 times further from the camera, or 5000 inches (416 feet, 8 inches). At this distance, the whole section of bars and spaces is covered by a single pixel. Recall that nothing fi ner than a pixel can be recorded. So no bars are seen by the camera. The frame is 3000 inches wide
FIGURE 6.5
Resolution of an Object for a Sensor Chip and Frame Width. The degree to which a digital camera can render detail in an object
depends on the number of
pixels and the width of the frame at the plane of the object. The sensor chip is shown on the right. T
o the left of the lens is the object as it would be seen by the chip due to
distance of the object from the camera. T
o the left of this is a representation of the chip as it would project through the len
s onto the object at its apparent size. The leftmost
column show a rendition of what the resulting image would be like. Up close and the bars are well reproduced. At some intermediary
, critical distance, the bars are distinguishable but blurry
. At greater distance there is just a faint blu
r.
C H A P T E R 6 : Resolution
(250 feet). We have shown that at a distance of fi ve inches from the camera, the image frame is three inches wide and the camera can easily resolve all three bars and both spaces. And, when the target is 5000 inches from the camera, the frame is 3000 inches wide and it is totally impossible for the camera to see any of the bars or spaces. It sees only a gray that is a bit darker than the white background. Somewhere between those extremes there is a point where the camera just resolves the bars and spaces, and beyond that point, the bars are no longer separate and distinct. The frame width associ- ated with just resolving the bars is called the critical frame width . The ques- tion is, where is that point? If we know that, we will be able to determine what width bars and spaces can be resolved for a given frame width.
There is a branch of mathematics known as Sampling Theory, and one of the key tenets of this fi eld is the Nyquist Sampling Frequency. This holds that if we take two samples per wavelength, and reconstruct using the sinc function, we will achieve adequate sampling. This does not hold for digital photography. The Nyquist requirement would amount to one pixel per dark bar and one per white space. It would also require that the printing system utilize the sinc function to make the print. Neither requirement works in photographic applications. Referring to Figure 6.6 , note that when the sam- pling frequency is twice the wavelength (one bar and one space) the bars are resolved only if the one pixel is over a bar and the next one is over a space,
FIGURE 6.6 Sampling at Two Pixels per Line Pair. The common assumption is that only two pixels are required per line pair. This will work if the lines are well lined up with the columns of pixels. If not, the approach will disintegrate to the point where the lines are aligned with the junctions between the columns of pixels, shown at the top. Since each pixel sees half line and half background, and since the pixels average what light falls on them, each pixel will see the same thing—the average of the overall frame. This is shown at the bottom.
or at least close to this situation. If one pixel is over the junction of a bar and its adjacent space, then the next pixel will be over the junction of the next bar and space. The result will be that both pixels will register some mid-level gray value and there will be virtually no reproduction of the separate bars. As for the sync function, no photographic system uses this function.
Note in Figure 6.5 , in which there are three pixels per wavelength, the bar and the space will be represented as separate no matter the alignment, and there is no sinc function requirement for the printer. Practical experience has shown that three pixels per line pair is a robust requirement for a digital cam- era being able to reproduce a line pair of one bar and one of the adjacent spaces. With this requirement known it is possible to calculate the resolution of a digi- tal camera. We will refer to this factor as the pixel conversion effi ciency ; for a wide range of digital cameras it has been found to be equal to three.
The camera just described has 3000 pixels per frame width. Since three pixels are required for robust resolution conditions, the camera is capable of 3000 pixels divided by 3, which is equal to 1000 line pairs per frame width. The test target assumed has bars and spaces of 0.2 inches each. One line pair will be 1 bar plus 1 space, or 0.2 ⫹ 0.2 ⫽ 0.4 inches. We require 3 pixels per 0.4 inches. Since there are 3000 pixels, this amounts to a frame width of 3000 * 0.4 ⫽ 1200 inches. If the width of the frame is W o (in distance units such as inches or millimeters) and the number of pixels per frame width is P w , then the size (in the same units as W o ) of a minimally resolved line pair, D will be given by:
D W P o w ⫽ ⫻3
Note that line pairs per unit length (LPL) is equal to the reciprocal of D, so
LPL P W w o ⫽ ⫻3
This clearly shows that as the number of pixels increases, the number of line pairs per unit length will increase, and as the frame width increases, it will decrease.
The level of detail that can be resolved depends upon the frame width, which in turn is governed by the distance from the camera to the object. With a camera that has 4000 pixels per frame width, we would be able pho- tograph a full hand print that is 5 ⫻ 9 inches and still be able to see ridge detail of 6/1000 of an inch. Since ridge detail is about 10/10,000 of an inch, the shot will work well. If the camera had only 2000 pixels per frame width, the same shot would be limited to detail of 12/1000 of an inch with less quality of reproduction of the detail. To get the detail, the photographer would be required to move the camera closer and settle for a smaller frame width.
C H A P T E R 6 : Resolution
Taking the lens formulae from Chapter 3, it is possible to perform some additional interesting calculations. We set up to take a photo with a cam- era with a given focal length, F, and the object frame width is W o , and the distance from the camera lens to the object is d o . Putting these into the lens formulae from Chapter 3, the effective width of the sensor chip, W i , must be: so ⫽ si* (do⫺F F,)/ or wo ⫽ wi* (do⫺F F)/ s s d F F W W d F F W W F d F o i o o i o i o o ⫽ ⫻ ⫺ ⫽ ⫻ ⫺ ⫽ ⫻ ⫺ ( ) , ( ), ( ) or or Finally, if the chip has P w pixels across its width, then the width of each pixel, W p , is given by: W W P W F P d F p i W o W o ⫽ ⫽ ⫻ ⫻( ⫺ )
Most digital cameras have square pixels, so the height of each pixel can be assumed to be the same as the width. The same process can be used again, working in the height dimension instead of the width. Some cameras have octagonal pixels, and so different methods must be used. Video cam- eras have rectangular pixels in which the width is 90% of the height. Again, modifi ed techniques must be used.