Data quantization results

To test the validity of reducing the data precission I simulated a reduced byte representation by first normalizing the dataset so its maximum value is one and then using various quantization intervals. Figure 1 shows three different quantization levels. The left collumn is the migration result. The center collumn is the difference between the full precission migration and the reduced precission image clipped at the same level as the migration. The right collumn is the same as the center panel but clipped at 1/10 of the value. The top row is quantizated at .25 (two bits)., the center row at .0625 (four bits), and the bottom row at .015625 (six bits). At two bits you can see some very minor differences in structure and a significant difference in amplitude the quanitized image. At four bits the error is only noticable at 1/10th the clip, and at six bits the error has almost completely disapeared.

 

bitinput
Figure 1 The result of migrating with a reduced precision input. The left collumn is the migration result. The center collumn is the difference between the full precission migration and the reduced precission image clipped at the same level is the migration. The right collumn is the same as the center panel, clipped at 1/10 the value. The top row is quantizated at .25 (two bits)., the center row at .0625 (four bits), and the bottom row at .015625 (six bits).

If we assume that our migration aperture is 200 samples with 50 offsets and 2 non-zero elements in $\bf G$ in time we get an error reduction of about 140 through the migration process. The image on the screen is an eight bit representation of the migration result. The data is clipped at $40\%$ of the maximum value. As a result we should need $\frac{256}{.4}=640$ quantization intervals without the migration process and a little more than $\frac{640}{140}\approx 5$ after. Which is confirmed by our test. Two bits wasn't sufficient four bits was. If we decrease our clip (right panel) by a factor of ten we should need a little more than $\frac{640*10}{140}\approx 46$ intervals which explains why the error is elimiated using 6 bits.

If we form angle gathers rather than just the zero offset image we are summing over very few points in the offset space to form each angle. If we assume each angle is on averge formed from two offsets we get an error reduction of $\sqrt{20*2*2}=30$.So an average should take two more bits to form a good angle domain image. Figure 2 confirms this hypothesis. The collumns follow the same structure as Figure 1. The top row is equivalent to 4 bit representation, the center row 6 bit, and to bottom row 8 bit. As predicted two more bits are needed. The top panel shows significant error at the migration clip, the center panel only shows error at $\frac{1}{10}$ the clip and the bottom panel shows no meaningful error.

 

angle
Figure 2 The result of migrating with a reduced precision input. The left collumn is the migration result. The center collumn is the difference between the full precission migration and the reduced precission image clipped at the same level is the migration. The right collumn is the same as the center panel, clipped at 1/10 the value. The top row is equivalent to 4 bit representation, the center row 6 bit, and to bottom row 8 bit.

For 3-D problems the number of bits that are needed should further reduce. Instead of summing over 15000 points to form a single image space in this small 2-D synthetic the number quickly jumps to more than 10000000, increasing the error reduction factor to more than 3000 for post-stack images and 300 for prestack images. As a result even less precision is needed to achieve the same result. A significant caveat is where to begin the quantization intervals. In areas with very strong noise (ground roll) or a strong reflector (salt) a clip before introducing the quantization intervals is a good idea.