DØ Run IIb L1Cal Et(miss) Algorithm

updated: 06-Jun-06

Index

1. Inputs to the algorithm in firmware
2. The calculation and results
3. Examples

Inputs to the Algorithm in the Firmware

Inputs to the procedure are:

1. Energies summed in eta for each phi (16 bit integers)
2. sine and cosine (13 bit integers) for phi 2 to 5
   The sine an cosine have range 0-1 multiplied by 1024. (We can change the number 1024 to something else if needed.)
   Example: sine(eta)=0.2 will be represented by 1024*(2/10)=204.8=204
3. one common energy threshold (16 bit integer)

The Calculation and Results

• Exsum() = (e2()-thr)*cos2()+(e3()-thr)*cos3()+ (e4()-thr)*cos4()+(e5()-thr)*cos5()
• Eysum() = (e2()-thr)*sin2()+(e3()-thr)*sin3()+ (e4()-thr)*sin4()+(e5()-thr)*sin5()
• Etsum = (e2()-thr) + (e3()-thr) + (e4()-thr) + (e5()-thr)

Your results are then: exsum[28:0] eysum[28:0], 29 bit long and etsum [16:0] is 17 bit long. Our agreement (Hal and others) was that our results which we send to GAB will be 12 bit number and if the overflow is detected the numbers will be all ones(0xFFF). We agreed also on the rounding off the numbers. We need also normalize the numbers for the maximum energy deposits. We simply divide the results by 16 at the moment.

Bit usage is summarized in the following table:

Variable	Full Sum	Output Subset	Used for Roundoff
Ex,ySum	[28..0]	[25..14]	[13..2]
EtSum	[16..0]	[15..4]	[3..0]

Details of it are in the procedure multiply.vhd.

Examples

An example of what we are doing at the moment in multiply.vhd is

• Sin2=0.2 -> 204
• Sin3=0.3 -> 307
• Sin4=0.4 -> 409
• Sin5=0.5 -> 512

Let say e2=e3=e4=e5=1000 and thr=0. Then

• Eysum = (204*1000 + 307*1000 + 409*1000 + 512*1000)/1024
  = (204000 + 307000 + 409000 + 512000) /1024
  = (1430000)/1024 = 1396.484375
• Etsum = 1000+1000+1000+1000 = 4000

• Ey = 1396.484375/16 = 87
• Et = 4000/16 = 250