The DIAGnostic:MEMory:ADDResses? command returns a list of 32-bit
integers for each segment in memory. In each 32-bit list, bit 1 is the aborted
flag and bit 0 is the memory wrapped flag. Bit 31 through
bit 2 is the value of the address counter for that segment. Thus, to obtain the
memory address only, a divide by 4 (right shift of 2) of the 32-bit list must
be done. Since the address counter points to the memory location where the
next reading will be stored, the address should be decremented by 1 before
use.
As an example, assume: ARM:COUN 5; SENS:SWE:POIN -20;
TRIG:COUN 35
Because of the ARM:COUN specified memory is divided into eight
segments (Table 3-5), each of which could contain up to 65536 readings.
For ARM:COUN 5, only five of the eight possible segments will be used,
starting with the segment that begins at address 0 and ends at address 65535.
The ending address for each of the five segments can be calculated from the
equation:
ending segment address = segment_number * 65536 - 1
This yields the following ending addresses for the five segments:
segment 1 = 65535
segment 2 = 131071
segment 3 = 196607
segment 4 = 262143
segment 5 = 327679
Assume you want to read the data from segment number 4. Then, the
number of readings = 35 + 1 = 36 (padded to make divisable by 4). The
first reading in the segment is at address:
first address = ending segment address - (number of readings - 1)
262143 - (36 - 1) = 262108
If the readings in the memory segment did not wrap around, the address
262108 and count 35 would be specified in the DIAGnostic:FETCh?
command.
If, for example, readings in the segment did wrap around, then the starting
address specified in the DIAGnostic:FETCh? command will not be the first
address in the segment where a reading is stored. To determine the starting
address, use the DIAGnostic:MEMory:ADDRess? command and get the 32
bits representing the 4th segment (this would be the 13th through 16th bytes
in the block of data returned).
144 Understandin
the HP E1429 Di
itizer Chapter 3
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