HP 1630A Operating And Programming Manual page 116

Model 1630A/D/G Using HP-113 or HP-IL Interface
Referring to the real format arrangement example, byte 18 is the exponent byte . Byte 18 is decoded as :
Since the binary point is to the left of byte 19, the first place to the right is 2-13 . The rest of the number (bytes 19
through 21) can be translated as follows:
Byte 19 Byte 20 Byte 21
Hexadecimal: 9 8 9 1 5 E
Binary :
1001 1000 1001 0001 0101 1110
L.-1
X 2-35
~1 X 2-34
1 X 2-33
1 X 2-32
1 X 2-30
1 X 2-28
1 X 2-24
1 X 2-21
1 X 2-17
1 X 2-16
1 X 2-13
The sum of the entire column to the right is 145.44 Asec total accumulated time .
Time (T) Format
Another example involving the Timing Trace Learn String illustrates the prior discussion pertaining to
multiplication by a standard time increment . In this example, the timing trace was acquired with a 10 nsec
sample period with a mean time between the x and o cursors of 1 .52 btsec .
Parameter: Sample Period (Timing Trace Learn String)
Byte Position : 11 12
Format : (T) (T)
Hexadecimal: 03
00
From the listing for time in the Data Format section of this chapter, this translates to 10 nsec .
Parameter: Mean Time Between Cursors (Timing Trace Learn String)
Byte Position :
Format :
Hexadecimal:
Binary :
24
(R)
1 0
0001 0000
25
(R)
9
8
.1 I 001 1
A
8 places
positive number
10011000=128+16+8=152
At 10 nsec : 152 X 10 nsec = 1 .52 usec mean time between cursors.
9-9
26 27
(R) (R)
9 4 7 B
1001 0010 0111 1011
Hexidecimal:
E 8
Binary : 1110
1000
L-indicates number is positive
indicates exponent is negative
7-bit 2's complement : 111
0100
Invert : 000 1011
Add one: +1
Exponent :
0001 1100 = -12