Agilent Technologies N5183A MXG User Manual
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Agilent N518xA, E8663B, E44x8C, and E82x7D Signal Generators Programming Guide
295
Creating and Downloading User-Data Files
User File Data (Bit/Binary) Downloads (E4438C and E8267D)
AUTOGEN_PRAM_1 file size. If you select different user files for the timeslots within a frame, the
user file that produces the largest number of frames determines the size of the AUTOGEN_PRAM_1
file.
Use this procedure to calculate the volatile memory usage for a GSM signal with two active timeslots
and two user binary files. One user file, 57 bytes, is for a normal timeslot and another, 37 bytes, is
for a custom timeslot.
1.
Determine the total number of bits per timeslot.
A GSM timeslot consists of 156.25 bits (control and payload data).
2.
Calculate the number of bits per frame.
A GSM frame consists of 8 timeslots: 8
× 156.25 = 1250 bits per frame
3.
Determine how many bytes it takes to produce one frame in the signal generator:
The signal generator creates a 32- bit word for each bit in the frame (1 bit equals 4 bytes).
4 x 1250 = 5000 bytes
Each GSM frame uses 5000 bytes of PRAM memory.
4.
Analyze how many timeslots the user file data will fill.
A normal GSM timeslot (TS) uses 114 payload data bits, and a custom timeslot uses 148 payload
data bits. The user file (payload data) for the normal timeslot contains 57 bytes (456 bits) and the
user file for the custom timeslot contains 37 bytes (296 bits).
Normal TS
456 / 114 = 4 timeslots
Custom TS
296 / 148 = 2 timeslots
NOTE
Because there is an even number of bytes, either a bit or binary file works in this scenario.
If there was an uneven number of bytes, a bit file would be the best choice to avoid data
discontinuity.
5.
Compute the number of frames that the signal generator will generate.
There is enough user file data for four normal timeslots and two custom timeslots, so the signal
generator will generate four frames of data.
6.
Calculate the AUTOGEN_PRAM_1 file size:
7.
Calculate the number of memory blocks that the AUTOGEN_PRAM_1 file will occupy:
Volatile memory allocates memory in blocks of 1024 bytes.
20000 / 1024 = 19.5 blocks
Number of Frames Bytes per Frame
4
5000
4 x 5000 = 20000 bytes