Hi
In the application report "Throughput Performance Guide for C66x KeyStone Devices"--July 2012, on the page 43 Table 30 showing below, it said for the Payload size 8 bytes, it has 1024 bits without overhead. From my point of view 8 bytes should be equal with 64 bits. How does 1024 bits without overhead calculated out? Thank You.
Hi Xining,
In the description of the DirectIO tests, step 5 mentions that 16 transactions are sent. So 8 byte payload x 8 bits/byte x 16 transactions yields the 1024 bits without taking into account overhead. I've included that portion of the document below for reference.
Throughput Performance Guide for C66x KeyStone Devices: http://www.ti.com/lit/pdf/sprabk5
Hope that helps. Let us know if you have any other questions.
Thanks,
Clinton
