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Could someone explain the attached table and how the values are calculated? (the table is taken from the official usb 2.0 specification page 42).

What I could understand is just the relation between the bytes/frame and bytes/second (as in Full Speed the second consists of 1000 frames). But how is the max transfers, the bytes remaining, frame bandwidth per transfer, and the bytes/frame calculated?

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But how is the max transfers,

The max transfers is the number of complete packets of the chosen size that can be sent in a frame.

the bytes remaining,

Once you've sent max transfers * frame size, this is the number of bytes that are left over in the frame.

and the bytes/frame calculated?

Bytes/frame are the number of payload bytes sent in the frame, or the number of bytes/transfer * max transfers.

So, if you send one byte, there's 46 bytes in the packet that carries that byte. The max transfers is the number of packets in a frame, which is 1495 bytes, or 1495/46 = 32.5. We're not counting the half-packet, so that becomes 32, which takes up 32*46=1472 bytes. 23 of the 1495 are unused, so that's bytes remaining. Your bytes/frame, since each transfer passed only one byte, is 32*1=32.

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  • \$\begingroup\$ Thank you for the answer! I almost got everything, just few questions yet. How did you get this 1495? Shouldn't it be Max Speed (12 MBit) / 8 (to get it in byte) / 1000 to get per frame, which gives 1500 byte/frame? Could you also show where did frame bandwidth per transfer come from? \$\endgroup\$ – Mohammed Noureldin Aug 23 at 20:14
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    \$\begingroup\$ I got 1495 from calculations from the table...I'm not sure where the other 5 bytes went, maybe into a frame sync. The frame bandwidth per transfer is the percentage of the 1495 bytes are used for each packet. \$\endgroup\$ – Cristobol Polychronopolis Aug 23 at 20:34
  • \$\begingroup\$ Thanks a lot! I guess the 5 bytes are disappeared because of the rounding of the fractions according to your calculations. \$\endgroup\$ – Mohammed Noureldin Aug 23 at 20:38

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