Burst transfer

SystemVerilog
1

following is the example of burst transfer in AHB…
my question is what happened in addresses 0x11. 0x13, 0x15…
I think that first byte is transfered to 0x10 and second in 0x11. is it right ?

example2:- WRAP16 - HALFWORD
steps:
1> count the size of transfer 16 * 2 = 32 bytes.
2> assume that the memory is divided in the segments of 8 bytes.
so trnsfers can be stored from 0 to 31 or 32 to 63 or 64 to 95 etc. address locations (here address locations are in decimal).
3>address 0x0E (starting address)
0x10
0x12
0x14
0x16
0x18
0x1A
0x1C
0x1E
0x00
0x02
0x04
0x06
0x08
0x0A
0x0C

2

In reply to keshav_chokshi:

As it is WRAP transaction, address will wrap at 32 byte boundaries i.e. 0x1F.
And each of 16 beat transactions requires two addresses for two bytes (as HSIZE represents half word).