-
Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students! -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
Cyclic Redundancy Check (1 Viewer)
- Thread starter Becky222
- Start date
samthebear
Member
Using a 16-bit CRC:
using the example of [17, 24, 6, 73]
Step 1: treat the entire message block as a single binary number. So you'll get:
(17) 00010001, (24) 00011000, (6) 00000110, (73) 01001011 = 286 787 147 in decimal (you need to convert the binary numbers into decimal for the next step to work.)
Step 2: Divide by 69 665 (because you're using a 16-bit CRC. If you're using a 36-bit CRC divide the number by 39 882 463) and the result should be 4116 with a remainder of 46 007.
Step 3: CRC-16 = 46 007 (the remainder)
I've taken most of this out of the Jacaranda HSC course IPT textbook.
using the example of [17, 24, 6, 73]
Step 1: treat the entire message block as a single binary number. So you'll get:
(17) 00010001, (24) 00011000, (6) 00000110, (73) 01001011 = 286 787 147 in decimal (you need to convert the binary numbers into decimal for the next step to work.)
Step 2: Divide by 69 665 (because you're using a 16-bit CRC. If you're using a 36-bit CRC divide the number by 39 882 463) and the result should be 4116 with a remainder of 46 007.
Step 3: CRC-16 = 46 007 (the remainder)
I've taken most of this out of the Jacaranda HSC course IPT textbook.
samthebear
Member
you work it from the bits so you've got:
[128, 64, 32, 16, 8, 4, 2, 1] (this is one byte - each individual piece is called a bit) and to get a number (say 17) you put a 1 under the bit to add up the numbers that make 17. in this case, [0(128)0(64)0(32)1(16)0(8)0(4)0(2)1(1)] everywhere else with a 0 indicated means that you dont add those numbers.
So 16 and 1 have a 1 under it so you add them together and you'll get 17. And you just keep doing that for all the numbers to get your binary numbers.
[128, 64, 32, 16, 8, 4, 2, 1] (this is one byte - each individual piece is called a bit) and to get a number (say 17) you put a 1 under the bit to add up the numbers that make 17. in this case, [0(128)0(64)0(32)1(16)0(8)0(4)0(2)1(1)] everywhere else with a 0 indicated means that you dont add those numbers.
So 16 and 1 have a 1 under it so you add them together and you'll get 17. And you just keep doing that for all the numbers to get your binary numbers.