GE GFK-0804B manual Example CRC±16 Calculation, Address Function CRC±16

C

Example CRC±16 Calculation

As an example we will calculate the CRC±16 for RTU message, Read Exception Status

07). The message format is:

Address	Function	CRC±16
01	07

The Micro PLC transmits the rightmost byte (of registers or discrete data) first. The first bit of the CRC± 16 transmitted is the MSB. Therefore, in this example the MSB of the CRC polynomial is to the extreme right. The X16 term is dropped because it affects only the quotient (which is discarded) and not the remainder (the CRC characters). The generating polynomial is therefore 1010 0000 0000 0001. The remainder is initialized to all 1s.

In this example we are querying device number 1 (address 01). We need to know the amount of data to be transmitted. This information can be found for every message type on page C-8, RTU Message Length. For this message the data length is 2 bytes.

	TRANSMITTER
	CRC±16 ALGORITHM
MSB2	LSB2	Flag MSB2
Initial Remainder		1111 1111 1111 1111
XOR 1st data byte		0000 0000 0000 0001
Current CRC		1111 1111 1111 1110
Shift 1		0111 1111 1111 1111
Shift 2		0011 1111 1111 1111
XOR Gen. Polynomial		1010 0000 0000 0001
Current CRC		1001 1111 1111 1110
Shift 3		0100 1111 1111 1111
Shift 4		0010 0111 1111 1111
XOR Gen. Polynomial		1010 0000 0000 0001
Current CRC		1000 0111 1111 1110
Shift 5		0100 0011 1111 1111
Shift 6		0010 0001 1111 1111
XOR Gen. Polynomial		1010 0000 0000 0001
Current CRC		1000 0001 1111 1110
Shift 7		0100 0000 1111 1111
Shift 8		0010 0000 0111 1111
XOR Gen. Polynomial		1010 0000 0000 0001
Current CRC		1000 0000 0111 1110
XOR 2nd data byte		0000 0000 0000 0111
Current CRC		1000 0000 0111 1001
Shift 1		0100 0000 0011 1100
XOR Gen. Polynomial		1010 0000 0000 0001
Current CRC		1110 0000 0011 1101
Shift 2		0111 0000 0001 1110
XOR Gen. Polynomial		1010 0000 0000 0001
Current CRC		1101 0000 0001 1111
Shift 3		0110 1000 0000 1111
XOR Gen. Polynomial		1010 0000 0000 0001
Current CRC		1100 1000 0000 1110
Shift 4		0110 0100 0000 0111
Shift 5		0011 0010 0000 0011
XOR Gen. Polynomial		1010 0000 0000 0001
Current CRC		1001 0010 0000 0010
Shift 6		0100 1001 0000 0001
Shift 7		0010 0100 1000 0000
XOR Gen. Polynomial		1010 0000 0000 0001
Current CRC		1000 0100 1000 0001
Shift 8		0100 0010 0100 0000
XOR Gen. Polynomial		1010 0000 0000 0001
Transmitted CRC		1110 0010 0100 0001
		E	2	4	1

		RECEIVER1
		CRC±16 ALGORITHM
		LSB2 Flag
	Rcvr CRC after data		1110 0010 0100 0001
	XOR 1st byte Trns CRC 0000 0000 0100 0001
	Current CRC		1110 0010 0000 0000
0	Shift 1		0111 0001 0000 0000	0
1	Shift 2		0011 1000 1000 0000	0
	Shift	3	0001 1100 0100 0000	0
	Shift	4	0000 1110 0010 0000	0
0	Shift 5		0000 0111 0001 0000	0
1	Shift 6		0000 0011 1000 1000	0
	Shift	7	0000 0001 1100 0100	0
	Shift	8	0000 0000 1110 0010	0
0	XOR 2nd byte trns CRC 0000 0000 1110 0010
1	Current CRC		0000 0000 0000 0000
	Shift 1±8 yields		0000 0000 0000 0000
			ALL ZEROES FOR RECEIVER
0			FINAL CRC±16 INDICATES
1			TRANSMISSION CORRECT!

1

1Remember, the receiver processes

1incoming data the same way as the

1transmitter. The example for the receiver begins when all data bits but not the

0transmitted CRC have been received

1correctly. Therefore, the receiver CRC

should be equal to the transmitted CRC at 01 this point. When this occurs, the output of the CRC algorithm will be zero indicating

1that the transmission is correct.

The transmitted message with CRC would then be:

		Address	Function		CRC±16
		01	07	41		E2
	2 The MSB and LSB references are to the data bytes only, not the CRC bytes. The CRC
	MSB and LSB order are the reverse of the data byte order.
GFK±0804B	Appendix C RTU Protocol						C-7