Aufgaben:Exercise 1.6: Cyclic Redundancy Check

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Formation of the CRC4 checksum

The synchronization happens at the primary multiplex connection in each case in channel  $0$  – the synchronization channel - of each frame:

• For odd time frames  (number 1, 3, ... , 15)  this transmits the so-called "frame password" with the fixed bit pattern  $\rm X001\hspace{0.05cm} 1011$.
• Each even frame  (with number 2, 4, ... , 16)  on the other hand contains the "message word"  $\rm X1DN\hspace{0.05cm}YYYY$.
• Error messages are signaled via the  $\rm D$ bit and the   $\rm N$ bit, and the four  $\rm Y$ bits are reserved for service functions.

The  $\rm X$ bit is obtained in each case by the CRC4 method, the implementation of which is shown in the diagram:

• From eight input bits each - in the entire exercise the bit sequence  $\rm 1011\hspace{0.05cm} 0110$ is assumed for this purpose - the four check bits  $\rm CRC3$, ... , $\rm CRC0$  are obtained by modulo-2 additions and shifts, which are added to the input word in this order.
• Before the first bit is shifted into the register, all registers are filled with zeros:

$${\rm CRC3 = CRC2 =CRC1 =CRC0 = 0}\hspace{0.05cm}.$$

• After eight shift clocks, the four registers  $\rm CRC3$, ... , $\rm CRC0$  contains the CRC4 checksum.

The taps of the shift register are  $g_{0} = 1, \ g_{1} = 1, \ g_{2} = 0, \ g_{3} = 0$  and  $g_{4} = 1$.

• The corresponding generator polynomial is:

$$G(D) = D^4 + D +1 \hspace{0.05cm}.$$

• The CRC4 checksum on the transmission side is also obtained as the  remainder  of the polynomial division

$$(D^{11} +D^{9} +D^{8}+D^{6}+D^{5})/G(D) \hspace{0.05cm}.$$

• The divisor polynomial results from the input sequence and four appended zeros:  $\rm 1011\hspace{0.09cm} 0110\hspace{0.09cm} 0000$.

The CRC4 check at the receiver according to subtask  (4)  can also be represented by a polynomial division. It can be implemented by a shift register structure in a similar way as the CRC4 checksum is obtained at the transmitting end.

Notes:

• The exercise belongs to the chapter  "ISDN Primary Multiplex Connection" .
• To solve the exercise, some basic knowledge of  "channel coding  is required.

Questions

1 Which result  $E(D)$  and which remainder  $R(D)$  results from the polynomial division $(D^{11} + D^{9} + D^{8} + D^{6} + D^{5}) : (D^{4} + D + 1)$ ?

	$E(D) = D^{5} + D^{3} + 1, \hspace{2.13cm}R(D) = D^{3} + D$,
	$E(D) = D^{7} + D^{5} + D^{3} + 1, \hspace{1cm}R(D) = D^{3} + D + 1$,
	$E(D) = D^{7} + D^{5} + D^{3} + 1, \hspace{1cm}R(D) = 0$.

2 What is the CRC checksum in the present case?

$\rm CRC0 \ = \ $

$\rm CRC1 \ = \ $

$\rm CRC2 \ = \ $

$\rm CRC3 \ = \ $

3 The following bit sequences arrive at the receiver,   $\text{eight information bits each plus (CRC3, CRC2, CRC1,CRC0)}$.
When is there no bit error?

	$1011 \hspace{0.1cm}0010\hspace{0.08cm} 1011$,
	$1011 \hspace{0.1cm}0110 \hspace{0.08cm}1011$,
	$1011 \hspace{0.1cm}0110\hspace{0.08cm} 1001$.

4 Which received bit sequences were falsified during transmission?

	$0000 \hspace{0.1cm}0111 \hspace{0.1cm}0010$,
	$0000 \hspace{0.1cm}1111\hspace{0.1cm} 0010$,
	$0000\hspace{0.1cm} 1111\hspace{0.1cm} 1010$.

Solution

Solution

(1)  Solution 2 is correct:

• Due to the largest numerator exponent $(D^{11})$ and the highest denominator exponent $(D^{4})$, the first suggestion $E(D) = D^{5} + D^{3} + 1$ can be excluded as the result   ⇒   $E(D) = D^{7} + D^{5} + D^{3} + 1$.

• Modulo-2 multiplication of $E(D)$ by the generator polynomial $G(D) = D^{4} + D + 1$ yields:

$$E(D) \cdot G(D) \ = \ (D^7+ D^5+D^3+1)\cdot (D^4+ D+1) \ = D^{11}+D^8+D^7+D^9+D^6+D^5+D^7+D^4+D^3+D^4+ D+1 \hspace{0.05cm}.$$

• It must be taken into account here that for modulo-2 calculations $D^{4} + D^{4} = 0$. This results in the following remainder:

$$R(D) = D^{11}+D^9+D^8+D^6+D^5- E(D) \cdot G(D) = D^3+D+1 \hspace{0.05cm}.$$

Register assignments for CRC4

(2)  From the result of subtask (1) follows:

$${\rm CRC0 = 1},\hspace{0.2cm}{\rm CRC1 = 1},\hspace{0.2cm}{\rm CRC2 = 0},\hspace{0.2cm}{\rm CRC3 = 1}\hspace{0.05cm}.$$

The table shows a second way of solution: It contains the register assignments of the given circuit at the times 0, ... , 8.
(3)  Only solution 2 is correct:

• The receiver divides the polynomial $P(D)$ of the receive sequence by the generator polynomial $G(D)$.
• If this modulo-2 division returns the remainder $R(D) = 0$, then all 12 bits were transmitted correctly.
• This is true for the second solution, as a comparison with subtasks (1) and (2) shows. It is valid without remainder:

$$(D^{11}+D^9+D^8+D^6+D^5+D^3+D+1) : (D^4+ D+1)= D^7+D^5+D^3+1 \hspace{0.05cm}.$$

• In solution 1 the 6th information bit was corrupted, in solution 3 the CRC1 bit.

(4)  Solutions 1 and 3 are correct:

• The diagram illustrates the modulo-2 divisions for the given received sequences in simplified form (with zeros and ones).
• You can see that only for sequence 2 the division without remainder is possible.
• In written form, the polynomial divisions are:

$$\ (1) \ \hspace{0.2cm}(D^6+D^5+D^4+1) : (D^4+ D+1)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Rest}\hspace{0.15cm}D^3+ D+1\hspace{0.05cm},$$

$$\ (2) \ \hspace{0.2cm}(D^7+D^6+D^5+D^4+1) : (D^4+ D+1)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm ohne \hspace{0.15cm}Rest}\hspace{0.05cm},$$

$$\ (3) \ \hspace{0.2cm}(D^7+D^6+D^5+D^4+D^3+1) : (D^4+ D+1) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Rest}\hspace{0.15cm}D^3\hspace{0.05cm}.$$

Polynomial division of the three received sequences