Question

1.Describe the effects of the FCS length on error detection performance

2.Describe the effect of frame length on the CRC performance

3.Summarize the burst error detection performance for a given FCS length in terms

of burst size and probabilities.

Answer 1

1)

Frame check sequence (FCS) errors indicate that frames of data are being corrupted during transmission. FCS error count is the number of frames that were transmitted/received with a bad checksum (CRC value) in the Ethernet frame. These frames are dropped and not propagated onto other ports.

FCS errors can be caused by any of these options:
1. Collisions
2. Duplex mismatch
3. Hardware (NIC, cable or port)
4. Frames with bad FCS being generated by the connected device

Resolution
To resolve this issue, perform these steps:
1. If possible, hard set the speed and duplex settings on both the port and the connected device.
2. Check the cable with a tester or replace it with a known good cable.
3. Check the NIC drivers on the end device to make sure they support the required duplex and speed settings.
4. Check the port by connecting the end device to a different port on the switch.

Naturally the FCS length is not a factor on error detection performance. The problem only with the damage of the cable. If cable damaged then we have to change it in to another one.

2)

Cyclic Redundancy Check (CRC) Error indicates when data is corrupted. Calculating from all data, CRC validates packets of information sent by devices and verifies it against the data extracted, ensuring its accuracy. ... If the two values do not exactly match a CRC error occurs.

Programmers have used the Cyclic Redundance Check (CRC) algorithm for years to uncover errors in a data transmission. It turns out that you can also use CRCs to correct a single-bit error in any transmission.

3)

a burst error or error burst is a contiguous sequence of symbols, received over a communication channel, such that the first and last symbols are in error and there exists no contiguous subsequence of m correctly received symbols within the error burst

There is more than one upper bound on the achievable code rate of linear block codes for multiple phased-burst correction (MPBC). One such bound is constrained to a maximum correctable cyclic burst length within every subblock, or equivalently a constraint on the minimum error free length or gap within every phased-burst. This bound, when reduced to the special case of a bound for single burst correction, is the Abramson bound (a corollary of the Hamming bound for burst-error correction) when the cyclic burst length is less than half the block length.[3]

Theorem (number of bursts). For {\displaystyle 1\leqslant \ell \leqslant {\tfrac {1}{2}}(n+1),} over a binary alphabet, there are {\displaystyle n2^{\ell -1}+1} vectors of length {\displaystyle n} which are bursts of length {\displaystyle \leqslant \ell }.[1]

Proof. Since the burst length is {\displaystyle \leqslant {\tfrac {1}{2}}(n+1),} there is a unique burst description associated with the burst. The burst can begin at any of the {\displaystyle n} positions of the pattern. Each pattern begins with {\displaystyle 1} and contain a length of {\displaystyle \ell }. We can think of it as the set of all strings that begin with {\displaystyle 1} and have length {\displaystyle \ell }. Thus, there are a total of {\displaystyle 2^{\ell -1}} possible such patterns, and a total of {\displaystyle n2^{\ell -1}} bursts of length {\displaystyle \leqslant \ell .} If we include the all-zero burst, we have {\displaystyle n2^{\ell -1}+1} vectors representing bursts of length {\displaystyle \leqslant \ell .}

Theorem (Bound on the number of codewords). If {\displaystyle 1\leqslant \ell \leqslant {\tfrac {1}{2}}(n+1),} a binary {\displaystyle \ell }-burst error correcting code has at most {\displaystyle 2^{n}/(n2^{\ell -1}+1)} codewords.

Proof. Since {\displaystyle \ell \leqslant {\tfrac {1}{2}}(n+1)}, we know that there are {\displaystyle n2^{\ell -1}+1} bursts of length {\displaystyle \leqslant \ell }. Say the code has {\displaystyle M} codewords, then there are {\displaystyle Mn2^{\ell -1}} codewords that differ from a codeword by a burst of length {\displaystyle \leqslant \ell }. Each of the {\displaystyle M} words must be distinct, otherwise the code would have distance {\displaystyle <1}. Therefore, {\displaystyle M(2^{\ell -1}+1)\leqslant 2^{n}} implies {\displaystyle M\leqslant 2^{n}/(n2^{\ell -1}+1).}

Theorem (Abramson's bounds). If {\displaystyle 1\leqslant \ell \leqslant {\tfrac {1}{2}}(n+1)} is a binary linear {\displaystyle (n,k),\ell }-burst error correcting code, its block-length must satisfy:

{\displaystyle n\leqslant 2^{n-k-\ell +1}-1.}

Proof: For a linear {\displaystyle (n,k)} code, there are {\displaystyle 2^{k}} codewords. By our previous result, we know that

{\displaystyle 2^{k}\leqslant {\frac {2^{n}}{n2^{\ell -1}+1}}.}

Isolating {\displaystyle n}, we get {\displaystyle n\leqslant 2^{n-k-\ell +1}-2^{-\ell +1}}. Since {\displaystyle \ell \geqslant 1} and {\displaystyle n} must be an integer, we have {\displaystyle n\leqslant 2^{n-k-\ell +1}-1}.

Remark. {\displaystyle r=n-k} is called the redundancy of the code and in an alternative formulation for the Abramson's bounds is {\displaystyle r\geqslant \lceil \log _{2}(n+1)\rceil +\ell -1.}