Cyclic Redundancy Check: burst error detection capability

Question

For a given polynomial generator $G(X) = 1 + \cdots + X^r$ of degree $r$, it can detect a burst of errors of length $r$.

So if the polynomial generator does not contain the $+1$ term, I guess it can detect a burst of errors of length $r - i$, where i is the degree of the least significant monomial (for example $G(X) = X^9 + X^7 + X^5$, could detect a burst of $9 - 5 = 4$ errors).

Is this statement correct? If so, how could we prove it mathematically?

Answer 1

Since a codeword is a multiple of $G(X)$ an error pattern $E(X)$ added to a codeword $C(X)$ is detected if and only if $C(X)+E(X)$ is not divisible by $G(X).$ This is equivalent to saying an error pattern $E(X)$ is not detected if and only if $G(X)$ divides $E(X).$

Write your proposed polynomial as $$ G(X)=X^5(1+X^2+X^4)=X^5(1+X+X^2)^2 $$ where the arithmetic is over $GF(2).$ This polynomial is divisible by $(1+X+X^2)$ so it will not detect burst errors of the form $\cdots 01110\cdots$ of length 3.

Answer 2

So if the polynomial generator does not contain the +1 term....

That cannot happen. The polynomial generator always has a not null independent term.

Elsewhere, you could factor $g(x) = x \, h(x)$ and then $h(x)$ (being a shift of $g(x)$) would be a valid codeword... which cannot be written as $g(x) u(x)$, and hence $g(x)$ would not be a valid polynomial generator .