CRC Evaluation for a Given Frame and Generator Polynomial

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Nov 4, 2026

To evaluate the CRC for a transmitted frame, we treat the bit string as a polynomial over GF(2) (so addition/subtraction are XOR operations). For a generator polynomial G(x) of degree $5$ , we append $5$ zero bits to the message, divide by $G(x)$ , and the remainder (degree $<5$ ) becomes the CRC checksum that is appended to form the final codeword.

Generator polynomial:
$G(x)=x^5+x^4+x^2+1$

Frame (message) bits (as given):

Data: $1010011011$
Extra bits in the prompt: $01$ (we will interpret the required message for CRC as the data field only, i.e., $1010011011$ ; the $01$ is commonly presented as an additional example suffix but CRC inputs must be explicit—here the CRC is computed for the frame/message $1010011011$ against the given $G(x)$ ).

We compute CRC-5 for data $1010011011$ using $G(x)$ .

Mermaid view of the standard CRC pipeline:

CRC Long Division (GF(2)) to get the 5-bit remainder

1
Step 1
Since $G(x)$ has highest power $x^5$ , the CRC degree is $k=5$ . The bit pattern of $G(x)=x^5+x^4+x^2+1$ is 110101 (because coefficients from $x^5$ to $x^0$ are: 1,1,0,1,0,1).
2
Step 2
Message data: 1010011011. Append 5 zeros → 1010011011 00000.
3
Step 3
We align the generator’s leading 1 with the leftmost 1 of the current dividend segment. Subtraction over GF(2) is XOR, so when a generator is applied, we XOR with 110101.
4
Step 4
Proceed left-to-right across the dividend. Each time the current leftmost bit is 1, XOR the next 6 bits with 110101; if it is 0, XOR is effectively skipped. After finishing all positions, the final 5 remaining bits are the remainder.
5
Step 5
Carrying out the XOR steps yields the remainder: 11101.
6
Step 6
Codeword = data + remainder = 1010011011 11101.

Verification by constructing the dividend remainder relationship

Let the augmented message be:

Data polynomial times $x^5$ corresponds to appending $5$ zeros: $M'(x)=M(x)\,x^5$

When we divide $M'(x)$ by $G(x)$ we obtain: $M'(x)=Q(x)\,G(x)+R(x)$ where:

$\deg(R)<5$ , so $R(x)$ corresponds to exactly 5 bits. The computed remainder bits are $R=11101$ .

Bit-level mapping from remainder to CRC bits

Remainder $R=11101$ means the CRC checksum to append is:

CRC bits: 11101

Thus the transmitted/codeword frame becomes:

1010011011 11101

Pro Tip: Polynomial bit ordering

When converting $G(x)$ to bits, use coefficients from highest power down to $x^0$ . For $x^5+x^4+x^2+1$ : bits are 1 (x^5), 1 (x^4), 0 (x^3), 1 (x^2), 0 (x^1), 1 (x^0) → 110101.

Warning: Clarify what the extra '01' refers to

CRC is computed for a specific message bitstring. If '01' is meant to be appended to the message before CRC, the dividend changes. In this section, the CRC is evaluated for the stated frame/message '1010011011' using $G(x)$ . If instead the full input is '1010011011 01', you must recompute with a different augmented dividend.

CRC Evaluation Roadmap

Generator setup

Step 1

Convert $G(x)$ to a bit pattern (MSB to LSB)."

Augment message

Step 2

Append $k=5$ zeros to the data."

GF(2) division

Step 3

Perform polynomial long division using XOR."

Remainder becomes CRC

Step 4

Take the final 5 remainder bits and append to form the codeword."

CRC Output Summary

Remainder length is 5 bits because deg( $G(x)$ )=5.

Common CRC evaluation details

CRC Quick Checks

1 / 5

Question · Term

For $G(x)$ with highest term $x^5$, how many CRC bits are produced?

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Answer · Definition

5 bits, because CRC size equals deg( $G(x)$ ).

Knowledge Check

Question 1 of 4

Q1Single choice

Given $G(x)=x^5+x^4+x^2+1$ , how many CRC bits are appended?

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