Solution:

• There are ' q ' stage shift registers to generate a ' q ' bit syndrome vector. Initially, all the shift register contents are zero $\&$ the switch is closed in position 1.

• The received vector $\mathrm{Y}$ is shifted bit by bit into the shift register. The contents of flip flops keep changing according to input bits of Y and values of $g 1, g 2$, etc.

• After all, the bits of Y are shifted, and the ' q ' flip flops of the shift register contain the q-bit syndrome vector. The switch is then closed to position $2 \&$ clocks are applied to the shift register.

• The output is a syndrome vector $\mathrm{S}=\left(\mathrm{S}_{\mathrm{q}-1}, \mathrm{~S}_{\mathrm{q}-2} \ldots \mathrm{S}_1, \mathrm{~S}_0\right)$