Digital Logic · Logic minimization and hazards

Even parity on a bit string means that the total number of 1s, counting an appended…

Problem

Even parity on a bit string means that the total number of $1$s, counting an appended parity bit, is even. The even-parity bit of an $n$-bit data word $D$ is therefore \[ PE(D)=d_{n-1}\oplus d_{n-2}\oplus\cdots\oplus d_0. \] An even-parity checker on an $(n+1)$-bit received string $D$ concatenated with $PE$ outputs $E=PE(D)\oplus PE_{\mathrm{received}}$; $E=1$ signals a parity error. Let the 8-bit data word be $D=11010010$. (a) Compute $PE(D)$. Write the 9-bit transmitted string with the parity bit on the left of the data (the MSB of the 9-bit string). (b) A single-bit inversion corrupts data bit $d_3$ (counting $d_0$ as the LSB of $D$) and leaves the parity bit unchanged. Compute the checker output $E$. (c) Two inversions corrupt $d_3$ and $d_6$, parity bit unchanged. Compute $E$. State whether the checker is reliable for double errors. (d) Realise $PE(D)$ as a balanced tree of two-input XOR gates (four XORs in the first rank, two in the second, one in the third). If each XOR has delay $3\,\mathrm{ns}$ and all eight data bits arrive at $t=0$, compute the delay until $PE(D)$ is settled.

Hint

Even-parity bit is the XOR of the data bits, which is $0$ iff the data already has even weight.

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