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Answer - B. Degree of generator polynomial is $3$ hence $3-bits$ are appended before performing division After performing division using $2$'$s$ complement arithmetic remainder is $011$ The remainder is appended to original data bits and we get $M' = 11001001\bf{011}$ from $M = 11001001.$ Courtesy, Anurag Pandey
posted Mar 16, 2019 in Preparation Experience Parth Shah 2,597 views
A serial transmission $T1$ uses $8$ information bits, $2$ start bits, $1$ stop bit and $1$ parity bit for each character. A synchronous transmission $T2$ uses $3$ eight-bit sync characters followed by $30$ ... /sec $80$ characters/sec, $136$ characters/sec $100$ characters/sec, $136$ characters/sec $80$ characters/sec, $153$ characters/sec
posted Mar 17, 2018 in 2018 aehkn 1,926 views
Consider the following program segment for concurrent processing using semaphore operators $P$ and $V$ for synchronization. Draw the precedence graph for the statements $S_1$ to $S_9$. var a,b,c,d,e,f,g,h,i,j,k : semaphore; begin cobegin begin S1; V(a); V(b) end; begin P(a); S2; V(c); V(d) end; ... end; begin P(g); S6; V(i) end; begin P(h); P(i); S8; V(j) end; begin P(j); P(k); S9 end; coend end;
posted Jan 16, 2018 in Preparation Advice Ashish Subscription 2,246 views
Consider a simple connected graph $G$ with $n$ vertices and $n$ edges $(n > 2)$. Then, which of the following statements are true? $G$ has no cycles The graph obtained by removing any edge from $G$ is not connected $G$ has at least one cycle The graph obtained by removing any two edges from $G$ is not connected None of the above
posted Oct 8, 2017 in Preparation Experience prateekkmr60 4,905 views
If you think of a $12\times 12$ grid (like a chess board of size $12\times 12$), then each each point $\left(i,j\right)$, which is in $i^{th}$ row and $j^{th}$ column, is a vertex $\left(i,j\right)$. Now we are allowed to connect only those points which are ... $2$ diagonals, so total diagonals $= 242$. So total edges $= 132 + 132 + 242 = 506.$
posted Jun 3, 2017 in Others Surajit 951 views
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