GATE Breaker

The truth table  represents the Boolean function  (A) $X$  (B) $X + Y$  (C) $X \bigoplus  Y$  (D) $Y$  (GATE 2012)  Answer: (A) $X$  Explanation:  From the given truth table, we get  $f(X, Y)$  $= XY' + XY$  $= X(Y' + Y)$  $=X$  Previous Post Next Post

 The minterm expansion of $f(P, Q, R) = PQ + QR' + PR'$ is  (A) $m_2 + m_4 + m_6 + m_7$  (B) $m_0 + m_1 + m_3 + m_5$  (C) $m_0 + m_1 + m_6 + m_7$  (D) $m_2 + m_3 + m_4 +m_5$  (GATE 2010)  Answer: (A) $m_2 + m_4 + m_6 + m_7$  Explanation:  $f(P, Q, R)$  $= PQ + QR' + PR'$  $= PQ(R + R') + (P + P')QR' + P(Q + Q')R'$, since $A + A' = 1$  $= PQR + PQR' + PQR' + P'QR' + PQR' + PQ'R'$  $= PQR + PQR' + P'QR' + PQ'R'$, since $A + A = A$  $= m_7 + m_6 + m_2 + m_4$   Previous Post Next Post

If $P$, $Q$, $R$ are Boolean variables, then $(P + Q')( PQ' + PR)( P'R' + Q')$ simplifies to  (A) $PQ'$  (B) $PR'$ (C) $PQ'$  (D) $PR' + Q$  (GATE 2008)  Answer: (A) $PQ'$  Explanation:  $(P + Q')( PQ' + PR)( P'R' + Q')$  $= (PQ' + PR + PQ' + PQR)(P'R' + Q')$, since $AA = A$  $= PQ' + PQ'R + PQ' + PQ'R$, since $AA' = 0$  $= PQ' + PQ'R$  $= PQ'(1 + R)$  $= PQ'$  Previous Post Next Post

What is the minimum number of gates required to implement the Boolean function $(AB + C)$ if we have to use only 2-input NOR gates?  (A) 2  (B) 3  (C) 4  (D) 5  (GATE 2009)  Answer: (B) 3  Explanation:  $(A + BC) = (A + C)(B + C) = \big((A + C)' + (B + C)\big)'$  Therefore, two NOR gates is requiredto implement the given Boolean function.  Previous Post Next Post

In this tutorial, we will learn how to compute the difference between two binary numbers using 1's complement and 2's complement.  Let us consider two binary numbers $X$ and $Y$.  We want to compute $X - Y$ here.  Method 1: Using 1's complement  Step 1: Find 1's complement of $Y$.  Step 2: Add $X$ and $Y$.  Step 3: If there is a carry, then add this carry to the answer. If there is no carry, then find the 1's complement of the answer and place a negative sign in front of the answer.  Example 1: Compute $(101)_2 - (011)_2$ using 1's complement method.   Solution:  Here, $X = (101)_2$ and $Y = (011)_2$.  Step 1: 1's complement of $Y$ is $(100)_2$.  Step 2: $X + Y = (101)_2 + (100)_2 = (1 001)_2$.  Step 3: There is a carry of 1, so now we will add this carry to the answer.  Therefore, the final result is $(1)_2 + (001)_2 = (010)_2$.  Example 2: Compute $(011)_2 - (101)_2$ using 1's complement method.  Solution...