A 2 $$ \times $$ 2 ROM array is built with the help of diodes as shown in the circuit below. Here W0
and W1 are signals that select the word lines and B0 and B1 are signals that are output of the
sense amps based on the stored data corresponding to the bit lines during the read operation.
During the read operation, the selected word line goes high and the other word line is in a
high impedance state. As per the implementation shown in the circuit diagram above, what
are the bits corresponding to Dij (where i = 0 or 1 and j = 0 or 1) stored in the ROM?
The logic gates shown in the digital circuit below use strong pull-down nMOS transistors for
LOW logic level at the outputs. When the pull-downs are off, high-value resistors set the
output logic levels to HIGH (i.e. the pull-ups are weak). Note that some nodes are
intentionally shorted to implement “wired logic”. Such shorted nodes will be HIGH only if
the outputs of all the gates whose outputs are shorted are HIGH.
The number of distinct values of X3X2X1X0 (out of the 16 possible values) that give
𝑌 = 1 is _______.
Your input ____
3
GATE ECE 2018
MCQ (Single Correct Answer)
+2
-0.67
A four-variable Boolean function is realized using
4 $$ \times $$ 1
multiplexers as shown in the
figure.
The minimized expression for F(U, V, W, X)
is
A
$$\left( {UV + \overline U \overline V } \right)\overline W $$
B
$$\left( {UV + \overline U \overline V } \right)\left( {\overline W \overline X + \overline W X} \right)$$
C
$$\left( {U\overline V + \overline U V} \right)\overline W $$
D
$$\left( {U\overline V + \overline U V} \right)\left( {\overline W \overline X + \overline W X} \right)$$
4
GATE ECE 2018
MCQ (Single Correct Answer)
+1
-0.33
A function F(A, B, C) defined by three Boolean variables A, B and C when expressed as sum
of products is given by
F = $$\overline A .\overline B .\overline C + \overline A .B.\overline C + A.\overline B .\overline C $$
where, $$\overline A $$, $$\overline B $$, and $$\overline C $$ are the complements of the respective variables. The product of sums
(POS) form of the function F is
A
F = (A + B + C)(A + $$\overline B $$ + C)($$\overline A $$ + B + C)
B
F = ($$\overline A $$ + $$\overline B $$ + $$\overline C $$)($$\overline A $$ + B + $$\overline C $$)(A + $$\overline B $$ + $$\overline C $$)
C
F = (A + B + $$\overline C $$)(A + $$\overline B $$ + $$\overline C $$)($$\overline A $$ + B + $$\overline C $$)($$\overline A $$ + $$\overline B $$ + C)($$\overline A $$ + $$\overline B $$ + $$\overline C $$)
D
F = ($$\overline A $$ + $$\overline B $$ + C)($$\overline A $$ + B + C)(A + $$\overline B $$ + C)(A + B + $$\overline C $$)(A + B + C)