1
GATE ECE 1995
MCQ (Single Correct Answer)
+1
-0.3
The output of the circuit shown in figure is equal to GATE ECE 1995 Digital Circuits - Combinational Circuits Question 12 English
A
0
B
1
C
$$\overline A B + A\overline B $$
D
$$\left( {\overline {A \oplus B} } \right) \oplus \left( {\overline {A \oplus B} } \right)$$
2
GATE ECE 1995
Subjective
+10
-0
A ROM is to be used to implement the Boolean functions given below:
$${F_1}$$$$(A,\,B,\,C,\,D) = ABCD + \bar A\,\overline B \,\bar C\,\bar D$$
$${F_2}(A,\,B,\,C,\,D) = (A + B)(\bar A\, + \overline B + C)$$
$${F_3}(A,\,B,\,C,\,D) = \sum {13,15 + \sum {3,5} } $$

(a) What is the minimum size of the ROM required?

(b) Determine the data in each location of the ROM.

3
GATE ECE 1995
MCQ (Single Correct Answer)
+1
-0.3
The minimum number of NAND gates required to implement the Boolean function $$A + A\overline B $$ $$ + A\overline B C$$ is equal to
A
Zero
B
1
C
4
D
7
4
GATE ECE 1995
Subjective
+10
-0
A 'code converter' is to be designed to convert from the BCD(5421) to the normal BCD (8421). The input BCD combinations for each digit are below. A block diagram of the converter is shown in figure. GATE ECE 1995 Digital Circuits - Boolean Algebra Question 6 English
(A) draw K-map for outputs, W, X, Y, and Z.
(B) Obtain minimized expression for the outputs W, X, Y, and Z.
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