1
GATE ECE 2007
+2
-0.6
The Boolean expression Y= $$\overline A \,\overline B \,\overline C \,D + \overline A BC\overline D + A\overline {B\,} \overline C \,D + AB\overline C \,\overline D$$
A
Y = $$\overline A \,\overline B \,\overline C \,D + \overline A B\overline C + A\overline C D$$
B
Y = $$\overline A \,\overline B \,\overline C \,D + BC\overline D + A\overline B \overline C \,D$$
C
Y=$$\overline A \,BC\,\overline D + \overline B \,\overline C D + A\overline B \overline C \,D$$
D
Y= $$\overline A \,BC\,\overline D + \overline B \,\overline C D + AB\overline C \,\overline D$$
2
GATE ECE 2006
+2
-0.6
The point p in the following figure is stuck- at-1. The output f will be A
$$\overline {AB\overline {C\,} }$$
B
$$\overline A$$
C
$$AB\overline C$$
D
A
3
GATE ECE 2004
+2
-0.6
The Boolean expression AC + B$$\overline C$$ is equivalent to
A
$$\overline A \,C + B\overline C + AC$$
B
$$\overline B C\, + AC + B\overline C + \overline A C\overline B$$
C
AC+$$B\overline C + \overline B C + ABC$$
D
$$\overline A \,B\,\overline C + A\,B\,\overline C + A\overline {\,B} \,C$$
4
GATE ECE 2004
+2
-0.6
A Boolean function 'f' of two variables x and y is defined as follows: f(0,0)=f(0,1)=f(1,1)=1;f(1,0)=0 Assuming complements of x and y are not available, a minimum cost solution for realizing F using only 2-input NOR gates and 2-input OR gates (each having unit cost) would have a total cost
A
1 unit
B
4 unit
C
3 unit
D
2 unit
GATE ECE Subjects
Signals and Systems
Network Theory
Control Systems
Digital Circuits
General Aptitude
Electronic Devices and VLSI
Analog Circuits
Engineering Mathematics
Microprocessors
Communications
Electromagnetics
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