Digital Logic Design
Exam Duration: 45 Mins Total Questions : 30
2's complement of a 16-bit number (one sign and 15 magnitude bits) in FFF, its magnitude in decimal representation is
- (a)
0
- (b)
1
- (c)
23767
- (d)
6453.5
In number FFF, all bits are 1. Thus, 1's complement of FFFF will be (0000) call 1's complemented, 2's complement will be 0000+0001=0001
The binary number (10.10001)2 is equivalent to
- (a)
(2.53125)10
- (b)
(1.7869)10
- (c)
(1.75)10
- (d)
(40.31)8
10.10001
=1x21 + 0x20 + 1x2-1 + 0x2-2 + 0x2-3 + 0x2-4 + 1x2-5
=\(2+\frac { 1 }{ 2 } +\frac { 1 }{ 32 } \)
=\(2+\frac { 17 }{ 32 } =2.53125\)
Gray code for (45)10 is
- (a)
110011
- (b)
101101
- (c)
111011
- (d)
111100
For gray code
(45)10=(101101)2
b5 b4 b3 b2 b1 b0
1 0 1 1 0 1
\({ g }_{ 5 }={ b }_{ 5 }=1\\ { g }_{ 4 }={ b }_{ 5 }\bigoplus { b }_{ 4 }=1\bigoplus 0=1\\ { g }_{ 3 }={ b }_{ 4 }\bigoplus { b }_{ 3 }=0\bigoplus 1=1\\ { g }_{ 2 }={ b }_{ 3 }\bigoplus { b }_{ 2 }=1\bigoplus 1=0\\ { g }_{ 1 }={ b }_{ 2 }\bigoplus { b }_{ 1 }=1\bigoplus 0=1\\ { g }_{ 0 }={ b }_{ 2 }\bigoplus { b }_{ 0 }=0\bigoplus 1=1\)
(101101)2=(111011)gray
(45)10=(111011)gray
What will be the expression obtained by reducing function (A + C + D)(A + C + D')(A + C' + D)(A + B')?
- (a)
A + ABC + BC
- (b)
AB + AC + BC
- (c)
A + ABC' + BC
- (d)
A + AB'C' + BC
(A + C + D)(A + C + D')(A + C' + D)(A + B')
=(A.A + AC + AD' + CA + C.C + CD' + DA + DC + DD')(A.A + AB' + C'A + C'B' + DA + D'B')
=(A + AC + AD' + C + CD' + AD + DC + 0)(A + AB' + AC' + B'C' + AD + B'D)
=(A(1+C) + AD' + C(1+D') + AD +DC)(A(1+B') + AC' + B'C' + AD + B'D)
=(A + C + A(D + D') + DC)(A(1+C') + B'C' + AD + BD)
=(A + C + DC)(A + AD +B'C' + BC)
=(A + C(1 + D))(A(1 + D) + B'C' + BC)
=(A + C)(A + B'C' + BC)
=A.A + AB'C' + ABC + AC + CB'C' + BC
=A + AB'C' + ABC + AC + CB'C' + BC
=A + AB'C' + AC(1 + B) + CB'C' + BC
=A + AB'C' + AC + CB'C' + BC
=A(1 + C) + AB'C' + 0 + BC
=A + AB'C' + BC
The Boolean function A + BC is the reduced form of
- (a)
AB + BC
- (b)
(A + B)(A + C)
- (c)
\(\bar { A } B+A\bar { B } C\)
- (d)
(A + C)B
(A + B)(A + C)
What are the maximum number of Boolean function involving n Boolean variable?
- (a)
n2
- (b)
2n
- (c)
22n
- (d)
\({ 2 }^{ { n }^{ 2 } }\)
For 1 binary half substractor having two inputs A and B, the correct set of logical expression for the output D=(A minus B) and X(=borrow) are
- (a)
\(D=AB+\bar { A } B,X=\bar { A } B\)
- (b)
\(D=\bar { A } B+A\bar { B } ,X=\bar { A } B\)
- (c)
\(D=\bar { A } B+A\bar { B } ,X=\bar { A } B\)
- (d)
\(D=AB+\bar { A } \bar { B } ,X=A\bar { B } \)
The truth table of half-subtrator will be as
A | B | D(A-B) | X(Borrow) |
0 | 0 | 0 | 0 |
0 | 1 | 1 | 1 |
1 | 0 | 1 | 0 |
1 | 1 | 0 | 0 |
Thus, D=A'B + AB'
X=A'B
An SR flip-flop can be converted into T flip-flop by connecting.........to Q and................to \(\bar { Q } \).
- (a)
S' ,R'
- (b)
S ,R'
- (c)
S' ,R
- (d)
S ,R
Out of S-R flip-flop is,
S | R | Qn |
0 | 0 | undefined |
0 | 1 | 0 |
1 | 0 | 1 |
1 | 1 | No change |
Output of T flip-flop is
T | Qn |
0 | Qn |
1 | \(\bar { { Q }_{ n } } \) |
Thus, if S is connected to Q and R is connected to \(\bar { { Q } } \) the equation will be as,
S | R | Qn+1 | \(\bar { { { Q }_{ n+1 } } } \) |
1 | 0 | 1 | 0 |
0 | 1 | 0 | 1 |
The present output Qn of an edge triggered J-K flip-flop is logic 0. If J=1 then Qn+1
- (a)
will be logic 0
- (b)
will be logic 1
- (c)
cannot be determined
- (d)
will race around
As the truth table of J-K flip-flop is
J | K | Qn+1 |
0 | 0 | Qn |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | \(\bar { { Q }_{ n } } \) |
When an odd number is converted into binary number the LSB is
- (a)
0
- (b)
1
- (c)
0 or 1
- (d)
None of these
An odd number is generated by add 1 to even number. The LSB is always zero for an even binary number because 20 =1. Thus, for odd number the LSB will be always 1.
If (211)x = (152)8 then the value of base x is
- (a)
6
- (b)
5
- (c)
9
- (d)
7
Converting both (211)x and (152)8 in decimal, we get
\(2\times { x }^{ 2 }+1\times x+1=1\times { 8 }^{ 2 }+5\times { 8 }^{ 1 }+2\times { 8 }^{ 0 }\\ \quad \quad \quad 2{ x }^{ 2 }+x+1=64+40+2\\ \quad \quad \quad 2{ x }^{ 2 }+x+1=106\\ \quad \quad 2{ x }^{ 2 }+x-105=0\)
An n-bit gray code can be obtained by rejecting an .............. bit code.
- (a)
n
- (b)
n + 1
- (c)
n -1
- (d)
None of these
As n-bit gray code are obtained by n-bit binary number.
The equivalent in decimal for the excess-3 code 843 is
- (a)
840
- (b)
510
- (c)
846
- (d)
None of these
(843)excess-3
8 4 3
-3 -3 -3
__________
5 1 0
Thus, (843)excess-3=(510)10
A positive logic AND operation will behave as negative logic
- (a)
AND-NOT
- (b)
OR
- (c)
NOR
- (d)
OR-NOT
The truth table for positive logic AND would be represented as
A | B | Output |
0 | 0 | 0 |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | 1 |
Negative logic AND-NOT will have truth table like
A | B | AND output | AND-NOT (output) |
0 | 0 | 0 | 1 |
0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 |
Thus, negative logic AND-NOT \(\neq \) positive logic AND.
For negative logic OR gate truth table will be
A | B | Output |
0 | 0 | 0 |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | 1 |
Thus, negative logic OR = positive logic AND
XOR is
- (a)
odd function
- (b)
even function
- (c)
Both a and b
- (d)
None of these
XOR is odd function because its output is high when number of 1's is odd.
\(\left( \left( A+\bar { A } B \right) \left( A+\bar { A } \bar { B } \right) \right) \left( CD+\bar { C } \bar { D } \right) +\left( C\bigoplus D \right) =\)
- (a)
A
- (b)
B
- (c)
0
- (d)
1
\(\left( \left( A+\bar { A } B \right) \left( A+\bar { A } \bar { B } \right) \right) \left( CD+\bar { C } \bar { D } \right) +\left( C\bigoplus D \right) \\ \left( \left( A+\bar { A } \right) \left( A+B \right) \left( A+\bar { A } \right) A+\bar { B } \right) \left( \left( CD+\bar { C } \bar { D } +\left( C\bar { D } +\bar { C } D \right) \right) \right) \\ \left\{ A+\bar { A } B=\left( A+\bar { A } \right) \left( A+B \right) distributive\quad law,C\bigoplus D=\bar { C } D+C\bar { D } \right\} \\ =\left( \left( A+B \right) \left( A+\bar { B } \right) \right) \left( CD+\bar { C } \bar { D } +CD+\bar { C } D \right) \\ =\left( A+A\bar { B } +AB+B\bar { B } \right) \left( CD+\bar { D } \left( \bar { C } +C \right) +\bar { C } D \right) \\ =\left( A+A\left( \bar { B } +B \right) \right) \left( CD+\bar { D } +\bar { C } D \right) \\ =\left( A+A \right) \left( D\left( C+\bar { C } \right) +\bar { D } \right) \\ =A\left( D+\bar { D } \right) \\ =A\)
A n-variable K-map contains
- (a)
n cells
- (b)
2n cells
- (c)
n2 cells
- (d)
None of these
A two variable K-map contains 2x2 =4 cells, similarly 3 variable K-map contains 3x3 =9 cells. Thus n variable K-map will contain nxn =n2 cells.
Decimal digits can be converted to binary format using
- (a)
Decoder
- (b)
Encoder
- (c)
MUX
- (d)
DE MUX
n bit binary numbers are used to represent 2n decimal digits, thus decimal to binary conversion means 2n to n-bit conversion which is done by encoder.
The m-bit adder contains........full-adder.
- (a)
m
- (b)
m+1
- (c)
m/2
- (d)
m-1
A X:Y decoder can be constructed using, how many A:B decoders with enable?
- (a)
\(\frac { Y }{ B } \)
- (b)
\(\frac { X }{ A } \)
- (c)
\(\frac { X }{ B } \)
- (d)
\(\frac { Y }{ A } \)
x:y decoder means, the decoder has x inputs and A:B decoder means decoder has A inputs and B outputs for constructing 4:16 decoder from 2:4 decoder. 4, 2:4 decoder will be needed to generate 16 digits output. Similarly, for developing 5:32 bit decoder 5 2:4 decoder will be needed. Thus the number of A:B decoder required for developing x:y decoder will be \(\frac { Y }{ B } \)
Determine the value of x if (211)x =(152)8
- (a)
-7.5
- (b)
-6.5
- (c)
-5.5
- (d)
0
(211)2 =(152)8
Converting both into decimal, we get
\(\quad \quad \quad \quad \quad \quad 2\times { x }^{ 2 }+1\times x+1=1\times { 8 }^{ 2 }+5\times 8+2\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =64+40+2\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =104+2\\ \quad \quad \quad \quad \quad \quad 2\times { x }^{ 2 }+1\times x+1=106\\ \quad \quad \quad \quad \quad 2\times { x }^{ 2 }+x-105=0\\ 2\times { x }^{ 2 }+15x-14x-105=0\\ x\left( 2x+15 \right) -7\left( 2x+15 \right) =0\\ \quad \quad \quad \quad \quad \left( x-7 \right) \left( 2x+15 \right) =0\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad x+15=0\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad x=-7.5\)
Multiplication of (296)12 and (57)12 is
- (a)
11706
- (b)
13706
- (c)
12706
- (d)
None of these
2 9 6
x 5 7
____
(296)12=2x1212+9x12+6
=(402)10
(57)12=5x12+7
=(67)10
(402)10x(67)10=(26934)10
12 | 26934 | |
12 | 2244 | 6 |
12 | 187 | 0 |
12 | 15 | 7 |
12 | 1 | 3 |
1 |
(13706)12
10th element of base-3 number system is
- (a)
22
- (b)
12
- (c)
100
- (d)
None of these
The 10th element of any number is (9)10 so,
3 | 9 | |
3 | 3 | 0 |
3 | 1 | 0 |
0 | 1 |
(9)10=(100)3
Convert (14.34)10 into binary.
- (a)
1011.1101
- (b)
1110.01010
- (c)
1110.1001
- (d)
1011.01001
(14.34)10
For integer part,
2 | 14 | |
2 | 7 | 0 |
2 | 3 | 1 |
2 | 1 | 1 |
0 | 1 |
(14)10=(1110)2
For fractional part,
(-34)10
Integer | Fraction | Coefficient | |
0.34x2 | 0 | 0.68 | a-1=0 |
0.68x2 | 1 | 0.36 | a-2=1 |
0.36x2 | 0 | 0.72 | a-3=0 |
0.72x2 | 1 | 0.44 | a-4=1 |
0.44x2 | 0 | 0.88 | a-5=0 |
Thus, (0.34)10=(0.01010)2
(14.340)10=(1110.01010)2
If (250)10=(X)12 the X is
- (a)
16A
- (b)
18A
- (c)
14A
- (d)
None of these
(250)10=(X)12
12 | 250 | |
12 | 20 | 10\(\Rightarrow \) A |
12 | 1 | 8 |
0 | 1 |
Thus, (250)10=(18A)12
\(\overline { \left( A+\bar { B } \right) \overline { \left( \bar { A } +B \right) } } =\)
- (a)
\(\bar { A } +B\)
- (b)
\(A+\bar { B } \)
- (c)
A
- (d)
B
\(\overline { \left( A+\bar { B } \right) \overline { \left( \bar { A } +B \right) } } \)
\(\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\overline { A+\bar { B } } +\overline { \bar { A } +\bar { B } } \\ \quad \quad \quad (Demorgans\quad law\quad \overline { A.B } =\bar { A } +\bar { B } )\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\left( \bar { A } +\bar { \bar { B } } \right) +\left( \bar { A } +B \right) \\ \quad \quad \quad (\bar { \bar { A } } =A\quad and\quad \left( \overline { A+B } \right) =\bar { A } .\bar { B } \quad Demorgans\quad law)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\left( \bar { A } .B \right) +\bar { A } +B\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\bar { A } B+\bar { A } +B\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\bar { A } \left( B+1 \right) +B\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\bar { A } +B\)
Express \(A+\bar { A } B+\bar { A } \bar { B } C+\bar { A } \bar { B } \bar { C } D+\bar { A } \bar { B } \bar { C } \bar { D } E\) would be simplified to
- (a)
\(A+\bar { A } B+CD+E\)
- (b)
A+B+C+D+E
- (c)
A+B+CDE
- (d)
A+BC+CD+DE
\(A+\bar { A } B+\bar { A } \bar { B } C+\bar { A } \bar { B } \bar { C } D+\bar { A } \bar { B } \bar { C } \bar { D } E\\ \quad \quad \quad \quad =\left( A+\bar { A } \right) \left( A+B \right) +\bar { A } \bar { B } C+\bar { A } \bar { B } \bar { C } \left( D+\bar { D } E \right) \\ \quad \quad \quad \quad =A+B+\bar { A } \bar { B } C+\bar { A } \bar { B } \bar { C } \left( \left( D+\bar { D } \right) \left( D+E \right) \right) \\ \quad \quad \quad \quad =A+B+\bar { A } \bar { B } C+\bar { A } \bar { B } \bar { C } \left( D+E \right) \\ \quad \quad \quad \quad =A+B+\bar { A } \bar { B } C+\bar { A } \bar { B } \bar { C } D+\bar { A } \bar { B } \bar { C } E\\ \quad \quad \quad \quad =A+\bar { A } \bar { B } C+B+\bar { A } \bar { B } \bar { C } D+\bar { A } \bar { B } \bar { C } E\\ \quad \quad \quad \quad =\left( A+\bar { A } \right) \left( A+\bar { B } C \right) +B+\bar { A } \bar { B } \bar { C } D+\bar { A } \bar { B } \bar { C } E\\ \quad \quad \quad \quad =A+\bar { B } C+B+\bar { A } \bar { B } \bar { C } D+\bar { A } \bar { B } \bar { C } E\\ \quad \quad \quad \quad =A+\left( B+\bar { B } \right) \left( B+C \right) +\bar { A } \bar { B } \bar { C } D+\bar { A } \bar { B } \bar { C } E\\ \quad \quad \quad \quad =A+B+C+\bar { A } \bar { B } \bar { C } D+\bar { A } \bar { B } \bar { C } E\\ \quad \quad \quad \quad =\left( A+\bar { A } \right) \left( A+\bar { B } \bar { C } D \right) +B+C+\bar { A } \bar { B } \bar { C } E\\ \quad \quad \quad \quad =A+\bar { B } \bar { C } D+B+C+\bar { A } \bar { B } \bar { C } E\\ \quad \quad \quad \quad =A+\left( B+\bar { B } \right) \left( B+\bar { C } D \right) +C+\bar { A } \bar { B } \bar { C } E\\ \quad \quad \quad \quad =A+B+\bar { C } D+C+\bar { A } \bar { B } \bar { C } E\\ \quad \quad \quad \quad =A+B+\left( C+\bar { C } \right) \left( C+D \right) +\bar { A } \bar { B } \bar { C } E\\ \quad \quad \quad \quad =A+B+C+D+\bar { A } \bar { B } \bar { C } E\\ \quad \quad \quad \quad =\left( A+\bar { A } \right) \left( A+\bar { B } \bar { C } E \right) +B+C+D\\ \quad \quad \quad \quad =A+\bar { B } \bar { C } E+B+C+D\\ \quad \quad \quad \quad =A+\left( B+\bar { B } \right) \left( B+\bar { C } E \right) +C+D\\ \quad \quad \quad \quad =A+B+\bar { C } E+C+D\\ \quad \quad \quad \quad =A+B+\left( C+\bar { C } \right) \left( C+E \right) +D\\ \quad \quad \quad \quad =A+B+C+D+E \)
The Boolean expression \(\bar { X } Y\bar { Z } +\overline { XY } Z+XY\bar { Z } +X\bar { Y } Z+XYZ\) can be simplified to
- (a)
\(X\bar { Z } +\bar { X } Z+YZ\)
- (b)
\(\bar { Z } Y+\bar { Y } Z+XZ\)
- (c)
\(XY+\bar { Y } Z+YZ\)
- (d)
\(\bar { X } \bar { Y } +Y\bar { Z } +X\bar { Z } \)
\(\bar { X } Y\bar { Z } +\overline { XY } Z+XY\bar { Z } +X\bar { Y } Z+XYZ\\ \quad \Rightarrow \quad \quad \quad \bar { X } Y\bar { Z } +\overline { XY } Z+XY\bar { Z } +XZ\left( Y+\bar { Y } \right) \\ \quad \quad \quad \quad \quad \quad \quad \bar { X } Y\bar { Z } +\overline { XY } Z+XY\bar { Z } +XZ\left( Y+\bar { Y } \right) \\ \quad \quad \quad \quad \quad \quad \quad \bar { X } Y\bar { Z } +\overline { XY } Z+XY\bar { Z } +XZ\\ \quad \Rightarrow \quad \quad \quad \quad \quad \quad Y\bar { Z } \left( X+\bar { X } \right) +\overline { XY } Z+XZ\\ \quad \Rightarrow \quad \quad \quad \quad \quad \quad Y\bar { Z } +\overline { XY } Z+XZ\\ \quad \Rightarrow \quad \quad \quad \quad \quad \quad Y\bar { Z } +Z\left( \overline { XY } +X \right) \\ \quad \Rightarrow \quad \quad \quad \quad \quad \quad Y\bar { Z } +Z\left( \left( \bar { Y } +X \right) \left( \bar { X } +X \right) \right) \\ \quad \Rightarrow \quad \quad \quad \quad \quad \quad Y\bar { Z } +Z\left( \bar { Y } +X \right) \\ \quad \Rightarrow \quad \quad \quad \quad \quad \quad Y\bar { Z } +Z\bar { Y } +ZX\)
A logical function of three variables is given as
\(f\left( A,B,C \right) =\left( A+BC \right) \left( B+\bar { C } A \right) \)
The canonical SOP form is
- (a)
\(\Sigma m\left( 2,4,8,10 \right) \)
- (b)
\(\Sigma m\left( 2,4,6,7 \right) \)
- (c)
\(\Sigma m\left( 3,5,8,9 \right) \)
- (d)
\(\Sigma m\left( 3,4,6,7 \right) \)
\(f\left( A,B,C \right) =\left( A+BC \right) \left( B+\bar { C } A \right) \\ \quad \quad =AB+A\bar { C } +BC+BC.\bar { C } A\\ \quad \quad =AB+A\bar { C } +BC+0\\ \quad \quad =AB+A\bar { C } +BC\\ \quad \quad =AB\left( C+\bar { C } \right) +A\bar { C } \left( B+\bar { B } \right) +BC\left( A+\bar { A } \right) \\ \quad \quad =ABC+AB\bar { C } +A\bar { B } \bar { C } +A\bar { B } \bar { C } +ABC+\bar { A } BC\\ \quad \quad =ABC+AB\bar { C } +A\bar { B } \bar { C } +\bar { A } BC\\ \quad \quad =\left( 7,6,4,3 \right) \left( ABC=7,\quad AB\bar { C } =6,\quad A\bar { B } \bar { C } =4,\quad \bar { A } BC=3 \right) \\ Thus,\quad f=\Sigma \left( 3,4,6,7 \right) \)
The number of 2x1 muxer needed to construct 8x1 mux is
- (a)
7
- (b)
3
- (c)
8
- (d)
none of these
\(\log _{ 2 }{ 8 } =3\)
2x1 mux will be needed for 8x1 mux.