MGU UGP BCA ,S1 DIGITAL FUNDAMENTALS SECOND INTERNAL EXAMINATION , OCTOBER 2025

Answer Key

Section A

1.    The 8421 code is a Binary Coded Decimal (BCD) code — one of the most common ways to represent decimal numbers in binary form. Binary Coded Decimal, or BCD, is used for converting decimal numbers into their binary equivalents

2.    The Hexadecimal Number System (often called Hex) is a base-16 number system.
It uses 16 unique symbols to represent all possible values for a single digit.

3.    Overflow occurs in binary addition when the result of an addition exceeds the range that can be represented with the available number of bits.

4.     

 

A	B	SUM	CARRY
0	0	0	0
0	1	1	0
1	0	1	0
1	1	0	1

 

 

 

 

 

5.    Final BCD Result = 1001 1000

6.     

7.    f=ABC+ABC′+AB′C+A′B′C

= 111 + 110 + 101 + 001

= m7+m6+m5+m1

=∑m (1,5,6,7)

8.    Combinational Circuit	Sequential Circuit

 

A circuit whose output depends only on the current inputs.

A circuit whose output depends on both      current inputs and past history (previous inputs).

9.     

 

X	Y	Z	F(x,y,z)
0	0	0	1
0	0	1	0
0	1	0	0
0	1	1	1
1	0	0	0
1	0	1	0
1	1	0	1
1	1	1	0

 

10.  XOR - Outputs 1 when the number of 1s at the input is odd (for 2 inputs, when inputs are different).

XNOR- Outputs 1 when the number of 1s at the input is even (for 2 inputs, when inputs are the same).

SECTION B

 

11.   483 – 0100 1000 0011 3marks explanation with steps 3 marks.

12.   Correct diagram 3 marks explanation 3 marks.

13.   A multiplexer (MUX) is a combinational circuit that selects one of many input lines and forwards it to a single output line.

It acts like an electronic “switch”.

The selection of the input line is controlled by selection lines (or control lines)

Explanation – 3 marks

Diagrams – 3 marks

 

14.   674​=37C   3 marks

Explanation 3 marks

 

15.  1’s complement = invert all bits:

A+B′=1010 1101+1010 0100

            = 0101 0001

01010001+1=01010010 , no overflow

2’s Complement Method

Step 1: Take 2’s complement of B

2’s complement = 1’s complement + 1:

1’s complement of B: 01011011  ⟹  10100100

Add 1:

10100100+1=10100101

A+B2′s​=1010 1101+1010 0101

            = Result = 0101 0010 

In 2’s complement, overflow occurs if carry into MSB ≠ carry out of MSB. No overflow

16.  (i) F=[AB’(C+BD) +A’B’]C = [AB’C+AB’BD+A’B’]C = AB’CC+AB’BCD+A’B’C

                                   =AB’C + 0 + A’B’C

                                   =B’C(A+A’)

=B′C

(ii) (ABC+DEF) ′=(ABC)′⋅(DEF)′= (A′+B′+C′) ⋅(D′+E′+F′)

 (iii) [AB′+C′D+EF] ′= (AB′) ′⋅(C′D) ′⋅(EF)′ =(A′+B) ⋅(C+D′) ⋅(E′+F′)

17.  Half adder – diagram 1 mark, truth table 1 mark, equation 1 mark

full adder – diagram 1 mark, truth table 1 mark, equation 1 mark

Section C

18.  4 number group, First simplified term( m0,m1,m8,m9) : B'C'

2 number group ( m7,m6) : A’BC

2 number group( m0,m2) : A’B’D’

2 number group ( m6, m14) : BCD’

 

Expression is B’C’ + A’BC + A’B’D’ +BCD’

            Plotting k map -2 mark

               Grouping – 2 mark

               Simplification – 2 mark

                Logic diagram – 4 marks

             

19.   Encoder with explanation – 5 marks

                       Decoder with explanation – 5 marks.

20. Basic laws of Boolean Algebra -5 marks. Simplification Steps 5marks

F=AB+A(B+C) +B(B+C)

= AB+AB+AC+BB+BC

= AB+AC+B+BC

=AB+AC+B(1+C)

 =AB+AC+B=B(A+1) +AC

=B+AC

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