Question

Q2: Implement F(A,B,C)=(A+B+C)(A’+C’)(B’+C’) using: (5 pts each)

A. A 3x8 active high decoder

B. A 3x8 active low decoder

C. A 2x1 multiplexer.

D. A 4x1 multiplexer.

Q3: Implement a Full Adder using: (5 pts each)

A. A 3x8 active high decoder

B. A 3x8 active low decoder

C. With two 2x4 Active high decoders.

Answer 1

answer 1:

truth table:

A   B   C   F
0   0   0   0
0   0   1   1
0   1   0   1
0   1   1   0
1   0   0   1
1   0   1   0
1   1   0   1
1   1   1   0

active high encoder implementation:

active low implementation:

now for 2:1 mux implementation

lets keep A as select line

hence from the above truth table, we see

when a=0 then output is B xor C

when a=1 then output is B'C'+BC'

hence:

4:1 mux implementation:

let's keep AB as select inputs

for the truth table we see that

AB=00 F=C

01 F=C'

10 F=C'

11 F= C'

hence circuit:

answer3:

full adder truth table:

A   B   cin   S   cout
0   0 0 0 0
0   0 1 1 0
0   1 0 1 0
0   1 1 0 1
1   0 0 1 0
1   0 1   0 1
1   1 0 0 1
1   1 1   1 1

implementation using 3*8 decoder

implementation using 2*8 active low decoder

implementation using two 2*4 decoders

BC are inputs to the decoder and A is given to the enable of the decoder to select the decoder