Question

Implement the following Boolean function with a 4 x 1 multiplexer and external gates. Connect
inputs A and B to the selection lines. The input requirements for the four data lines will be a furca
tion of variables C and D. These values are obtained by expressing F as a function of C and D for
each of the four cases when AB = 00, 01, 10, and 11. These functions may have to be implemented
with external gates.
F(A, B, C, D) = {(1, 3, 4, 11, 12, 13, 14, 15)
​

Answer 1

Answer:

The input requirements for the four data lines will be a function of varibles C and D. These values are obtained by expressing F as a function of C and D of the four cases AB = 00, 01, 10 and 11. These functions may have to be implemented with external gates.

Explanation:

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Answer 2

The given multiplexer will have the 2 Input Bits (A,B) connected to the 2 select lines and the other 2 bits (C,D) attached to the 4 input lines using the NOR and AND gates.

The given expression F(A, B, C, D) requires 4 variable bits A, B, C, and D to represent the numbers 0 to 15 using it.

But we need to implement this, using a 4*1 Multiplexer. A 4*1 Multiplexer has :

• 4 Input Lines : $I_{0} ,I_{1} ,I_{2} ,I_{3}$ which means it can represent only 4 numbers i.e. 0 to 3.
• 2 Select Lines : $S_{0} ,S_{1}$ to select the bits. Numbers 0 to 3 require only 2 bits for representation.
• 1 Output Line : The Multiplexer returns a single-bit output (0 or 1) based upon the values in the Input and Select Lines.

We have 16 combinations (0 to 15) which require 4 bits to represent using 2 select lines corresponding to 4 values only. So, we group the 16 values into 4 groups, each group having the same bits for A and B. Thus, we use the bit A and B in the select lines and C and D as external inputs.

Group 1 : A = 0, B = 0 ( Numbers = 0, 1, 2, 3 ) :

The tabular form of Input (C, D) and Output(Y) for this group is :

0 : C(0)  D(0)   Y(0)

1 :  C(0)  D(1)    Y(1)

2 : C(1)   D(0)   Y(0)

3 : C(1)   D(1)    Y(1)

Thus, we see that the output value Y is the exact same as the Input bit D.

Expression for Group 1 : Y = D

Group 2 : A = 0, B = 1 ( Numbers = 4, 5, 6, 7 ) :

The tabular form of Input (C, D) and Output(Y) for this group is :

4 : C(0)  D(0)   Y(1)

5 : C(0)  D(1)    Y(0)

6 : C(1)   D(0)   Y(0)

7 : C(1)   D(1)    Y(0)

Thus, we see that the output value Y is the negation of the Boolean OR of the two inputs C and D.

Expression for Group 2 : Y = NOT( C OR D) = NOR(C,D)

Group 3 : A = 1, B = 0 ( Numbers = 8, 9, 10, 11 ) :

The tabular form of Input (C, D) and Output(Y) for this group is :

8 :  C(0)  D(0)   Y(0)

9 :  C(0)  D(1)    Y(0)

10 : C(1)   D(0)   Y(0)

11 :  C(1)   D(1)    Y(1)

Thus, we see that the output value Y is the Boolean AND of the two inputs C and D.

Expression for Group 3 : Y =AND(C,D)

Group 4 : A = 1, B = 1 ( Numbers = 12, 13, 14, 15 ) :

The tabular form of Input (C, D) and Output(Y) for this group is :

12 :  C(0)  D(0)   Y(1)

13 :  C(0)  D(1)    Y(1)

14 :  C(1)   D(0)   Y(1)

15 :  C(1)   D(1)    Y(1)

Thus, we see that the output value Y is 1 irrespective of any combination of the bits C and D.

Expression for Group 4 : Y = 1

So, the final implementation of the 4*1 Multiplexer for the given SOP function of minterms is :

• Input line 0 ( Group1 ) will connect to Input Line D
• Input line 1 ( Group 2 ) will connect to NOR gate for C and D
• Input line 2 ( Group 3 ) will connect to AND gate for C and D
• Input line 3 will directly pass 1
• Select line 0 will connect the input bit A
• Select line 1 will connect the input bit B

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