Use K-Map to Simplify a 4-Variable Function with Dont-Care Conditions

In this part of the procedure, you will use K-Map to simplify a 4-variable function and verify your results following the same flow of Verilog.

1. Write the function F in the form of sum of proper minterms.
2. Make a truth table to evaluate the function F for all input combinations use Boolean Algebraic.
3. Write a Verilog module called original to implement the original circuit with all the minterms.
4. Use 4-Variable K-Map to simplify the function. You need to include the details of this step in your lab report.
5. Write a Verilog module called simplified to implement the simplified circuit you obtained from the K-Map simplification process.

Given that solution :- F (A, B, C, D) - & m (0,3,5,7, 11, 13) d (A, B, C, D) - um (4,6, 14, 15) don't care conditions So F (A, B, C, D) - & m (0,3,5, 7, 11, 13 & &d (4, 6, 14, 15) the function. K-maps to cp TD AB 00 B oo solve and minimize TD CD co 0 10 10 A o + BD + CD AB 0 With ABC AB 100 Simplified minterm ; TO + BD + CD : F (A, B, C, D) = TO + BD + CD circuit : ACD + BD+CD. TI BD B- Toruth table i= = TO+BD + CD ACO BD CD LA B 0 0 0 0 0 0 0 0 0 1 Looooooo OOOO C D 0 0 0 1 1 0 1 1 0 0 0 I 10 0 1 1 1 1 0 0 0 -0-0000-0-00 000 OOOOO OOO 000-000- --eo- -oo-- -ooo ooo - ooo-oool -0-0-000-0---00- 1 0 o-o-o o 1 - Here the above truth table if A,C and I are o then to i I same like Bo is I then BD is I and co i 1 in following possible inputs. then CD is 1 finally in the three and gates, any one value is I then the total value is 1. Given that solution :- F (A, B, C, D) - & m (0,3,5,7, 11, 13) d (A, B, C, D) - um (4,6, 14, 15) don't care conditions So F (A, B, C, D) - & m (0,3,5, 7, 11, 13 & &d (4, 6, 14, 15) the function. K-maps to cp TD AB 00 B oo solve and minimize TD CD co 0 10 10 A o + BD + CD AB 0 With ABC AB 100 Simplified minterm ; TO + BD + CD : F (A, B, C, D) = TO + BD + CD circuit : ACD + BD+CD. TI BD B- Toruth table i= = TO+BD + CD ACO BD CD LA B 0 0 0 0 0 0 0 0 0 1 Looooooo OOOO C D 0 0 0 1 1 0 1 1 0 0 0 I 10 0 1 1 1 1 0 0 0 -0-0000-0-00 000 OOOOO OOO 000-000- --eo- -oo-- -ooo ooo - ooo-oool -0-0-000-0---00- 1 0 o-o-o o 1 - Here the above truth table if A,C and I are o then to i I same like Bo is I then BD is I and co i 1 in following possible inputs. then CD is 1 finally in the three and gates, any one value is I then the total value is 1