![6. [8 marks] For the Boolean functions E and F, as given in the following truth table XİYİZ E F 0 1 10 0 1 1 10 a. List the minterms and maxterms of each function b. List the minterms of E and F. c. List the minterms of F and E.F d. Express E and Fin sum-of-minterms form e. Simplify E and F to expressions with a minimum of literals.](https://media.cheggcdn.com/media%2F870%2F8701bcde-5f7a-47cc-b72a-7849b1ad66b7%2Fimage)
6. [8 marks] For the Boolean functions E and F, as given in the following truth table XİYİZ E F 0 1 10 0 1 1 10 a. List the minterms and maxterms of each function b. List the minterms of E’ and F. c. List the minterms of F and E.F d. Express E and Fin sum-of-minterms form e. Simplify E and F to expressions with a minimum of literals.
(Homework Solution): 6. [8 marks] For the Boolean functions E and F, as given in the following truth table XİYİZ E
Expert Answer
Answer.
Note: ‘ ` ‘ is used to represent not operation.
a.
| X | Y | Z | E | Min tem of E | Max Term of E | F | Min term of F | Max term of F |
| 0 | 0 | 0 | 0 | X`+Y`+Z` | 1 | X`Y`Z` | ||
| 0 | 0 | 1 | 1 | X`Y`Z | 0 | X+Y+Z` | ||
| 0 | 1 | 0 | 1 | X`YZ` | 1 | X`YZ` | ||
| 0 | 1 | 1 | 0 | X+Y`+Z` | 0 | X+Y`+Z` | ||
| 1 | 0 | 0 | 1 | XY`Z` | 1 | XY`Z` | ||
| 1 | 0 | 1 | 0 | X`+Y+Z` | 0 | X`+Y+Z` | ||
| 1 | 1 | 0 | 1 | XYZ` | 0 | X`+Y`+Z | ||
| 1 | 1 | 1 | 0 | X`+Y`+Z` | 1 | XYZ |
Note: Min term is written when the function gives ‘1’ as output. While, max term is written, only when the function gives ‘0’ as output.
b.
| X | Y | Z | E | E` | Min tem of E` | F | F` | Min term of F` |
| 0 | 0 | 0 | 0 | 1 | X`Y`Z` | 1 | 0 | |
| 0 | 0 | 1 | 1 | 0 | 0 | 1 | X`Y`Z | |
| 0 | 1 | 0 | 1 | 0 | 1 | 0 | ||
| 0 | 1 | 1 | 0 | 1 | X`YZ | 0 | 1 | X`YZ |
| 1 | 0 | 0 | 1 | 0 | 1 | 0 | ||
| 1 | 0 | 1 | 0 | 1 | XY`Z | 0 | 1 | XY`Z |
| 1 | 1 | 0 | 1 | 0 | 0 | 1 | XYZ` | |
| 1 | 1 | 1 | 0 | 1 | XYZ | 1 | 0 |
Note: E` is complement of E i.e. if E = 0 then E`= 1 and if E =1 then E`= 0.
c.
| X | Y | Z | E | F | E+F | Min tem of E+F | E.F | Min term of E.F |
| 0 | 0 | 0 | 0 | 1 | 1 | X`Y`Z` | 0 | |
| 0 | 0 | 1 | 1 | 0 | 1 | X`Y`Z | 0 | |
| 0 | 1 | 0 | 1 | 1 | 1 | X`YZ` | 1 | X`YZ` |
| 0 | 1 | 1 | 0 | 0 | 0 | 0 | ||
| 1 | 0 | 0 | 1 | 1 | 1 | XY`Z` | 1 | XY`Z` |
| 1 | 0 | 1 | 0 | 0 | 0 | 0 | ||
| 1 | 1 | 0 | 1 | 0 | 1 | XYZ` | 0 | |
| 1 | 1 | 1 | 0 | 1 | 1 | XYZ | 0 |
Note: ‘+’ is OR operation- it gives ‘1’ as output if any of the two input is 1. ‘.’ is AND operation- it gives ‘1’ as output if both the input is ‘1’ else it gives ‘0’ as output.
d.
Sum of min terms form of E (using the result from part a.)-
= X`Y`Z+X`YZ`+XY`Z`+XYZ`
Note: all the min terms of E which are calculated in part one are added and written.
Sum of min terms form of F (using the result from part a.)-
= X`Y`Z`+X`YZ`+XY`Z`+XYZ
Note: all the min terms of F which are calculated in part one are added and written.
e.
Simplifying the result of sum of min term of E –
= X`Y`Z+ X`YZ`+XY`Z`+XYZ` using the min term of ‘E’ from part d.
= X`Y`Z+ X`YZ`+ (XY`Z`+XYZ`)
= X`Y`Z+ X`YZ`+ (Y`+Y)XZ` rearranging and taking XZ` common
= X`Y`Z+ X`YZ`+ 1.XZ` using complementary law
= X`Y`Z+ (X`YZ`+ XZ`) using property of 1: 1.X=X
= X`Y`Z+ (X`Y+ X)Z` rearranging terms and taking Z` common
= X`Y`Z+ (Y+ X)Z` using X+X`Y = X+Y
= X`Y`Z+ YZ`+ XZ`
Hence this is the most simplified form of the sum of min term of E.
Simplifying the result of sum of min term of F –
= X`Y`Z`+X`YZ`+XY`Z`+XYZ using the min term of ‘F’ from part d.
= (X`Y`Z`+X`YZ`)+XY`Z`+XYZ
= (Y`+Y)X`Z`+XY`Z`+XYZ rearranging and taking X`Z` common
= 1.X`Z`+XY`Z`+XYZ using complementary law
= (X`Z`+XY`Z`) +XYZ using property of 1: 1.X=X
= (X`+XY`)Z` +XYZ rearranging terms and taking Z` common
= (X`+Y`)Z` +XYZ using X+X`Y = X+Y
= X`Z` +Y`Z` +XYZ
Hence this is the most simplified form of the sum of min term of F.
Note: X+X`Y = X+Y is also called as third distributive law.
