# (Homework Solution): 6. [8 marks] For the Boolean functions E and F, as given in the following truth table XİYİZ E

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.

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.

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