Boolean Expression Minimisation Using Boolean Algerbric Law and K-MAP.
Summary of Grouping Rules
-
A group must only contain
1s, noOs -
A group can only be
horizontalorvertical,not diagonal -
A group must contain $2 ^n 1s$ (1, 2, 4, 8, etc.)
-
Each group should be as large as possible
-
Groups may
overlap -
Groups may wrap around a table
-
Every
1must be inat least one group
Exercises:
For each of the following Truth Tables:
-
Produce the
Standard Sum of the ProductTerms for the outputS. -
Then
reducethe Boolean Expression to a simpler expression usingK-MAPand compare your answers to the provided soluation using Boolean algebric Laws.
Task-1:
| A | B | S |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
Click for Solution
\(S = \overline{A}.\overline{B} + \overline{A}.B + A.\overline{B}\)
\(S = \overline{A}.(\overline{B} + B) + A.\overline{B}\)
\(S = \overline{A}.(1) + A.\overline{B}\)
\(S = \overline{A} + A.\overline{B}\)
\(S = \overline{A} +\) A}\(.\overline{B}\)
\(S = \overline{A} + \overline{B}\)
Task-2:
| A | B | S |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
Click for Solution
\(S = \overline{A}B + A\overline{B}\)
Task-3
| A | B | C | S |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 0 |
Click for Solution
\(S = \overline{A}.\overline{B}.\overline{C} + \overline{A}.\overline{B}.C\)
\(S = \overline{A}.\overline{B}.(\overline{C} + C)\)
\(S = \overline{A}.\overline{B}.(1)\)
\(S = \overline{A}.\overline{B}\)
Task-4
| A | B | C | S |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
Click for Solution
\(S = \overline{A}.\overline{B}.C + \overline{A}.B.C + A.\overline{B}.C + A.B.C\)
\(S = \overline{B}.C.(\overline{A} + A) + B.C.(\overline{A} + A)\)
\(S = \overline{B}.C.(1) + B.C.(1)\)
\(S = \overline{B}.C + B.C\)
\(S = C.(\overline{B} + B)\)
\(S = C.(1)\)
\(S = C\)
Task-5
| A | B | C | S |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
Click for Solution
\(S = \overline{A}.\overline{B}.\overline{C} + \overline{A}.\overline{B}.C + \overline{A}.B.C + A.B.C\)
\(S = \overline{A}.\overline{B}.(\overline{C} + C) + B.C.(\overline{A} + A)\)
\(S = \overline{A}.\overline{B}.(1) + B.C.(1)\)
\(S = \overline{A}.\overline{B} + B.C\)
Task-6
| A | B | C | S |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
Click for Solution
\(S = \overline{A}.B.\overline{C} + \overline{A}.B.C + A.\overline{B}.C + A.B.\overline{C} + A.B.C\)
\(S = \overline{A}.B.(\overline{C} + C) + A.\overline{B}.C + A.\overline{B}.(\overline{C} + C)\)
\(S = \overline{A}.B.(1) + A.\overline{B}.C + A.\overline{B}.(1)\)
\(S = \overline{A}.B + A.\overline{B}.C + A.\overline{B}\)
\(S = B.(\overline{A} + A) + A.\overline{B}.C\)
\(S = B.(1) + A.\overline{B}.C\)
\(S = B + A.\overline{B}.C\)
\(S = B + A.\) B \(.C\)
\(S = B + A.C\)
Task-7
| A | B | C | S |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
Click for Solution
\(S = \overline{A}.B.C + A.\overline{B}.\overline{C} + A.\overline{B}.C + A.B.\overline{C} + A.B.C\)
\(S = \overline{A}.B.C + A.\overline{B}.(\overline{C} + C) + A.B.(\overline{C} + C)\)
\(S = \overline{A}.B.C + A.\overline{B}.(1) + A.B.(1)\)
\(S = \overline{A}.B.C + A.\overline{B} + A.B\)
\(S = \overline{A}.B.C + A.(\overline{B} + B)\)
\(S = \overline{A}.B.C + A.(1)\)
\(S = \overline{A}.B.C + A\)
\(S =\) A \(B.C + A\)
\(S = B.C + A\)