![Q3) (10 points] Tabulate the truth table of an 8x4 ROM that implements the following functions: F. (A,B,C) = AB + AC F1 (A,B](http://img.homeworklib.com/questions/840c5ec0-cbca-11eb-884a-133924478877.png?x-oss-process=image/resize,w_560)
Homework Answers
F0 = A’B + AC’
F1 = A’B’C’ + C
F2 = A’B’ + AB
F3 = AB + AC + BC
|
Address |
F0 |
F1 |
F2 |
F3 |
|
0 0 0 |
0 |
1 |
1 |
0 |
|
0 0 1 |
0 |
1 |
1 |
0 |
|
0 1 0 |
1 |
0 |
0 |
0 |
|
0 1 1 |
1 |
1 |
0 |
1 |
|
1 0 0 |
1 |
0 |
0 |
0 |
|
1 0 1 |
0 |
1 |
0 |
1 |
|
1 1 0 |
1 |
0 |
1 |
1 |
|
1 1 1 |
0 |
1 |
1 |
1 |
F0: A’B + AC’
0 0 0: 1.0 + 0.1 = 0
0 0 1: 1.0 + 0.0 = 0
0 1 0: 1.1 + 0.1 = 1
0 1 1: 1.1 + 0.0 = 1
1 0 0: 0.0 + 1.1 = 1
1 0 1: 0.0 + 1.0 = 0
1 1 0: 0.1 + 1.1 = 1
1 1 1: 0.1 + 1.0 = 0
F1: A’B’C’ + C
0 0 0: 1.1.1 + 0 = 1
0 0 1: 1.1.0 + 1 = 1
0 1 0: 1.0.1 + 0 = 0
0 1 1: 1.0.0 + 1 = 1
1 0 0: 0.1.1 + 0 = 0
1 0 1: 0.1.0 + 1 = 1
1 1 0: 0.0.1 + 0 = 0
1 1 1: 0.0.0 + 1 = 1
F2 : A’B’ + AB
0 0 0: 1.1 + 0.0 = 1
0 0 1: 1.1 + 0.0 = 1
0 1 0: 1.0 + 0.1 = 0
0 1 1: 1.0 + 0.1 = 0
1 0 0: 0.1 + 1.0 = 0
1 0 1: 0.1 + 1.0 = 0
1 1 0: 0.0 + 1.1 = 1
1 1 1: 0.0 + 1.1 = 1
F3: AB + AC + BC
0 0 0: 0.0 + 0.0 + 0.0 = 0
0 0 1: 0.0 + 0.1 + 0.1 = 0
0 1 0: 0.1 + 0.0 + 1.0 = 0
0 1 1: 0.1 + 0.1 + 1.1 = 1
1 0 0: 1.0 + 1.0 + 0.0 = 0
1 0 1: 1.0 + 1.1 + 0.1 = 1
1 1 0: 1.1 + 1.0 + 1.0 = 1
1 1 1: 1.1 + 1.1 +1.1 = 1
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