From the following tables
Two bits equality detector:
| x1 |
x2 |
y1 |
y2 |
f |
| 0 |
0 |
0 |
0 |
1 |
| 0 |
0 |
0 |
1 |
0 |
| 0 |
0 |
1 |
0 |
0 |
| 0 |
0 |
1 |
1 |
0 |
| 0 |
1 |
0 |
0 |
0 |
| 0 |
1 |
0 |
1 |
1 |
| 0 |
1 |
1 |
0 |
0 |
| 0 |
1 |
1 |
1 |
0 |
| 1 |
0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
1 |
0 |
| 1 |
0 |
1 |
0 |
1 |
| 1 |
0 |
1 |
1 |
0 |
| 1 |
1 |
0 |
0 |
0 |
| 1 |
1 |
0 |
1 |
0 |
| 1 |
1 |
1 |
0 |
0 |
| 1 |
1 |
1 |
1 |
1 |
Min-term expresson:
| x |
y |
f |
min-term |
| 0 |
0 |
0 |
------------- |
| 0 |
1 |
1 |
x'$\cdot$y |
| 1 |
0 |
1 |
x$\cdot$y' |
| 1 |
1 |
0 |
------------- |
$f=\sum all the midterms = \bar{x}\cdot y+x\cdot \bar{y}$
Max-term expression:
| x |
y |
f |
max-term |
| 0 |
0 |
0 |
$x + y$ |
| 0 |
1 |
1 |
-------------- |
| 1 |
0 |
1 |
-------------- |
| 1 |
1 |
0 |
$\bar{x} + \bar{y}$ |
$f=\prod all the maxterms=(x+y)(\bar{x}+\bar{y})$
Write down the min-term expression of the f output of two bits equlity detector.
Write down the max-term expression of the f output of the truth table below:
| x |
y |
f |
| 0 |
0 |
0 |
| 0 |
1 |
1 |
| 1 |
0 |
0 |
| 1 |
1 |
1 |
