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Design a circuit to compare two-2 bit numbers. Implement this circuit using only NOR gate
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written 3.0 years ago by
teamques10
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Let A and B are the two 2-bit numbers. The outputs of this circuit when we compare "A" with "B".
- G (greater) should be 1 only when A>B
- E (equal) should be 1 only when A=B
- L (lesser) should be 1 only when A<B</li>
| A | B | G | E | L | ||
|---|---|---|---|---|---|---|
| A1 | A0 | B1 | B0 | |||
| 0 | 0 | 0 | 0 | 0 | 1 | 0 |
| 0 | 0 | 0 | 1 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 1 | 0 | 0 | 1 |
| 0 | 1 | 0 | 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 1 | 0 | 0 | 1 |
| 1 | 0 | 0 | 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 1 | 1 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 1 | 0 |
| 1 | 0 | 1 | 1 | 0 | 0 | 1 |
| 1 | 1 | 0 | 0 | 1 | 0 | 0 |
| 1 | 1 | 0 | 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 0 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 | 0 | 1 | 0 |
To obtain the simplified Boolean expression, let's use k-maps.
There are three functions (each with the same inputs A1, A0, B1, B0), so we need three k-maps.
G(A1, A0, B1, B0)=A1A0B0'+A0B1B0'+A1B1'
E(A1, A0, B1, B0)=A1'A0'B1'B0'+A1'A0B1'B0'+A1A0B1B0+A1A0'B1B0'
L(A1, A0, B1, B0)=A1'A0'B0+A0'B1B0+A1'B1
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