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Design 2 bit comparator

written 3.6 years ago by teamques10 ★ 68k

modified 3.1 years ago by apsareena • 670

digital electronics

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1 Answer

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written 3.6 years ago by teamques10 ★ 68k

A1	A0	B1	B0	A>B	A=B	A
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	1	0	0
0	1	0	0	0	1	0
0	1	0	1	0	0	1
0	1	1	0	0	0	1
0	1	1	1	1	0	0
1	0	0	0	1	0	0
1	0	0	1	0	1	0
1	0	1	0	0	0	1
1	0	1	1	1	0	0
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

for A=B

$=\bar{A_1}\bar{A_0}\bar{B_1}\bar{B_0}+\bar{A_1}A_0\bar{B_1}B_0+A_1A_0B_1B_0+A_1\bar{A_0}B_1\bar{B_0}$

$=\bar{A_1}\bar{B_1}(A_0 \odot B_0) + A_1B_1(A_0 \odot B_0)$

$=(A_0 \odot B_0) . (A_1 \odot B_1)$

for A>B

$A_1\bar{B_1}+A_0\bar{B_1}\bar{B_0}+A_1A_0\bar{B_0}$

for A < B

$\bar{A_1}B_1+B_1B_0\bar{A_0}+\bar{A_1}\bar{A_0}B_0$

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