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NPP Number System, Boolean Algebra and Logic Circuits 279

10's

10's

3.33 1’s and 2’s Complement 3.33 1’s VWm 2’s H$m°påßb_|Q>
1’s complement: 1’s complement of a 1’s H$m°påßb_|Q>… {H$gr Xr JB© ~mBZar g§»`m Ho$ àË`oH$
binary number is a Binary number obtained {~Q> H$mo CëQ>m H$aZo na Omo ~mBZar g§»`m àmá hmoVr h¡
after complementing each bit. Suppose Cgo nhbr H$m 1’s H$m°påßb_|Q> H$hVo h¡ Ÿ& O¡go EH$ 4-{~Q>
A A A A is a 4-bit binary number. Its 1’s ~mBZar g§»`m A A A A H$m 1’s H$m°påßb_|Q> h¡Ÿ:
1
2
0
3
complement is A A AA . For example 3 2 1 0
2
3
0
1
1’s complement of 1010 is 0101. A 3 A 2 A 1 A 0 . CXmhaU Ho$ {bE, 1010 H$m 1’s H$m°påßb_|Q>
0101 h¡ Ÿ&
2’s Complement : 2’s complement can be 2’s H$m°påßb_|Q> … {H$gr g§»`m Ho$ 1’s H$m°påßb_|Q> H$mo
found after incrementing 1’s complement by EH$ go ~‹T>mZo na 2’s H$m°påßb_|Q> àmá hmoVm h¡ AV…
one. Thus
2’s complement = 1’s complement + 1 2’s H$m°påßb_|Q> = 1’s H$m°påßb_|Q> + 1
Take one example to find 2’s complement h_ Bgo EH$ CXmhaU go g_PVo h¢…
of a binary number.
Given number = 1110, 1’s complement = 0001 Xr JB© g§»`m =1110, 1’s H$m°påßb_|Q> = 0001
2’s complement = 0001 + 1 = 0010 2’s H$m°påßb_|Q> = 0001 + 1 = 0010