Page 317 - FUNDAMENTALS OF COMPUTER

NPP Number System, Boolean Algebra and Logic Circuits 317

Start from right side and add A and 1. gd©àW_ h_ A VWm 1 H$mo Omo‹S>|JoŸ& AV… EH$ hm\$
0
0
Therefore, one Half Adder is needed. Next we ES>a Amdí`H$ hmoJmŸ& BgHo$ níMmV² A VWm hm{gb H$mo
add A and previous carry C generated from 1
1
1
Half Adder. Therefore we need Four Half Omo‹S>|JoŸ& nwZ… hm\$ ES>a Amdí`H$ hmoJmŸ& Bg àH$ma Hw$b Mma hm\$
Adders and one controlled invertor: ES>a Amdí`H$ hm|JoŸ& BgH$m g§nyU© n[anW Bg àH$ma hmoJm…

COMPLEMENT

H.A. NPP H.A. H.A.
H.A.
C 4 C 3 C 2 C 1

X 3 X 2 X 1 X 0

As soon as COMPLEMENT = 1 is given, O¡go hr COMPLEMENT = 1 {X`m OmVm h¡ Mmam|
the outputs of XOR gates give 1's complement XOR JoQ> NOT JoQ> H$m H$m`© H$aVo h¢ Am¡a
A 3 A 2 A 1 A . This is applied to four Half A A A A àXmZ H$aVo h¢Ÿ& àW_ hm\$ ES>a A
0
0
3
2
1
0
Adders. Addition of 1 is done with the help of VWm 1 H$mo Omo‹S>Vm h¡ Ÿ& Bg àH$ma ~mH$s hm\$ ES>a H$m`©
COMPLEMENT input. Whenever COMPLE- H$aVo h¢ Ÿ& VWm h_| X X X X na H$m 2'g H$m°påßb_|Q
1
0
3
2
MENT=0, The output becomes A A A A . {_bVm h¡ Ÿ& bo{H$Z O~ COMPLEMENT = 0 N>moQ>m h¡,
2
3
1
0
Thus above circuit is a controlled 2's V~ X X X X = A A A A {_bVm h¡ Ÿ& AV… Cnamo³V
Complement Generator as shown: 3 2 1 0 3 2 1 0
n[anW {Z`§{ÌV 2's H$m°påßb_|Q> OZaoQ>a h¡…
A 3 A 2 A 1 A 0
Controlled COMPLEMENT
2's Complement
Generator
X 3 X 2 X 1 X 0