Question

Using single bit Full Adder (FA) blocks (as shown below) and required gates, construct a 6-bit Adder/Subtractor for signed numbers.

Use the signed two’s complement system for the signed numbers.

Verify your design for the following addition and subtraction by specifying A as A5A4A3A2A1A0 and B as B5B6B3B2B1B0, determining the inputs to the FAs and their outputs and showing that the outputs correspond to the correct results:

a) A-B with A = -13, B = +20 (5 points)

b) A+B with A = -23, B = +9 (5 points)

Ci in i out FA Block diagram of a single bit Full Adder

Answer 1

to = So before designing the Corcuit ale must know that to add a cicuit are add A+B, but looo substract two number LA-B We adel B's any compliment with A. To find 2's Compliment ale toke is compliment of the number and add I to it So the given Circuit shows the same... When M=0 Coso and B. O = B: So we get addition Ait Bet (i = 1; and when we need to substract we make M=1 so when Mal Go=1 Bio 1 = B so out our $i = Ait Bitli and ule have co= L We are basically - 25 Compliment of B to A. since adding ramod with CamScanner

The output circuit is givan below.

1 LEAF CScannot wth Cam$canner

Here we will get si bits but we should know whether our answere is positive or negetive. If the answere is positive it is in its normal form but when the answere is negetive it is in 2s compliment form so below are some rules to check the answere.

Here are some Rules to check if the no. is in its compliment foom00_Normal form- Rule 1 In the int fig Co is input Cooray C is the next Coory and so on A So CAD is last wory and clk 2 is ----Second last ceerd. Vüthe-Overflow bit. So if u = Cot CK-D C(K-2) = 1 We have a over flow. If kth 66th in this case bit =0 then the two ne numbers all negetive and awr aus were is in a's compliment form if kth bit 1 and 1=1 then we have colded two non-negetive numbers hence answere is in normal form. n i h Rule & IY v=0 then two honumbers are of different sign & hence iy 6th Corfithibit) - is I then this is a's Compliment Representation emdiy 6th bit is oo then it is normal Representating, CScanned with CamScanner

Part A answere

honce validating our circuit ill get so So= not Bot Go = 10 % C) = 1 on SC Ait Bit a = 1 Cal E s A + B t C = 1 C₂ = 0 53 Azt Bait Co=1 alig Ca Fort She Aut But cu = 1 9 cs=o. Sga Ast Bs +(c = o CG=la U= ( 0 Cs=1 liter so owe cool ens is O LULL (25 Compliment foun CScanned with CamScanner

Answere B.

a Atb ATB with A=-23 L. But 1 23 = hence OTOT -23 = . lolooo -A3-23 = 101001 B = BADO 01.00 hence A & in our So circuit M= 0 since we are adding B Aos 1. A = 0 A2 = 0 Az = 1 Auzo Asal Bo = 1 L B = ofl B2 - 0 B 1 B4 = 0 Bs=0 Co = 0 CScanned with CamScanner

END.