Question

(a) Convert 10100011 from binary to decimal (b) Convert (5789) 10 to Octal (c) Convert (2576), to binary (d) Convert (ABDF).. to decimal (e) Convert (241823), to Hexadecimal Q2. (a) Subtract 1100110 from 0110001 using 2's complement (CLO 1: 2 MARKS) (b) Add 79 and (-98) using 10's complement (CLO 1: 2 MARKS) Q 3. Derive the Boolean expression for the logic circuit shown below: (CLO 1: 3 MARKS) A Q 4. Simplify the given Boolean expression using Boolean laws and implement the result using gate. (CLO 1: 3 MARKS) A= xy'z' + x'y'z + xyz + x'yz'+x'y'z' Q 5. Simplify the following functions using K map and implement the result using logic gates. (CLO 2: 5 MARKS) (a) F(A,B,C,D)= m(0,1,3,4,5,6,8,10,12,14,15) (b) F(A,B,C) = (A’+B’+C') (A’+B)(A’+C')

Answer 1

(10100011) - Decimal x 26 + 1897 1x90 + 1x2 + 0x2? + 0x2 3 + 0x2 + 1x 95 1+ 2+ 32 + 128 - 163 (163) 10 Ans. (5789) Lo to octal 85789 8 723 5 1 (13235) 8 go 3 81 13 (57.89)40 — (13235), I & (2576)8 to binay Each octal digits represents with 3 bit binay (2576)8 - (010 101 110)2 * 2 b 7 6 Answer (010 101111 110). decimal (ABDF) I A-10 B 11 D-13 F 15 16! 15x 16 + 13x16' + 11x16? + 10x 183 = (43999) 10

(941823) Hexadecimal (2 4 18 23710 - Hexadecimal 161 941823 1615713 715 We know that 16 guy Ig 15 represendes F 16 59 To 1 represent — B 163 TO 3 (3B09F) 16 (241823)70 - (38097)16 (1100110) - (olloool) using 28 11 2 First 28 Complement of B 0110001 10011102 I Stomplement Add can 1 100 110 of 10 1001114 2.8 Complement Addition of A with B's 28 Complement gives desired result My0orio - Rotschrit 1 rooito otorola do 1001111 to li o Xo1101.01 1+ Tol Indo carry dix cand Hence answer is 0110101

79+ (-98) - 79-98. Firs we had the lor - 1 - los complement of go gs complemen Add to LSB 99 98 01 t g's Complement le's complement 02 Now add 79+ 10's complement of 98 79 02 H A Creprenste) 1 (17) but 10²8 Complement 8 A ΤΟ ΤΤ Τ ΡΙΓΕ A+B Given CCA+B) COLA+B) 7. (A+B) C DE ya (A+B) CDE Gate · © 3 Gate (4) OR (A+B) AND (1+1) C M43)CD AND (A&B) CDE -

Ф А- 15 и+ 7 = 1 в = н+өв. (в+а) (+3) хў і няўz + y + 1 + йўt ці (+Х) + ў24 x 2 +хуі і я ўz my24 яу? ( 8+ яу) - z(муу я) £( +%) (843) + ( »y+%8) 78 %я + xyz йў7 У ( z+1) +7 x + Яyz ( z+z) (Я) + 2х + лу, Jžtų atžă tnya (x+2) + (х+у) - (y+1) + xyz. - Е - & A+B= A.B = Ч+2 (+) + (19) + (37) ИЧЕ + уу?

F(A,B,C,D) - Em(0,3,4,5,6, 8, 10, 12, 14, 15) CO TY AB INIT An of f lot = Ã č + Ao + ABD & BCD + ABC a 1007 1h совет plod MA

FCA,B,C). (+3 +7) (Ā +B)( +7) B BC BEB POS A latalo Sop form (A+B+0 A B C O O O A+B+C) 0 10 LATRICT 0o 0 1 A+äte] loola+B+C) lo1 A +13 to ] IL O +18+ ILI [5+1340] Eimplified FF Act AB TF ACHAB Ã +B to ABC FF AC & AB Pos A + B - Бti — AB Ас OR A B C + AB & AG ABC HC) & AC A B + AC AND