Question

6. (20 Points) Find the minterms of the function realized by the circuit below. DO וס 2-to-4 mo Decoder ml A (LSE) m2 B m3 b fab.c) 2-10-4 mo Decoder mi GA (LSB) m2 с B m3 D2 8-to-1 D3 MUX D4 Y D5 DO D7 (MSB) с в А obc

Answer 1

Go through the solution given below.

CIRCUIT DIAGRAM mo a | ACLSB) m2 B m3 b fla,b,c) Do Di D2 D3 IDY 105 D6 (P7CB A mo mi ALLSB m2 m3 с abc MSB) TRUTH TABLE OF 2 to 4 DECODER bb b a mo mi m₂ m3 - 0 0 Do =mo = b D1 = y =ba D2 D3 = m₂ = ba 0 0 m₂ =ba O 1 o 0 Similiarly for and 2 to 4 decoder Dy = my = ł Ds DG Dz mz ca m, =ła m2 =ca

FOR 8 to I HUX MSB LSB ។ 4 b 0 in оооо OO--OO-- Do DU D2 D₃ 0 - 0 - - Therefore, 4- ābē (Do] + ābo CD) abECD) + abc CD3) abc (04) + abc Cos) abc (Do) tabc (D7) Du. DS - D6 D7 put the value of Do,.. Dz as obtained from #2 to 4 Decoder a.az a.a = a Y = abchat a toc ba + ābībā tā bcha + abocat abocat abocat abcca ។ =āb + abc final muin-term obtained.