Question

11. Refer to figure 2 to complete the table given. A3-AO B3-BO x S4-SO (5 bits) BCD sum (4 bits) 0101 0010 0111 0100 1000 1001 1001 0011 B, B2, Bo BCD code group CA 4-bit parallel dder (74LS93) Ca, carry from + lower position adder SA S; S S, S. Az Az A, A BCD code group Carry to next BCD adder X 74LS32 74LSOB Correction Logic 74LS32 Co = 0 CA not Leed 4-bit parallel addar 174L983) TIL 1 Zz 21 Σο BCD Sum Correction Adder Figure 2 A BCD adder contains two 4-bit adders and a correction-detector circult.

Answer 1

B3-Bo Az-Ao @ 0101 0010 BCD olol 0010 OE 0111 Sy sz 2 si 50 100 x = (Sa+s,). S3+ Sy BCD sum oll) oo oo -> Bob sum @ O C4 Ez & Eco BCP 0101 0100 olll 0100 or lot sy S3 525.so Table: x= $2+5) •S3 + Sy Az-Ao = (ori. I to 0101 - 0111 BCD som 1000 1011 olo Goool Cy Ez E₂ EEO 1001 1000 lool 100 o 100) Goool sy S3 S2 S1 so x = 625) - Sztsy (oto), 0+1

BCD som 1 ooo! Ol10 okoll Ez E2h, co 1001 ool) =) 1001 ool a Too Sy S3425150 x= Setsi). S3+ Sy 11 (to). I to 1 BCP sum - 11oo ollo Itolo Ezez E, ko Table: x cuns ang Bz-Bo sy-so (5-bits) (4 bils) Az-Ao 1 OO10 olol 00111 ololl 1 000! 000 / oll 001 0 0 100 0111 lool 1 0-100 1000 cool ool