When I read it, I think a \$2\$-digit BCD number is something between \$0\$ and \$99\$ in decimal. So, for example, \$0100 1001\$ is a \$2\$-digit BCD number and decimal equivalent of it is \$49\$ in my opinion.
However, this question confused me. I want to interpret \$A_0\$ or \$A_1\$ as a \$4\$ binary digits, representing a decimal number(BCD). And the circuit will add two \$2\$-digit binary numbers then. However, in such a scenario, I would expect to see an output labeling either as \$S_0\$ and \$S_1\$, which will also correspond to \$4\$ binary digits, but will represent BCD. Or, I would expect to see \$S_0\$ to \$S_7\$, each representing one binary digit.
In short, I am stuck in this question. Either I have some misunderstandings about wording, or I just don't get it.
I would appreciate if someone explains what this question wants to tell.
Best Answer
One 2-digit BCD number is 8-bit wide and is made from
High nibble = \$A_1\$ and
Low nibble = \$A_0\$
Each nibble consists of 4 bits.
So e.g. \$A_1 = 1000, A_0 = 0110\$ gives \$A = 86\$ in BCD.
Similarly \$B = B1 + B0\$.
\$S\$ is \$16\$ bits
\$= S_3 S_2 S_1 S_0\$ \$= 4 \times 4\$-bit digits
So e.g. \$S_4 S_3 S_2 S_1= 0001 \space 0010 \space 0100\space 1000 \$,
Gives \$S = 0001001001001000 = 1248\$ in BCD.