Consider the digital circuit shown below with two input lines A and B, two select lines S0 and S1, and an output line Y. The blocks Q and M represent an active high 2:4 decoder and a 4-to-1 multiplexer, respectively. Out of 16 possible input combinations, the number of combinations that produce Y=1 is ______.
Note: One input combination is an instance of [A B S1 S0].
Show Hint
Write the decoder outputs D0=A'B' and D3=AB, note that mux inputs 1 and 2 are hardwired to 0 and 1, then count, separately for each S1S0 setting, how many of the 4 A,B combinations give Y=1.
Step 1: Read the connections from the circuit.
The decoder Q takes inputs A and B and produces four active high outputs D0, D1, D2, D3, where
\[
D0=\overline{A}\,\overline{B},\quad D1=\overline{A}B,\quad D2=A\overline{B},\quad D3=AB
\]
Looking at the wiring into the multiplexer M: input 0 of M is connected to D0, input 1 of M is tied to a fixed 0, input 2 of M is tied to a fixed 1, and input 3 of M is connected to D3. Outputs D1 and D2 of the decoder are not used anywhere in the circuit.
Step 2: Write the multiplexer output in terms of the select lines.
A 4-to-1 mux with select lines S1 (MSB) and S0 (LSB) picks its output as:
\[
Y = \begin{cases}
D0, & S1S0=00 \\
0, & S1S0=01 \\
1, & S1S0=10 \\
D3, & S1S0=11
\end{cases}
\]
Step 3: Check each S1S0 case against A, B.
Case S1S0=00 (Y=D0): \(Y=1\) only when \(A=0,B=0\). Out of the 4 combinations of A, B, exactly 1 gives Y=1.
Case S1S0=01 (Y=0): Y is fixed at 0 regardless of A and B, so 0 out of 4 combinations give Y=1.
Case S1S0=10 (Y=1): Y is fixed at 1 regardless of A and B, so all 4 combinations give Y=1.
Case S1S0=11 (Y=D3): \(Y=1\) only when \(A=1,B=1\). Out of the 4 combinations of A, B, exactly 1 gives Y=1.
Step 4: Add up the winning combinations.
\[
1 + 0 + 4 + 1 = 6
\]
Final Answer:
Out of the 16 possible [A B S1 S0] combinations, the number that produce Y=1 is
\[ \boxed{6} \]
