Question

The figure shows a 4-to-1 multiplexer. The inputs are connected as: 
$I_0 = 1$,  $I_1 = 0$,  $I_2 = 1$,  $I_3 = y$. 
The select lines are $S_1 = x$ and $S_0 = z$. 
Find the Boolean function $f(x,y,z)$. 
 

Hint:

For MUX problems, always write the general output expression first, then substitute input values and simplify using Boolean identities like $x + \bar{x} = 1$.

Answer 1

Step 1: Write the MUX output expression.
For a 4-to-1 MUX, output is:
\[ f = I_0 \bar{S_1}\bar{S_0} + I_1 \bar{S_1}S_0 + I_2 S_1\bar{S_0} + I_3 S_1 S_0. \]
Here, $S_1 = x$ and $S_0 = z$.
Thus,
\[ f = I_0 \bar{x}\bar{z} + I_1 \bar{x}z + I_2 x\bar{z} + I_3 xz. \]
Step 2: Substitute given inputs.
Given:
\[ I_0 = 1, \quad I_1 = 0, \quad I_2 = 1, \quad I_3 = y. \]
Substituting:
\[ f = 1(\bar{x}\bar{z}) + 0(\bar{x}z) + 1(x\bar{z}) + y(xz). \]
\[ f = \bar{x}\bar{z} + x\bar{z} + xyz. \]
Step 3: Simplify expression.
Factor $\bar{z}$ from first two terms:
\[ f = \bar{z}(\bar{x} + x) + xyz. \]
Since, \[ \bar{x} + x = 1, \]
\[ f = \bar{z} + xyz. \]
Step 4: Final Answer.
\[ \boxed{f(x,y,z) = \bar{z} + xyz}. \]