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 $f(x,y,z)$.
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For MUX problems, always write the general output expression first, then substitute input values and simplify using Boolean identities like $x + \bar{x} = 1$.
Step 1: Understanding the Question The question asks to derive the simplified Boolean expression for the output `f` of a 4-to-1 multiplexer (MUX) given its specific data inputs and select line connections. Step 2: Key Formula or Approach The general Boolean expression for the output `f` of a 4-to-1 MUX with data inputs $I_0, I_1, I_2, I_3$ and select lines $S_1, S_0$ is: \[ f = I_0 \cdot \overline{S_1} \cdot \overline{S_0} + I_1 \cdot \overline{S_1} \cdot S_0 + I_2 \cdot S_1 \cdot \overline{S_0} + I_3 \cdot S_1 \cdot S_0 \] This equation selects one of the four inputs based on the binary value of $S_1S_0$. Step 3: Detailed Explanation Substitute the given inputs and select lines into the general formula: The given connections are: Data Inputs: $I_0 = 1, I_1 = 0, I_2 = 1, I_3 = y$ Select Lines: $S_1 = x, S_0 = z$
Substituting these into the MUX equation gives: \[ f(x,y,z) = (1) \cdot \bar{x} \cdot \bar{z} + (0) \cdot \bar{x} \cdot z + (1) \cdot x \cdot \bar{z} + (y) \cdot x \cdot z \] Simplify the resulting expression: The term with $I_1 = 0$ becomes zero and can be removed. \[ f = \bar{x}\bar{z} + 0 + x\bar{z} + xyz \] \[ f = \bar{x}\bar{z} + x\bar{z} + xyz \] We can factor out $\bar{z}$ from the first two terms using the distributive law: \[ f = \bar{z}(\bar{x} + x) + xyz \] Using the Boolean identity $\bar{A} + A = 1$, we get $\bar{x} + x = 1$. \[ f = \bar{z}(1) + xyz \] \[ f = \bar{z} + xyz \] Step 4: Final Answer The simplified Boolean function for the given MUX circuit is: \[ \boxed{f(x,y,z) = \bar{z} + xyz} \]
