Function Realization with an 8-to-1 Multiplexer: 25-Comp-A2 Digital Systems Design, Undated paper, Q2

Question 2: Function Realization with an 8-to-1 Multiplexer (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A 3-variable function $F(A,B,C)$ specified by the K-map-style table above (Gray-coded columns $AB=00,01,11,10$; rows $C=0,1$).

Find. (a) $F$ as a Boolean expression in $A,B,C$; (b) a realization of $F$ using one 8-to-1 multiplexer.

Approach. Read the eight cells off the table as minterms of $(A,B,C)$, write the canonical sum and simplify it, then — because a 3-variable function maps onto an 8-to-1 MUX one-for-one — drive the select lines directly from $A,B,C$ and tie each data input to the truth-table value of its own minterm.

Part (a) — read the eight minterms and spot the pattern. Expanding the table cell-by-cell (order $A,B,C$): $F=1$ at $(0,1,0),(1,1,0),(0,0,1),(1,0,1)$ and $F=0$ at $(0,0,0),(1,0,0),(0,1,1),(1,1,1)$. Grouping the four $1$-cells on a K-map, $A$ never fixes a group (both $A{=}0$ and $A{=}1$ appear in each pair), while $B$ and $C$ are always opposite on a $1$-cell: $$F=\boxed{A'BC'+ABC'+A'B'C+AB'C = B\overline C+\overline BC = B\oplus C}$$ $F$ is independent of $A$ — it is simply the XOR of $B$ and $C$.
Part (b) — map $A,B,C$ onto the MUX select lines. An 8-to-1 MUX has three select lines $S_2,S_1,S_0$ and eight data inputs $D_0\ldots D_7$, with output $=D_i$ where $i=4S_2+2S_1+S_0$. Since $F$ has exactly three variables and the MUX has exactly $2^3=8$ data inputs, connect $S_2{=}A,\ S_1{=}B,\ S_0{=}C$ directly — every possible $(A,B,C)$ combination then selects exactly one data input, so no reduction/Shannon-expansion trick is required.
Tie each data input to $F$'s own minterm value. Reading $F=B\oplus C$ (independent of $A$) at every one of the 8 select combinations: $$D_0{=}0,\ D_1{=}1,\ D_2{=}1,\ D_3{=}0,\ D_4{=}0,\ D_5{=}1,\ D_6{=}1,\ D_7{=}\boxed{0}$$ (each pair $D_{2k},D_{2k+1}$ for fixed $A,B$ repeats the same $0,1,1,0$ pattern since $F$ never depends on $A$).

Fig. Q2-b — 8:1 MUX realizing $F(A,B,C)=B\oplus C$: select lines $S_2S_1S_0=ABC$ chosen directly; data inputs tied to the truth-table value of their own minterm ($D_1{=}D_2{=}D_5{=}D_6{=}1$, the rest $0$).

Final Results — Question 2
Part	Result
(a) $F(A,B,C)$	$F=B\oplus C$ (independent of $A$)
(b) MUX select mapping	$S_2{=}A,\ S_1{=}B,\ S_0{=}C$
(b) Data inputs	$D_1{=}D_2{=}D_5{=}D_6{=}1$; $D_0{=}D_3{=}D_4{=}D_7{=}0$

Topic: Combinational function realization with a multiplexer.

Key relations: $n$-variable function $\to$ $2^n{:}1$ MUX: connect all $n$ variables to the select lines, wire $D_i=F(\text{minterm }i)$; MUX output $=D_{4S_2+2S_1+S_0}$ (3-select case).

Why this works: A MUX is a physical realization of a truth table — each select combination is a unique address that steers exactly one data line to the output, so tying $D_i$ to the desired output value for minterm $i$ reproduces the table exactly, with no minimization needed even though $F$ happens to reduce to a 2-variable expression.

Common pitfall: Assuming a function that simplifies to fewer variables (here $F$ dropped $A$ entirely) needs a smaller MUX — the exam specifies an 8-to-1 MUX, so all three select lines are still wired even though two adjacent data inputs always repeat ($D_{2k}=D_{2k+1}$'s pattern for every fixed $A$).

Source: Mano & Ciletti, Digital Design, Multiplexers / combinational-circuit implementation with MSI.

Question 2 of 6: Function Realization with an 8-to-1 Multiplexer

Question 2: Function Realization with an 8-to-1 Multiplexer (20 marks)