Foundations — Multiplexers (2 - 1, 4 - 1, n - 1)

Before you can enjoy the parent note on multiplexers, you must be able to read it. The parent throws around bars, subscripts, plus signs that aren't addition, ${\log }_2$, minterms, and a big $\sum$. This page picks up each of those marks, one at a time, and hands you a plain-words meaning and a picture for it. Nothing here assumes you've seen digital logic before.

0. What a "signal" even is (the atoms)

Look at the two wires in the figure: one sits at the bottom rail ($0$), one at the top rail ($1$). There is no in-between in this world — that is what makes the maths so clean. We need the bit because a multiplexer's inputs, output, and selects are all bits.

1. Naming many wires: $I_0,I_1,\dots$ (subscripts)

So "$I_3$" is read aloud as "I-three" and simply means the fourth data wire. The letter $Y$ (no subscript) is reserved for the single output wire.

2. The overline: $\bar{S}$ means "NOT" (inversion)

In the figure, follow the wire through the little triangle-with-a-bubble (that shape is the inverter). A $1$ goes in the left, a $0$ comes out the right. The bubble on the tip is the picture of the bar. We need inversion because a multiplexer must be able to say "do this when the switch is 0" — and "$S=0$" is exactly the situation where $\bar{S}=1$.

3. The dot / touching letters: AND (logical multiply)

Here is the entire behaviour in four lines:

$0\cdot 0=0,0\cdot 1=0,1\cdot 0=0,1\cdot 1=1$

4. The plus sign: OR (logical add, but capped at 1)

$0+0=0,0+1=1,1+0=1,\boxed{1+1=1}$

5. Binary numbers & ${\log }_2$ (why n selects address $2^n$ inputs)

The select wires don't act alone — read together, high wire first, they spell a number in base-2.

The figure is an odometer: 2 wheels each with digits $0,1$ can show $00,01,10,11$ — that's 4 distinct settings from 2 wheels. In general $n$ wheels give $2^n$ settings.

Check: for $N=4$ inputs, ${\log }_{2}4=2$ exactly, so $2$ selects. For $N=8$, ${\log }_{2}8=3$. For $N=5$, ${\log }_{2}5\approx 2.32$, round up to $3$.

6. Minterm $m_k$ — the "spotlight" AND-term

The recipe: for each select bit, use the plain letter where you want that bit to be $1$, and the barred letter where you want it to be $0$. Then AND them.

Now the parent's scary formula reads like plain English: $Y={\sum }_{k=0}^{2^n-1}{m}_k{I}_k=({\text{spotlight}}_0)I_0+({\text{spotlight}}_1)I_1+\cdots$ "Light exactly one spotlight, and pass the input under it."

7. The big $\sum$ — shorthand for a long OR

Because in our topic $+$ means OR, this $\sum$ is a "big OR". Nothing new — it just saves you from writing out $2^n$ terms when $n$ is large.

8. Putting the marks together — reading the 2:1 formula

You can now read the parent's first boxed equation cold: $Y=\bar{S}{I}_0+S{I}_1$ Word for word: "the output $Y$ equals (NOT-S AND input-0) OR (S AND input-1)." Test both dial positions:

• $S=0$: $\bar{S}=1$, so $Y=1\cdot I_0+0\cdot I_1=I_0$. ✅
• $S=1$: $Y=0\cdot I_0+1\cdot I_1=I_1$. ✅

Every symbol you needed was defined above, in order.

Prerequisite map

This feeds directly into Boolean Algebra Minterms and Shannon Expansion, and the minterm-generator idea becomes the decoder inside the parent's multiplexer note. The inverse routing device is the demultiplexer; the output-gating idea connects to Tri-state Buffers and the whole thing lives inside Combinational Logic Design.

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