Multiplexers (2 - 1, 4 - 1, n - 1) — Hinglish notes

3.3.5 · Hardware › Combinational Circuits

MUX ki ZAROORAT kyun hai?

Multiplexer KYA hai — definition

$N$ inputs ke liye kitni select lines chahiye: kyunki $n$ bits $2^{n}$ inputs ko address karti hain, $n = ⌈ lo g_{2} N ⌉ .$

Boolean expression kaise derive karein (scratch se)

Hum equation us tarah banate hain jaise ek decoder karta hai: har input ko select variables ke us minterm ke saath AND kiya jaata hai jo uski taraf "point" karta hai, aur ye saare products OR kar diye jaate hain.

2:1 MUX derivation

Hum chahte hain $Y = I_{0}$ jab $S = 0$ aur $Y = I_{1}$ jab $S = 1$ . " $S = 0$ hone par" ko term $\overset{ˉ}{S}$ capture karta hai; " $S = 1$ hone par" ko $S$ capture karta hai. Toh:

$Y = \overset{ˉ}{S} I_{0} + S I_{1}$

4:1 MUX derivation

Do selects $S_{1} S_{0}$ chaar minterms dete hain. Input $I_{k}$ tab pass hota hai jab $S_{1} S_{0}$ ki value $k$ ho:

$S_{1}$	$S_{0}$	minterm	passes
0	0	$\overset{ˉ}{S}_{1} \overset{ˉ}{S}_{0}$	$I_{0}$
0	1	$\overset{ˉ}{S}_{1} S_{0}$	$I_{1}$
1	0	$S_{1} \overset{ˉ}{S}_{0}$	$I_{2}$
1	1	$S_{1} S_{0}$	$I_{3}$

General n:1 MUX

$Y = k = 0 \sum 2^{n} - 1 m_{k} I_{k}$ jahaan $m_{k}$ select variables ka wo minterm hai jo $k$ se correspond karta hai. Ye sum-of-products literally ek decoder (minterms) hai jo AND–OR logic ko feed kar raha hai.

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

MUX as a universal function generator (80/20 gem)

Trick (Shannon expansion): ek $n$ -variable function ko chhote $2^{n - 1} : 1$ MUX se banane ke liye, $n - 1$ variables ko selects pe rakho; har data input baaki bache variable $v$ ke ${0, 1, v, \overset{v}{ˉ}}$ mein se ek ban jaata hai.

Enable line

Real MUX chips mein ek enable $E$ add hoti hai. Typically $Y = E \cdot (MUX expression)$ (active-high), toh $E = 0$ force karta hai $Y = 0$ (ya tri-state parts pe high-Z). Enable se tum cascade kar sakte ho aur shared bus pe tri-state kar sakte ho.

Common mistakes

Recall Feynman: ek 12-saal ke bacche ko samjhao

Socho 4 paani ke pipes hain aur ek nal jo ek baar mein sirf ek pipe se pi sakta hai. Tum ek chhota dial ghuma te ho jo 2 clicks mein chalta hai (wo hain 2 select bits = 4 positions). Dial jis number pe hai, usi pipe ka paani bahar aata hai. MUX electricity ke liye wohi magic nal hai — woh sirf ek chosen input ko single output wire pe copy karta hai, baaki sab ko ignore karta hai.

Connections

Decoders — har MUX ke andar minterm generator (ek MUX = decoder + OR).
Demultiplexers — exact inverse operation (one to many).
Boolean Algebra Minterms — wo SOP form jo humne derive ki.
Shannon Expansion — functions ko chhote MUXes pe shrink kaise karein.
Tri-state Buffers — enable line aur bus sharing.
Combinational Logic Design — MUX as universal function block.

Flashcards

Number of select lines for an N-input MUX?

⌈ lo g_{2} N ⌉

Boolean expression of a 2:1 MUX?

Y = \overset{ˉ}{S} I_{0} + S I_{1}

Boolean expression of a 4:1 MUX?

Y = \overset{ˉ}{S}_{1} \overset{ˉ}{S}_{0} I_{0} + \overset{ˉ}{S}_{1} S_{0} I_{1} + S_{1} \overset{ˉ}{S}_{0} I_{2} + S_{1} S_{0} I_{3}

Which data input is selected when

S_{1} S_{0} = 10

I_{2}

(since

1 0_{2} = 2

General n:1 MUX equation?

Y = \sum_{k} m_{k} I_{k}

, minterm of selects times input.

Why can a

2^{n} : 1

MUX implement any n-variable function?

Wire each input to the truth-table value of its row; MUX selects the correct row.

Implement XOR(A,B) on a 4:1 MUX (selects A,B). Inputs?

I_{0} = 0, I_{1} = 1, I_{2} = 1, I_{3} = 0

Implement XOR(A,B) on a 2:1 MUX with A on select. Inputs?

I_{0} = B, I_{1} = \overset{ˉ}{B}

Difference between MUX and DEMUX?

MUX = many inputs to one output; DEMUX = one input to many outputs.

Role of the enable line?

Gates output;

E = 0

forces output 0 / high-Z, enabling cascading and bus sharing.

How to build an 8:1 MUX from smaller ones?

Two 4:1 MUXes (select

S_{1} S_{0}

) feeding a 2:1 MUX (select

S_{2}