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Which of the following sets of component(s) is/are sufficient to implement any arbitrary Boolean function?

1. XOR gates, NOT gates

2. $2$ to $1$ multiplexers

3. AND gates, XOR gates

4. Three-input gates that output $(A.B) + C$ for the inputs $A, B$ and $C$.

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Sir, what do you mean by xor is linear? Can you please explain a bit ?

1. XOR and NOT gates can only make XOR and XNOR which are not functionally complete- $a \oplus \bar a = 1, a \oplus a = 0.$

2. 2-1 multiplexer is functionally complete provided we have external 1 and 0 available. For NOT gate, use $x$ as select line and use 0 and 1 as inputs. For AND gate, use $y$ and 0 as inputs and $x$ as select. With {AND, NOT} any other gate can be made.

3. XOR can be used to make a NOT gate ($a \oplus 1 = \bar a$) and {AND, NOT} is functionally complete. Again this requires external 1.

4. We have $AB + C$. Using $C = 0$, we get an AND gate. Using $B = 1$ we get an OR gate. But we cannot derive a NOT gate here.

So, options B and C are true provided external 1 and 0 are available.
by

Multiplexers Are Functionally Complete

Can you please explain point no. 3??

What does external 1 mean?
If we say that "with the help of 0 and 1 B and C will become functionally complete obviously partial" then why with the same help of 0 and 1 we can not have A also as partially functionally complete.
Functionally complete operation set is a set of Logic from which any arbitrary Boolean function can be realized .

Exp of Functionally complete set operation are

PC 1 {OR ,AND  NOT}

PC 2 { NAND }

PC 3 ( NOR )

PC 4 { XOR  AND }

Note - By Heart  all.

MUX too
XOR,OR

NOT,OR

NOT,AND too

{XOR, AND} are partially functionally complete. You need external 1 to get the NOT gate.

Answer is B because Multiplexer is the universal logic circuit that Comprise All the boolean function
by

I think B ,C

can you tell me please why not B????
2:1 MUX cant implement binary boolean functions

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