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Let $f(x, y, z)=\bar{x} + \bar{y}x + xz$ be a switching function. Which one of the following is valid?

1. $\bar{y} x$ is a prime implicant of $f$

2. $xz$ is a minterm of $f$

3. $xz$ is an implicant of $f$

4. $y$ is a prime implicant of $f$

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Ans is 3 or 4 ?
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This question needs a better Explanation

In sum of terms, any term is an implicant because it implies the function. So, $xz$ is an implicant and hence C is the answer. Still, lets see the other options.

If no minimization is possible for an implicant (by removing any variable) it becomes a prime implicant.

If a prime implicant is present in any possible expression for a function, it is called an essential prime implicant. (For example in K-map we might be able to choose among several prime implicants but for essential prime implicants there won't be a choice).

So, $f = x' + y'x + xz$
$= y' + x' + z$ (could be also derived using algebraic rules as in http://www.ee.surrey.ac.uk/Projects/Labview/boolalgebra/ )

So, the prime implicants are $x', y'$ and $z$. Being single variable ones and with no common variables, all must be essential also.

Now, a choice is false, as $y'$ is a prime implicant and hence, $y'x$ is just an implicant but not prime.
b choice - $xz$ is not a minterm. A minterm must include all variables. So, $xyz$ is a minterm so, is $xy'z$, but not $xz$.
d choice - $y'$ is a prime implicant not $y$.

by
edited
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Sir

any term is an implicant because it implies the function.Can you explain it a bit.Not able to understand
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Whenever a term in SOP form is 1, the function value becomes 1. i.e., any term implies the function.
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@Arjun sir

In above K-map,number of prime implicant $(PI) = 3$

and number of essential prime implicant$(EPI) = 3$

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@Lakshman Patel RJIT

Yap.

Just wanted to confirm, Is Implicants is equal to Number of minterms or is it product terms in SOP?

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A normal product term that implies $Y.$

Example: For the function $Y = AB + ABC + BC,$ the implicants are $AB, ABC,$ and $BC$ because if any one of those terms are true, then $Y$ is true.

Reference:

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@Lakshman Patel RJIT

What if we minimize the function? You can take this example where f is reduced to $\bar{x}+\bar{y}+z$. Can we say implicants are $\bar{x}, \bar{y}$ and z?

Edit:- Got it. Thanks anyway.

f(x,y,z) = x' + y'x + xz

An implicant of a function is a product term that is included in the function.

so x', y'x and xz ,all are implicants of given function.

A prime implicant of a function is an implicant that is not included in any other implicant of the function.

option a)   y'x is not a prime implicant as it is included in xz [ xy'z+ xyz]

option d) y is not a prime implicant as it include in both x' and xz.

a product term in which all the variables appear is called a minterm of the function

option b) xz is not a minterm

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so which is the prime implicant?
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Draw kmap then derive .... Any subcube which is not fully part of other sub cube in Kmap is Prime implicant ....
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@Praveen sir.
How $xy'$ is included in $xz?$
$xy' \leftrightarrow \color{RED}{xy'z} + xy'z'$
$xz \leftrightarrow xyz + \color{RED}{xy'z}$
$xy'z'$ is not included in $xz.$

Similarly, how $y$ is included in $x' \ and \ xz?$
$y$ is not even implicant.
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y'x is not a prime implicant as it is included in xz [ xy'z+ xyz]

Indeed, y'x is not a prime implicant But the given reason(explanation) for it is Wrong.

y is not a prime implicant as it include in both x' and xz.

Again, Wrong reason.

@Soumya,

xy′z′ is not included in xz.

Similarly, how y is included in x′ and xz?
y is not even implicant.

You are absolutely right.

Explanation:

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Why not group last row instead of last two as xz?

$xz$ is an implicant and $\neg y$ is both prime and essential prime implicant. The sop would be $z+\neg x+\neg y$.

Implicant: Something that implies a function is its implicant
Prime implicant: The most reduced (minimal) implicant
Essential prime implicant: The prime implicant which cannot be avoided in any SOP

edited by
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thanks @Marv
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need better explanation

let suppose Y=f(A,B,C)

the expression for y=AB+ABC+BC

1)Implicant : for y the implicants are AB,ABC,BC basically in simple words implicant is 'for any SOP , the product term that implies our y is basically implicant ' i mean when your implicant would be 1 then y would be 1.

for given sop set any product term (AB or ABC or BC) 1 and it will make y=1.

2)Prime implicant : Implicants whose removal don't now imply y (mean their presence is prime)are called PI.

for y PI are AB,BC as if they are removed and here we lost y. ABC is not PI as if you remove it from y then still our y is not lost.

3)Essential PI : An EPI of a function y is one that cover a minterm of F , not covered by any other PI of y.

Minterm for given function are x'(y+y')(z+z') and y'z(x+x') and xz(y+y')

When you would solve given function using K-Map you will get f=x'+y'+z

For f if any one of (x' , y'z , xz) is 1 then it would make f=1 so these all are implicants.

as f= x'+y'+z and if we remove anyone of x' or y' or z then here we lost f so all three are PI.

http://web.cecs.pdx.edu/~mcnames/ECE171/Lectures/Lecture10.html

prime implicant isthe biggest subcube possible . if you minimize Z+~X+~Y ,~Y is one of the prime implicant of this not Y .

if you see z is a prime implicant so xz is an implicant of this function .

so option 3 .

Implicant- A product term that covers one or more minterms.

Prime Implicant- Not subset of higher implicant

EPI- There are many PIs but an EPI is one which has atleast one cell that is not covered by any other PI

Although C is the answer but I have a doubt that if we expand xz as x(y+y')z         xyz + xy'z are minterms
annswer - C
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y'x is not a prime implicant as it is included in implicant xz

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