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39 votes
39 votes

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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9 Answers

5 votes
5 votes

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

0 votes
0 votes

 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 . 

0 votes
0 votes

after converting into canonical form f= x'+y'x'+xz becomes

f=x'yz+x'yz'+x'y'z+x'y'z'+xyz+xy'z its k-map will be as

clearly y and y'x' are not prime implicant hence choice a and d are out but xz(xy'z+xyz) is an implicant not prime though hence choice b should be correct

0 votes
0 votes
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
Answer:

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