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+17 votes

Let $f(w, x, y, z) = \sum {\left(0,4,5,7,8,9,13,15\right)}$. Which of the following expressions are NOT equivalent to $f$?

P: $x'y'z' + w'xy' + wy'z + xz$

Q: $w'y'z' + wx'y' + xz$

R: $w'y'z' + wx'y' + xyz+xy'z$

S: $x'y'z' + wx'y'+ w'y$

  1. P only
  2. Q and S
  3. R and S
  4. S only
asked in Digital Logic by Veteran (59.5k points)
edited by | 1.6k views
can somebody remove that cursor from option A.

3 Answers

+16 votes
Best answer
  w'x' w'x wx wx'
y'z' 1 1   1
y'z   1 1 1
yz   1 1  

So, minimized expression will be 

$xz  + w'y'z' + wx'y'$ which is Q. From the K-map, we can also get P and R. So, only S is NOT equivalent to $f$.

answered by Veteran (355k points)
selected by
I am not understanding how to check P and R. one way is that we can draw the 4bit truth table but that is time consuming. How to derive it(P and R) using k-map?
+19 votes

Let me show u a very simple method 

Let w =1 ,x =1 ,y=1 ,z=1 then the value of f is 1

consider each statement

x'y'z' + w' x y' +w y' z +x z = 0.0.0 + 0.1.0 + 1.0.1 + 1.1 =1

w' y' z' +w x' z' y' +x z = 0.0.0 +1.0.0 +1.1  =1

w' y' z' +w x' y' +x y z +x y' z =0.0.0 +1.0.0 +1.1.1 +1.0.1 =1

x' y' z' +w x' y' + w' y = 0.0.0 + 1.0.0 + 0.1 =0

So statement (d) is false because w=1 x=1 y=1 z=1 the value of f is 0 .

(d) does not contain the essential Minterms .



answered by Boss (45.1k points)
Is this method correct for generalisation? If yes,why?
is this a perfect method to solve such question, sometimes I got the error  by this method ?

need help @shekhar_chauhan
@arjun sir confirm it please
Here it works by only putting 0 and 1 values. But this is not a general way to solve this problem. For e.g. for $x.z$ we get $1.1=1$ though it's not a correct representation of function.
+5 votes
Minterms of P, Q and R are same as f.

So, S is not equivalent to f

Ans (D)
answered by Boss (13.5k points)

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