GATE CSE 2002 | Question: 7

Question

• A. Express the function $f(x,y,z) = xy' + yz'$ with only one complement operation and one or more AND/OR operations. Draw the logic circuit implementing the expression obtained, using a single NOT gate and one or more AND/OR gates.
• B. Transform the following logic circuit (without expressing  its switching function) into an equivalent logic circuit that employs only $6$ NAND gates each with $2$-inputs.

  

Answer 1

$f(x,y,z)=xy'+yz' =xy'z'+xy'z+x'yz'+xyz'$

$f(x,y,z)=\sum_m(2,4,5,6)$

K-map

 

By pairing of $1's$, we get two pairs $(2,6),(4,5)$ resulting in same expression $F= xy'+yz'$

But by pairing of $0's$, we get two pairs $(0,1),(2,7)$, we get $F'= yz+x'y'$

Take complement, $F= \overline{(yz)}.(x+y)$

so we can implement the function with $1$ NOT , $1$ OR and $2$ AND gates.

For the second part , we need to implement given circuit using NANDs only.

so best way is to replace OR with Invert NAND, $A+B = \overline{(\bar A\bar B)}$

 

 

 

 

 

Comments

second last diagram is the key

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Yes it does but, xy'+xz'+yz' == xy'+yz'

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Salute you @praveen sir

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sir why it is (0,1),(2,7) while pairing by 0's ?. Is it (0,1),(3,7) and F= xy+y'z' ?

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the second answer is just lit

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@Amal

Yes , it must be  (3,7).

Answer 2

This is answer for part b of this Questions. Drawing the given circuit with 6 NAND gates.

Answer 3

For 1st part make equation using AND and OR

like give  x AND  gate y => to not gate u get x'+ y'

now give x OR gate  y= x+y now do

X' + Y' AND gate (X+Y)= x'y + y'x

Comments

We have got three variable, x & y & z !

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Where is the variable z?

Answer 4

First part of question

xy'+yz'

(xy'+y). (xy'+z')

(x+y). (xy'+z')

(xy'+x+y). (z'+x+y)

(x+y). (z'+x+y)

1 NOT gate

1 AND gate

2 OR gates

Comments

You applied the distributive law wrongly.

(x+y). (xy'+z') != (xy'+x+y). (z'+x+y).

If possible, then how? Explain.

Answer 5

THE question is in SOP form get the POS form and then try to construct the gates