Unit -1

Minimization Technique

Q1. Simplify f(X,Y,Z)=∏M(0,1,2,4)f(X,Y,Z)=∏M(0,1,2,4)using K-map.

Therefore, the simplified Boolean function is

f = (X + Y).(Y + Z).(Z + X)

Q2. Simplify: F(P,Q,R,S)=∑(0,2,5,7,8,10,13,15)

F = P’Q’R’S’ + PQ’R’S’ + P’Q’RS’ +PQ’RS’ + QS

F = P’Q’S’ + PQ’S’ + QS

F = Q’S’ +QS

Q3. Simplify: F(A,B,C)=π(0,3,6,7)

F = A’BC +ABC +A’B’C’ +ABC’

F = BC + C’ ( A’B’ + AB )

Q4. Simplify, f(W,X,Y,Z)=∑m(2,6,8,9,10,11,14,15) and f(W,X,Y,Z)=∑m(2,6,8,9,10,11,14,15) using Quine-McClukey tabular method.

A4:

Therefore, the prime implicants are YZ’, WX’ & WY.

The prime implicant table is shown below.

The reduced prime implicant table is shown below.

Hence, f(W,X,Y,Z) = YZ’ + WX’ + WY.

Q5. Simplify Y(A,B,C,D) =∑ m(0,1,3,7,8,9,11,15)

A5: Groups are made with respect to the no. of one’s present.

Now, comparing with the above table wherever we have a different bit present we put a ’-‘ there.

The same thing is done by comparing it from the above table.

The table for prime implicants is:

Rounding the min terms which has X in the column and hence the final answer is B’C’ + CD.

Q6. Explain 1's complement

A6:

Fig.: 1's complement

Q7. Explain 2’s complement.

A7:

Fig.: 2's complement

Q8. Convert the Boolean function into Standard SoP form.

f = p’qr + pq’r + pqr’ + pqr

A8:

Step 1 – By using the Boolean postulate, x + x = x and also writing the last term pqr two more times we get

⇒ f = p’qr + pq’r + pqr’ + pqr + pqr + pqr

Step 2 – By Using Distributive law for 1st and 4th terms, 2nd and 5th terms, 3rdand 6th terms.

⇒ f = qr(p’ + p) + pr(q’ + q) + pq(r’ + r)

Step 3 – Then Using Boolean postulate, x + x’ = 1 we get

⇒ f = qr(1) + pr(1) + pq(1)

Step 4 – hence using Boolean postulate, x.1 = x we get

⇒ f = qr + pr + pq

⇒ f = pq + qr + pr

This is the required Boolean function.

Q9. Convert the Boolean function into Standard PoS form.

f = (p + q + r).(p + q + r’).(p + q’ + r).(p’ + q + r)

A9:

Step 1 – By using the Boolean postulate, x.x = x and writing the first term p+q+r two more times we get

⇒ f = (p + q + r).(p + q + r).(p + q + r).(p + q + r’).(p +q’ + r).(p’ + q + r)

Step 2 – Now by using Distributive law, x + (y.z) = (x + y).(x + z) for 1st and 4thparenthesis, 2nd and 5th parenthesis, 3rd and 6th parenthesis.

⇒ f = (p + q + rr’).(p + r + qq’).(q + r + pp’)

Step 3 − Applying Boolean postulate, x.x’=0 for simplifying of the terms present in each parenthesis.

⇒ f = (p + q + 0).(p + r + 0).(q + r + 0)

Step 4 − Using Boolean postulate, x + 0 = x we get

⇒ f = (p + q).(p + r).(q + r)

⇒ f = (p + q).(q + r).(p + r)

This is the simplified Boolean function.

Hence, both Standard SoP and Standard PoS forms are Dual to one another.

Q10. Simplify  f(X,Y,Z)=∏M(0,1,2,4)f(X,Y,Z)=∏M(0,1,2,4)using K-map.

A10:

Therefore, the simplified Boolean function is

f = (X + Y).(Y + Z).(Z + X)

Q11. Simplify:  F(P,Q,R,S)=∑(0,2,5,7,8,10,13,15)

A11:

F = P’Q’R’S’ + PQ’R’S’ + P’Q’RS’ +PQ’RS’ + QS

F = P’Q’S’ + PQ’S’ + QS

F = Q’S’ +QS

Q12. Simplify:  F(A,B,C)=π(0,3,6,7)

A12:

F = A’BC +ABC +A’B’C’ +ABC’

F = BC + C’ ( A’B’ + AB )