Digital Logic Test 2

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Digital Logic Test 2

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Question 1

1 / -0

If function f(A, B) = ∑ m(0, 1, 2, 3) is implemented using SOP form, the resultant Boolean function would be:

A + B

A + B̅

AB

1 Solution

Laws of Boolean Algebra:

Name	AND Form	OR Form
Identity law	1.A = A	0 + A = A
Null Law	0.A = 0	1 + A = 1
Idempotent Law	A.A = A	A + A = A
Inverse Law	AA’ = 0	A + A’ = 1
Commutative Law	AB = BA	A + B = B + A
Associative Law	(AB)C	(A + B) + C = A + (B + C)
Distributive Law	A + BC = (A + B)(A + C)	A(B + C) = AB + AC
Absorption Law	A(A + B) = A	A + AB = A
De Morgan’s Law	(AB)’ = A’ + B’	(A + B)’ = A’B’

Application:

f(A, B) = ∑ m(0, 1, 2, 3)

= A̅ B̅ + A̅ B + A B̅ + AB

= A̅ ( B + B̅) + A (B̅ + B)

= A̅ + A = 1

Question 2
1 / -0

Consider the below given Boolean function
\(f\left( a,b,c \right)=\overline{\left( \overline{a+b}~+c \right).\bar{a}}+\left( \bar{b}~+c \right)~\)
What is the number of literal in the simplified expression of f(a, b, c)?

A
0-0

B
04-01

C
07-06

D
09-02

Solution

Concepts:
A literal is any Boolean variable x or its complement x’.
De Morgan’s Law:
\(\overline{a+b}=\bar{a}.\bar{b}\)
Calculation:
\(f\left( a,b,c \right)=\overline{\left( \overline{a+b}~+c \right).\bar{a}}+\left( \bar{b}~+c \right)\)
\(f\left( a,b,c \right)=\overline{\overline{a+b}+c}+a+\bar{b}+c\)
\(f\left( a,b,c \right)=\left( a+b \right).\bar{c}+a+\bar{b}+c\)
\(f\left( a,b,c \right)=a\bar{c}+b\bar{c}+a+\bar{b}+c\)
\(f\left( a,b,c \right)=a\left( \bar{c}+1 \right)+\left( \bar{b}+b \right)\left( \bar{b}+\bar{c} \right)+c\)
\(f\left( a,b,c \right)=a+\bar{b}+\bar{c}+c\)
\(f\left( a,b,c \right)=a+\bar{b}+1\)
\(f\left( a,b,c \right)=1\)
Question 3
1 / -0

What is the function f(A, B, C) = A̅ + B̅C in product of sum in which A is MSB and C is LSB?

A
(A̅ + B̅ + C̅) (A̅ + B̅ + C) (A̅ + B + C̅) (A̅ + B + C) (A + B̅ +C) (A̅ + B̅ + C)

B
A̅BC + A̅BC̅ + A̅B̅C + A̅B̅C̅ + AB̅C + A̅B̅C

C
(A̅ + B + C) (A̅ + B̅ + C) (A̅ + B̅ + C̅)

D
(A̅ + B̅ + C̅) (A̅ + B̅ + C) (A̅ + B + C̅) (A̅ + B + C) (A + B̅ +C)

Solution

f(A, B, C) = A̅ + B̅C
f(A, B, C) = A̅(B + B̅)(C + C̅) + (A + A̅)B̅C
f(A, B, C) = A̅(B + B̅)(C + C̅) + (A + A̅)B̅C
f(A, B, C) = A̅BC + A̅BC̅ + A̅B̅C + A̅B̅C̅ + AB̅C + A̅B̅C
f(A, B, C) = A̅B̅C̅ + A̅B̅C + A̅BC̅ + A̅BC + AB̅C
f(A, B, C) = ∑ m (0, 1, 2, 3, 5)
f(A, B, C) = Π M (4, 6, 7)
In POS
f(A, B, C) = (A̅ + B + C) (A̅ + B̅ + C) (A̅ + B̅ + C̅)