Digital Logic Test 1

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Digital Logic Test 1

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Question 1
1 / -0

If \(X\;\# \;Y = \overline {\overline X.Y} \;then\;\overline {X\# Y} \;\# \;\overline X\) is equivalent to

A
X

B
Y

C
X + Y

D
X̅ + Y

Solution

\(X\;\# \;Y = \overline {\overline X.Y} = X + \overline Y\)
\(\overline {X\# Y} \;\# \;\overline X = \overline {X +\overline Y} \;\# ̅ X = \left( {\overline {X\;} .Y} \right)\;\# \;\overline X\)
\( =\overline{ \overline {\overline X.Y\;} .\overline X} \)
= (X̅.Y + X) \\apply deMorgan's law
= X + Y

Question 2

1 / -0

If X = 0 in the logic equation, \(\left[ {X + Z\left\{ {Y + \left( {Z + \bar XY} \right)} \right\}} \right]\left[ {\bar Y + \bar X\left( {Z + Y} \right)} \right] = 0\) then

Z = 0

Z = Y

Z = 1

Z = Y̅

Solution

Laws of Boolean Algebra:

All Boolean algebra laws are shown below

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:

By putting X = 0, the above Boolean equation becomes

\( \Rightarrow \left[ {0 + Z\left\{ {Y + \left( {Z + Y} \right)} \right\}} \right]\left[ {\bar Y + \left( {Z + Y} \right)} \right] = 0\)

\( \Rightarrow \left[ {Z\left\{ {Y + Z} \right\}} \right]\left[ {1 + Z} \right] = 0\)

⇒ YZ + Z = 0

⇒ Z = 0

Question 3
1 / -0

Consider the following Boolean expression:
\(\left( {\bar A + \bar B} \right)\left[ {\overline {A\left( {B + C} \right)} } \right] + A\left( {\bar B + \bar C} \right)\)
It can be represented by a single three-input logic gate. Identify the gate.

A
AND

B
OR

C
XOR

D
NAND

Solution

\(\begin{array}{l}\left( {\bar A + \bar B} \right)\left[ {\overline {A\left( {B + C} \right)} } \right] + A\left( {\bar B + \bar C} \right)\\= \left( {\bar A + \bar B} \right)\left[ {\bar A + \left( {\overline {B + C} } \right)} \right] + A\left( {\bar B + \bar C} \right)\end{array}\)
\(\begin{array}{l}= \left( {\bar A + \bar B} \right)\left[ {\bar A + \bar B\bar C} \right] + A\bar B + A\bar C\\= \bar A + \bar A\bar B\bar C + \bar A\bar B + \bar B\bar C + A\bar B + A\bar C\end{array}\)
\(\begin{array}{l}= \bar A\left( {1 + \bar B\bar C + \bar A\bar B} \right) + \bar B\bar C + A\bar B + A\bar C\\= \bar A + \bar B\bar C + A\bar B + A\bar C\end{array}\)
\(\begin{array}{l}= \bar A + A\bar B + A\bar C\\= \bar A + A\left( {\bar B + \bar C} \right)\end{array}\)
\(\begin{array}{l}= \bar A + \bar B + \bar C\\= \overline {ABC} \end{array}\)
The given Boolean expression represents NAND gate.
Question 4
1 / -0

What is the probability that the below-given f(x, y, z) will return 1?
\(f\left( {x,\;y} \right) = \;\bar x + \bar y{\rm{\;}} + x.z\left( {\overline {x.\left( {y + z} \right)} } \right)\)
where f(x, y, z) is a Boolean function. (answer upto 2 decimal places)

A
0.75-0.75

B
0.754-0.751

C
0.757-0.756

D
0.759-0.752

Solution

\({\rm{f}}\left( {{\rm{x}},{\rm{y}}} \right) = \bar x + \bar y{\rm{\;}} + {\rm{x}}.{\rm{z}}\left( {\overline {{\rm{x}}.\left( {{\rm{y}} + {\rm{z}}} \right)} } \right)\)
De Morgan's Law
\({\rm{f}}\left( {{\rm{x}},{\rm{y}}} \right) = \overline {{\rm{x}}.{\rm{y}}} + {\rm{x}}.{\rm{z}}\left( {\bar x + \overline {\left( {{\rm{y}} + {\rm{z}}} \right)} } \right)\)
De Morgan's Law
\({\rm{f}}\left( {{\rm{x}},{\rm{y}}} \right) = \overline {x.y} + {\rm{x}}.{\rm{z}}\left( {{\rm{\bar x}} + \left( {{\rm{\bar y}}.{\rm{\bar z}}} \right)} \right)\)
\({\rm{f}}\left( {{\rm{x}},{\rm{y}}} \right) = \overline {x.y} + {\rm{x}}.{\rm{z}}.{\rm{\bar x}} + {\rm{xz}}.{\rm{\bar y}}.{\rm{\bar z}}\)
\({\rm{f}}\left( {{\rm{x}},{\rm{y}}} \right) = \overline {x.y} + {\rm{x}}.{\rm{z}}.{\rm{\bar x}} + {\rm{x}}.{\rm{z}}.{\rm{\bar y}}.{\rm{\bar z}}\)
\({\rm{f}}\left( {{\rm{x}},{\rm{y}}} \right) = \overline {x.y} \; + 0 + 0\)
\({\rm{f}}\left( {{\rm{x}},{\rm{y}}} \right) = \overline {x.y} \)
NAND Gate Truth Table:
x
y
\(\overline {x.y} \)
0
0
1
0
1
1
1
0
1
1
1
0

\(Probability\;of\;getting\;1 = \frac{3}{4} = 0.75\)
Question 5
1 / -0

The minimum Boolean expression for the following circuit is

A
AB + AC + BC

B
A + BC

C
A + B

D
A + B + C

Solution

A(B + C) + AB + (A + B)C
= AB + AC + AB + AC + BC
= AB + AC + BC