Digital Logic Test 2

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Digital Logic Test 2

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Weekly Quiz Competition

Question 1
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Consider a quadratic equation x2 – 13x + 36 = 0 with coefficients in a base b. The solutions of this equation in the same base b are x = 5 and x = 6. Then b = _______.

A
7

B
8

C
9

D
10

Solution

Formula:
In quadratic equation, ax²+ bx + c = 0
Sum of root = \(- \frac{b}{a}\)
Product of roots = \(\frac{c}{a}\)
Calculation:
Quadratic equation: x² – 13x + 36 = 0
Given base is b.
Roots of the equation with base b is 5 and 6
\({5_{b\;}} + \;{6_b} = \;{\left( {13} \right)_b}\)
\(11 = b + 3\)
b = 8
Also, by considering the product of roots,
\({5_{b\;}} \times {6_b} = {\left( {36} \right)_b}\)
30 = 3b + 6,
24 = 3b, b = 8
In both cases, base b = 8
Question 2
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Simplify Y = AB’ + (A’ + B)C

A
AB' + C

B
AB' + A'C + BC

C
A’B + AC’

D
AB + A

Solution

The expression Y = AB' + (A' + B)C can be simplified using Boolean algebra rules. First, distribute C across the terms in the parentheses: Y = AB' + (A' + B)C = AB' + A'C + BC. This expression AB' + A'C + BC represents the fully simplified form according to Boolean algebra.
Question 3
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A function (A, B, C) defined by three boolean variables A, B, and C when expressed as the sum of products is given by:
F = A̅.B̅.C̅ + A̅.B.C̅ + A.B̅.C̅
where, A̅, B̅, and C̅ are the complements of the respective variables. The product of sums (POS) form of the function F is

A
F = (A + B + C) . (A + B̅ + C) . (A̅ + B + C)

B
F = (A̅ + B̅ + C̅) . (A̅ + B + C̅) + (A + B̅ + C̅)

C
F = (A + B + C̅) . (A + B̅ + C̅) . (A̅ + B + C̅) . (A̅ + B̅ + C) . (A̅ + B̅ + C̅)

D
F = (A̅ + B̅ + C) . (A̅ + B + C) . (A + B̅ + C) . (A + B + C̅) . (A + B + C)

Solution

F = A̅.B̅.C̅ + A̅.B.C̅ + A.B̅.C̅
In terms of minterms, this can be represented as:
F = ∑m (0, 2, 4)
The equivalent maxterm will contain the terms not present in the minterm representation, i.e.
F = ∑m (0, 2, 4) = π(1, 3, 5, 6, 7) = M₁. M₃. M₅. M₆. M₇
⇒ (A + B + C̅) (A + B̅ + C̅) (A̅ + B + C̅) (A̅ + B̅ + C) (A̅ + B̅ + C̅)
Question 4
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Find values of Boolean variables A,B,C which satisfy the following equations:

A
1,0,0

B
0,1,0

C
0,0,1

D
0,1,1

Solution

From A + B = 1 and AB = 0
We get either of A,b is 1 and another is 0.
Now AC = BC ,here C has to be 0 (because A,b has different values)
C = 0
Now A + C = 1
So, A = 1 and Ab = 0, So, B = 0
So we get
A = 1, B = 0 , C = 0
These are the values.
Question 5
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What are the first eight bits of mantissa part and the decimal value stored in exponent if (-28.375)10 is represented in single-precision IEEE 754 format?

A
Mantissa = 11000110

B
Exponent = 132

C
Mantissa = 11100011

D
Exponent = 131

Solution

Single precision IEEE 754 format
Signed (S)
Mantissa (M)
Exponent (E)
1 bit
23 bits
8 bits
1 (negative)
11000110
131 (decimal)
Since (-28.375)10 is negative therefore S = 1.
(28.375)10 = (11100.0110)2
(11101.011)2 = 1.11000110 × 24
Mantissa = 11000110
Since it is excess 127
Exponent = 4 + 127 + 131
Question 6
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Complement of the expression A’B + CD’ is:

A
(A’ + B)(C’ + D)

B
(A + B’)(C’ + D)

C
(A’ + B)(C’ + D)

D
(A + B’)(C + D’)

Solution

(A’B + CD’)’ = (A’B)'(CD’)’ = (A” + B’)(C’ + D”) = (A + B’)(C’ + D).
Question 7
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In boolean algebra, the OR operation is performed by which properties?

A
Associative properties

B
Commutative properties

C
Distributive properties

D
All of the Mentioned

Solution

The expression for Associative property is given by A+(B+C) = (A+B)+C & A*(B*C) = (A*B)*C.

The expression for Commutative property is given by A+B = B+A & A*B = B*A.

The expression for Distributive property is given by A+BC=(A+B)(A+C) & A(B+C) = AB+AC.
Question 8
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A

B
xyz

C

D
None of the above

Solution
Question 9
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A

B

C

D

Solution

x'y' + x'y = x'(y+y') = x'

x' + xy = x' + y
Question 10
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The clock frequency of 12 MHz is applied to a cascaded counter of modulus-4 counter, modulus-5 counter and modulus-6 counter.

A
The overall modulus of the cascaded counter is 60

B
The lowest output frequency of the cascaded counter is 100 kHz

C
The overall modulus of the cascaded counter is 120

D
The lowest output frequency of the cascaded counter is 200 kHz

Solution

The overall modulus of the cascaded counter is the multiplication of the modulus of the individual counters.

Overall modulus (N) = 4 × 5 × 6 = 120

The clock frequency (f_clk) = 12 MHz

The lowest output frequency \(= \frac{{{f_{clk}}}}{N}\)

\(= \frac{{12 \times {{10}^6}}}{{120}} = 100\;kHz\)
Question 11
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The simplified Sum of Product form of the following Boolean expression:
\((P + \bar Q + \bar R )(P + Q+R)(P+Q+\bar R )\) is

A
\(\bar P Q+R\)

B
\(P+Q\bar R \)

C
\(P\bar Q +R\)

D
\(PQ+ R\)

Solution

\((P+\bar Q+ \bar R).(P+Q+R).(P+Q+\bar R)\)
\((P+\bar Q+\bar R).(P+(Q+R)(Q+\bar R))\)
\((P+\bar Q+\bar R).(P+Q)\)
\((P+(\bar Q+\bar R)Q)\)
\(P+Q\bar R\)

Question 12

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In the circuit shown below, X and Y are digital inputs, and Z is a digital output. The equivalent circuit is a

NAND gate

NOR gate

XOR gate

XNOR gate

Solution

Concept:

XOR GATE

Symbol:

Truth Table:

Input A	Input B	Output Y = A ⊕ B
0	0	0
0	1	1
1	0	1
1	1	0

Z = X̅Y + XY̅

The output represents the equation of the XOR gate.

Question 13
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Consider the unsigned 8-bit point binary number representation below:
X₇X₆X₅X₄ .X₃X₂X₁X₀
where the position of the binary point is between X₄ and X₃. X₇ is the MSB. Which of the decimal numbers cannot be represented exactly in the above representation:
(a) 14.1875
(b) 16.125
(c) 3.03125
(d) 1.6875

A
Only a and c

B
Only b and c

C
Only b and d

D
All (a), (b), (c) and (d)

Solution

(a) 14.1875 = 1110. 0011
Therefore, it can be represented
(b) 16.125 = 10000.0010
Since 5 bits is consumed by 16 ∴ it cannot be executed
(c) 3.03125 = 0011.00001
Since 5 bits is consumed by 0.3125 ∴ it cannot be executed
(d) 1.6875 = 0001.1011
Therefore, it can be represented
Hence (b) and (c) cannot be represented
Question 14
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The function AB’C + A’BC + ABC’ + A’B’C + AB’C’ is equivalent to

A
AC’ + AB + A’C

B
AB’ + AC’ + A’C

C
A’B + AC’ + AB’

D
A’B + AC + AB’

Solution

So, the equivalent expression will be A'C + AC' + AB'
Question 15
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Which of the following expressions is equivalent to

A

B

C

D
None of these

Solution
Question 16
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A
Only P and Q are valid.

B
Only Q and R are valid.

C
Only P and R are valid.

D
All P, Q, R are valid.

Solution

P:
Y * Z = Y * (X*Y)
= Y * (XY + X'Y')
= Y (XY + X'Y') + Y' (XY + X'Y')'
= XY + Y' ((X' + Y') (X+ Y))
= XY + Y' (X'Y + XY')
= XY + XY'
= X(Y + Y')
= X
So, P is valid.

Q:
X * Z = X * (X*Y)
= X * (XY + X'Y')
= X (XY + X'Y') + X' (XY + X'Y')'
= XY + X' ((X' + Y') (X+ Y))
= XY + X' (X'Y + XY')
= XY + X'Y
= Y(X + X')
= Y
So, Q is also valid.

R:
X * Y * Z = (X * Y) * (X * Y)
= (XY + X'Y') * (XY + X'Y')
= (XY + X'Y') (XY + X'Y') + (XY + X'Y')' (XY + X'Y')'
= (XY + X'Y') + (XY + X'Y')' (Since, AA = A) = 1 (Since A + A' = 1)

So, R is also valid.

Hence, D choice.
Question 17
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If P,Q,R are Boolean variables, then

A

B

C

D

Solution
Question 18
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In boolean algebra, the OR operation is performed by which properties?

A
All of the Mentioned

B
Associative properties

C
Commutative properties

D
Distributive properties

Solution

The expression for Associative property is given by A + (B + C) = (A + B) + C & A * (B * C) = (A * B) * C.

The expression for Commutative property is given by A + B = B + A & A * B = B * A.

The expression for Distributive property is given by A + BC = (A + B)(A + C) & A(B + C) = AB + AC.

Question 19

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Consider the function F = A + B̅C, where F is a three-variable Boolean function in which A is MSB and C is LSB. Which of the following is/are false among the given options?

F = ∑ (1, 4, 5, 6, 7)

F = ∑ (1, 3, 5, 6, 7)

F = ∏ (0, 2, 4)

F = ∏ (0, 2, 3)

Solution

Concept:

Sum of product (SOP) – The concept of sum of products mainly includes min-term. Min-term is a Boolean expression resulting in 1 for the output of a single cell and 0’s for all other cells in K-Map. Min-term is all the 1’s in the truth table.

Product of sum (POS) – Product of sum mainly includes max term. Max term is a Boolean expression resulting in 0 for the output of a single cell and 1 for all other cells in K-Map. Maxterm is all the 0’s in the truth table.

Explanation:

Function: X = A + B̅C

Construct the truth table for this function:

A	B	C	A + B̅C
0	0	0	0
0	0	1	1
0	1	0	0
0	1	1	0
1	0	0	1
1	0	1	1
1	1	0	1
1	1	1	1

Here F = ∑ (1, 4, 5, 6, 7) means sum of product which is all the 1’s in output of truth table.

F = ∏ (0, 2, 3) means product of sum which is all the 0’s in the output of truth table.

Option 2 and 3 are false.

Question 20
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Consider the 64-bit register which stops floating point number in IEEE Double precision format
What is the value of the number if 64 bits are given below?
1
10000000111
010111000…

A
-348--348

B
-3484--3481

C
-3487--3486

D
-3489--3482

Solution

In IEEE 64-bit format,
signed(s)
Exponent(E)
Mantissa(M)
1 bit
11 bits
52 bits

(-1)^s × 1.M × 2^{E – 1023}
(-1)¹ × 1.010111 × 2^{1031 – 1023}
(-1)¹ × 1.010111 × 2⁸
-1×(1 + 2^-2 + 2^-4 + 2^-5 + 2^-6)×2⁸
-1(256 + 64 + 16 + 8 + 4) = -348
Question 21
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Which one of the following expressions does NOT represent exclusive NOR of x and y?

A

B

C

D

Solution

A: means both are either true OR both are false. then it will be true = ExNOR

B & C: Whenever any one of the literal is complemented then ExOR can be turned to ExNOR and complement sign on the literal can be removed. So these two also represents ExNOR operation of x and y.

Hence, the correct answer is option D it is the ExOR operation b/w the two.
Question 22
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P, Q, and R are the decimal integers corresponding to the 4-bit binary number 1100 considered in signed magnitude, 1’s complement, and 2’s complement representations, respectively. The 6-bit 2’s complement representation of (P + Q + R) is

A
110101

B
110010

C
111101

D
111001
Solution
Concept:

1. Signed magnitude representation uses the most significant bit (MSB) a sign bit.
- If the sign bit is ‘0’ then the number is positive.
- If the sign bit is ‘1’ then the number is negative.
The remaining bits represent the magnitude of the binary number.

2. 1’s complement representation:

It is a representation of a binary number obtained by toggling all bits in it i.e. transforming the 0 bit to 1 and the 1 bit to 0.

3. 2’s complement representation:

It is obtained by simply adding 1 to the 1’s complement of that binary number.

Calculation:

Given: 4-bit binary number = 1100

Signed magnitude representation (P): - 4

1’s complement representation (Q): - 3

2’s complement representation (R): - 4

So,

P + Q + R = (-4) + (-3) + (-4) = -11

(-11)₁₀ is represented in 2’s complement as:

-(11)₁₀ = 10101

Since the Options are in 6 bits so, we copy sign bit once towards left.

So 6-bit representation of (-11)₁₀ is 110101
Question 23
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denote the exclusive OR (XOR) operation. Let '1' and '0' denote the binary constants. Consider the following Boolean expression for F over two variables P and Q:

The equivalent expression for is

The equivalent expression for F is

A

B

C

D

Solution
Question 24
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How many functions does f1.f2 and f1 + f2 represents respectively if the given function is
\({\rm{f}}1{\left( {{\rm{a}},{\rm{b}},{\rm{c}}} \right)_{}} = \mathop \sum \nolimits_M \;\left( {0,{\rm{\;}}1,{\rm{\;}}4} \right) + {\rm{\;}}\mathop \sum \limits_\emptyset \left( {3,{\rm{\;}}5,{\rm{\;}}7} \right)\)
\({\rm{f}}2\left( {{\rm{a}},{\rm{b}},{\rm{c}}} \right) = \mathop \sum \nolimits_M \;\left( {2,3} \right) + {\rm{\;}}\mathop \sum \limits_\emptyset \left( {1,{\rm{\;}}6,{\rm{\;}}7} \right)\)
where ∑_m represents sum of min-terms and ∑_ϕ represents and don’t care

A
2, 4

B
8, 8

C
4, 8

D
4, 4

Solution

Given:
\(f1.f2\left( {a,b,c} \right) = \mathop \sum \limits_\emptyset \left( {1,3,7} \right)\)
\(f1 + f2\left( {a,b,c} \right) = \sum \left( {0,1,2,3,4} \right) + \mathop \sum \limits_\emptyset \left( {5,6,7} \right)\)
It is given in min-term form with the don’t care.
In table form it is represented as :
a
b
c
f1
f2
f1.f2
f1+f2
0
0
0
1
0
0
1
0
0
1
1
Ø
Ø
1
0
1
0
0
1
0
1
0
1
1
Ø
1
Ø
1
1
0
0
1
0
0
1
1
0
1
Ø
0
0
Ø
1
1
0
0
Ø
0
Ø
1
1
1
Ø
Ø
Ø
Ø

Now,
\(f1\left( {a,b,c} \right) = \sum \left( {0,\;1,\;4} \right) + \;\mathop \sum \limits_\emptyset \left( {3,\;5,\;7} \right)\)
\(f2\left( {a,b,c} \right) = \sum \left( {2,3} \right) + \;\mathop \sum \limits_\emptyset \left( {1,\;6,7} \right)\)
As, in f1.f2(a ,b ,c), there are 3 don’t cares .So, number of functions possible in this are : 2³ = 8
Similarly, in f1 + f2(a, b, c), there are 3 don’t cares. So, number of functions possible are : 2³= 8
Question 25
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D input of a clocked D-flip-flop receives and input A ⊕ Q_n where A is an external logic input and Q_n is the output of the n^th D-FF before the clock appears. The circuit works as

A
EXOR gate

B
T-FF

C
D-FF

D
JK-FF

Solution

Characteristic equation of a D flip flop is, Q (n + 1) = D
Characteristic equation of a T flip flop is, Q (n + 1) = T ⊕ Qn
The input of the D flip flop = A ⊕ Q_n
The output of the D flip flop will be, Q (n + 1) = A ⊕ Q_n
It represents, the characteristic equation of T flip flop.