Digital Electronics Test 1

Result

Digital Electronics Test 1

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Weekly Quiz Competition

Question 1
1 / -0

Let us consider the following equation in a 6-bit binary number system X = A + B.
A is given as (001010)₂ in 1's complement binary number system, B is given as (111010)₂ in signed number system. What would be X in the 2’s complement number system?

A
(110000)₂

B
(010000)₂

C
(101011)₂

D
(110101)₂

Solution

A in 1’s complement binary number system = (001010)₂
As the MSB is ‘0’, the number is positive.
Therefore, the number A = (001010)₂ = (10)₁₀
B in signed number system
B = (111010)₂
As the MSB is ‘1’, the number is negative.
Therefore the number B = (-26)₁₀
Now, X = A + B = (10)₁₀ + (-26)₁₀ = (-16)₁₀
The two’s complement representation of (-16)₁₀ is the two’s complement of (+16)₁₀
(+16)₁₀ = 010000
2’s complement = 110000
X in 2’s complement representation = (110000)₂

Question 2

1 / -0

Match the Boolean expression with its minimal realization

	Boolean expression		Minimal realization
P	X̅ Y̅ Z̅ + X̅ Y Z̅ + X̅ Y Z	K	X (Y + Z)
Q	X Y Z + X Y̅ Z + X Y Z̅	L	X̅ (Y + Z̅)
R	X Y + X Y Z + X Y Z̅ + X̅ Y Z	M	Z
S	X̅ Y̅ Z + X̅ Y Z + X Y̅ Z + X Y Z	N	Y (X + Z)

P – K, Q – L, R – N, S - M

P – L, Q – K, R – N, S - M

P – L, Q – N, R – M, S - K

P – M, Q – K, R – L, S - N

Solution

(P) X̅ Y̅ Z̅ + X̅ Y Z̅ + X̅ Y Z

= X̅ Y̅ Z̅ + X̅ Y̅ Z̅ + X̅ Y Z̅ + X̅ Y Z

= X̅ Z̅ (Y̅ + Y) + X̅ Y (Z̅ + Z)

= X̅ Z̅ + X̅ Y

= X̅ (Y + Z̅) = L

(Q) X Y Z + X Y̅ Z + X Y Z̅

= X Y Z + X Y̅ Z + X Y Z + X Y Z̅

= X Z (Y + Y̅) + X Y (Z + Z̅)

= XZ + X Y

= X (Y + Z) = K

(R) X Y + X Y Z + X Y Z̅ + X̅ Y Z

= XY (1 + Z + Z̅) + X̅ Y Z

= X Y + X̅ Y Z

= Y (X + X̅ Z) = Y (X + Z) = N

(S) X̅ Y̅ Z + X̅ Y Z + X Y̅ Z + X Y Z

= X̅ Z (Y̅ + Y) + XZ (Y̅ + Y)

= X̅ Z + X Z = Z (X̅ + X)

= Z = M

Question 3
1 / -0

The following two numbers are converted into desired base x and y receptively.
(6.FE5)₁₆ = (6.7745)_x
(4563)₁₀ = (11D3)_y
Find the X + Y = _____

A
24-24

B
244-241

C
247-246

D
249-242

Solution

For (6.7745)x , x could be anything greater than or equal to 8.
The standard number systems with base greater than or equal to 8 are decimal and octal.
Converting it into decimal and checking will involve large powers of x. So we start by checking it with the octal number system.
6.FE5)₁₆ = (0110.111111100101)₂
Now the binary number is converted into an octal number.
\(\underbrace {000}_0\;\underbrace {110}_6\;.\underbrace {111}_7\;\underbrace {111}_7\;\underbrace {100}_4\;\underbrace {101}_5\)
Thus, (6.FE5)₁₆ = (6.7745)₈
X = 8
Similarly,
Now, converting the decimal number (4563)₁₀ into a hexadecimal number.
16
4563

16
285
3
16
17
13
16
1
1

1
(4563)₁₀ = (11D3)₁₆
Y = 16
So the required value of X + Y = 8 +16 = 24