Digital Electronics Test 1

Concept:

1’s complement representation of a binary number is obtained by toggling all the bits, i.e. replacing 1 with 0, and 0 with 1.

2’s complement representation of a binary number is obtained by adding 1 to the 1’s complement representation.

Application:

(127)₁₀ = (01111111)₂

1’s complement representation will be:

1’s complement = 10000000

Number of 1’s is the 1’s complement is, n = 1

Now, the two (2’s) complement representation will be:

2’s complement = 10000000 + 1

= 10000001

Number of 1’s in 2’s complement is, m = 2

∴ The required ratio is m : n = 2 : 1

Concept:

Let a number is represented in a base 'b' number system as b₃ b₂ b₁ b₀

The decimal equivalent of this number will be:

b₃ × 2³ + b₂ × 2² + b₁ × 2¹ + b₀ × 2⁰

We observe that the decimal equivalent will be minimum when the base of the number system 'b' is minimum.

Application:

The given number is 11C.0

The minimum decimal equivalent for this number is possible for a valid minimum base system for this number.

Since the largest digit in the given number is C which is expressed by 12, the minimum valid base will be 13.

∴ The minimum decimal equivalent of the given number will be:

(11 C)₁₃ = 1 × 13² + 1 × 13¹ + 12 × 13⁰

= 169 + 13 + 12

= (194)₁₀

Concept:

• If two functions are represented in terms of max terms, the product will include all the max terms that are present in either of F₁ or F₂. This is because max terms are the values for which the function, gives a 0 output.
• If two functions are represented in terms of minterms, the product will include all the common min terms present in both F₁ and F₂. This is because minterms are the values for which the function gives an output of 1.

Application:

F₁ (a, b, c) = Π M (0, 4, 5, 6)

And F₂ (a, b, c) = Π M (0, 3, 4, 6, 7)

The product F₁F₂  will include all the max terms that are present in either F₁ or F₂

∴ F(a, b, c) = F₁.F₂  = Π M (0, 3, 4, 5, 6, 7)

Alternate Method:

In the form of minterms, the given function can be expressed as:

F₁(a, b, c) = ∑ m (1, 2, 3, 7)

F₂ (a, b, c) = ∑ m (1, 2, 5)

The product F₁F₂ will include minterms that are present in both F₁ and F₂ (common minterms), i.e.

F(a, b, c) = ∑ m (1, 2)

Now, the function can be expressed in the form of max terms as:

F(a, b, c) = ∑ m(1, 2) = Π M (0, 3, 4, 5, 6, 7)

Concept:

A quadratic equation with factor α and β can be written as:

x² – (α + β) x + (αβ) = 0

Application:

Given the quadratic equation:

x² – 11x + 22 = 0         ---(1)

Also, the solutions to the given quadratic equation are 3 and 6.

∴ We can write the quadratic equation as:

x² – (6 + 3) x + (6 × 3) = 0       ---(2)

Comparing the two-equations, we can write:

(6)_b + (3)_b = (11)_b i.e. for some base ‘b’, this equation must hold true.

Converting both the RHS and LHS to their respective decimal equivalent, we can write:

(6 × b⁰) + (3 × b⁰) = (1 × b¹ + 1 × b⁰)

6 + 3 = b + 1

b = 8

∴ The base of the number is 8 that satisfies the above quadratic equation.

According to the given problem, we represent the fruits as

A = Apple, B = Banana; M = Mango and O = Orange

So the logical expression that specifies the fruit available for dessert is

f (A, B, M, O) = (1^st choice) + (2^nd Choice) + (3^rd Choice)

= (O + A + OA) + (MA) + (OB)

= O (1 + A + B) + A + AM

= O + A (1 + M)