Test: Code Converters - Electronics and Communication Engineering (ECE) MCQ

Concept:
The general procedure for calculating binary number to Gray is as shown:

The same procedure can be extended for any number of bits.
Analysis:
The hexadecimal number (A5)16 can be written in Binary format as:
1 0 1 0 0 1 0 1

Concept:
Resolution: It is the minimum change at the DAC output.

Full-Scale Output: It is the total voltage level given to the DAC. It is calculated as:
FSO = Total steps × Step size
FSO = (2^N − 1)×stepsize

Calculation:
Given output voltages are 4.5 V and input (1001)₂
Converting 1001 into decimal form, we get:
(1001)₂ = 1 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰
(1001)₂ = 8 + 0 + 0 + 1
(1001)₂ = (9)₁₀
Resolution = 4.5/9 = 1/2
New input given is (0110)₂
Converting (0110)₂ into decimal
(0110)₂ = 0 × 2³ + 1 × 2² + 1 × 2¹ + 0 × 2⁰
(0110)₂ = 0 + 4 + 2 + 0
(0110)₂ = (6)₁₀
Now the output will be:

Output = 3 V

Concept:
Resolution
It is the minimum change at the DAC output.
Resolution = Step size 

Full-Scale Output
It is the total voltage level given to the DAC or in simple terms, it is calculated as:
FSO = Total steps × Step size
FSO = (2^N − 1) × step size
Calculation:
Given output voltages are 5 V and input (1111)₂
Converting 1111 into decimal form we get
(1111)₂ = 1 × 2³ + 1 × 2² + 1 × 2¹ + 1 × 2⁰
(1111)₂ = 8 + 4 + 2 + 1
(1111)₂ = (15)₁₀
Resolution = 5/15
Resolution = 1/3
New input given is (1100)₂
Converting (1100)₂ into decimal
(1100)₂ = 1 × 2³ + 1 × 2² + 0 × 2¹ + 0 × 2⁰
(1100)₂ = 8 + 4 + 0 + 0
(1100)₂ = (12)₁₀
Now the output will be

Output = 4 V

Decimal to BCD
BCD is represented with four digits.
The valid decimal numbers for BCD conversion are from 0 to 9 only.

Calculation:
According to the given statement, 4 inputs and 5 outputs are there.
Inputs are D₀, C₀, B₀, A₀
Outputs are D, C, B, A, Valid
Logic implementation is shown in the below table
For valid output logic 1 is taken.

The K-map for Valid is
Valid = Σ (0 to 9)

Output equation is

The K-map for D is
Valid = Σ (8, 9)

Output equation is

The K-map for C is
Valid = Σ (4, 5, 6, 7)

Output equation is

The K-map for B is
Valid = Σ (2, 3, 6, 7)

Output equation is

The K-map for A is
Valid = Σ (1, 3, 5, 7, 9)

Output equation is

Implementing the above logic circuit we require 8 minimum gates as shown:

Multiplier

• A multiplier is a combinational logic circuit that we use to multiply binary digits.
• We use a multiplier in several digital signal processing applications.
• We use it to design calculators, mobiles, processors, and digital image processors.

Process
1) The first product obtained from multiplying A₀ with the multiplicand is called as partial product 1.
2) And the second product obtained from multiplying A₁ with the multiplicand is known as the partial product 2.

The structure of multiplication is explained below

There are 6 product terms so to get all those 6 AND gates are required.
To get the addition of B₁A₀ and B₀A₁ we require one-half adder and this produces a carry also.
To get the addition of B₂A₀, B₁A₁ and C₁ we require Full adder because of 3 inputs and this also produces a carry.
To get the B₂A₁ and C₂ we require one-half adder because of 2 inputs.

Conclusion:

Total 6 AND gates, 2 Half adders and 1 Full adder are required.

Let the input to the circuit be 1010. The circuit is redrawn as:

∴ For an input 1010, we get the output as 1100.
Similarly, let the input be 0110. The circuit is drawn as:

The output for 0110 input is 0100
Observations:

∴ The given circuit converts Gray code to Binary code

Concept:
Gray to Binary
Gray codes are widely used to facilitate error correction in digital communications such as digital terrestrial television and some cable TV systems.
Conversion from Gray Code to Binary Code:
Let Gray Code be G₃  G₂  G₁  G₀ and Binary Code be B₃ B₂ B₁ B₀. Then the respective Grey Code can be obtained as follows:

Binary to Gray
It is given by

y₁ = x₁
y₂ = x₁ ⊕ x₂
y₃ = x₂ ⊕ x₃
y₄ = x₃ ⊕ x₄

Analysis:
Assume X₂X₁X₀ = (111)₂ = 7₁₀
Then the output of the first decoder = OP₇ which is connected to IP5 of the encoder
(5)₁₀ = (101)₂
∴ output of encoder Y₂Y₁Y₀ = 101
∴ input to the circuit = 1 1 1 and output = 1 0 1
Consider Gray to Binary converter if input = 111 then output is 101.

∴ So the given circuit acts like a Gray to Binary converter
Hence option (3) is correct

The reflected binary code is also known as gray code because one digit reflected to the next bit.

The given BCD number 00101001 has three 1s.
So, it can be rewritten as 0000001-1, 0001000-8, 0010100-20 and after addition, we get 0011101 as output.

Gray code is useful because only one bit changes at a time, which is implemented easily in Coded representation of a shaft’s mechanical position.