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

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

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## 10 Questions MCQ Test Digital Circuits - Test: Code Converters

Test: Code Converters for Electronics and Communication Engineering (ECE) 2024 is part of Digital Circuits preparation. The Test: Code Converters questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: Code Converters MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Code Converters below.
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Test: Code Converters - Question 1

### The Grey code for (A5)16 is equivalent to

Detailed Solution for Test: Code Converters - Question 1

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

Test: Code Converters - Question 2

### A 4-bit D/A converter gives an output voltage of 4.5 V for an input code of 1001. The output voltage for an input code of 0110 is

Detailed Solution for Test: Code Converters - Question 2

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 = (2N − 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

Test: Code Converters - Question 3

### A 4 bit Digital to Analog converter (DAC) gives an output voltage of 5 V for an input code of 1111. What is the output voltage for an input code of 1100?

Detailed Solution for Test: Code Converters - Question 3

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− 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

Test: Code Converters - Question 4

A binary-to-BCD encoder has four inputs D0 C0, B0, and A0 and five outputs D, C, B, A, and VALID. The outputs D, C, B and A give the proper BCD value of the input and the VALID output is 1 if the input combination is a valid decimal code. If the input combination is an invalid decimal code, the VALID output becomes 0, and all of the D, C, B, and A outputs show 0 values. If only NOT gates and 2-input OR and AND gates are available, the minimum number of gates required to implement the above circuit is

Detailed Solution for Test: Code Converters - Question 4

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:

Test: Code Converters - Question 5

Minimum number of Half adders, Full adders, AND gates required to implement 2 × 3 multiplier is given as

Detailed Solution for Test: Code Converters - Question 5

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.

Test: Code Converters - Question 6

The circuit shown below converts

Detailed Solution for Test: Code Converters - Question 6

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

Test: Code Converters - Question 7

Identify the circuit below.

Note: Change in the above connection
OP4 is connected to IP7
OP5 is connected to IP6
Rest connections are same

Detailed Solution for Test: Code Converters - Question 7

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

Test: Code Converters - Question 8

Reflected binary code is also known as

Detailed Solution for Test: Code Converters - Question 8

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

Test: Code Converters - Question 9

The binary representation of BCD number 00101001 (decimal 29) is

Detailed Solution for Test: Code Converters - Question 9

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.

Test: Code Converters - Question 10

The primary use for Gray code is

Detailed Solution for Test: Code Converters - Question 10

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.

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