GATE > Test: Code Converters

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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 = (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 × 2^{3} + 0 × 2^{2} + 0 × 2^{1} + 1 × 2^{0}

(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 × 2^{3} + 1 × 2^{2} + 1 × 2^{1} + 0 × 2^{0}

(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

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

Given output voltages are 5 V and input (1111)

Converting 1111 into decimal form we get

(1111)

(1111)

(1111)

Resolution = 5/15

Resolution = 1/3

New input given is (1100)

Converting (1100)

(1100)

(1100)

(1100)

Now the output will be

Output = 4 V

Test: Code Converters - Question 4

A binary-to-BCD encoder has four inputs D_{0} C_{0}, B_{0}, and A_{0} 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 D_{0}, C_{0}, B_{0}, A_{0}

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 A_{0} with the multiplicand is called as partial product 1.

2) And the second product obtained from multiplying A_{1} 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 B_{1}A_{0} and B_{0}A_{1} we require one-half adder and this produces a carry also.

To get the addition of B_{2}A_{0}, B_{1}A_{1} and C_{1} we require Full adder because of 3 inputs and this also produces a carry.

To get the B_{2}A_{1} and C_{2} we require one-half adder because of 2 inputs.

**Conclusion:**

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

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.

Let Gray Code be G

It is given by

y

y

y

y

Analysis:

Assume X_{2}X_{1}X_{0} = (111)_{2} = 7_{10}

Then the output of the first decoder = OP_{7} which is connected to IP5 of the encoder

(5)_{10} = (101)_{2}

∴ output of encoder Y_{2}Y_{1}Y_{0} = 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

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.

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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