###### Operating-Systems

October 11, 2023###### 2017 November NTA UGC NET Paper 1

October 11, 2023# Digital-Logic-Design

Question 14 |

Consider the circuit given below which has a four bit binary number b_{3}b_{2}b_{1}b_{0} as input and a five bit binary number d_{4}d_{3}d_{2}d_{1}d_{0} as output. The circuit implements:

Binary of Hex conversion
| |

Binary to BCD conversion | |

Binary to grey code conversion | |

Binary to radix-12 conversion |

Question 14 Explanation:

Here ф means 0.

Whenever, b

Note that the last 4 combinations leads to b

1100 is 12

1101 is 13

1110 is 14

1111 is 15

in binary unsigned number system.

1100 + 0100 = 10000

1101 + 0100 = 10001, and so on.

This is conversion to radix 12.

Whenever, b

_{2}= b_{3}= 1, then only 0100, i.e., 4 is added to the given binary number. Lets write all possibilities for b.Note that the last 4 combinations leads to b

_{3}and b_{2}as 1. So, in these combinations only 0010 will be added.1100 is 12

1101 is 13

1110 is 14

1111 is 15

in binary unsigned number system.

1100 + 0100 = 10000

1101 + 0100 = 10001, and so on.

This is conversion to radix 12.

Correct Answer: D

Question 14 Explanation:

Here ф means 0.

Whenever, b

Note that the last 4 combinations leads to b

1100 is 12

1101 is 13

1110 is 14

1111 is 15

in binary unsigned number system.

1100 + 0100 = 10000

1101 + 0100 = 10001, and so on.

This is conversion to radix 12.

Whenever, b

_{2}= b_{3}= 1, then only 0100, i.e., 4 is added to the given binary number. Lets write all possibilities for b.Note that the last 4 combinations leads to b

_{3}and b_{2}as 1. So, in these combinations only 0010 will be added.1100 is 12

1101 is 13

1110 is 14

1111 is 15

in binary unsigned number system.

1100 + 0100 = 10000

1101 + 0100 = 10001, and so on.

This is conversion to radix 12.

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