Exercise :: Digital Electronics - Section 1
- Digital Electronics - Section 1
- Digital Electronics - Section 2
- Digital Electronics - Section 3
- Digital Electronics - Section 4
- Digital Electronics - Section 5
- Digital Electronics - Section 6
- Digital Electronics - Section 7
- Digital Electronics - Section 8
- Digital Electronics - Section 9
- Digital Electronics - Section 10
- Digital Electronics - Section 11
- Digital Electronics - Section 12
- Digital Electronics - Section 13
- Digital Electronics - Section 14
- Digital Electronics - Section 15
- Digital Electronics - Section 16
- Digital Electronics - Section 17
- Digital Electronics - Section 18
- Digital Electronics - Section 19
- Digital Electronics - Section 20
- Digital Electronics - Section 21
- Digital Electronics - Section 22
- Digital Electronics - Section 23
- Digital Electronics - Section 24
| 16. | The access time of a word in 4 MB main memory is 100 ms. The access time of a word in a 32 kb data cache memory is 10 ns. The average data cache bit ratio is 0.95. The efficiency of memory access time is |
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Answer: Option B Explanation: Access time = 0.95 x 10 + 0.05 x 100. |
| 17. | The expression Y = pM (0, 1, 3, 4) is |
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Answer: Option A Explanation: This is product of sums expression. |
| 18. | An 8 bit DAC has a full scale output of 2 mA and full scale error of ± 0.5%. If input is 10000000 the range of outputs is |
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Answer: Option A Explanation: 10000000 = 128, 11111111 = 255 If there is no error, output = Maximum error = Hence range of output 994 to 1014 μA. |
= 1004μA. 
| 19. | Decimal 43 in hexadecimal and BCD number system is respectively. |
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Answer: Option B Explanation: (43)10 = (2B)16 (43)10 = (01000011)2 . |
| 20. | The circuit of the given figure realizes the function |
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Answer: Option A Explanation:
Y = (A + B)C + DE. |
or