Electronics and Communication Engineering - Digital Electronics
Exercise : Digital Electronics - Section 1
- Digital Electronics - Section 13
- Digital Electronics - Section 24
- Digital Electronics - Section 23
- Digital Electronics - Section 22
- Digital Electronics - Section 21
- Digital Electronics - Section 20
- Digital Electronics - Section 19
- Digital Electronics - Section 18
- Digital Electronics - Section 17
- Digital Electronics - Section 16
- Digital Electronics - Section 15
- Digital Electronics - Section 14
- Digital Electronics - Section 1
- Digital Electronics - Section 12
- Digital Electronics - Section 11
- Digital Electronics - Section 10
- Digital Electronics - Section 9
- Digital Electronics - Section 8
- Digital Electronics - Section 7
- Digital Electronics - Section 6
- Digital Electronics - Section 5
- Digital Electronics - Section 4
- Digital Electronics - Section 3
- Digital Electronics - Section 2
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
Answer: Option
Explanation:
Access time = 0.95 x 10 + 0.05 x 100.
17.
The expression Y = pM (0, 1, 3, 4) is
Answer: Option
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
Answer: Option
Explanation:
10000000 = 128, 11111111 = 255
If there is no error, output = = 1004μA.
Maximum error =
Hence range of output 994 to 1014 μA.
19.
Decimal 43 in hexadecimal and BCD number system is respectively.
Answer: Option
Explanation:
(43)10 = (2B)16
(43)10 = (01000011)2 .
20.
The circuit of the given figure realizes the function


Answer: Option
Explanation:
or
Y = (A + B)C + DE.
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