By R.S. Sedha

ISBN-10: 8121923786

ISBN-13: 9788121923781

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Additional info for A Textbook of Digital Electronics

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A) 142 (b) 221 (c) 435 Solution. 25 × 8 = 2 LSD MSD Bottom Since the quotient is zero, therefore we will stop the diversion-by-8. Reading the remainders from bottom to top, we find that the octal equivalent of the decimal number 142 is 216.  14210 = 2168 Ans. (b) Given: The decimal number = 221 We know that in order to correct the given number 22110 to its octal equivalent, we have to use the repeated diversion-8 method. 375 × 8 = 3 MSD Bottom Since the quotient is zero, therefore we will stop the diversion by 8.

Com/ 60 A TEXTBOOK OF DIGITAL ELECTRONICS Example 2-45. , 2008) Solution. )2 We know that integral part can be expressed by the sum-of-weight as follows 129 = 128 + 1 = 27 + 20 Placing 1s in the appropriate weight positions 27 and 20 and 05 in the 26, 25, 24 23, 22, and 21 position determines the binary number for 129 thus, 27 26 25 24 23 22 21 20 1 0 0 0 0 0 0 1 12910= 100000012 The fraction part of decimal number is converted to binary number by repeated multiplication by 2 method. 1001)2 Ans.

Reading the remainders from bottom to top, we find that the hexadecimal equivalent of the decimal number 650 is 28A.  65010 = 28A16 Ans. Example 2-31. Convert the number 15110 to its equivalent hexadecimal number. Solution. Given : The number = 15110 Created with Print2PDF. com/ 50 A TEXTBOOK OF DIGITAL ELECTRONICS We know that we can use the repeated division by 16 for converting a decimal number to its haxadecimal equivalent. Using division by hand, we get, Top 151 ÷ 16 = 9 with a remainder 7 LSD 9 ÷ 16 = 0 with a remainder 9 MSD Bottom Since the quotient is zero, so we will stop the further division.

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A Textbook of Digital Electronics by R.S. Sedha

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