By R.S. Sedha

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Additional resources for A Textbook of Digital Electronics

Sample text

45 = 32 + 8 + 4 + 1 25 + 23 + 22 + 20 Placing 1s in the appropriate weight positions 25, 23, 22 and 20 05 in the 24 and21 positions determine the binary number for decimal 45 25 24 23 22 21 20 1 0 1 1 0 1  4510 = 1011012 The fractional part of decimal number is converted in binary form by multiplication by 2 method. 976 with a carry with a carry with a carry with a carry with a carry with a carry with a carry 0 1 0 1 1 1 0 LSB MSB Bottom And so on …. 010111)2 Ans. Created with Print2PDF. com/ 60 A TEXTBOOK OF DIGITAL ELECTRONICS Example 2-45.

Com/ NUMBER SYSTEMS AND CODES 61 (iii) Given: (11001010 . )8 Binary to hexadecimal: Step 1. 1100 1 0 1 0. 1001 Step 2. 1100 1010 1001 Step 3. 9)16 Ans. Binary to octal 1100 adding a leading zero Step 2. 01 1 3 1 0 1 0. 1001 00 1 01 0 1 4 adding zero } Step 1. 10 0 4  (11001010 . 44)8 Ans. Example 2-46. ,2007) Solution. (i) Given: (650)10  ()16 We know that we can use the repeated division 16 for converting a decimal number to its hexadecimal equivalent. Using division by hand. 125 × 16 = 2 MSD Bottom Since the quotient i1 zero, so we will stop the further division.

Step 2. Divide each resulting quotient by 16 until there is a zero whole number quotient. Step 3. The remainders generated by each division form the hexadecimal number. The first remainder to be produced is the least-significant digit (LSD) in the hexadecimal number and the last remainder to be produced is the most-significant digit (MSD). In other words, reading the remainders from bottom-to-top constitutes the hexadecimal number. Note that any remainders that are greater than 9 are represented by the letters A through F.