Download Foundation of Switching Theory and Logic Design: (As Per by Arun Kumar Singh PDF

Download Foundation of Switching Theory and Logic Design: (As Per by Arun Kumar Singh PDF

By Arun Kumar Singh

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Extra info for Foundation of Switching Theory and Logic Design: (As Per JNTU Syllabus)

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Result Manipulation The way result is manipulated is different for signed and unsigned arithmetic. (a) UNSIGNED 1. If a carry is generated in step 4 then the result is positive(+) and the digits in the result shows the correct magnitude of result. NUMBER SYSTEMS AND CODES 2. 27 If there is no carry from MSB in step 4 then the result is negative (–) and the digits in result is not showing the correct magnitude. So, we must go for a post processing of result (Step 7) of result to determine the correct magnitude of the result.

I) Replace each 1 by 0 and each 0 by 1 0 1 1 0 0 1 ↓ ↓ ↓ ↓ ↓ ↓ 1 0 0 1 1 0 So, 1’s complement of 011001 is 100110. (ii) Subtract each binary bit from 1. 1 1 1 1 1 1 1 1 –0 0 1 0 0 1 1 1 1 1 0 1 1 0 0 0 ← 1’s complement one can see that both the method gives same result. 2’s Complement 2’s complement of a binary number can be obtained by adding 1 to its 1’s complement. Example 1. Find 2’s complement of (i) 011001 (ii) 0101100 16 FOUNDATION OF SWITCHING THEORY AND LOGIC DESIGN Solution. (i) (ii) 0 1 1 0 1 0 0 1 0 1 ← Number 1 0 ← 1’s complement + 1 ← Add 1 to 1’s complement 1 0 0 1 1 1 ← 2’s complement 0 1 0 1 1 0 0 ← Number 1 0 1 0 0 1 1 ← 1’s complement + 1 ← Add 1 to 1’s complement 0 0 ← 2’s complement 1 0 1 0 1 There is an efficient method to find 2’s complement based upon the observation made on the above 2 examples.

9’s complement of N = (10n–1) – N. Example 1. Find out the 9’s complement of following decimal numbers. (i) 459 (ii) 36 (iii) 1697 n Solution. (i) By using (10 –1) – N; But, n = 3 in this case So, (10n–1) – N = (103 – 1) – 459 = 540 Thus, 9’s complement of 459 = 540 (ii) By subtracting each digit from 9 9 9 –3 6 6 3 So, 9’s complement of 36 is 63. (iii) We have N = 1697, so n = 4 n Thus, 10 –1 = 104 – 1 = 9999 So, (10n–1) – N = (104–1) – 1697 = 9999 – 1697 = 8302 Thus, 9’s complement of 1697 = 8302 10’s complement: 10’s complement of a decimal number is defined as 10n – N.

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