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hw1sol - CSE140 Spring 2010 Homework#1 Solution 1 a F(A,B,C...

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CSE140 Spring 2010, Homework #1 Solution 1. a. F (A,B,C) = A’ B’ C’ + A’ B’ C + A B’ C’ + A B’ C + A B C’ + A B C = A’B’ (C’ + C) + AB’ (C’ + C) + AB (C’ + C) [ Complementarity] = A’B’ + AB’ + AB = A’B’ + AB’ + AB’ +AB [ Idempotency] = B’ (A’ + A) + A (B’ + B) [ Complementarity] = B’ + A b. F(A,B) = (A ’+B) ( A+B) = AA ’ + A’B+ AB+ BB [Complementarity ] = A ’B+AB+BB = B ( A ’+A+1) [Complementarity , Idempotency ] = B . 1 [ identity ] = B 2. 3.
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4. a. (F23) 16 --> ( ) 10 3 *16^0=3 2* 16^1=32 F*16^2= 3840 (F23) 16 --> (3875 ) 10 = 3+32+3840 = 3875 b. (1E0) 16 --> ( ) 2 0 * 16^0=0 E * 16^1=224 1 * 16^2=256 =224+256 = 480 10 2|480 2|240 - 0 2|120 0 2|60 0 (1E0) 16 --> (111100000 ) 2 2|30 0 2|15 0 2|7 1 2|3 1 1 1
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c. (10011111) 2 --> ( ) 8 1*2^0= 1 1*2^1=2 1*2^2=4 1*2^3=8 1*2^4=16 0*2^5=0 0*2^6=0 1*2^7=128 = 159 10 8|159 8|19 7 |2 3 (10011111) 2 --> (237 ) 8 d. (104) 10 --> ( ) 2 2|104 2|52 0 2|26 0 2|13 0 hence (104) 10 = ( 1101000) 2 2|6 1 2|3 0 1 1 e. (300) 10 --> ( ) 16 16|300 16|18 C Hence (300) 10= (12C ) 16 16|1 2
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5. a. 1 1 1 0 1 1 0 0 1 0 1 0 1 1 1 0 0 1 0 0 1 1 0 1 0 1 Method 1 : (-) ve number + (+) ve number No overflow Method 2 : Cin XOR Cout = 0 No overflow b. 0 1 0 0 1 1 0 0 1 1 0 1 1 1 0 1 1 0 1 0 1 0 1 0 0 1
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