sol5

# sol5 - Chapter 5 5.1(a(b(c(d e 478 743 2025 41567 61680...

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Chapter 5 5.1. ( a ) 478 ( b ) 743 ( c ) 2025 ( d ) 41567 ( e ) 61680 5.2. ( a ) 478 ( b ) - 280 ( c ) - 1 5.3. ( a ) 478 ( b ) - 281 ( c ) - 2 5.4. The numbers are represented as follows: Decimal Sign and Magnitude 1’s Complement 2’s Complement 73 000001001001 000001001001 000001001001 1906 011101110010 011101110010 011101110010 - 95 100001011111 111110100000 111110100001 - 1630 111001011110 100110100001 100110100010 5.5. The results of the operations are: ( a ): 00110110 54 ( b ): 01110101 117 ( c ): 11011111 ( - 33) +01000101 +69 +11011110 - 34 +10111000 +( - 72) 01111011 123 01010011 83 10010111 ( - 105) ( d ): 00110110 54 ( e ): 01110101 (117) ( f ): 11010011 ( - 45) - 00101011 - 43 - 11010110 - ( - 42) - 11101100 - ( - 20) 00001011 11 10011111 (159) 11100111 ( - 25) Arithmetic overFow occurs in example e ; note that the pattern 10011111 represents - 97 rather than +159. 5-1

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5.6. The associativity of the XOR operation can be shown as follows: x ( y z ) = x ( yz + y z ) = x ( yz + y z ) + x ( y · z + yz ) = x · yz + xy z + x y · z + xyz ( x y ) z = ( xy + x y ) z = ( x · y + xy ) z + ( xy + x y ) z = x · yz + xyz + xy z + x y · z The two SOP expressions are the same. 5.7. In the circuit of Figure 5.5b, we have: s i = ( x i y i ) c i = x i y i c i c i +1 = ( x i y i ) c i + x i y i = ( x i y i + x i y i ) c i + x i y i = x i y i c i + x i y i c i + x i y i = y i c i + x i c i + x i y i The expressions for s i and c i +1 are the same as those derived in Figure 5.4 b . 5.8. We will give a descriptive proof for ease of understanding. The 2’s complement of a given number can be found by adding 1 to the 1’s complement of the number. Suppose that the number has k 0s in the least- signi±cant bit positions, b k - 1 ...b 0 , and it has b k = 1 . When this number is converted to its 1’s complement, each of these k bits has the value 1. Adding 1 to this string of 1s produces b k b k - 1 b k - 2 ...b 0 = 100 ... 0 . This result is equivalent to copying the k 0s and the ±rst 1 (in bit position b k ) encountered when the number is scanned from right to left. Suppose that the most-signi±cant n - k bits, b n - 1 b n - 2 ...b k , have some pattern of 0s and 1s, but b k = 1 . In the 1’s complement this pattern will be complemented in each bit position, which will include b k = 0 . Now, adding 1 to the entire n -bit number will make b k = 1 , but no further carries will be generated; therefore, the complemented bits in positions b n - 1 b n - 2 ...b k +1 will remain unchanged. 5.9. Construct the truth table x n - 1 y n - 1 c n - 1 c n s n - 1 (sign bit) Over²ow 0 0 0 0 0 0 0 0 1 0 1 1 0 1 0 0 1 0 0 1 1 1 0 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 0 1 0 1 1 1 1 1 1 0 Note that over²ow cannot occur when two numbers with opposite signs are added. From the truth table the over²ow expression is Overflow = c n c n - 1 + c n c n - 1 = c n c n - 1 5-2
5.10. Since s k = x k y k c k , it follows that x k y

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## This note was uploaded on 02/16/2011 for the course EE 4745 taught by Professor Rai during the Spring '07 term at LSU.

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sol5 - Chapter 5 5.1(a(b(c(d e 478 743 2025 41567 61680...

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