Chapter 5 - Chapter 5 Problem Solutions 5.1(a X = 23(10 Y =...

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- 5.1 - Chapter 5 Problem Solutions 5.1. (a) X = 23 (10) , Y = 6 (10) c 5 s 4 s 3 s 2 s 1 s 0 = 011101 (2) / 29 (10) (b) X = 26 (10) , Y = 13 (10) c 5 s 4 s 3 s 2 s 1 s 0 = 100111 (2) c 5 =1 indicates an overflow since 39 (10) cannot be represented with five sum bits. (c) X = 25 (10) , Y = 5 (10) c 5 s 4 s 3 s 2 s 1 s 0 = 110100 (2) c 5 =1 is ignored in subtraction using 2's-complements and s 4 s 3 s 2 s 1 s 0 = 10100 (2) / 20 (10) is the true difference. (d) X = 19 (10) , Y = 26 (10) c 5 s 4 s 3 s 2 s 1 s 0 = 011001 (2) c 5 =0 indicates a negative result since subtraction was performed and s 4 s 3 s 2 s 1 s 0 = 11001 (2) is the difference in 2's-complements representation, i.e., the 2's-complement of 11001 is 00111 which corresponds to 7 (10) . Hence, the result indicates -7 (10) . 5.2. (a) As indicated in Sec. 2.8, an overflow condition occurs when the signs of the operands entering the high order stage of the binary full adder are the same, but the sign of the result of the addition is opposite. Thus, the high order stage of the binary adder/subtracter becomes (where V=1 indicates an overflow condition):
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- 5.2 - 5.2. (continued) (b) As indicated in Sec. 2.8, an overflow condition exists when (1) there is a carry into the sign digit position and no carry from the sign digit position or (2) there is no carry into the sign digit position and a carry from the sign digit position. This can be realized by use of an exclusive-or gate having the two carries as inputs.
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- 5.3 - 5.2. (continued) 5.3. Let n be the number of bits in an operand and that n is divisible by 4. A carry generated in the least significant adder stage requires one level of logic, two levels of logic are needed to propagate the carry to the end of the first
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This note was uploaded on 10/26/2008 for the course ENEE 244 taught by Professor Petrov during the Spring '08 term at Maryland.

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Chapter 5 - Chapter 5 Problem Solutions 5.1(a X = 23(10 Y =...

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