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EECS452_HW#2

# 3 answer a 2n bits will be needed for the

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Unformatted text preview: 1 + 0_0100 = 0_1111 = 15 (should be 19, saturate occurs) The final result is incorrect. For b:  ­15  ­12 = 1_0001 + 1_0100 = 1_0000 =  ­16 (should be  ­27, saturate occurs)  ­16  ­7 = 1_0000 + 1_1001 = 1_0000 =  ­16 (should be  ­23, saturate occurs)  ­16 + 14 = 1_0000 + 0_1110 = 1_1110 =  ­2  ­2 + 11 = 1_1110 + 0_1011 = 0_1001 = 9 9 + 10 = 0_1001 + 0_1010 = 0_1111 = 15 (should be 19, saturate occurs) 15 + 8 = 0_1111 + 0_1000 = 0_1111 = 15 (should be 23, saturate occurs) 15 + 4 = 0_1111 + 0_0100 = 0_1111 = 15 (should be 19, saturate occurs) 15 + 1 = 0_1111 + 0_0001 = 0_1111 = 15 (should be 16, saturate occurs) The final result is incorrect. 3. Answer: a. 2n bits will be needed for the representation. The Q value of the result would be Q(2x). b. No. A counterexample is that 1.00(3 ­bit Q2) times itself gets 1 in decimal and is out of the range of 3 ­bit Q2 representation (0.75 ~  ­1). c. 2.25 = 010.010,  ­1.75 = 110.010 Calculated as follows: 0010.100 * 1110.010  ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ 0000000 0010100 0000000 0000000 0010100 0010100 – 0010100 + 1101001  ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ 111100000100 The answer is 111100.000100 =  ­3.9375 = 2.25 *  ­1.75. 4. Answer: a. As !! = 125!"#, and ! ! = !"# ∗ !! !!"...
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