hw05sol - HW#5 SOLUTIONS ECSE-2610 Computer Components and...

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HW #5 SOLUTIONS 1/7 ECSE-2610 Computer Components and Operations SOLUTIONS HW # 5 Due March 3, 2010 1. (10 points) Number Systems, Codes, Computer Arithmetic a. (1 point each, 6 points total) i. Convert 0010.1111 2 to a decimal number. ____2.9375 10 __________ 10.11110000 ii. Convert 01101010 2 to a hexadecimal number. ____006A 16 ______________ 01101010 iii. Convert AB 16 to a binary number. ____10101011 2 _________ AB iv. Convert 83.76 10 to a binary number. ____01010011.11000010 2 _ 83.7578 v. Convert 83 10 to a BCD number . ____1000 0011 BCD _______ 8 3 vi. Convert 01010101 2 to its 2s-complement negative. ____10101011 2s-comp ______ 81 -81 b. (3 points) Add using 2s-complement arithmetic 01000111 2 s-comp - 10101110 2 s-comp . Does this result in an overflow? Explain your answer. Result = ________10011001 2 s-comp ________________________________ Overflow = Yes. Subtracting a negative from a positive should not result in negative. c. (1 point) Convert the numbers of part (b) to decimal and add them. What answer do you get? S u m = ________71 – (-82) = 153 Overflow = because 153 > 127, the largest 8-bit 2’s comp number/ .
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HW #5 SOLUTIONS 2/7 2. (20 points) a. (10 points) Simplify the following Boolean function to sum-of-products form using the laws of Boolean algebra. Show each step. You do not need to identify the name of the property you use. i = X’Z + Y’(X’Z)’ = X’Z + Y’ by Rule 11D (Jan 28) A more pedestrian approach is to find the SOP canonical form and recombine minterms (as with K-map): X’Z + Y’Z’ + XY’ = X’Y’Z + X’YZ + XY’Z’ +X’Y’Z’ + XY’Z + XY’Z X’Y’Z + X’Y’Z’ = X’Y’, XY’Z’ + XY’Z = XY’, X’Y’Z + X’YZ = X’Z and X’Y + XY’ = Y’ Therefore G = X’Z + Y’ TT#1 has columns for each of the OR-ed terms, TT#2 has columns for each AND term.
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hw05sol - HW#5 SOLUTIONS ECSE-2610 Computer Components and...

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