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ECE314sp06_HW3_solutions

# ECE314sp06_HW3_solutions - ECE/CS 314 Spring 2006 Homework...

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1 ECE/CS 314 Spring 2006 Homework 3 Solutions Problem 1. (5 points) Suppose register \$8 contains 0x00004000 and register \$9 contains 0xDEAFBEEF. (a) Suppose we execute the instruction sw \$9,0(\$8) on a big-endian MIPS computer. At what address will each byte of the value in register \$9 be stored. DE at 0x00004000, AF at 0x00004001, BE at 0x00004002, and EF at 0x00004003. (b) Suppose we execute the instruction sw \$9,0(\$8) on a little-endian MIPS computer. At what address will each byte of the value in register \$9 be stored. DE at 0x00004003, AF at 0x00004002, BE at 0x00004001, and EF at 0x00004000 (c) Suppose we execute the instruction sh \$9,0(\$8) on a big-endian MIPS computer. Which bytes of the value in register \$9 will be stored and at what address? BE at 0x00004000, and EF at 0x00004001. (d) Suppose we execute the instruction sh \$9,0(\$8) on a little-endian MIPS computer. At what address will each byte of the value in register \$9 be stored. BE at 0x00004001, and EF at 0x00004000 (e) Suppose we execute the instruction sb \$9,0(\$8). Does it matter if the machine is big-endian or little-endian? Why? No, byte EF gets stored at address 0x00004000 in either case.

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2 Problem 2. (10 points) Draw the Karnaugh map, circle the prime implicants, and derive the sum of product form for F= Σ ABCD (0,2,4,6, 8,9,10,13) 00 01 11 10 00 0 1 4 1 12 0 8 1 01 1 0 5 0 13 1 9 1 11 3 0 7 0 15 0 11 0 10 2 1 6 1 14 0 10 1 F = B’•D’ + A’•D’ + A•C’•D Problem 3. (10 points) Draw the Karnaugh map, circle the prime implicants, and derive the sum of product for G= Σ ABCD (1,3,5,7,11,12,14,15) 00 01 11 10 00 0 0 4 0 12 1 8 0 01 1 1 5 1 13 0 9 0 11 3 1 7 1 15 1 11 1 10 2 0 6 0 14 1 10 0 G = A’•D + A•B•D’ + C•D A B C D A B C D
3 Problem 4 . (10 points) Show that F=G’ using Boolean algebra. For each step of the derivation, mention which law of Boolean algebra you applied. G’ = (A’D + ABD’ + CD)’ = (A + D’)(A’ + B’ +D)(C’ + D’) [DeMorgan’s Laws] = (AA’ + AB’ + AD + A’D’ + B’D’ + DD’ )(C’+D’) [identity: a a’ = 0] = AB’C’ + AB’D’ + AC’D + ADD’ + A’C’D’ + A’D’D’ + B’C’D’ + B’D’D’ [identities: a a’ = 0, a a = a] = AB’C’ (1)

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ECE314sp06_HW3_solutions - ECE/CS 314 Spring 2006 Homework...

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