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NAU EE 110 - Octal solution

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EE 110 Spring 2017 CQUPT HW 7 Solution Mlsna EE 110 Spring 2017 CQUPT Homework 7 Solution 1. (10 points) In octal, perform the following calculation: 2404.38 + 156.28 Show all steps. Check your results by converting all numbers to decimal and making sure addition in decimal gives the same result. 2 4 0 4 . 3 + 0 1 5 6 . 2 (Carrying occurs by 8 since we’re working in octal) _________________ 2 5 6 2 . 5 Check: 2404.38 = 2x83 + 4x82 + 0x81 + 4x80 + 3x8–1 = 1024 + 256 + 4 + 3/8 = 1284.37510 156.28 = 1x82 + 5x81 + 6x80 + 2x8–1 = 64 + 40 + 6 + 2/8 = 110.2510 2562.58 = 2x83 + 5x82 + 6x81 + 2x80 + 5x8–1 = 1024 + 320 + 48 + 2 + 5/8 = 1394.62510 This checks, since in decimal: 1284.375 + 110.25 = 1394.625 2. (10 points) In octal, perform the following calculation: 2404.38 – 156.28 Show all steps. Check your results by converting all numbers to decimal and making sure subtraction in decimal gives the same result. 2 4 0 4 . 3 – 0 1 5 6 . 2 (Borrowing occurs by 8) _________________ 2 2 2 6 . 1 Check: 2226.18 = 2x83 + 2x82 + 2x81 + 6x80 + 1x8–1 = 1174.12510 This matches the decimal: 1284.375 – 110.25 = 1174.125EE 110 Spring 2017 CQUPT HW 7 Solution Mlsna 3. (10 points) In hexadecimal, perform the following calculation: A230C.7F16 – EDB.C16 Consider these to be positive, unsigned numbers. Show all steps. Show the decimal equivalent to all of the numbers involved. I’ll show this one column at a time A 2 3 0 C . 7 F – 0 0 E D B . C 0 (Borrowing occurs by 16 since we’re working in hex) ____________ F ! “F – 0” or 15 – 0 “7 – C” or 7 – 12. Need to borrow 16 from the left. B ! This produces 16+7–12 = 11 0 ! “B – B” (remember, we borrowed 16 from here) 3 ! “0 – D” Need to borrow 16 from left, producing 16+0–13 4 ! “2 – E” (we borrowed from here). Borrow: 16+2–14 1 ! “1 – 0” (we borrowed from here) A ! “A – 0” ____________ A 1 4 3 0 . B F Now let’s look at the decimal equivalent values. A230C.7F16 = 10 x 164 + 2 x 163 + 3 x 162 + 12 + 7 x 16-1 + 15 x 16-2 = 655360 + 8192 + 768 + 12 + 0.4375 + 0.05859375 = 664332.49609375 EDB.C16 = 14 x 162 + 13 x 16 + 11 + 12 x 16-1 = 3584 + 208 + 11 + 0.75 = 3803.75 A1430.BF16 = 10 x 164 + 163 + 4 x 162 + 3 x 16 + 11 x 16-1 + 15 x 16-2 = 655360 + 4096 + 1024 + 48 + 0.6875 + 0.05859375 = 660528.74609375EE 110 Spring 2017 CQUPT HW 7 Solution Mlsna 4. (10 points each) Perform the following calculations in 8-bit two’s complement binary and show all steps. Also check your results by showing the calculation in decimal and showing the answers are correct. a. 5510 + 11410 First convert numbers to 8-bit 2’s complement form: 5510 = 0 0 1 1 0 1 1 1 11410 = 0 1 1 1 0 0 1 0 ------------------------------------ now add 1 0 1 0 1 0 0 1 = -8710 This answer is negative! We added two 8-bit 2’s complement positive numbers and should have obtained a positive answer. The problem is sign-bit overflow because the correct answer cannot be represented in 8-bit two’s complement form. b. 5510 + (-11410) 5510 = 0 0 1 1 0 1 1 1 -11410 = 1 0 0 0 1 1 1 0 ------------------------------------ now add 1 1 0 0 0 1 0 1 = - 00111011 = -5910 correct! c. (-5510) + 11410 -5510 = 1 1 0 0 1 0 0 1 11410 = 0 1 1 1 0 0 1 0 ------------------------------------ now add 0 0 1 1 1 0 1 1 = 5910 correct!EE 110 Spring 2017 CQUPT HW 7 Solution Mlsna 5. (20 points each) Use the Quine-McCluskey method to fully minimize the following Boolean expressions, producing SOP expressions. Show all steps! You may check your results with a K-map, but don’t use the K-map as a substitute for doing Quine-McCluskey. a. € f (A, B,C, D) = m(1,2, 3, 6, 7,8,9,12,14)∑ First make a minterm list arranged by number of 1’s in the binary pattern Individual Minterms Number of 1’s in binary pattern Minterms covered Binary pattern Prime Implicants ( * = covered ) 1 1 2 8 0001 0010 1000 * * * 2 3 6 9 12 0011 0110 1001 1100 * * * * 3 7 14 0111 1110 * * Pairs of minterms Minterms covered Binary pattern Prime Implicants ( * = covered ) 1,3 1,9 2,3 2,6 8,9 8,12 00–1 –001 001– 0–10 100– 1–00 PI2 PI3 * * PI4 PI5 3,7 6,7 6,14 12,14 0–11 011– –110 11–0 * * PI6 PI7 Groups of 4 minterms (quads) Minterms covered Binary pattern Prime Implicants ( * = covered ) 2,3,6,7 0–1– PI1EE 110 Spring 2017 CQUPT HW 7 Solution Mlsna Prime Implicant Chart 1 2 3 6 7 8 9 12 14 PI1 X X X X PI2 X X PI3 X X PI4 X X PI5 X X PI6 X X PI7 X X Minterms 2 and 7 can only be covered by PI1, therefore PI1 is an essential prime implicant. That means PI1 must be included in the final answer. Reduced Prime Implicant Chart 1 8 9 12 14 PI2 X PI3 X X PI4 X X PI5 X X PI6 X PI7 X X Now we must choose a minimal (smallest) set of remaining prime implicants in order to cover all minterms in this chart. Possible solutions: PI1 + PI3 + PI5 + PI6 = € A C + B C D + C AD + BCD PI1 + PI3 + PI5 + PI7 = € A C + B C D + C AD + ABD PI1 + PI2 + PI4 + PI7 = € A C + A B D + AB C + ABDEE 110 Spring 2017 CQUPT HW 7 Solution Mlsna b. € f (A, B,C, D, E) = m( 5, 6, 7, 8,10,11,13,18,20,21,23,24,26)∑ First make a minterm list arranged by number of 1’s in the binary pattern Individual Minterms Number of 1’s in binary pattern Minterms covered Binary pattern Prime Implicants ( * = covered ) 1 8 01000 * 2 5 6 10 18 20 24 00101 00110 01010 10010 10100 11000 * * * * * * 3 7 11 13 21 26 00111 01011 01101 10101 11010 * * * * * 4 23 10111 * Pairs of minterms Minterms covered Binary pattern Prime Implicants ( * = covered ) 8,10 8,24 010–0 –1000 * * 5,7 5,13 5,21 6,7 10,11 10,26 18,26 20,21 24,26 001–1 0–101 –0101 0011– 0101– …


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