HW5_solutions_last

HW5_solutions_last - Homework 5 Solutions 2012 ECE 15A 1....

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Homework 5 Solutions ECE 15A 2012 1. (10p) For this function, ±nd a minimum sum-of-products solution, using the Quine-McCluskey method: F ( a, b, c, d, e )= m (0 , 2 , 3 , 7 , 10 , 11 , 16 , 17 , 18 , 20 , 23 , 31) + d (4 , 15 , 21) Column I Column II Column III group 0 0 00000 ± (0,2) 000-0 ± (0,2,16,18) -00-0 (0,16) -0000 ± (0,16,4,20) -0-00 (0,4) 00-00 ± group 1 2 00001 ± (2,3,10,11) 0-01- 4 00100 ± (2,3) 0001- ± (16,17,20,21) 10-0- 16 01000 ± (2,10) 0-010 ± (2,18) -0010 ± (16,17) 1000- ± (16,18) 10-00 ± (16,20) 01-00 ± (4,20) -0100 ± group 2 3 00011 ± (3,7) 00-11 ± (3,11,7,15) 0- -11 10 01010 ± (3,11) 0-011 ± 17 10001 ± (10,11) 0101- ± 18 10010 ± (17,21) 10-01 ± 20 10100 ± (20,21) 1010- ± group 3 7 00111 ± (7,23) -0111 ± (7,23,15,31) - -111 11 01011 ± (21,23) 101-1 21 10101 ± (7,15) 0-111 ± (11,15) 01-11 ± group 4 15 01111 ± (15,31) -1111 ± 23 10111 ± (23,31) 1-111 ± group 5 31 11111 ± 1
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02371 01 11 61 71 82 02 33 1 (0,2,16,18)* b 0 c 0 e 0 XX X X (0,16,4,20) b 0 d 0 e 0 X (2,3,10,11)* a 0 c 0 d X X (16,17,20,21)* ab 0 d 0 X X X (3,11,7,15) a 0 de X (7,23,15,31)* cde X (21,23) ab 0 ce X * indicates essential prime implicant f ( a, b, c, d, e )= a 0 c 0 d + ab 0 d 0 + b 0 c 0 e 0 + cde 2
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2. (10p) Find all prime implicants of the following function using the Quine-McCluskey method and then ±nd all minimum sum-of-products solutions using Petrick’s method. F ( a, b, c, d, e )= m (0 , 2 , 4 , 5 , 10 , 11 , 13 , 15 , 16 , 18 , 20 , 21 , 26 , 27 , 29 , 31) Column I Column II Column III group 0 0 00000 ± (0,2) 000-0 ± (0,2,16,18) -00-0 (0,4) 00-00 ± (0,4,16,20) -0-00 (0,16) 000-0 ± group 1 2 00010 ± (2,10) 0-010 ± (2,10,18,26) - -010 4 00100 ± (2,18) -0010 ± (4,5,20,21) -010- 16 10000 ± (4,5) 0010- ± (4,20) -0100 ± (16,18) 100-0 ± (16,20) 10-00 ± group 2 5 00101 ± (5,13) 0-101 ± (5,13,21,29) - -101 10 01010 ± (5,21) -0101 ± (10,11,26,27) -101- 18 10010 ± (10,11) 0101- ± 20 10100 ± (10,26) -1010 ± (18,26) 1-010 ± (20,21) 1010- ± group 3 11 01011 ± (11,15) 01-11 ± (11,15,27,31) -1-11 13 01101 ± (11,27) -1011 ± (13,15,29,31) -11-1 21 10101 ± (13,15) 011-1 ± 26 11010 ± (13,29) -1101 ± (21,29) 1-101 ± (26,27) 1101- ± group 4 15 01111 ± (15,31) -1111 ± 27 11011 ± (27,31) 11-11 ± 29 11101 ± (29,31) 111-1 ± group 5 31 11111 ± 02451 01 11 31 51 61 82 02 12 62 72 93 1 P0 (0,2,16,18) b 0 c 0 e 0 XX X X P1 (0,4,16,20) b 0 d 0 e 0 X X P2 (2,10,18,26) c 0 de 0 X X P3 (4,5,20,21) b 0 cd 0 P4 (5,13,21,29) cd 0 e X X P5 (10,11,26,27) bc 0 d P6 (11,15,27,31) bde P7 (13,15,29,31) bce This table can be reduced noting that the table repeats itself starting at column 16: 3
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02451 01 11 31 5 P0 (0,2,16,18) b 0 c 0 e 0 XX P1 (0,4,16,20) b 0 d 0 e 0 P2 (2,10,18,26) c 0 de 0 P3 (4,5,20,21) b 0 cd 0 P4 (5,13,21,29) cd 0 e P5 (10,11,26,27) bc 0 d P6 (11,15,27,31) bde P7 (13,15,29,31) bce Applying Petrick’s method: (P0 + P1)(P0 + P2)(P1 + P3)(P3 + P4)(P2 + P5)(P5 + P6)(P4 + P7)(P6 + P7)
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This note was uploaded on 04/04/2012 for the course ECE 15A taught by Professor M during the Spring '08 term at UCSB.

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HW5_solutions_last - Homework 5 Solutions 2012 ECE 15A 1....

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