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    Download CS302 Assignment Solution
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    @zareen said in CS302 Assignment 3 Solution and Discussion: Design the final Circuit diagram. A circuit diagram is a graphical representation of an electrical circuit. A pictorial circuit diagram … Circuit diagrams are used for the design (circuit design), construction (such as PCB layout), and maintenance of electrical and electronic … This results in the final layout artwork for the integrated circuit or printed circuit board. Reff [image: L3kBbVH.png]
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    ASSIGNMENT NO:2 Course: CS302 1: Write the SOP expression for the given sum. Sol: A B C D E 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 1 1 0 0 0 1 1 0 1 0 1 1 1 0 0 1 1 1 1 1 0 0 0 0 1 0 0 0 1 1 0 0 1 0 1 0 0 1 1 1 0 1 0 0 1 0 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1 0 0 0 1 1 0 0 1 1 1 0 1 0 1 1 0 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 A B C D E OUTPUT (F) 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 1 1 0 0 0 1 0 0 1 0 0 1 0 1 0 0 0 1 1 0 1 0 0 1 1 1 0 0 1 0 0 0 1 0 1 0 0 1 0 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 1 1 0 1 0 1 1 1 1 0 1 0 0 0 0 1 1 0 0 0 1 0 1 0 0 1 0 1 1 0 0 1 1 0 1 0 1 0 0 1 1 0 1 0 1 0 1 0 1 1 0 1 1 0 1 1 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 0 1 0 1 1 1 0 1 1 0 1 1 1 0 0 1 1 1 1 0 1 0 1 1 1 1 0 1 1 1 1 1 1 0 FOR SOP WE FOCUS ON 1 VALUE. A B C D E MINTERM 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 1 0 1 0 0 1 1 1 0 0 1 1 0 0 0 1 1 0 1 0 1 1 1 0 0 1 1 1 1 1 ABCDE SOP EXPRESSION: SUM OF PRODUCT EXPRESSION 2: Find Prime Implicant of minterm using QuineMcculsky method. Step-1 00010 2 00100 4 01000 8 10000 16 00110 6 01010 10 01100 12 10010 18 10100 20 11000 24 01110 14 10110 22 11010 26 11100 28 11110 30 Step-2 2,6(00-10) 2,10(0-010) 2,18(-0010) 4,12(0-100) 4,6(001-0) 4,20(-0100) 8,10(010-0) 8,24(-1000) 16,18(100-0) 16,20(10-00) 16,24(10-00) 6,14(0-110) 6,22(-0110) 10,14(01-10)10,26(-1010) 12,14(011-0) 12,28(-1100) 18,26(1-010) 18,22(10-10) 20,22(101-0) 20,22(1-100) 24,26(110-0) 24,28(11-00) 14,30(-1110) 22,30(1-110) 26,30(11-10) 28,30(111-0) Step-3 2,6,18,22(-0-10) 2,6,10,14( 0–10)2,10,18,26(–010)2,18,6,22(-0-10) 2,18,10,26(–010)4,12,6,14(0-1-0) 4,6,12,14 (0-1-0) 4,6,20,22(-01-0)4,20,6,22(-01-0) 4,20,12,28(–100) 8,10,12,14(01–0) 8,24,10,26(-10-0) 8,24,12,28 (-1-00) 6,14,22,30 (–110) 6,22,14,30(–110) 10,14,26,30(-1-10) 10,26,14,30(-1-10) 12,14,28,30(-11-0) 12,14,28,30(-11-0) 18,26,22,30(1–10) 18,22,26,30(1–10) 20,22,28,30(1-1-0) 20,22,22,30(1-1-0) 24,26,28,30(11–0) 24,28,26,30(11–0) Step-3 2,6,18,22(-0-10) 4,6,20,22(-01-0) 2,18,10,22(-0-10) 4,20,6,22(-01-0) 2,10,18,26(–010) 4,12,6,14(0-1-0) 2,18,6,22(–010) 4,6,12,14 (0-1-0) 4,20,12,28(–100) 10,14,26,30(-1-10) 6,14,22,30 (–110) 10,26,14,30(-1-10) 12,14,28,30(-11-0) 18,26,22,30(1–10) 12,14,28,30(-11-0) 18,22,26,30(1–10) 20,22,28,30(1-1-0) 24,26,28,30(11–0) 20,22,22,30(1-1-0) 24,28,26,30(11–0) 6,22,14,30(–110) 8,10,12,14(01–0) 8,24,12,28 (-1-00) 8,24,10,26(-10-0) Step-4 2,6,8,22(-0-10) 4,6,20,22(-01-0) 2,10,18,26(–010) 4,12,6,14(0-1-0) 4,20,12,28(–100) 10,14,26,30(-1-10) 12,14,28,30(-11-0) 18,26,22,30(1–10) 20,22,28,30(1-1-0) 24,26,28,30(11–0) 6,22,14,30(–110) 8,10,12,14(01–0) 8,24,12,28 (-1-00) 8,24,10,26(-10-0) These are the prime implicates cs302-assign 2.docx
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    The large energy cost of memory fetches limits the overallefficiency of applications no matter how efficient the ac-celerators are on the chip. As a result the most importantoptimization must be done at the algorithm level, to reduce off-chip memory accesses, to createDark Memory. The algorithmsmust first be (re)written for both locality and parallelism beforeyou tailor the hardware to accelerate them.Using Pareto curves in theenergy/opandmm2/(op/s)spaceallows one to quickly evaluate different accelerators, memorysystems, and even algorithms to understand the trade-offsbetween performance, power and die area. This analysis isa powerful way to optimize chips in the Dark Silicon era.
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