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CS302 - Digital Logic Design

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    cyberianC

    20200614_190524.jpg 20200614_190550.jpg

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    zaasmiZ

    Download CS302 Assignment Solution

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    zareenZ

    @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

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    zareenZ

    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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    M

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