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Assignment || Conversion Of Binary Number

   

Added on  2020-03-02

4 Pages724 Words181 Views
Task 1.a.Convert the binary number to floating number0Number is positive+01111110Exponent(0*27) + (1*26) + (1*25) + (1*24) + (1*23) + (1*22) + (1*21) + (1*20)= 126= 126 – 127= -110100000000000000000000Mantissa1.1011.101 * 10-1= 0.1101= 0.1101= (1*2-1) + (1*2-2) + (0*2-3) + (1*2-4)= 1/2 + 1/4 + 1/16= 0.8125b.1. Signed magnitude: 01111 to 11111-15 to +152. One's complement: 10000 to 01111-16 to +153. Two’s complement: 10001 to 01111-17 to +15Task 2.a)Let Clock timing denoted by ALet the Switch denoted by BLet the Gate denoted by CBoolean Expression:C = A.BMinimisation:For minimisation, let the switch (B) be always on, then the status for gate is always dependent on the clock timing (A).C = A

Diagram:b)X’Y + XYZ’ + Y’ + XZ (Y+Y’) = 1Consider the L.H.S to prove the Boolean expression.X’Y + XYZ’ + Y’ + XZ (Y+Y’)= X’Y + XYZ’ + Y’ + XZ.1= X’Y + (XZ’ + Y’) (Y + Y’) + XZ= X’Y + (XZ’ + Y’).1 + XZ= X’Y + XZ’ + Y’ + XZ= X’Y + Y’ + X (Z’ + Z)= (X’ + Y’) (Y + Y’) + X.1= (X’ + Y’).1 + X= X’ + Y’ + X= 1 + Y’= 1Hence Proved.Identities usedInverse law: A + A’ = 1Distributive law: A + BC = (A + B) (A + C)Identity law: A.1 = ANull law: A + 1 = 1Task 3.1.Memory – 256M * 128RAM chips – 1024K * 32Memory is word addressablea)No. of Rows 256∗210∗210210∗210 = 256No. of Columns 12832 = 4b)4 RAM chips (equal to the no. of the columns)c)1204K = 22020 bits are needed for each RAM chip.d)256 banks (equal to the no. of the rows)AC = A.BB

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