Boolean Algebra Answer and Question

Added on - 26 Nov 2019

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Running head: BOOLEAN ALGEBRABoolean AlgebraName of the Student:Name of the University:Author Note
1BOOLEAN ALGEBRAAnswer to question 1a)Single precision IEEE 754 format- 0 01111110 10100000000000000000000Sign bit is 0Exponent is 01111110Mantissa is 10100000000000000000000Decimal value -.8125Hence the value is 8.125x10^(-1)b)5-bit word- 00110Signed magnitude- (-6)One’s complement- 11001Two’s complement- 11010Answer to question 2a)The table below is used for the representation of the clockHoursClockDecimal RepresentationABCDE1000012000103000114001005001016001107001118010009HIGH(1)0100110HIGH(1)01010
2BOOLEAN ALGEBRA11HIGH(1)0101112HIGH(1)0110013HIGH(1)0110114HIGH(1)0111015HIGH(1)0111116HIGH(1)1000017100011810010191001120101002110101221011023101112411000In the above table the 24-hour clock is represented by the 5 bit binary values.Now for the High or 1 represents that the door is open.Hence the values that represents that the door is open are:The logic of the diagram is:A'BC'D'E+A'BC'DE'+A'BC'DE+A'BCD'E' = A’B(C’D’E+C’DE’+C’DE+CD’E’)= A’B(D(C’E’+C’E) +D’(C’E+CE’)) = A’B(D(C’E’+C’E) +D’(C’E+CE’)) =A’B (D’C’+D (Cxor E))A'BCD'E’+A'BCDE'+A'BCDE+AB'C'D'E' = A’BCD + A'BCD’E’+AB'C'D'E' = A’BCD + D’E’(A’BC + AB’C’)Hence, the expression of the problem is minimized and the diagram is provided below. Thisdiagram represents the clock and the output represents the intervals in the which the doors areopen.
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