LOGIC CIRCUITS AND SWITCHING THEORY FOR ELECTRICAL ENGINEERS
Convert the following: (Show your solutions.)
rarr FACE_(H)It can be written as 1111101011001110 in binaryTo convert octal, group 3 digits from right to left1111101011001110write the decimal equivalent of the groups(175316)_(8)" FACEH "rarr1753168" FACEH "rarr(15 xx16^(3))+(10 xx16^(2))+(12 xx16)+(14 xx16^(0))rarr(64206)10{:[rarr(7635)8],[(7xx8^(3))+(6xx8^(2))+(3xx8)+5=(3997)10]:}(7635)_(8)rarr write the each digit in 3 bit binary number(7635)_(8)rarr(111110011101)_(2)rarr587To convert hexadecimal((16)/(587)){:[16quad36quad B],[(587)_(10)rarr(24 B)_(16)],[(24 B)_(16)rarr001001001011],[(11113)_(8)],[(587)10 rarr(1113)_(8)],[rarr(101101)_(2)],[(1xx2^(5))+(0xx2^(4))+(1xx2^(3))+(1xx2^(2))+(6xx2)+(1xx2^(0))]:}32+8+4+1=(45)10{:[(101101)= rarr(2D)16],[rarr" gray code of "1010111100101_(2)]:}In binary to gray conversion mSB is sameFor remaining gray code of that digit isXOR operation of that digit and previous digito+o+o+o+o+o+(o+o+ox o+o+o+o+1010111100101011111000101111gray code of (10101111001010)_(2) is11111000101111rarr ABF_(5)ABF5_(H)-9CDF_(H)obrace(AB)^(+16)* obrace(S)^(+16)(-CDF)/(F 16)ABFS_(H)-9CDF_(H)=F 16 Hrarr2^(') s complement of 9FDB5_(H)GFDB5 in binary can be written as 10011111110110110101one's compliment of above binary nomber is011000000010010010102's compliment is01100000001001001010+(1)/(01100000001001001011)2's com ... See the full answer