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the boolean expression aolineb aboline ab is equivalent to ans the boolean expression aoline b a boline ab is equivalent to a baoline b a boline
eprom contents can be erased by exposing it to ans by exposing eprom contents to ultraviolet rays it can be erasedthe ultraviolet uv light passes
q--gtthe program simulates a student management system having the followingthe interface uses command buttons to i addeditdeleteupdate and cancel the
can u please tell me the assembly language program for carry look ahead adder that can run in 8086 emulator its
add 648 and 487 in bcd codeans in bcd code addition of 648 and 4876 4 8 0 1 1 0 0 1 0 0 1 0 0 04 8 7 0
conversioin of the decimal number 8267 into octal number ans the binary equivalent is 1010010101010112 of decimal number 8267 after that convert each
conversion of the decimal number 8267 into hexadecimal ans 1010010101010112 is the binary equivalent of decimal number 8267 now convert each 4-bit
simplify the given expressions using boolean postulates xy xolinezoline xyolinez xy zans xy xolinezoline xyolinez xy z xy xolinezoline
simplify the given expressions using boolean postulatesy a baoline cb cans y a baoline cb c a aoline ac b aoline bc b c ac b aoline bc
subtraction of 01000-01001 using 2s complement methodans firstly 1s complement of 01001 is 10110 and 2s complement is 10110 1 10111thus 01000
simplify the boolean expression f cb ca b cans simplification of the given boolean expression f c b c abc given asf c bc abc cb cc abc cb c
perform 2s complement subtraction of 710 - 1110 ans 2s complements subtraction of 710 - 1110firstly convert the decimal numbers 7 and 11 to there
which device consume minimum power ans minimum power consume by cmos as in its one p-mos and one n-mos transistors are connected in complimentary
which of the memories stores the most number of bits ans most number of bits stores in 32m x 8as 25 x 220 225therefore 1m 220 1k x 1k 210 x
which of the logic gates are known as universal gates ans nand and nor are termed as universal gates since any digital circuit can be
for jk flipflop j 0 k1 the output after clock pulse will be ans j0 and k1 such inputs will reset the flip-flop after the clock pulse therefore
how many address bits are required to represent 4k memory ans 12 address bits are required for representing 4k memory as4k 22 x 210
how many select lines will a 321 multiplexer will have ans 5 select lines will be required for 32 inputs as 25
how many two input and gates and two input or gates are required to realizey bdceab ans here three product terms therefore three and gates of two
the decimal equivalent of binary number 10101 is ans 1x24 0x23 1x22 0x21 1x20 16 0 4 0 1
the 2s complement of the number 1101110 is ans 1s complement of 1101110 is 0010001 ans hence 2s complement of 1101110 is 0010001 1
a weighted resistor digital to analog converter using n bits requires a total of ans digital to analog converter a weighted resistor using n bits
the conversation speed of an analog to digital converter is maximum with which techniqueans with parallel comparator ad converter technique the
the information in rom is stored ans by the manufacturer throughout fabrication of the
the output of a jk flipflop with asynchronous preset and clear inputs is 1 the output can be changed to 0 with which conditions ans through applying