RAKESH SINGH BADAL IFTM UNIVERSITY

What’s difference between 1’s Complement and 2’s Complement? 1’s complement  of a binary number is another binary number obtained by toggling all bits in it, i.e., transforming the 0 bit to 1 and the 1 bit to 0. Examples: Let numbers be stored using 4 bits 1's complement of 7 (0111) is 8 (1000) 1's complement of 12 (1100) is 3 (0011) 2’s complement  of a binary number is 1 added to the 1’s complement of the binary number. Examples: Let numbers be stored using 4 bits 2's complement of 7 (0111) is 9 (1001) 2's complement of 12 (1100) is 4 (0100) These representations are used for signed numbers. The  main difference  between 1′ s complement and 2′ s complement is that 1′ s complement has two representations of 0 (zero) – 00000000, which is positive zero (+0) and 11111111, which is negative zero (-0); whereas in 2′ s complement, there is only one representation for zero – 00000000 (+0) because if we add 1 to 11111111 (-1), we get 00000000 (+0) which is t

K-Map (Karnaugh Map) In many digital circuits and practical problems we need to find expression with minimum variables. We can minimize Boolean expressions of 3, 4 variables very easily using K-map without using any Boolean algebra theorems. K-map can take two forms Sum of Product (SOP) and Product of Sum (POS) according to the need of problem. K-map is table like representation but it gives more information than TRUTH TABLE. We fill grid of K-map with 0’s and 1’s then solve it by making groups. Steps to solve expression using K-map- Select K-map according to the number of variables. Identify minterms or maxterms as given in problem. For SOP put 1’s in blocks of K-map respective to the minterms (0’s elsewhere). For POS put 0’s in blocks of K-map respective to the maxterms(1’s elsewhere). Make rectangular groups containing total terms in power of two like 2,4,8 ..(except 1) and try to cover as many elements as you can in one group. From the groups made in step 5 find the p

Calculate the maximum packing fraction of the unit cell volume that can be filled by hard spheres in the SC, BCC, FCC, Diamond and Hexagonal structures.  Which structure most efficiently fills space?  SC :  It is easy to determine that there is 1 lattice point per unit cell.  The maximum radius that a hard sphere can have is  , where  a  is the lattice constant.  When the radius is this value, then the spheres on the corners of the unit cell just touch each other.  Hence the packing fraction is: B.C.C :  It is easy to determine that there is 2 lattice points per unit cell.  The maximum radius that a hard sphere can have is  , where  a  is the lattice constant.  When the radius is this value, then the spheres on the corners of the unit cell just touch the body-centered sphere.  Hence the packing fraction is: F.C.C :  It is easy to determine that there is 4 lattice points per unit cell.  The maximum radius that a hard sphere can have is  , where  a  is the lattice