DIGITALS

Question

Answer 1

Let us see one by one :

a) For generator and propogator functions :

n AND and n XOR gates..So 2n gates required the output of which will be fed to carrylookahead circuit..

b) For carry lookahead circuit :

It is a 2 level implementation with n(n+1)/2 AND gates and n OR gates.

So total gates required in carry lookahead circuit = n(n+1)/2  + n

                                                                      = ( n² + 3n) / 2

c) Finally for generating sum bits :

For sum bits which will be n in no , hence we need n XOR gates..

So total number of gates required for carry lookahead adder  = 2n + (n² + 3n) / 2 + n

                                                                                        =  (n² + 3n) / 2 + 3n

                                                                                        =  (n² + 9n) / 2

Hence the correct answer is option B)..