04-24-2011, 04:41 AM
[size=14]full solution in attachment [/SIZE]
Answers
Question No.1 (Marks = 10)
a) How many combinations of bits are exist when we make a truth table of 9 inputs?
2n are the total combinations of bits in Truth table, where n is number of variables, so
29=512
512 bits will there
b) Prepare a truth table for the following I = (w + x) (y . z) where w, x, y and z are four binary inputs in the truth table.
Note :symbol represent XOR gate)
Solution:
Remember in XOR, output will high only if only all inputs are mismatch to each others.
Answers
Question No.1 (Marks = 10)
a) How many combinations of bits are exist when we make a truth table of 9 inputs?
2n are the total combinations of bits in Truth table, where n is number of variables, so
29=512
512 bits will there
b) Prepare a truth table for the following I = (w + x) (y . z) where w, x, y and z are four binary inputs in the truth table.
Note :symbol represent XOR gate)
Solution:
Remember in XOR, output will high only if only all inputs are mismatch to each others.