Discussion in 'Homework Help' started by kuxz2008, Dec 26, 2009.
i would like to know is it possible to simplify further the circuit that i have attached?
I'm not sure about simplifying, because i haven't checked it myself. but in order to do so, you'll need to put the bulean expression into karnough map maybe just to make sure, although it's not a very long expression.
on the other hand, you can usde less gates for the same circuit (it's important if you want to save money for who you'll work for in the future..), instead of using the 2 NOT gates at the lower part, remove them and change the AND that's attached to them to NAND. also in the start, you can remove the first NOT and change the following AND to NAND. hmmm.. that's it.. you can make it with 3 gates less... i think that's all
This is a good problem for applying DeMorgan's Theorems to solve.
Using these, you can do this problem in your head. You'll see that the operation is a simple 4 input OR-gate with the top input inverted.
Yes, it is. Your circuit implements the logic expression ((AB')(C'D'))' , where A is the noninverted input . You can manipulate the expression by using DeMorgan's theorem and Boolean algebra to simplify. Or, you can download the program in the link below and plug in the above logic expression. You will see that all minterms are covered except minterm number 8, corresponding to logic term AB'C'D' . AB'C'D' represents the negative of your original expression ((AB')(C'D'))' . AB'C'D' can be implemented by a 4 input AND gate. To restore it to its positive value, a NAND gate can be used. Therefore, your circuit can be implemented by a single 4-input NAND gate with 3 inverters or whatever is equivalent.