I have a problem getting this following boolean equation into to using only nand gates. AB+DA+CA+DCB ... just can't figure it out with the basic rules of boolean algebra.

 

Have another look at DeMorgan's Theoram. (Theorem? Theorom?) Have another look at DeMorgan's stuff.

And remember... nobody restricted anyone to the use of two-input gates...

 

That was given to me in the assignment that I can you more than 2 inputs. But my problem is how to get complenment lines in order to form nand gates? please elaborate your answer

 

But my problem is how to get complenment lines in order to form nand gates?

You must use NAND gates exclusively, yes? Your outputs get inverted, yes?

If AB=X, then (AB)'=X'.

That's where DeMorgan's Theorem comes in.

The circuit can be made with three gates from a 7400, one from a 7410, and one from a 7420.

If you are still having trouble, try making a truth table for the problem.

 

See attached:

 

I have a problem getting this following boolean equation into to using only nand gates. AB+DA+CA+DCB ... just can't figure it out with the basic rules of boolean algebra.

Please note [' = NOT] and I am not simplifying the expression in the following:

AB+DA+CA+DCB

Double NOT the expression (because double NOT does not alter the expression):

(AB+DA+CA+DCB)''

Using DeMorgans (A+B)' = A'.B'

Therefore:

((A.B)'.(D.A)'.(CA)'.(DCB)')'

Which can be implemented using purely NAND gates.

You may be advised to look at simplifying the expression, although that depends on what your assignment requires.

Dave

 

I see your point dave but the way you did it requires a not gate and nand gates...And my assignment was to use only nand gates.

 

I see your point dave but the way you did it requires a not gate and nand gates...And my assignment was to use only nand gates.

It doesn't, it requires:

- 1x4-I/P NAND
- 1x3-I/P NAND
- 3x2-I/P NAND

Incidently, you can also make a NOT gate out of a 2-I/P NAND gate by tying the two inputs together.

Dave