minimise a logic gate..? boolean algebra q

x in the city

hi guys

can anyone look at this and see if its right and can be minimised more, it seems the a and > a bar, b > b bar cancel off and left with just c bar, d bar , but not sure....

Eliot Rosewater

You've made a mistake on the last line with the A's and B's. You haven't applied De Morgan's law properly in that you didn't change the gate. I just clarified the xor notation, that's the root of it really.

EDIT: The 2nd line from the bottom of mine has a typo; the sign between A' and B'' should be "+", not ".".

x in the city

Im useles with de morgans...

Its late, will be up in the morning to look into it. thanks bud.!

ps: what do you make of this one....?! (tricky....)



* in case your wondering, exam practise you kno....!

Eliot Rosewater

I think you've made a few mistakes again.

Take the third last line. You have lost the dot between the A and the D, so when you apply DeMorgans law in the future you do it wrong because your treating "AD" as one thing. But it's supposed to be "A.D".

[LATEX]\displaystyle F=\overline{\left(C+\left(A.B\right)\right).\left(\overline{\overline{D}.B}\right)}[/LATEX]

Your next move here was correct, but you dropped some of the dots so it added confusion for later on. Leaving in the dots:

[LATEX]\displaystyle F=\left(\overline{C+\left(A.B\right)\right)}+\left(\overline{\overline{\overline{D}.B}}\right)}[/LATEX]

In the right hand brakets, the two overlines cancel. The rule is:

[LATEX]\displaystyle \overline{\overline{A}}=A[/LATEX]

So the equation becomes

[LATEX]\displaystyle F=\left(\overline{C+\left(A.B\right)\right)}+\left(\overline{D}.B\right)}[/LATEX]

The right hand bracket is done now. Applying the law to the left hand bracket:

[LATEX]\displaystyle F=\left(\overline{C}.\left(\overline{A.B}\right)\right)}+\left(\overline{D}.B\right)}[/LATEX]

The only thing left is the A.B part:

[LATEX]\displaystyle \overline{A.B} = \overline{A}+\overline{B}[/LATEX]

So in the end:

[LATEX]\displaystyle F=\left(\overline{C}.\left(\overline{A}+\overline{B}\right)\right)}+\left(\overline{D}.B\right)}[/LATEX]

I *think* :pac:

I think the main thing you're doing wrong is you keep losing dots, which causes slip ups later one.

x in the city wrote: »

* in case your wondering, exam practise you kno....!

I've actually an exam in 2 or 3 weeks on this stuff too, so your helping me if anything!

x in the city

I think I got it now, thanks.

what do you think of these, I need the inverse function (Fn) and the inverse Input of the 'F' which was calculated.

Fn is just 'F.


If it makes sense..:pac:

x in the city

Eliot Rosewater wrote: »

I think you've made a few mistakes again.

Take the third last line. You have lost the dot between the A and the D, so when you apply DeMorgans law in the future you do it wrong because your treating "AD" as one thing. But it's supposed to be "A.D".

[LATEX]\displaystyle F=\overline{\left(C+\left(A.B\right)\right).\left(\overline{\overline{D}.B}\right)}[/LATEX]

Your next move here was correct, but you dropped some of the dots so it added confusion for later on. Leaving in the dots:

[LATEX]\displaystyle F=\left(\overline{C+\left(A.B\right)\right)}+\left(\overline{\overline{\overline{D}.B}}\right)}[/LATEX]

In the right hand brakets, the two overlines cancel. The rule is:

[LATEX]\displaystyle \overline{\overline{A}}=A[/LATEX]

So the equation becomes

[LATEX]\displaystyle F=\left(\overline{C+\left(A.B\right)\right)}+\left(\overline{D}.B\right)}[/LATEX]

The right hand bracket is done now. Applying the law to the left hand bracket:

[LATEX]\displaystyle F=\left(\overline{C}.\left(\overline{A.B}\right)\right)}+\left(\overline{D}.B\right)}[/LATEX]

The only thing left is the A.B part:

[LATEX]\displaystyle \overline{A.B} = \overline{A}+\overline{B}[/LATEX]

So in the end:

[LATEX]\displaystyle F=\left(\overline{C}.\left(\overline{A}+\overline{B}\right)\right)}+\left(\overline{D}.B\right)}[/LATEX]

I *think* :pac:

I think the main thing you're doing wrong is you keep losing dots, which causes slip ups later one.



I've actually an exam in 2 or 3 weeks on this stuff too, so your helping me if anything!

** I think you meant A+D on that diagram I showed, no..?

(rather than A+B)