Anyone here know about Boolean Algebra?

[Kirby]

I've got to find the sum-of-products for a timing diagram. I made a truth table and copied down the data.

Now, for SOP, the functions that you are supposed to use are the ones that are true, correct?

If the aforementioned is correct, then I would need to find the SOP for m(1,2,3,4,7).

So this is the equation I got: x'y'z + x'yz' + x'yz + xy'z' + xyz

Now comes the part that I'm confused on, simplification.
x'(y'z+yz' +yz) + x(y'z' + yz) = x'(1 + yz) + x(1)

OR

x'(y'z + yz') +yz(x' +x) + xy'z' = x' + yz + xy'z

I would think that this would simplify this way, but that doesn't make sense. The book gives the answer: x'z + x'y + yz + xy'z'

Can anyone help?

 

[shortylickens]

I thought I did but looking at your issue I guess its been a while for me.

 

[PottedMeat]

you got a truth table and need SOP for a 3 variable eqn?

you try putting it into a kmap?

 

[CraKaJaX]

Originally posted by: PottedMeat
you got a truth table and need SOP for a 3 variable eqn?

you try putting it into a kmap?

That's what I would do. Did you try putting it into a K-map?

 

[Kirby]

We haven't done K-map yet. I believe it's in Ch. 4, were just in Ch. 2.

 

[j00fek]

had to do it for CS in college thats all i know

 

[Kirby]

Originally posted by: j00fek
had to do it for CS in college thats all i know

Well, thats what I'm doing. The professors's accent is really thick (asian) so I'm having a hard time understanding him.

 

[CalvinHobbs]

try the k map then try simplifying again, i can feel your pain, it never got in my head this boolean algebra however easy it seemed for others

 

[Kirby]

Blegh. Does anyone know what (y'z + yz') simplifies to? Is it one? (y' + y)=1 and (z+z')=1, so I'm going off that assumption.

 

[Kyteland]

y'z+yz' doesn't simplify in to anything. That's as simple as it gets.

One thing to keep in mind in boolean algebra is that x+x = x = x+x+x+.... This isn't like adding numbers, so you can use terms multiple times in your simplifications.

You had x'(y'z+yz' +yz) + x(y'z' + yz). Start with y'z+yz' +yz. y'z+yz = z and yz'+yz = y so y'z+yz'+yz = (y'z+yz)+(yz'+yz) = z+y. It's perfectly valid to use the yz term twice in that reduction.

Plug this back in to your equation and you have x'(y+z) + x(y'z' + yz) = x'y+x'z+x(y'z' + yz) which is halfway to the answer from your book.

You're left with the xy'z' and xyz terms. These two terms don't simplify with each other, but remember you can reuse the other terms. xyz+x'yz can be simplified to yz. unfortunately the xy'z' term can't be combined with any of the other 4 parts of the equation to get anything simpler.

This leaves you with x'z + x'y + yz + xy'z' which is what you were looking for.

 

[Kirby]

Originally posted by: Kyteland
y'z+yz' doesn't simplify in to anything. That's as simple as it gets.

One thing to keep in mind in boolean algebra is that x+x = x = x+x+x+.... This isn't like adding numbers, so you can use terms multiple times in your simplifications.

You had x'(y'z+yz' +yz) + x(y'z' + yz). Start with y'z+yz' +yz. y'z+yz = z and yz'+yz = y so y'z+yz'+yz = (y'z+yz)+(yz'+yz) = z+y. It's perfectly valid to use the yz term twice in that reduction.

Plug this back in to your equation and you have x'(y+z) + x(y'z' + yz) = x'y+x'z+x(y'z' + yz) which is halfway to the answer from your book.

You're left with the xy'z' and xyz terms. These two terms don't simplify with each other, but remember you can reuse the other terms. xyz+x'yz can be simplified to yz. unfortunately the xy'z' term can't be combined with any of the other 4 parts of the equation to get anything simpler.

This leaves you with x'z + x'y + yz + xy'z' which is what you were looking for.

I understand everything except how you got yz. So far I have x'z + x'y + xy'z' +xyz. You said that xyz + x'yz = yz, but where did the x'yz come from?

I got my first homework back today. 56% 🙁

 

[neutralizer]

I'm taking a similar class, but we learned all this stuff with K-maps on the same time so you can check your work multiple times.

 

[Kirby]

Originally posted by: neutralizer
I'm taking a similar class, but we learned all this stuff with K-maps on the same time so you can check your work multiple times.

We actually started K-maps today, but we didn't have time to do any with 3 inputs. From what I understand, you can get rid of compliments by going up, down and across the diagram if they have the same output value (1 for minterm and 0 for maxterm).

 

[Kyteland]

Originally posted by: nkgreen
I understand everything except how you got yz. So far I have x'z + x'y + xy'z' +xyz. You said that xyz + x'yz = yz, but where did the x'yz come from?

I got my first homework back today. 56% 🙁

It comes from the original equation: x'y'z + x'yz' + x'yz + xy'z' + xyz

It's easier to pick out when you use the entire expansion, but you can also get that reduction simply from 'xz+xyz = 'xz+yz. Check the truth tables for these to see that they're equivalent.