Question

hi all: wood love a hand with a bit of a harder q, not understanding it sáll

S-O-P's (sum of Products)

Answer 1

@thomaster ur a genius at these...yeah?

Answer 2

@AccessDenied any ideas man...?

Answer 3

At the moment, no. I'm reading some stuff on the topic since I am not familiar with it. :)

Answer 4

First do the distributive property, right?

Answer 5

sorry @wio could u check my answers man?
i think a) is AB + CD
b) is ABD
and c) is A + BD
but im not that certain about part b)...

Answer 6

sorry, let me think it out for a moment.

Answer 7

are you using karnot?

Answer 8

a) is correct.

Answer 9

b) is correct.

Answer 10

sweet, cheers man

Answer 11

yeah, the tables

Answer 12

c) is also correct.

Answer 13

I'm doing it using boolean algebra. I'm a bit used to it. Want me to show you?

Answer 14

yes please, if u have time that'd be awesome thanks =D

Answer 15

Okay for first one: \[
\begin{split}
AB+CD(A\overline{B}+CD)
&= AB+CDA\overline{B}+CDCD\\
&= AB + CDA\overline{B}+CD\\
&= AB + CDA\overline{B}+CD1\\
&= AB + CD(A\overline{B} + 1)\\
&= AB + CD1\\
&= AB + CD
\end{split}
\]I am using \(1\) for true here.

Answer 16

You can always append \(1\) into any and group. You can always \(+0\) as well, where \(0\) is false.

Answer 17

ahhhhh, yes, i've seen this, these are from the ...was it 8 or 10 rules... from a few chapters back

Answer 18

To be honest, I all the lines with \(1\) I would normally do in my head in one step, but I wanted to show it explicitly to be clear.

Answer 19

yep that makes sense, and i've got the table of rules written down somewhere previousely, so cool man, thanks for that, i didn't know b4 that i could apply to this, i thought we had to use the other mmethod
cheers @wio

Answer 20

\[
\begin{split}
AB(\overline{B}\overline{C}+BD)
&= AB\overline{B}\overline{C} + ABBD \\
&= ABBD \\
&= ABD
\end{split}
\]

Answer 21

In this case, any duplicates such as \(BB\) can be made into a single \(B\).
Any terms with that exist with their not form, disappear, in this case \(B\overline B\).

Answer 22

sorry, can u explain where the NOT C disappeared to in that one? plz

Answer 23

\[
AB\overline B \overline C = A0\overline C = A0 = 0
\]

Answer 24

of course: 0
and anthing by 0 = 0

Answer 25

sweeeeeeet, cheers again @wio
dude r u any good at explaining truth tables?
next topic coming up is them, i've had a look, but just not getting how to go backwards from the numbers to the letters...

Answer 26

got an example?

Answer 27

well... to the *right* letters anyway ;D

Answer 28

if u cold talk me through a) dude, i'll see if i can hit b to d on my lonesum...?

Answer 29

What are you trying to do? Turn it into an expression?

Answer 30

yep, need to find the SOP (which i think means the expressions we were just doing above)... and the POS... which i have no idea what they are...?

Answer 31

okay so let's start with table a).

Answer 32

We ignore every row where \(x=0\).

Answer 33

ok

Answer 34

So the only rows that matter are 101|1, 110|1, 111|1

Answer 35

Now, for the row 101|1, what we have is A=1, B=0, C=1. Does that make sense?

Answer 36

We can write this as \(A\overline B C\). In other words, we write out the terms that are 1 normally, and we write out the ones that are 0 with a not.

Answer 37

yep
so not A, not B, C = 1
A, not B, not C = 1
a, not B, c = 1
A,B,C = 1
...yeah?

Answer 38

No

Answer 39

\(101\to A\overline B C\), \(110 \to AB\overline C\), and \(111\to ABC\)

Answer 40

ok... what about
\[001\to \overline A \overline B C\] tho?

Answer 41

oh, that is true.

Answer 42

ok, so how do i make all of those into some sort of ABC sum ? add them all together or times them all together... or....?

Answer 43

is it just a matter of
X = sum of the above ones u typed and the 1 that i typed...?

Answer 44

So you only translate the rows that have a 1, don't translate the rows with a 0.

Answer 45

Then you just sum up the translations

Answer 46

do we need to reduce from that?
coz now i have: \[x = \overline A \overline BC + A \overline B \overline C + A \overline BC + ABC\]

Answer 47

Yeah. Reduce from that.

Answer 48

when you have a lot of terms, karnot maps are easier than algebra.

Answer 49

sooo... how does that reduce if it's "add"and not "times"...?

Answer 50

Well, for example with \[
A\overline BC + ABC = AC(B+\overline B) = AC1 = AC
\]

Answer 51

It is easier to do it with karnot maps though.

Answer 52

no worries, look all good, cheers for your time and your help @wio i'm going to read the book again and see if it makes any more sense thanks bro