Boolean Algebra Hard

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Boolean Algebra Hard!!!

hey all, got stuck on some boolean algebra just wondering if you all can check my working out thanks :)
basically the ` represent bar's and in the example no. 1 the p is barred, r is barred,and the overall function is barred hope you get the gist of things thanks

1. [(p` + qr)(pq+r`)]`

my working out is, using de morgan's rule
= [(p` + qr)]` + [(pq + r`)]
= p``*(qr)` + r``*(pq)`
=p*(qr)` + r*(pq)` \\ De morgan's rule again
= P( q` + r` ) + r(P` + q`) \\ expanded out
= pq` + r`p + rp` + q`r
\\ i use the rule that r`p + rp` = `
thus = q`(R + p)

is that right hopefully i did it correctly :)

2. (z + (x*y`)) + yx + (x*(y` + z))
thats the function my working out is i expanded it out.
thus
= (z +xy`) + yx + xy` + zx
\\ then i left it as it is as use many of the boolean rules and got
xy` +Z + YX
X(Y` + y)+ z
=X + z
\\i used the karnaugh maps and got x + z to be the simpliest function as well but was wondering ifanyone can check this out for me thanks.


3. [x` + (y`*z`)][yz` + x`][y`+z`]
= [x` + z`y`][x` + yz`][y` + z`]
\\then i used the rule that P(P+Q)=P so that means taking x` as a common factor takinga look at the first 2 functions out ofthe 3
= x`(yz` + x`)
=x` \\ now we have (y` + z`) left as a function thus
= x`(y`+z`)
\\ using de morgan's rule
=(x+yz)`

is that the simpliestform and how would i draw this as a simplified switchinig circuit because i belive the whole function barred you cant draw it???

thats all hope this isnt a load of gibberish :)
 

Answers and Replies

AKG
Science Advisor
Homework Helper
2,559
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\\ i use the rule that r`p + rp` = `
I'm not sure of this rule. Suppose r = 0, p = 1, then r'p = 1*1 = 1, so 1 + rp' = 1 which is not equal to 0 or '.

Number 2 is correct. You will just use the fact that x + xA where A can be anything will just be x, and you'll find you have an x swallowing up everything, there's no need for Karnaugh maps. 3 is right. You can always bar a whole circuit. How you bar it depends on what gates you have available to you, but the following two ways work:

Use an AND and an OR gate to make x+yz, then use an inverter gate to bar it. Otherwise, use a NAND gate and NAND x+yz with itself. If you can bar a single thing like x, y, or z, then you can bar any larger expression. And I believe this way will be better then doing x'(y' + z') because you'd need to bar 3 things.
 

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