KayS Posted July 17, 2014 Posted July 17, 2014 Could someone help me or guide me through this quesiton, and does someone have any good resource where I could learn how ot get my head around boolean equations and truth tables, I have the following quesiton:
ydoaPs Posted July 17, 2014 Posted July 17, 2014 Do you know what all of the operators do? Break it down into parts.
Endy0816 Posted July 17, 2014 Posted July 17, 2014 "help me or guide me" 0 = false, 1= true help me guide me result 0 0 0 0 1 1 1 0 1 1 1 1 So if someone helps you or guides you, you see a result. Now these Or's and And's also equate to logic gates. So logic.ly if there were a logic simulator you could use the appropriate combination of gates and find your answer that way as well. 1
studiot Posted July 17, 2014 Posted July 17, 2014 (edited) Courses these days seem to plunge straight in to DeMorgan without spelling out the simple relationships. Here are all you will ever need. The last two between them make up De Morgan's Theorem. [math]0*0 = 0[/math][math]0 + 0 = 0[/math][math]1*1 = 1[/math][math]1 + 1 = 1[/math][math]1*0 = 0[/math][math]1 + 0 = 1[/math][math]\bar 0 = 1[/math][math]\bar 1 = 0[/math][math]A + 0 = A[/math][math]A + 1 = 1[/math][math]A*0 = 0[/math][math]A*1 = A[/math][math]A + A = A[/math][math]A*A = A[/math][math]A + \bar A = 1[/math][math]A*\bar A = 0[/math][math]\overline{\overline A} [/math] = [math]A[/math][math]nA = A[/math][math]{A^n} = A[/math][math]A + B = B + A[/math][math]A + B+C = A + (B + C) = (A + B) + C[/math][math]ABC = A(BC) = (AB)C[/math][math]A(B + C) = AB + AC[/math][math]A + AB = A[/math][math]A(A + B) = A[/math][math]A\left[ {\left( {\overline A } \right) + \left( B \right)} \right][/math] = [math]AB[/math][math]A + \overline A B = A + B[/math] [math]A + BC = (A + B)(A + C)[/math] [math]\overline {(AB)} [/math] = [math]\left( {\overline A } \right) + \left( {\overline B } \right)[/math] [math]\left( {\overline A } \right)*\left( {\overline B } \right)[/math] = [math]\overline {(A + B)} [/math] Edited July 17, 2014 by studiot 1
KayS Posted July 17, 2014 Author Posted July 17, 2014 (edited) I understand and gates or gates etc that for an both must b true in order to give true and for the rest. I know the rules however, its the equation I do not know what they express that is whats confusing me. for example what does /a.b <<--- stand for i know a and b are the gates representation but the division slash and the dot im sure the express a special type of gate but im just not sure Edited July 17, 2014 by KayS
Endy0816 Posted July 17, 2014 Posted July 17, 2014 And = A.B Or = A+B Not = / or a line above http://www.dragonwins.com/domains/getteched/binary/BooleanLogic.htm I read your example as ((Not A) And B). You may want to check with someone local on that one though. Not 100% sure if they meant to negate just the 'A' or the '(A And B)' portion. lot of logic symbols, very little standardization.
studiot Posted July 17, 2014 Posted July 17, 2014 (edited) I understand and gates or gates etc that for an both must b true in order to give true and for the rest. I know the rules however, its the equation I do not know what they express that is whats confusing me. Try to forget 'true' and false', they can be misleading. OK so we use capital letters for boolean variables, eg A, B, C, D, etc. Any boolean variable can take one of two values 0 or 1. Any boolean constant can take one of two values, 0 or 1. Since they are so simple we do not bother with letters for constants. Just as with ordinary algebra we can form equations or expressions combining (boolean) variable. (boolean) constants, We may use powers and multiples these may be integers as in ordinary algebra. eg 1+A ; B2; C(A+B); 4D and so on. We manipulate these using the rules of boolean algebra I listed So your expression is made up of the sum of two terms and Now = and [math]\left[ {\overline {\left( {\overline A } \right) + \left( {\overline B } \right)} } \right][/math]=[math]AB[/math] So your expression can be simplified to [math]\overline A [/math] + [math]\overline B [/math] + [math]AB[/math] You can simplify this further using more rules from my list. I note you mention gates and it is not clear whether you are studying maths or electronics here. If you like we can look at this suing the simplest electronic implementation, which is actually switches, not gates. For switches, variables A and B represent switches, either off or on (negated). The product AB is represented by switches in series The sum (A+B) is represented by switches in parallel Edited July 17, 2014 by studiot 1
Janus Posted July 17, 2014 Posted July 17, 2014 (edited) [math] \overline{A} [/math] means "not A" (the inverse of A) [math] \overline {A.B}[/math] means "not (A AND B)" (the inverse of the operation: A AND B.) ergo: [math]\overline {\overline{A}+ \overline{B}}[/math] is "not ( "not A" OR "not B") (the inverse of operation: 'inverse of A' OR the 'inverse of B') Edited July 17, 2014 by Janus
KayS Posted July 17, 2014 Author Posted July 17, 2014 thanks for clearing that up what about the plus sign in the middle and the question mark because thats what is really confusing me
studiot Posted July 17, 2014 Posted July 17, 2014 (edited) I'm sorry, why is the question mark a problem, don't you usually have one at the end of a question? The plus sign between the two terms is because you can have either plus or times as operations. You questioner has chosen plus. If you like the output (boolean variable C) is given by your expression between boolean variables A and B. So C = some function of A as well as B Edited July 17, 2014 by studiot
KayS Posted July 17, 2014 Author Posted July 17, 2014 from what I have understood the answer I came up with was D because : the inverse of A or B is is only shown in truth table D. for example whenever a or b is 1 it outputs 1, and when it is 0 it remains 0. Am I correct in thinking that or Am i still reading the equation wrong. Im reading the equations not(a and B) + not(a or b)
studiot Posted July 17, 2014 Posted July 17, 2014 (edited) try not A and B or not(notA or notB) You need to decide whether you are going for a mathematical solution using boolean algebra or an electronic solution using and and or gates. Do not mix them. I did ask this before? Edited July 17, 2014 by studiot
KayS Posted July 17, 2014 Author Posted July 17, 2014 (edited) if its not A and B or not(notA or notB) then it would just return 0 unless both are 1 or both are 0? Edited July 17, 2014 by KayS
ydoaPs Posted July 17, 2014 Posted July 17, 2014 if its not A and B or not(notA or notB) then it would just return 0 unless both are 1 or both are 0? Nope. This expression will give the same answer regardless of the values of A and B.
Janus Posted July 17, 2014 Posted July 17, 2014 if its not A and B or not(notA or notB) then it would just return 0 unless both are 1 or both are 0? It returns a 1 if either not(A AND B) or Not (not A OR not B) returns a 1. So first look at not(A AND B), this is just the inverted output of an and gate (A NAND gate), which means that it only returns a 0 when A and B are both 1 Not( not A OR not B) Is the inverted output from a OR gate (a NOR gate), which also has had its inputs inverted. Thus it only returns a 1 when not A and not B are both zero. ( IOW when A and B are both 1.) You may notice that this is the same truth table as an AND gate, and thus not(not A OR not B) = A AND B. Since not(A AND B) returns a 1 for everything but A AND B, and not(not A OR not B) returns a 1 for A AND B, One of the two returns a 1 for every combination of A and B. And if either of them returns a 1 then the final output is a 1.
KayS Posted July 17, 2014 Author Posted July 17, 2014 It returns a 1 if either not(A AND B) or Not (not A OR not B) returns a 1. So first look at not(A AND B), this is just the inverted output of an and gate (A NAND gate), which means that it only returns a 0 when A and B are both 1 Not( not A OR not B) Is the inverted output from a OR gate (a NOR gate), which also has had its inputs inverted. Thus it only returns a 1 when not A and not B are both zero. ( IOW when A and B are both 1.) You may notice that this is the same truth table as an AND gate, and thus not(not A OR not B) = A AND B. Since not(A AND B) returns a 1 for everything but A AND B, and not(not A OR not B) returns a 1 for A AND B, One of the two returns a 1 for every combination of A and B. And if either of them returns a 1 then the final output is a 1. thank you I think I get it now..let me explain the answer is A btw am I right? becausee.. (A AND B) or Not (not A OR not B) the OR in the middle of the equation allows both sides to allow the 1 to be the output for every scnerio for example if A and B do not fulfill the requirement then the other part of the equation allow the scnerio ot be complete because the OR requires either the left or right to fulfill the boolean criteria thats why all the outputs are 1. all this time i have been thinking that the plus sign means that both have to satisfy the criteria in order to deliver the 1. So in this case A is the right answer because the equation either left or right always allows use to get an output of 1? Am i correct?
studiot Posted July 17, 2014 Posted July 17, 2014 Am i correct? Yes OK so here is a boolean algebra solution I am ging to use ' instead of the overbar for ease of typing. expression = AB + A' + B' = A(expression) =AAB + AA' + AB' =A2B + 0 + AB' =AB + AB' =A(B + B') =A(1) = A (expression) Hence expression = 1
KayS Posted July 17, 2014 Author Posted July 17, 2014 Yes OK so here is a boolean algebra solution I am ging to use ' instead of the overbar for ease of typing. expression = AB + A' + B' = A(expression) =AAB + AA' + AB' =A2B + 0 + AB' =AB + AB' =A(B + B') =A(1) = A (expression) Hence expression = 1 Thank you very much solving it =] the way u explained it was so much better than I learnt this year made so much more sense about think of the gates when inverted sometimes tehy turn into simple ands. =]
studiot Posted July 18, 2014 Posted July 18, 2014 (edited) Formally to solve it algebraically we would assign a boolean variable to our expression and work on that. So I would start Let E = "the expression" Then AE =A(.......).. finally A(.......) = A(1) hence E = 1 But the trick of multiplying through by A is not obvious. Edited July 18, 2014 by studiot
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