coach94 Posted September 23, 2013 Share Posted September 23, 2013 I have this diagram I am trying to find the boolean equation for this. I know (d+c)(ba) I dont know how to write the equation to include the box X anyone give me a hand? Link to comment Share on other sites More sharing options...
EdEarl Posted September 23, 2013 Share Posted September 23, 2013 Why are you concerned only about X? The expression "(d+c)(ba)" doesn't mention W, V, X or f. Perhaps you will realize the answer if you write down separately what W and V do (label the lines between W and X and V and X), and the value of f. Link to comment Share on other sites More sharing options...
coach94 Posted September 23, 2013 Author Share Posted September 23, 2013 Why are you concerned only about X? The expression "(d+c)(ba)" doesn't mention W, V, X or f. Perhaps you will realize the answer if you write down separately what W and V do (label the lines between W and X and V and X), and the value of f. woops forgot a key piece of information haha! W is an OR gate, X is an AND gate, V is an XOR gate Link to comment Share on other sites More sharing options...
EdEarl Posted September 24, 2013 Share Posted September 24, 2013 (edited) Then the expression "(d+c)(ba)" cannot represent the circuit as shown; although, it may be equivalent (I haven't worked it out). Again, write each of the subexpressions for W, V and X separately. Edited September 24, 2013 by EdEarl Link to comment Share on other sites More sharing options...
coach94 Posted September 24, 2013 Author Share Posted September 24, 2013 Then the expression "(d+c)(ba)" cannot represent the circuit as shown; although, it may be equivalent (I haven't worked it out). Again, write each of the subexpressions for W, V and X separately. I think ive figured it out. (D+C) <- would be W (B.A) <- would be V and to put them together tying in the xor the final expression would be (D+C) (xor symbol) (B.A) ? i think that would be right? Link to comment Share on other sites More sharing options...
EdEarl Posted September 24, 2013 Share Posted September 24, 2013 Either you are guessing or you have given me incorrect information. You said V was .xor., what are the inputs to V? You said X was .and., what are the inputs to X? Link to comment Share on other sites More sharing options...
coach94 Posted September 24, 2013 Author Share Posted September 24, 2013 (edited) Either you are guessing or you have given me incorrect information. You said V was .xor., what are the inputs to V? You said X was .and., what are the inputs to X? Sorry im working on multiple things at once i mis-read my own question (D+C).(B(xor symbol)A) should be correct now id think Edited September 24, 2013 by coach94 Link to comment Share on other sites More sharing options...
EdEarl Posted September 24, 2013 Share Posted September 24, 2013 Yes, grats. You can write the equations as follows: W(d,c) = d .or. c = w V(b,a) = b .xor. a = v X(w, v) = w .and. v = (d .or. c) .and. (b .xor. a) = f Link to comment Share on other sites More sharing options...
coach94 Posted September 24, 2013 Author Share Posted September 24, 2013 Yes, grats. You can write the equations as follows: W(d,c) = d .or. c = w V(b,a) = b .xor. a = v X(w, v) = w .and. v = (d .or. c) .and. (b .xor. a) = f cool thanks! one last question though, would the truth tables look something like this? W's DC|X <- x because it goes through X right 00|0 01|1 10|1 11|1 X's WV|F <- F because this is where F is derived? 00|0 01|0 10|0 11|1 V's BA|X <- x because it goes through X right? 00|0 01|1 10|1 11|0 my professor never taught us how to construct them so im trying my best to go from the book but im not too sure Link to comment Share on other sites More sharing options...
EdEarl Posted September 24, 2013 Share Posted September 24, 2013 Those are correct for each of W, V, and X, but they aren't combined. Link to comment Share on other sites More sharing options...
coach94 Posted September 24, 2013 Author Share Posted September 24, 2013 Those are correct for each of W, V, and X, but they aren't combined. what do you mean by not combined? Link to comment Share on other sites More sharing options...
EdEarl Posted September 24, 2013 Share Posted September 24, 2013 One big logic matrix with all inputs, gates and the output. Link to comment Share on other sites More sharing options...
coach94 Posted September 24, 2013 Author Share Posted September 24, 2013 One big logic matrix with all inputs, gates and the output. oh okay so its fine how i have it written then Link to comment Share on other sites More sharing options...
EdEarl Posted September 24, 2013 Share Posted September 24, 2013 Yes, if that is what your instructor wants. Link to comment Share on other sites More sharing options...
coach94 Posted September 24, 2013 Author Share Posted September 24, 2013 Yes, if that is what your instructor wants. alright cool thanks for the help man appreciate it Link to comment Share on other sites More sharing options...
EdEarl Posted September 24, 2013 Share Posted September 24, 2013 YW 1 Link to comment Share on other sites More sharing options...
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