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Posted

I have this diagram

post-100409-0-48062600-1379978038_thumb.png

 

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?

Posted

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.

Posted

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

Posted (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 by EdEarl
Posted

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?

Posted

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?

Posted (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 by coach94
Posted

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

Posted

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

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