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?

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.

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

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, 201312 yr by EdEarl

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?

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?

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, 201312 yr by coach94

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

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

Those are correct for each of W, V, and X, but they aren't combined.

Those are correct for each of W, V, and X, but they aren't combined.

what do you mean by not combined?

One big logic matrix with all inputs, gates and the output.

One big logic matrix with all inputs, gates and the output.

oh okay so its fine how i have it written then

Yes, if that is what your instructor wants.

Yes, if that is what your instructor wants.

alright cool thanks for the help man appreciate it

YW