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help me on inverting t FF to d FF

hi~~~ there first of all let's define as

------------------------------------------------------------ XOR = ^

------------------------------------------------------------ you know that Q+ = T ^ Q in a t Flip-flip. That is. Q+ = T ^ Q in order to convert a T flip-flop to a D flip-flop. I've got some errors to do it doing boolean algebra. this is the procedure that I did. and the next is the compleltely correct boolean munipulation.

Mine is

--------------------------------------------------------------- Q+ = T ^ Q = TQ' + T'Q T = D ^ Q = DQ' + D'Q

Then, Q+ = (DQ' + D'Q)Q' + (DQ' + D'Q)'Q = DQ'Q' + D'QQ' + (DQ')'(D'Q)'Q = DQ' + (D' + Q)(D + Q')Q = DQ' + (D'Q' + DQ)Q = DQ' + D'Q'Q + DQQ = DQ' + DQ T = DQ' + D'Q isn't equal to DQ' + DQ

would tell me what is wrong?

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and another way to solve it is bellow

----------------------------------------------------------------- Q+ = T ^ Q T = D ^ Q

then, Q+ = (D ^ Q) ^ Q = D ^ ( Q ^ Q ) by associative law = D ^ 0 = D the this implies that I can convert a t FF to a D FF. and you can verify easilly the XOR operations.

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please help me I don't have much time .. and I'm sorry to bother you~~ Thanks. I'm going to appreciate all of your efforts

Reply to
dondora
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I feel your pain but I know nothing about the subject.

This looks good to me.....

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Have a click about upon it, you also get one of these....

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And lots of other stuff.

It might be too late.

Give your answer and your reasoning as you have done here and ask teacher person for the proper answer.

Either the teacher will go ballistic (idiot) or the teacher will explain it to you. It was, after all his fault in the first place.

DNA

Reply to
Genome

It's right all the way to =DQ' + DQ, which is equivalent to D(Q+Q'), which is equivalent to D. This is the same result as below. Your last statement is also correct, "T = DQ' + D'Q isn't equal to DQ' + DQ", but it shouldn't be, because the left-hand side of your series of equations was Q+, which is not equal to T, but to T^Q. Ok?

-- john

Reply to
John O'Flaherty

Ohh I'm so glad to receive your answers. Also I must to say thank you for your efforts. and I now understood. what you mean and that is Q+ = DQ' + DQ = D(Q' + Q) = D and 1 = D is it right? thanks~~~ I won't forget this kindness.

Reply to
dondora

No, D(Q'+Q) = D. It isn't equal to 1 unless you define D as 1.

-- john

Reply to
John O'Flaherty

To get the 2nd "OR" term of you r 2nd equation, you used deMorgans theorem to convert an "OR" to and "And". You correctly complemented the DQ', D'Q term so get (DQ')'(D'Q)'. However you forgot the last step, which is to invert the "New "And" term, which should be ((DQ')'(D'Q)')'.

Reply to
Jon

I think he did it correctly- the parenthesized part of the second term of the first equation is negated, so the transform of that part from the first to the second equation has the form (a+b)' = a'b', which is correct.

-- john

Reply to
John O'Flaherty

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