# Waveform result of an equation

I have this waveform and I would like to have the Y.

I know that the equation is Y = D' + E' + F'.

Where it has ' it means NOT.

As I see it, the only thing I did, and I am not sure if it is right, was this:

If I got it right, this is it:

• At the very beginning, D' is True, so Y should also be True. – ErikR Jun 22 at 0:20
• you mean I am right @ErikR ? – Yun.kon Jun 22 at 0:21
• You've drawn Y as being False at the beginning and only True near the end (where the bump is) – ErikR Jun 22 at 0:22
• I've updated my answer. – ErikR Jun 22 at 0:55
• @Yun.kon - your last graph is correct. – ErikR Jun 22 at 1:28

Do the analysis graphically.

Any red section in D, E or F will cause Y to be true.

You will also see that D is not required to generate Y.

• D is not required to generate Y WITH this example. – StainlessSteelRat Jun 22 at 16:39

The exercise shows the application of DeMorgan's Theorem. Stated simply,

• A'+ B' is equivalent to (AB)' (negative-in OR into positive-in NAND)

Likewise,

• A'B' is equivalent to (A+B)' (negative-in AND into positive-in NOR)

In this case, you have three inputs. Doesn't matter, DeMorgan's Theorem extends to any number of inputs:

• Y = D'+E'+F'

converts to..

• Y = (DEF)'

that is, the NAND of the three inputs.

Y will be low when all three inputs are high. Based on the sequence, Y will pulse low only at the end, when DEF are all high.

More about DeMorgan's Theorem here: https://www.electronics-tutorials.ws/boolean/demorgan.html

Do a truth table for both (D' + E' + F') and (DEF)'. They will be the same.

The other answers are already good, but since you have all inputs negated, then it might help you to simply flip the picture up-side down:

All that's left is to see where all of the three traces are false.

Since Y is equal to the sum of (not D), (not E) and (not F), it will be true whenever any one of D, E and F is not true. For then at least one of the three terms will be true. Thus, in your diagram, Y will be true or high whenever any of D, E and F is not true or low. It will only be not true or low whenever all of D, E and F are true or high. You should be able to construct Y from this information.

• I am trying to understand it but it is confusing.May cause english is not my first language – Yun.kon Jun 22 at 0:37

The easiest way to read the boolean equation Y=D'+E'+F' is D is false OR E is false OR F is false.

If you understand how to translate from your boolean to my English equivalent, the correct answer should be obvious. If any of D, E, or F are false, the output will be true. So it's always high, except when all three inputs are true.

Times 1, 2 and 3 are indicated by the green lines.

At time 1, D, E and F are all False.

At time 2, D and E are False and F is True.

At time 3, D is False, E is True and F is False.

• This is incorrect. – hacktastical Jun 22 at 0:47
• why it is incorrect – Yun.kon Jun 22 at 0:48
• Missed the inversion for (D'+E'+F') = (DEF)'. – hacktastical Jun 22 at 0:48
• @hacktasical - Y is from the OP's answer. I was just explaining how the logic levels are determined for D, E, and F. – ErikR Jun 22 at 0:51
• You still don't show the correct logic or waveform. – hacktastical Jun 22 at 1:02