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Enter image description here

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:

Enter image description here

If I got it right, this is it:

Enter image description here

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

enter image description here

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    \$\begingroup\$ D is not required to generate Y WITH this example. \$\endgroup\$ – StainlessSteelRat Jun 22 at 16:39
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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.

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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:

flip

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

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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.

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

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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.

enter image description here

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

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