I have the expression: x'y+xy'+y'z . I want to express this with only OR, NOT gates, but the issue is I have no idea how to remove the AND functions. I was thinking of using demorgans law, but I am unsure how and if it can be used on only part of the expression. Is there a way to simplify it and remove the ANDS?
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- 1\$\begingroup\$ Hint: Can you make an AND gate out of an OR gate and three NOT gates? \$\endgroup\$John Dvorak– John Dvorak2018-02-15 00:31:14 +00:00Commented Feb 15, 2018 at 0:31
- \$\begingroup\$ x'y+xy'+y'z === (x'y)+(xy')+(y'z). Now you can use De Morgan's laws inside each parathesis until all the ANDs are gone. \$\endgroup\$Dampmaskin– Dampmaskin2018-02-15 00:47:05 +00:00Commented Feb 15, 2018 at 0:47
- \$\begingroup\$ Oh I see. So (x'y) would be (x+y') correct? @Dampmaskin \$\endgroup\$Kytex– Kytex2018-02-15 00:50:17 +00:00Commented Feb 15, 2018 at 0:50
- 1\$\begingroup\$ The whole paranthesis also gets inverted. If De Morgan's laws are not clear to you, you should study them more closely. \$\endgroup\$Dampmaskin– Dampmaskin2018-02-15 00:55:37 +00:00Commented Feb 15, 2018 at 0:55
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6 Q: \$\overline{x}y+x\overline{y}+\overline{y}z\$, Turn this into NOT and OR gates only.
Just apply De Morgan's law on each of them individually.
Here's a refreshener:
\$\overline{AB}=\overline{A}+\overline{B}\$
\$\overline{A+B}=\bar{A}\bar{B}\$
\$\overline{\overline{A}}=A\$
First add some not gates.
\$\overline{\overline{\overline{x}y}}+\overline{\overline{x\overline{y}}}+\overline{\overline{\overline{y}z}}\$
Then De Morgan them to the maximum.
Here's the continuation in hidden format, I do encourage you to apply De Morgan's law on your own.
\$\overline{\overline{\overline{x}}+\overline{y}}+\overline{\overline{x}+\overline{\overline{y}}}+\overline{\overline{\overline{y}}+\overline{z}}\$
And after you've applied De Morgan's Law, then you might want to remove unnecessary gates. Use these two yellow boxes as a way to control that you've calculated correctly.
\$\overline{x+\overline{y}}+\overline{\overline{x}+y}+\overline{y+\overline{z}}\$
This is how it looks like in a schematic.
- \$\begingroup\$ Thank you, that makes much more sense. Now since the parenthesis have a complement, how would I go about making an AND gate? Wouldn't I need to have the complement somehow removed? \$\endgroup\$Kytex– Kytex2018-02-15 02:30:00 +00:00Commented Feb 15, 2018 at 2:30
- 1\$\begingroup\$ "how would I go about making an AND gate?", from what? And wasn't the OR + NOT gates the goal? - I am not following you. \$\endgroup\$Harry Svensson– Harry Svensson2018-02-15 02:32:02 +00:00Commented Feb 15, 2018 at 2:32
- \$\begingroup\$ sorry, misworded. I meant the or gate. How would I deal with the complement on the parenthesis since I need just the or gate \$\endgroup\$Kytex– Kytex2018-02-15 02:34:50 +00:00Commented Feb 15, 2018 at 2:34
- \$\begingroup\$ Are you talking about inverting the OR gate? \$\overline{x+y}=\$ NOT(OR(x,y)). - Just add a NOT gate at the output of the OR gate. \$\endgroup\$Harry Svensson– Harry Svensson2018-02-15 02:42:14 +00:00Commented Feb 15, 2018 at 2:42
- 1\$\begingroup\$ @Kytex Added schematic for clarity. \$\endgroup\$Harry Svensson– Harry Svensson2018-02-15 02:51:43 +00:00Commented Feb 15, 2018 at 2:51