Hi ... the green loop is just "A" because of the aread of the map that the loop occupies. On the top of the map, the "0" represents NOT_A and the "1" represents "A". So, the loop covers all of the "A" column, while the "BC" variables are 00, 01, 10, and 11, so regardless of what "B" and "C" are, "A" is still one. Because of that, the "B" and "C" variables have no effect on what "A" is ... so they are simplified out. Hope that helps! Cheers, Jim.
Very concise and easy to understand. My only question is...how did you know it was a OR b? Instead of a AND b? I guess my question pertains to what you would call the "gate" I think?
Hi Steve, I got the values from a truth table I made up ... but normally, they would come from the truth table that defines the circuit. Cheers and thanks, Jim.
What if my Y outputs aren't binary? I have outputs 0, 1, 2, and 3. The latter two are singletons, or special cases, so the other six boxes would be filled with just 0's and 1's. Thank you!
It seems that I still apply the same rules... If I label each box as 0, 1, 2, or 3, I won't find as many groupings of like outputs, but where I find my powers of two, I still have an opportunity to optimize.
+Axial LP Yes, the truth table just represents the inputs and outputs of a digital circuit. In this case, I just made up values, but when you have a digital circuit, there will be an output for all of the different input combinations ... the purpose of the video was to use an existing truth table, with the inputs and the outputs, and show you how to do the Boolean simplification using a Karnaugh Map rather than by using Boolean Algebra.
+Axial LP Also I believe since this an SOP Minimization from truth table to Karnaugh Map, SOP expressions output needs to be 1 because you are ANDing product variables in each term then ORing them. POS would be the opposite and you would group the 0's rather than the 1's.My issue in this simplification process is actually deriving the what needs to be pulled out of your K-map as your final result.
+shineunbyul95 Hi, so the confusion is that this video shows you how to solve a three-variable Karnaugh Map when you have the truth table ... this video doesn't show how you come up with the truth table. In this particular example, the truth table is just arbitrary or random values. In this case, the truth table shows the three inputs (A, B, C) and the output of the circuit (y). Once you have that, then you can use the Karnaugh Map to do your Boolean simplification using the Karnaugh Map rather than using Boolean algebra. So the "y" in this case just represents the output of a particular logic circuit ... we're just using that arbitrary output to show the technique used when using a Karnaugh Map to do Boolean simplification. Hope that helps.
You are literally the best. Holy cow. Thank you so muchhh
You're welcome! Glad you found it helpful ... good luck on the exam! Cheers, Jim.
Thank you very much for the tutorial!!
i find this very helpfull and simple to understand...and i'm not english. So all the haters are wrong!! great job subscribing!
You are welcome. Thanks for the comments and glad you found it useful! Cheers, Jim.
Thank you!
Cheers,
Jim.
Hi ... the green loop is just "A" because of the aread of the map that the loop occupies. On the top of the map, the "0" represents NOT_A and the "1" represents "A". So, the loop covers all of the "A" column, while the "BC" variables are 00, 01, 10, and 11, so regardless of what "B" and "C" are, "A" is still one. Because of that, the "B" and "C" variables have no effect on what "A" is ... so they are simplified out. Hope that helps! Cheers, Jim.
You're welcome ... glad you found it helpful! Cheers, Jim.
Thanks! Haven't used the Quine McClusky method, but I will have a look! Thanks again, Jim.
excellent video sir....... quite helpful..... thanks
Thanks for the feedback! You're welcome! Cheers, Jim.
thank you sir, it's very helpful
Thank you very much. Great video! Helped me a lot !
Hi,
You're weeeeelcome! :)
Glad you found it helpful!
cheers,
Jim.
Very concise and easy to understand. My only question is...how did you know it was a OR b? Instead of a AND b? I guess my question pertains to what you would call the "gate" I think?
Brian Deaver it's A OR B because of the columns they are in. If the c variable changed then it will become A OR (B AND C)
You've been veeeeery helpfull .. thanks alot
thanks man ,it really helped!!
Hi Steve, I got the values from a truth table I made up ... but normally, they would come from the truth table that defines the circuit. Cheers and thanks, Jim.
The thing about adjacency helped. Thanks!
It helped alot thanks
JUST PERFECT, I WAIT YOUR NEXT VIDEOES SOON///// THANX A LOT
Great explanation! Thank you!
Great video, simple and clear
I'm glad ... thanks for the feedback!
Great video! Do you think you can do one covering Quine mcclusky method? Have a great day!
Very helpful, thank you.
amazing explain and voice :D
Thank you sir, I will come again
Thank you 🫡🫶🏻🇮🇳
What if my Y outputs aren't binary? I have outputs 0, 1, 2, and 3. The latter two are singletons, or special cases, so the other six boxes would be filled with just 0's and 1's. Thank you!
It seems that I still apply the same rules... If I label each box as 0, 1, 2, or 3, I won't find as many groupings of like outputs, but where I find my powers of two, I still have an opportunity to optimize.
hmm i didn't catch why the green loop is just 'A'.
How do you make the truth table of this
Hello Sir, Could you please suggest me how do i implement function which outputs the number of 1's in the 3-bit
Wow, thank you sir!
I have a question! How did you get the y values? did your or the C, B, and A values?
Don't understand..that much, i don't understand why the example has to be that way, how that example come to exist?
excellent ! :) thanks !
Thank you!
Thanks
Thank u
Could you help me? Could you tell me, why is "y" 0 or 1?
+Gyula Fedor it's random...nothing more!
yah he just selected random outputs for y, you can try your own y values and create your own kmap
yes, can you please explain why y is 0 or 1?
+Axial LP Yes, the truth table just represents the inputs and outputs of a digital circuit. In this case, I just made up values, but when you have a digital circuit, there will be an output for all of the different input combinations ... the purpose of the video was to use an existing truth table, with the inputs and the outputs, and show you how to do the Boolean simplification using a Karnaugh Map rather than by using Boolean Algebra.
+Axial LP Also I believe since this an SOP Minimization from truth table to Karnaugh Map, SOP expressions output needs to be 1 because you are ANDing product variables in each term then ORing them. POS would be the opposite and you would group the 0's rather than the 1's.My issue in this simplification process is actually deriving the what needs to be pulled out of your K-map as your final result.
You gotta make more tutorials man.
Dzięki
***** Proszę bardzo
should have problem with more outcomes. hard to understand.
Jiwan N I have another video with 4 variables ... did you see that one?
So hard to understand
rnmln95 anything in particular I can help with?
+DigitalLogicAnalysis Can you just explain how we get y, there are so many question about that.. everything else is clear..
+Aida Ljajic probably just chose a random set of boolean values.
+Jo S (asdf071) No, it cant be like that. A, B, C are all possible combinations and than there is y.. and all I need is formula for y..
+shineunbyul95 Hi, so the confusion is that this video shows you how to solve a three-variable Karnaugh Map when you have the truth table ... this video doesn't show how you come up with the truth table. In this particular example, the truth table is just arbitrary or random values. In this case, the truth table shows the three inputs (A, B, C) and the output of the circuit (y). Once you have that, then you can use the Karnaugh Map to do your Boolean simplification using the Karnaugh Map rather than using Boolean algebra. So the "y" in this case just represents the output of a particular logic circuit ... we're just using that arbitrary output to show the technique used when using a Karnaugh Map to do Boolean simplification. Hope that helps.
Poorly explained