I'm having a heck of a time at this. I think the answer is either A or B but leaning towards B. I've mapped this out, its two variables that require 4 squares. Just can't get it, please any and all help is appreciated.
Thanks in advance for any and all help.
Corey Esson

 

Since this is Homework Help and not Homework Done For You, you need to post YOUR best attempt to solve YOUR homework. That gives us a starting point from which to make observations and suggestions. With nothing to work from, about the best we can do is to tell you to Google "Karnaugh map".

 

Since this is Homework Help and not Homework Done For You, you need to post YOUR best attempt to solve YOUR homework. That gives us a starting point from which to make observations and suggestions. With nothing to work from, about the best we can do is to tell you to Google "Karnaugh map".

Hi WBahn, sorry but this is my best attempt. I've mapped it, and using another example of a two variable with different minterms I don't understand how that answer is achieved either. I just don't understand it and have Googled till I'm blue in the face. Reading it is tougher than having it explained on a black board as well.
I'll just pick B and see if I'm right tomorrow. Thanks
Corey

 

Hi WBahn, sorry but this is my best attempt. I've mapped it, and using another example of a two variable with different minterms I don't understand how that answer is achieved either. I just don't understand it and have Googled till I'm blue in the face. Reading it is tougher than having it explained on a black board as well.
I'll just pick B and see if I'm right tomorrow. Thanks
Corey

Another thing that just came to me. I think that because (in my map) B cancels C-not, therefore, B would be the first Letter.
Then, AB makes D logic 1. Therefore letter B for BD.
Am I making any sense? Or am I still completely out in left field?
Thanks. Corey

 

Hi WBahn, sorry but this is my best attempt. I've mapped it, and using another example of a two variable with different minterms I don't understand how that answer is achieved either. I just don't understand it and have Googled till I'm blue in the face. Reading it is tougher than having it explained on a black board as well.
I'll just pick B and see if I'm right tomorrow. Thanks
Corey

So... are we supposed to use a crystal ball to view this map that you are talking about? Perhaps remote viewing? Or if you leave it out on your front porch we can steal some time on an NRO satellite and snap a picture of your work?

WE ARE NOT MIND READERS!

How can we possibly tell you where you went wrong unless you show us what you did?

 

So... are we supposed to use a crystal ball to view this map that you are talking about? Perhaps remote viewing? Or if you leave it out on your front porch we can steal some time on an NRO satellite and snap a picture of your work?

WE ARE NOT MIND READERS!

How can we possibly tell you where you went wrong unless you show us what you did?

Sorry, kind of stressed and forgot to show it. I'll take a pic here. I'm. Really trying my best. A simple "need a pic of map would do"
Map coming soon.

 

See if I might of mapped this right?

Corey

 

Um... I am not downloading 4.7 megabytes.

 

How about posting something a bit more reasonable than a 4.7 MB image when a 33 kB file is more than sufficient.



With four inputs, you have sixteen possible input combinations, so your K-map needs 16 boxes. This is normally done as follows:



Now try mapping each box to one of the expressions like you tried to do above.

 

So you have 4 inputs, A, B, C, D.

2^4=16
So your truth table will have 16 rows.

Of those 16 rows only four will have desired output.
A'BCD is same as 0111
ABC'D is same as 1101
ABCD is same as 1111
A'BC'D is same as 0101

This wiki has example of four input karnaugh map: https://en.wikipedia.org/wiki/Karnaugh_map

I hate to ask this, but... Can you even write the truth table using information that is given to you in the exercise?

 

A simple "need a pic of map would do"

I kinda thought, "you need to post YOUR best attempt to solve YOUR homework," was sufficiently suggestive.

 

How about posting something a bit more reasonable than a 4.7 MB image when a 33 kB file is more than sufficient.

View attachment 90919

With four inputs, you have sixteen possible input combinations, so your K-map needs 16 boxes. This is normally done as follows:

View attachment 90920

Now try mapping each box to one of the expressions like you tried to do above.

Oh my gosh! Sorry to both of you! Didn't mean to, nor realize I sent such a huge file. Sorry about this.
Okay, so WBahn and shteii01, didn't have a clue I needed 16 boxes, wow! So now I'll try to work this out that way and go from here. Thank you for the map to fill in as well.

 

Oh my gosh! Sorry to both of you! Didn't mean to, nor realize I sent such a huge file. Sorry about this.
Okay, so WBahn and shteii01, didn't have a clue I needed 16 boxes, wow! So now I'll try to work this out that way and go from here. Thank you for the map to fill in as well.

That is why I did the whole step by step setup.
Step 1.
# of inputs.

Step 2.
The size of the truth table.

Step 3.
Fill up the truth table.
(technically you don't need this part)

Step 4.
Draw up the Karnaugh Map and fill it up.

Step 5.
Get/derive the solution.


You can double check your work using Boolean Algebra rules.

 

If you get a problem like this on an exam and don't understand how to work it using K-maps, remember that it is sufficient to find the correct answer by eliminating all but one of the choices.

You KNOW that C is incorrect because there are (at least) FOUR possible input combinations that will produce a HI output.

You can determine that D is incorrect as long as you can perform ANY reduction. Well, we can combine the 1st and 3rd terms to get BCD (then ORed with the 2nd and 4th), so that rules out D.

Now you are down to the two you had identified.

The first one, CA, can ONLY be HI if both A is HI and C is HI. Well, look at the very first term, A'BCD. As long as B, C, and D are HI, then the output is HI even A is LO. So that eliminates A from consideration.

That's enough to allow you to choose answer B. But let's look at B anyway. It requires that B and D both be HI. A simple look shows that each term requires that B and D both be HI. So that is necessary -- but it is not sufficient since for that to be the correct answer it also requires that A and C cannot influence the output. If we limit ourselves to when B and D are both HI, we see that each term covers one of the four possible combinations of A and C, so one of them will always be HI.

 

So you have 4 inputs, A, B, C, D.

2^4=16
So your truth table will have 16 rows.

Of those 16 rows only four will have desired output.
A'BCD is same as 0111
ABC'D is same as 1101
ABCD is same as 1111
A'BC'D is same as 0101

This wiki has example of four input karnaugh map: https://en.wikipedia.org/wiki/Karnaugh_map

I hate to ask this, but... Can you even write the truth table using information that is given to you in the exercise?

Hi shteii01, sort of. But with yours and WBahns help I'm mapping out in 16 squares. I hope I get it in the next while or so.
Thank you
Corey

 

Hi shteii01, sort of. But with yours and WBahns help I'm mapping out in 16 squares. I hope I get it in the next while or so.
Thank you
Corey

You really only need 4 squares. The equation give in the problem tells you which squares will have 1 in them. The rest are zeros so you don't really care about them.

 

You really only need 4 squares. The equation give in the problem tells you which squares will have 1 in them. The rest are zeros so you don't really care about them.

Be a bit careful there. A zero is not the same as a don't care.

 

Be a bit careful there. A zero is not the same as a don't care.

Hello again, still working this problem (thanks for helping me with the B answer) but I still need to understand this. Like I said above, without "blackboard" learning, reading is tougher for me and you (WBahn) and shteii01 are my blackboards of sorts. This REALLY helps me! I hope the next file/pic is not too big. In this next one as the example of the book it gives 8 variables and this is where I thought I needed only 4 squares and now I see that I do really only need 8 squares. Pic to follow.

 

Dang! Still a big file! Sorry.

 

You really only need 4 squares. The equation give in the problem tells you which squares will have 1 in them. The rest are zeros so you don't really care about them.

Okay, I think I'm picking this up now. Now, with more practice I'll get sharper at it. Thank VERY MUCH WBahn and shteii01!!!!
Pic to follow, hope not too big a file.

Corey