I have attempted this several times and come up with different answers. Just can't see where I am going wrong
the truth table is all high except for a=1 b=0 c=0 d=1 then f=0;
using a K map i simplified it to f=CD+C+A+B.
Then converting it to nand gates i get F=(CD+C+A+B)"
=((CD)'C'A'B')'

 

I have attempted this several times and come up with different answers. Just can't see where I am going wrong
the truth table is all high except for a=1 b=0 c=0 d=1 then f=0;

...

For c=0 d=1 ( c.d )' = 1 and then f = 1
Is the original expression right? Not much sense if f is always 1

 

Don't forget that the 1's in each of the four corners can also be grouped together.

I think you may be missing a few other groups as well. It's been a while since I've used Karnaugh maps though.

 

JJW nailed it. If you populate the map using the original expression, all 16 boxes have a 1.

If you had a map with only a single zero, a 4 input NAND gate would implement the function.

 

I have attempted this several times and come up with different answers. Just can't see where I am going wrong
the truth table is all high except for a=1 b=0 c=0 d=1 then f=0;
using a K map i simplified it to f=CD+C+A+B.
Then converting it to nand gates i get F=(CD+C+A+B)"=((CD)'C'A'B')'

As others pointed out, "f=CD+C+A+B" does not agree with "all high except for a=1 b=0 c=0 d=1 then f=0".

Your K-map needs to be properly labeled. There is no indication of what each row and column represents. Don't make people guess or reverse engineer your work -- the grader will give you a low grade and your boss will simply fire you (sooner or later).

Also, your schematic implies that you think that you can build up a 4-input NAND gate by cascading a bunch of 2-input NAND gates the same way that you can with OR and AND gates. Doesn't work that way.

(ABC)' is NOT equal to ((AB)'C)'

(ABC)' is LO only if A, B, and C are ALL HI.

But for ((AB)'C)' to be HI, both (AB)' and C have to be HI. However, (AB)' can only be HI if at least one of A or B is LO.

And also as already pointed out, be sure to consider groups that wrap around the edges of the map. You missed one (and it's not the corners, although that is a valid group).

 

Thanks for all the replies,
I have reassessed the k map and got F=A+B+C+D+B.C+C.D
but that still does not satisfy a=1 b=0 c=0 d=1 then f=0,

To simulate a 4 input nand gate should the output be inverted before cascading into the next 2 input nand gate?

 

I have managed to come right thanks once again for all the help.
F= !A+B+C+!D reduced with K-map

and to nand gates
F=!(A.!B.!C.D)

 

Why do you want the condition f=0 when a=1 b=0 c=0 d=1, when the original expression is always 1.
It can be found easily without k-map by converting the term (c.d)' to equivalent c' + d' and then f has sum term c' + c which is always 1.

 

As far as i evaluated the original expression is F=A.B+(C.D)'+(A'+C) and that is F=0 when A=1 B=0 C=0 D=1 the rest is all high
and I wanted to reduce the equation and build it with nand gates only.

 

As far as i evaluated the original expression is F=A.B+(C.D)'+(A'+C) and that is F=0 when A=1 B=0 C=0 D=1 the rest is all high
and I wanted to reduce the equation and build it with nand gates only.

This should have been where your original post started. It always works best if you tell us what the actual problem is instead of starting where you happened to get part way through it.

 

I don't understand why you didn't start with MikeML's observation that when there's only 1 zero in the K-map, a single 4-input NAND will implement that. Was there some other criteria?

F is zero only when A=1, B=0, C=0, D=1. This notates directly as F = (AB'C'D)'. Done, with no circles drawn.

 

Problem here is that there are no zeros in the k-map of the original expression, which has been mentioned at least three times, but OP won't realize this.
Or have I missed something?

 

The big question is what is the original problem? Was it a truth table, a K-map, a Boolean expression, what?

We need the TS to tell us what the original problem is.

 

It is a Boolean expression in the title of this thread.
I guess, that there is an error in the expression, but who knows

 

It is a Boolean expression in the title of this thread.
I guess, that there is an error in the expression, but who knows

But IS that the original problem? Or is that the Boolean expression that he came up with from the truth table that he talks about in the original post? As you say, who knows?

 

Why can't you just see what's in his mind and help him? Lol.

 

@WBahn
F=A.B+(C.D)'+(A'+C) to nand gates only was the original problem as in the title.
I posted my attempt in the first post to show me where I was going wrong, and that was
the K-map and the logic gate error as was suggested.

@Russmax
Its obvious when you get the answer but I was frustrated and couldn't see it directly and i went with the method I was learning. But that is a straight forward answer and i didn't see it at first.

@jjw
There is one 0 in the truth table at 1001 so they arn't all high

and as for what MikeML said "If you had a map with only a single zero, a 4 input NAND gate would implement the function."
That is only true for 1111 would give a low and the rest high.

I hope this answers all your questions.

 

Two questions and then I give up:
1. was the k-map in your first post given in the exercise or
2. did you make the k-map from the Boolean expression in the title: f= a.b + (c.d)' + ( a' + c )

If 1. then there is an error in the Boolean expression
If 2. then you have made a mistake in the k-map

 

Alexander,
It cannot be both. If the truth table is correct, then the answer I gave is correct. Implement the single minterm and it's done.
Your k-map DOES NOT match the boolean expression. If the original expression is correct, the k-map is ALL ones.

Here's how to fill out the k-map:
a.b fills in row 3
(c.d)' fills in columns 1, 2, 4.
a'+c fills in rows 3, 4 plus columns 3, 4.

The last two terms (c.d)' + (a' + c) fill the table with ones, so the (a.b) term is redundant. The logic can be implemented with a single nand, all inputs tied to zero and a b c d ignored.

 

@WBahn
F=A.B+(C.D)'+(A'+C) to nand gates only was the original problem as in the title.

There is one 0 in the truth table at 1001 so they arn't all high

Where is this truth table -- that I don't recall ever seeing you post -- coming from? It does not match the equation that you give in the post title.

F = AB + (CD)' + (A' + C)

In order for F to be LO, A must be HI and C must be LO (otherwise the third term would be HI and F would then be HI). Since A must be HI, the first term means that B must be LO in order for F to be LO. In order for the second term to be LO, CD must be HI, which requires that both C and D be HI. However, we've already established that C must be LO.

Bottom line: If C is HI, then the third term is HI and F is HI. If C is LO, then the second term is HI and F is HI, hence F is always HI.