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Product of Sums Design Problem

shamieh

Active member
Sep 13, 2013
539
Need someone to check my answer please.

Consider a 4 input, 1 output digital system (W,X,Y,Z, and f respectively) . Design a POS circuit with any number of inputs such that f(W,X,Y,Z) = M(0,2,4,9,13) + D(6,14). First fill in the Truth Table, then find the minimum product of sums equation using a karnaugh map, and finally draw a schematic of a minimized circuit implementing the logic using NOR Gates.


My Answer: (y + z + \(\displaystyle \bar{w}\))(\(\displaystyle \bar{z}\) + w)(\(\displaystyle \bar{y}\) + \(\displaystyle \bar{w}\))
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
I got $(w+z)(\bar{w}+y+\bar{z})$.
 

shamieh

Active member
Sep 13, 2013
539
I got $(w+z)(\bar{w}+y+\bar{z})$.
Um..how?

I grouped the four zeros in the row above the last in the bottom left.. I also grouped the four zeros and the d in the bottom right. Finally, I grouped the two zeros in the top right. I don't understand how you got that solution.

See.
photo.JPG Apologies in advance for the horizontal placement of my image.(Bandit)
 
Last edited:

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
The product-of-maxterms notation like ΠM(0,2,4,9,13) lists rows where the function is 0, not 1 (the opposite of the sum-of-minterms notation). So, in your Karnaugh map you should flip zeros and ones.

Once again, a number, say, 6₁₀ = 110₂ denotes the minterm $xy\bar{z}$, but it denotes the maxterm $\bar{x}+\bar{y}+z$. Also, Σm(1, 2, 3) (lowercase m for "minterm"), by definition, denotes the sum of minterms 1, 2, 3, i.e., $\bar{x}\bar{y}z+\bar{x}y\bar{z}+\bar{x}yz$. This function returns 1 in rows 1, 2 and 3. Similarly, ΠM(1, 2, 3) (uppercase M for "maxterm"), by definition, denotes the product of maxterms 1, 2, 3. However, the expression now is $(x+y+\bar{z})(x+\bar{y}+z)(x+\bar{y}+\bar{z})$. This function returns 0 in rows 1, 2 and 3.

Thus, the only common thing is that both Σm(1, 2, 3) and ΠM(1, 2, 3) refer to min/maxterms 1, 2, 3. Everything else is the opposite: binary 0 in a minterm means the presence of negation, binary 0 in a maxterm means the absence of negation. Minterms are products, maxterms are sums. The function corresponding to Σm(1, 2, 3) has 1 in rows 1, 2, 3; the function corresponding to ΠM(1, 2, 3) has 0 in those rows.

See again the section in Wikipedia about indexing minterms and maxterms, as well as these slides (PDF) (the first result in Google).

I keep forgetting these conventions, so I am going to make a bookmark for this post. I recommend that you also make a note with this information.
 

shamieh

Active member
Sep 13, 2013
539
So my $f$ should consist of these values then correct?

$f$
0
1
0
1
0
1
d
1
1
0
1
1
1
0
d
1
 

shamieh

Active member
Sep 13, 2013
539
I see. But I believe your final answer is still incorrect. I got \(\displaystyle (y + z + w)(y + \bar{z} + \bar{w})(z + w + x)
\)
does that look correct?
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
So my $f$ should consist of these values then correct?

$f$
0
1
0
1
0
1
d
1
1
0
1
1
1
0
d
1
Yes.

I got \(\displaystyle (y + z + w)(y + \bar{z} + \bar{w})(z + w + x)\)
My expression $(w+z)(\bar{w}+y+\bar{z})$ and your \(\displaystyle (y + z + w)(y + \bar{z} + \bar{w})(z + w + x)\) are equivalent up to row 6 (0 1 1 0), where mine equals 0 and yours equals 1. But row 6 is "don't care", and my expression is shorter.

In fact, I got mine from the same Karnaugh map generator. Here is how to use it to create a minimal POS (rather than SOP) from an expression like ΠM(0, 2, 4, 9, 13) + D(6, 14). Enter it as if it is Σm(0, 2, 4, 9, 13) + D(6, 14), i.e., enter 1 in rows 0, 2, 4, 9, 13 and x in rows 6 and 14. This is the dual function to the one we need (we need 0's in rows 0, 2, etc.), but the program only knows how to group 1's in the Karnaugh map, so we have to adjust. Then take the dual of the minimal SOP it produces, i.e., switch + and * and all negations. This way you get the original function. In this example, the program produces $\bar{A}\bar{D}+A\bar{C}D$. Using $w, x, y, z$ instead of $A, B, C, D$, this is $\bar{w}\bar{z}+w\bar{y}z$. The dual of this is $(w+z)(\bar{w}+y+\bar{z})$. In any case, you can see by hovering the mouse over minterms how the program grouped 1's in the Karnaugh map.

Hint: write variables in min- and maxterms in the alphabetical order. This makes checking your answers easier.